Prior Knowledge Answers

Prior knowledge Answers

Test yourself (page 2)

1 The more reactive metal displaces the less reactive metal from its compound, e.g. Zn(s)CuSO4(aq)

→ ZnSO4(aq) + Cu(s) 2 A metal or hydrogen 3 Carbon dioxide is produced. acid + carbonate → metal compound (salt) + carbon dioxide + water 4 They are all white crystalline solids.

Test yourself (page 6)

5 Oxides of nitrogen split into oxygen and nitrogen on heating. Any ionic salt, such as sodium chloride or aluminium oxide, can be split into elements by electrolysis when molten. 6

Test yourself (page 7)

7 a) H–Cl , HCl

b) S=C=S, CS2

8 a) Hydrogen and sulfur, H2S

b) Chlorine and oxygen, Cl2O

c) Hydrogen and nitrogen, H3N or more usually NH3

Test yourself (page 8)

9 a) Copper, steel b) Salt, limestone c) Ice, sugar

10 a) C2H6O2

b) CH3O

11 Because Al(OH)3 contains one Al atom combined with 3 O atoms and 3 H atoms, whereas

AlOH3 would contain one Al atom combined with only one O atom and 3 H atoms.

12 a) K2SO4 b) Al2O3 c) PbCO3 b) PbCO3 b) Fe2(SO4)3

Test yourself (page 9)

13 Molecules a) Copper, steel c) Salt, limestone e) Ice, sugar Ions b) copper (i) oxide d) lithium fluoride 14

Test yourself (page 10)

15 a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) b)

16 a) H2(g) + Cl2(g) → 2HCl(aq)

b) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

c) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)

d) 2Fe(s) + 3Cl2(g) → 2FeCl3(s)

Test yourself (page 11)

17 a) Zn(s) + 2HCl(aq)→ ZnCl2(aq) + H2(g)

b) CaO(s) + 2HCl(aq) → CaCl2(aq) + H2O(l)

c) KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)

d) ZnCO3(s) + 2HCl(aq) → ZnCl2(aq) + CO2(g) + H2O(l) 18 a) Potassium nitrate b) Calcium chloride c) Copper(ii) sulfate d) Copper(ii) sulfate

Test yourself (page 13)

1 a) The third and fourth points are still thought to be correct. b) The first point is incorrect because atoms can be split into smaller particles. The second and fifth points are incorrect because of the existence of isotopes.

2 a) H2O, CO2, CuO b) c) Water

Test yourself (page 15)

3 a) Most of the atom is empty space, so most alpha particles miss the small central nucleus. b) Some positive alpha particles pass near a positive nucleus, which repels them and causes them to be deflected from their original path. c) A few alpha particles approach positive nuclei of the gold atoms ‘head-on’ and are repelled backwards, so they appear to rebound from the foil. 4 a) They must be very small, occupying relatively little space. b) Thomson’s plum pudding model suggested a more solid structure for atoms with little or no empty space. This would not have produced the scattering of alpha particles which Geiger and Marsden observed. 5 Scientists publish their work so that their results and interpretations can be checked by other scientists before being accepted as reliable new knowledge. This process of peer review adds support to valid claims; it also helps to detect false results that cannot be repeated and to test new theories. Scientists also compete to be the first to publish new discoveries.

Test yourself (page 16)

the hydrogen atom the helium atom (1p, 0n, 1e) (2p, 2n, 2e)

Atomic model Evidence

Discovery that compounds have a definite composition and Dalton’s solid atoms that atomic masses can be used to find formulae.

Discovery of beams of negative electrons coming from many Thomson’s plum pudding different substances. Neutral atoms made up of negative model ‘plums’ in a positive ‘pudding’. Rutherford’s ‘solar Alpha particle scattering experiment suggests that the positive system’ atom charge of atoms must be concentrated in a tiny nucleus. Chadwick’s model of a nuclear atom with Uncharged particles produced by bombarding beryllium with protons, neutrons and alpha particles accounts for the relative masses of atoms. electrons Test yourself (page 17)

8 The spacing between centres of atoms is about 7 mm = 7 × 10−3 m. The magnification is times 35 × 106. The diameter of one gold atom is about 2 × 10−10 m = 0.2 nm. 9 a) 4p, 5n, 4e b) 19p, 20n, 19e c) 92p, 143n, 92e d) 9p, 10n, 10e e) 20p, 20n, 18e 16 O 10 a) 8 40 Ar b) 18 23 Na + c) 11 32S 2− d) 16

Test yourself (page 18)

11 a) 4 b) 204, 206, 207 and 208 c) 1 : 16 : 11 : 34

Test yourself (page 21)

12 a) 12 b)2 c) 8 d) 3 e) 4 13 a) silicon-28: 14p, 14n; silicon-29: 14p, 15n; silicon-30: 14p, 16n b) Relative atomic mass = (.93 0×+ 28 )(. 5 0 ×+ 29 )(. 2 0 × 30 ) 100 = 28.1 14 a) The boiling temperature of neon-20 is lower than that of neon-22. Neon is a noble gas containing single atoms held together by weak intermolecular forces (here interatomic forces). On average, the energy required to separate and vaporise the lighter atoms of neon- 20 is slightly less than that for neon-22, so its boiling temperature is lower. ×+× b) Relative atomic mass of neon = (90 20 )( 10 22 ) 100 = 20.2 15 Isotopes have the same atomic number and therefore the same number of electrons and the same electronic structure. Chemical properties are determined by electronic structures, so isotopes have the same chemical properties. Isotopes have different mass numbers and therefore different masses. Physical properties, such as density, melting temperature and boiling temperature, are determined by mass, so isotopes have slightly different physical properties.

Test yourself (page 22)

16 a) 71.0 b) 256.8 c) 46.0 d) 154.0 17 a) 95.3 b) 159.6 c) 249.6 18 a) 58

b) CH3CH2− and CH3CH2CH2−

c) CH3CH2CH2CH3

Test yourself (page 25)

19 a) Ca+(g) → Ca2+(g) + e− b) Al2+(g) → Al3+(g) + e− 20 a) 4

b) For the first ionisation energy, an electron is being removed from a neutral atom. For the second and subsequent ionisation energies, an electron is being removed from an ion with an increasing positive charge. The process of removing an electron becomes more difficult and therefore more endothermic. c) The diagram should show two electrons in a lower level and two in a higher level. d) Group 2

Test yourself (page 28)

21 The sketch graph should show the first five electrons removed at low but slowly rising values of log ionisation energy, then eight electrons at significantly higher but slowly rising values of log ionisation energy, and finally two electrons at a significantly higher value, again with the final value of log ionisation energy slightly higher than the penultimate one.

22 a) 2, 1 b) 2, 6 c) 2, 8 d) 2, 8, 4

Test yourself (page 29)

23 a) 1s22s1 b) 1s22s22p4 c) 1s22s22p6 d) 1s22s22p63s23p2 24 a) b) c) d) 25 a) helium b) carbon c) magnesium d) sulfur

Test yourself (page 31)

26 a) 1s22s22p63s23p63d14s2 b) 1s22s22p63s23p63d54s2 c) 1s22s22p63s23p63d104s2

d) 1s22s22p63s23p63d104s24p2 27 a) calcium b) nickel c) germanium 28 a) 1s22s22p6 b) 1s22s22p63s23p6 c) 1s22s22p63s23p63d10 d) 1s22s22p63s23p63d104s24p6

Test yourself (page 32)

29 Because they have the same number of electrons in the outermost shell. 30 Noble gases have completely filled sub-shells of electrons and, in the case of helium and neon, completely filled shells. Filled shells and sub-shells are associated with higher stability and this results in the unreactive nature of the noble gases 31 a) Fluorine: 2, 7 and 1s22s22p5; chlorine: 2, 8, 7 and 1s22s22p63s23p5 b) Both fluorine and chlorine require one electron to fill their outer shell. When metals react they give up electrons. So, fluorine and chlorine react vigorously with metals taking the electrons which metals are prepared to give up in order to fill their outer shells. c) When fluorine and chlorine react with metals, they form ions, F– and Cl− both of which have one negative charge. These ions combine with positive metal ions and form compounds with

similar formulae such as NaF and NaCl, CaF2 and CaCl2.

Test yourself (page 34)

32 As atomic number increases in any group of the periodic table, the nuclear charge increases (tending to increase the first ionisation energy), the diameter of the atoms increase (tending to reduce the first ionisation energy) and shielding from filled inner shells of electrons increases (tending to reduce the first I.E.). Overall, the influence of the larger diameter and increased shielding outweighs that of larger nuclear charge

Activity: Mass spectrometry in sport (page 23)

1 301

2 Yes, because the relative molecular mass of dihydrocodeine (C18H23O3N) = (18 × 12) + (23 × 1) + (3 × 16) + 14 = 301 3 Relative mass of fragment lost = 17

4 The fragment lost with a relative mass of 17 could be −OH, and the fragment lost with a relative

mass of 31 (leaving a fragment of relative mass 270) could be CH3O−.

Activity: Evidence for subshells of electrons (page 26)

1 The graph should be like that in Figure 1.29 in the Student’s Book. The graph should have axes with scales and labels, points plotted accurately and clean straight lines drawn from one point to the next. 2 The points on the graph should be labelled with the symbols of the corresponding elements as in Figure 1.29. 3 a) In the lowest troughs of the graph. b) At the peaks on the graph. 4 There is a 2-3-3 pattern with secondary peaks after the second and fifth element in each period. 5 There are 2, 3 and 3 elements in the three sub-groups in each case.

Exam practice questions (pages 36-8)

1 a) The relative atomic mass of an element is the average mass of an atom of the element [1] relative to one-twelfth the mass [1] of a carbon-12 atom. [1] b) Relative atomic mass of antimony 57.. 3 42 7 = ×+ 121 × 123 [1] 100 100 = 69.33 + 52.52 = 121.85 [1] = 122 [1] to 3 significant figures c) Similarity: either same number of protons or same number of electrons. [1] Difference: different number of neutrons. [1] 2 a) F– and Ne [2] b) O2– and Ne [2] c) Na [1] 3 a) i) s-block [1] ii) p-block [1] iii) d-block [1]

b) i) [1]

ii) [1] [1]

iii) [1]

4 a) Isotopes are atoms with the same atomic number [1] but different mass numbers. [1] b) Protons Neutrons Electrons 24Mg 12 12 12 [1] 26Mg 12 14 12 [1]

c) 1s22s22p63s2 [1] 5 a) Melting temperatures increase towards Group 3 or 4 [1] and then fall to low values in Groups 5, 6, 7 and 0. [1] b) The elements change from metals on the left to non-metals on the right [1] of the periodic table. (Note that a more detailed answer is possible in terms of the ideas about structure and bonding in Chapter 2.) c) A property which has a repeating pattern [1] from left-to-right across each period of the periodic table. [1] d) 2 marks for any two properties, e.g. boiling temperature, atomic radius, ionisation energy, formulae of compounds, and so on. 6 a) Na+(g) → Na2+(g) + e– [2] b) The single outer electrons of lithium and sodium atoms are shielded from the full attraction of the atomic nucleus by inner full shells. [1] So the effective nuclear charge is 1+ and the first electron is relatively easy to remove. [1] The second electron has to come from a full shell and there is little or no shielding effect for electrons in the same shell so the nuclear attraction for the electrons in this shell are much larger. [1] c) Sodium has an additional electron shell compared to lithium. [1] The nuclear charge is greater in sodium [1], but the outermost electron is further from the nucleus [1] and the shielding effect is greater in sodium. [1] The increased distance of the electron from the nucleus and the increased shielding effect outweigh the increased nuclear charge [1] and therefore the ionisation energies in sodium are less than those of lithium. (Any 4 of the above 5 points.) d) i) Element X is Al. [1] ii) The increases in ionisation energy from first to second and second to third ionisations are relatively small compared with the very large increase from the third to the fourth ionisation. [1] This suggests there are 3 electrons in the outermost shell of element X. [1] e) Helium. Ionisation energies rise from left to right across each period [1] and fall down each group [1], suggesting that the element that is top-right in the table has the highest first ionisation energy. [1] (Electrons in a helium atom are in the innermost shell with no shielding from the 2+ nuclear charge.)

7 a) The general trend from Na to Ar is an increase [1] in first ionisation energies. From Na to Ar, the nuclear charge is increasing [1], the atomic radius is decreasing [1] and the shielding of inner shells is almost constant. [1] The increasing nuclear charge and decreasing atomic radius result in a stronger attraction on the outermost electron [1] and therefore an increase in the first ionisation energy. (Any 3 of the above points.) b) Electron configurations: Mg – 1s22s22p63s2; Al – 1s22s22p63s23p1 [1] In Al the 3p electron that is lost is further from the nucleus than the 3s electrons and slightly shielded from the nuclear charge by the 3s electrons. [1] c) In a sulfur atom there are two electrons in a p orbital whereas in phosphorus they are all unpaired. [1] Repulsion between the paired p electrons slightly reduces the first ionisation energy despite the increasing nuclear charge. [1] d) Value at around, or below 450 kJ mol–1. [1] First ionisations energies fall down each group so the value for K must be below the value for Na. [1] e) The first very large increase in ionisation energy comes between the second and third ionisations. This means that the first two electrons are relatively easy to remove. [1] So, the element is in Group 2, but the element cannot be Be, which has only four electrons. [1] So, the first possible element with these ionisation energies would be Mg with an atomic number of 12. [1] 8 a) i) The two lines are for ionised bromine atoms, Br+, [1] so the m/z values correspond to the mass numbers. [1] The two isotopes are equally abundant so the two peaks are the same height. [1] ii) These peaks correspond to ionised bromine molecules. [1] There are four possible combinations: 79Br79Br (158), 79Br81Br, 81Br79Br (two ways to get 160) and 81Br81Br (162). [1] Thus the 160 combination is twice as probable as the other two given that the isotopes are equally abundant. [1] b) i) The two lines are for ionised chlorine atoms, Cl+, so the m/z values correspond to the mass numbers. The ratio of abundances for 35Cl : 37Cl = 3 : 1. [1] So the two peaks are in the same ratio and separated by two units. [1] ii) These peaks correspond to ionised chlorine molecules. There is a 3 in 4 chance that any particular chlorine atom is 35Cl, so there is a ¾ × ¾ = 9/16 chance that both atoms will be 35Cl. There is a ¾ × ¼ = 3/16 chance that the molecules will be 35Cl37Cl and the same chance that it will be 37Cl35Cl; and a ¼ × ¼ = 1/16 chance that both atoms will be 37Cl. Hence the 9 : 6 : 1 ratio.

c) i) In a molecular ion of C2H2Cl2 with two chlorine atoms the possibilities are: most likely two atoms of chlorine-35 (96); next one of chlorine-35 and one of chlorine-37 (98); and

least likely two atoms of chlorine-37 (100). [1] The adjacent peaks differ by two mass units [1] and the expected ratio is 9 : 6 : 1. [1] ii) These two peaks are fragments formed by the loss of one chlorine atom [1] leaving a single chlorine atom in the fragment [1], so the ratio of abundances corresponds to the ratio of the abundance of the isotopes. [1] 9 a) In Period 4, the sub-shells fill in the order: 4s > 3d > 4p. So the 3d orbitals do not start to fill until there are two electrons in the s orbital. The same is true in Period 5. [1] Hence the d- block elements lie between the s- and p-blocks for Periods 3 and above. [1] b) Sn: 1s22s22p63s23p63d104s24p64d105s25p2, or [Kr]4d105s25p2 [1] c) Element 88 is radium with the electron configuration [Rn]7s2. [1] The next orbitals to fill are the 6d and 5f, which are not included in the diagram. [1] d) 7 orbitals [1] Either: the periodic table shows that there are 14 lanthanide elements, which corresponds to adding 2 electrons to each of 7 orbitals [1]. Or: The pattern is 1 × s, 3 × p, 5 × d [1] so next in the sequence is 7f. e) The higher the main shell, the more the energy levels converge. [1] The configurations suggest that the energies of the 4f and 5d orbitals are almost equal. [1] However, the diagram shows that in the main the 4f orbital fills before the 5d. [1] Hence the 4f fill from Ce to Lu placing the lanthanides between La and Hf in the periodic table. [1] 10 a) Neon and argon are the Group 0 elements in Periods 2 and 3. These elements are both noble gases [1] which have full outer shells of electrons so that they do not form chlorides. [1] b) i) From left to right across each period, the number of Cl atoms combining with one atom of the element rises from 1 in Group 1 to 4 in Group 4 [1] and then falls to 1 in Group 7. [1] ii) Metals to the left of the period form ionic chlorides by losing the outer electrons of their

atoms to form 1+, 2+ and then 3+ ions, hence the formulae NaCl, MgCl2 and AlCl3. [1] The non-metals to the right form covalent bonds. [1] The number of shared electrons is such that the outer shell is full with bonding or lone pairs. [1] Halogens form one covalent bond, Group 6 elements two and Group 5 elements three. [1] c) Because the formulae of binary compounds are written with the more electronegative element second. [1] d) i) There is a general pattern in the boiling temperatures of the chlorides of elements in Periods 2 and 3 in that those of the elements in Groups 1 and 2 in each period have high boiling temperatures [1], whereas those of the elements in Groups 4 to 7 have low boiling temperatures. [1]

ii) Group 1 and 2 metals form ionic chlorides [1] with strong bonding throughout the giant structures. [1] Groups 4–7 form molecular chlorides [1] with weak intermolecular forces between the molecules. [1] e) The outer electrons of a nitrogen atom are in the second main shell which only has one s orbital and three p orbitals. [1] This shell can accommodate a maximum of four pairs of

electrons. [1] In NCl3 there are three bonding pairs of electrons and one lone pair. [1] The outer electrons of a phosphorus atom are in the third main shell which has d orbitals in addition to s and p orbitals. [1] The five 3d orbitals can accommodate further bonding pairs of electrons and there can be more than eight electrons in the outer shell of a phosphorus atom. [1] 11 This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answers taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). a) Indicative content: • Atomic number shows the number of protons in the nucleus; hence the number of electrons and the electron configuration. • The atomic number determines the position of the element in the periodic table. • The chemistry of an element is strongly related to the periodic group it belongs to and also the row in the table in which it belongs. • However, the atomic number does not show how many neutrons there are in the nucleus. • The relative atomic mass is only a rough guide to the number of neutrons because this varies depending on the number of isotopes. • Relative atomic mass is an important quantity when determining formulae and reacting masses but it is of less value than atomic number as a guide to chemistry. • There are examples in the periodic table where ordering elements in terms of relative atomic mass puts elements in the wrong group (for example Te and I). b) Indicative content: • Elements in the same group have similar properties because they have the same outer electron configuration. • Elements in the same group do not have identical properties. • There are trends down groups – the number of inner full shells affects the chemical characteristics of an element. • Across a period there is shift in properties from metals on the left to non-metals on the right.

• This reflects the effect of the increasing nuclear charge as the number of electrons in inner full shells stays the same. • This increasing effective nuclear charge across a period means that electrons are held more strongly to the atom. • Across a period the atoms have a declining tendency to lose electrons and form positive ions but greater tendency to gain electrons and form negative ions.

Test yourself (page 40)

1 a) Giant structures: boron, silicon, manganese Simple molecular structures: fluorine, sulfur, iodine b) Simple molecular structures: fluorine, sulfur, iodine 2 a) Cubic b) In a cubic pattern

Test yourself (page 42)

3 a) b)

c) d)

4 Cs+, Sr2+, Ga3+, Se2−, At− 5 Metals have only 1, 2 or 3 electrons in their outer shell. It is much easier to lose a few electrons and obtain a more stable electron structure than to gain 7, 6 or 5 electrons respectively to achieve a structure like the next noble gas. By losing electrons metals form positive ions. Many non-metals have 5, 6 or 7 electrons in their outer shell. In forming ions, it is easier for them to gain 3, 2 or 1 electrons respectively and form a stable ion rather than to lose 5, 6 or 7 electrons. By gaining electrons, the non-metals form negative ions.

Test yourself (page 43)

6 a) 4 b) 6 c) 6 d) Because there are 6 Cl− ions nearest to (around) each Na+ ion and 6 Na+ ions nearest to each Cl− ion. e) The sodium and chloride ions are touching, but the sodium ions are not touching each other and the chloride ions are not touching each other. The strength of electrostatic forces depends on the distance between the ions, so the attractive forces are stronger than the repulsive forces.

Test yourself (page 45)

7 a) Na+F−, Mg2+O2− b) The electron structures of Na+ and Mg2+ are the same and the two ions have approximately the same radii. Similarly, the electron structures of F− and O2− are the same and these two ions also have approximately the same radii. However, the charges on Mg2+ and O2− are twice as large as those on Na+ and F−. This means that the ionic bonding in MgO is much stronger than that in NaF and so its melting temperature is higher. 8 a) Cathode (−) 2K+(l) + 2e− → 2K(l)

− − Anode (+) 2Br (l) → 2e + Br2(g) b) Cathode (−) Mg2+(l) + 2e− → Mg(l)

− − Anode (+) 2Cl (l) → 2e + Cl2(g)

− 9 The purple colour is due to the manganate(vii) ion, MnO4 . This ion moves in the solution on the filter paper towards the oppositely charged (+) terminal. The potassium ion moves in the opposite direction but is colourless and is not seen.

Test yourself (page 49)

10 a) A covalent bond b) A high density of electrons in the space between the nuclei of the two hydrogen atoms. This is clear evidence for a covalent bond. 11 a) b) c)

12 a) b) c)

d) e)

13 a) N has one lone pair. b) O has 2 lone pairs. c) F has 3 lone pairs. d) Each O has 2 lone pairs.

14 a)

b) BF readily reacts with ammonia because the B 3 atom in BF has only 6 electrons in its outer shell. 3 NH molecules have a lone pair of electrons which 3 form a dative covalent bond to the B atom in BF 3 and give it a more stable electron structure. The product is H N → BF . 3 3

Test yourself (page 50)

15 Iodine molecules in are in layers. In each layer the molecules are angled in an alternating pattern. 16 a) Bromine atoms are larger than chlorine atoms. The larger bromine has more electrons which shield the nuclei and reduce the attraction between the nuclei and the bonding electrons. b) The stronger, shorter bond between the chlorine atoms will require more energy to break. 17 Two pairs of shared electrons in O=O bond reduce the repulsion between the nuclei, which can then remain closer together.

Test yourself (page 53)

18 In covalent bonds, atoms share a pair of electrons between their nuclei. This means that atoms always remain in the same position relative to each other and therefore molecules have a definite shape. So, the covalent bonds between atoms point in a fixed direction and they are described as directional. 19 a) b) c)

20 a) Trigonal bipyramid; 90° and 120° b) Tetrahedral; 109.5° c) Trigonal planar; 120°

21 a) b) c)

Tetrahedral; 109.5° Tetrahedral; 109.5° Octahedral; 90° and 180°

Test yourself (page 55)

22 a) b)

Pyramidal; 107° ‘See-saw’ (distorted tetrahedron); 173° and 102° c) d)

Linear; 180° Linear; 180° 23 a)

Tetrahedral; 109.5°

− + b) AlH4 and PH4

Test yourself (page 56)

24 a) b)

Linear Linear c) d)

V-shaped (bent) Tetrahedral 25 a)

b) Tetrahedral c) The C−H bond is much shorter than the C−Cl bond. Consequently, the electron density in the C−H bond is greater than that in the C−Cl bond and repulsion from the electrons is greater than that in the C−Cl bond. So, the C−H bonds push each other further apart and they each move closer to the C−Cl bond.

Test yourself (page 58)

26 a) H−S N−O C−Cl I−Cl b) From most to least polar: C−F > C−O > C−Cl > C−Br > C−I > C−H

27 a) Na2O > MgO > Al2O3 > SiO2 b) CsI > KI > NaI > LiI

28 a) 2Fe + 3Cl2 → 2FeCl3 b) Iron 3+ is a highly charged ion so will polarise the chloride ion leading to covalency. In dry conditions iron(iii) chloride forms covalent molecules so will sublime when heated. c) Iron 2+ is less polarising so iron(ii) chloride is essentially ionic and cannot sublime. 29 This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answers taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Points to make: Strengths Strong electrostatic forces between oppositely charged ions, therefore: • high melting temperature

• solubility in polar solvents such as water • conductivity when molten or dissolved. Weaknesses Polarisation by small highly charged positive ions leads to partial covalency, so that: • some metal compounds, e.g. aluminium chloride, sublime/easily vaporise • some compounds, e.g. silver halides, are insoluble.

Test yourself (page 60)

30 Polar: HBr, CHBr3 and SO2, the rest non-polar.

31 CCl4 and CO2 both have polar bonds but the symmetrical shape of the molecules means that the bond polarities cancel and overall the molecules are non-polar. Drawing the shapes of the other molecules and highlighting the polar bonds shows that the polar bonds do not cancel each other out. There is a net polarity. See Figures 2.29 and 2.32 in the Student’s Book. 32

Oxygen is more electronegative than hydrogen so in water, O is δ− and H is δ+. Fluorine is more electronegative than oxygen so in oxygen difluoride, O is now δ+.

Test yourself (page 62)

33 The molecules in noble gases are single atoms. The number of electrons in a noble gas atom is given by its atomic number. Boiling temperatures rise from helium (2 electrons) to xenon (54 electrons), suggesting that intermolecular forces increase from He to Xe. 34 a) Chlorine, bromine and iodine consist of non-polar, diatomic molecules. The number of electrons per molecule increases from Cl to I so iodine molecules are much more polarisable than chlorine molecules. This results in melting and boiling points which rise down the group. b) Astatine has more electrons than iodine so will be a solid at room temperature.

Test yourself (page 64)

35 a) Both ethane and fluoromethane are molecules with 18 electrons. Ethane is non-polar and the only intermolecular forces are London forces. Fluoromethane has a permanent dipole because of the polarity of the C−F bond. The extra attraction between permanent dipoles accounts for the higher boiling temperature of fluoromethane.

b) Butane (34 electrons) and propanone (32 electrons) molecules each contain about the same number of electrons. Butane molecules are non-polar and held together by London forces. Propanone molecules include the polar C=O bond. The extra attraction between permanent dipoles accounts for the higher boiling temperature. 36 ICl is polar so will have intermolecular dipole–dipole forces in addition to London forces. Bromine will only have London forces so will have a lower boiling temperature. 37 a)

b) Heptane 98.4 °C; 3-methylhexane 92.0 °C; 2,2-dimethylpentane 79.2 °C The molecules, as isomers, have the same number of electrons. The differences lie in the extent of chain branching. The more branching the more compact the molecule and the smaller the area over which London forces can act, hence the lower the boiling temperature.

Test yourself (page 66)

38 a)

39 a) Boiling temperatures of hydrogen halides

b) HF is the only one of the compounds affected by hydrogen bonding. Its boiling temperature is much higher than expected by the general trend. HCl, HBr and HI all have permanent dipoles. The polarity of the H–X bond decreases down the group so dipole–dipole forces decrease down the group. This would suggest that the boiling temperatures should fall from HCl to HI. However, the number of electrons per molecule increases down the group, so HI is more polarisable than HBr and HBr more than HCl and therefore London forces increase from HCl to HI. The increase in London forces has more effect on the boiling temperatures than the decrease in dipole–dipole forces. 40 None of the hydrides in Group 4 is affected by hydrogen bonding. The trend in boiling points down the group is as expected for a series of compounds with similar structures but increasing numbers of electrons per molecule. The molecules become more polarisable down the group and so London forces increase. There is a similar trend for Group 6 hydrides apart from water, which is out of line because of the much stronger hydrogen bonding between water molecules. 41 a) Hydrogen bromide: London forces and permanent dipole–dipole attractions b) Propane: London forces c) Methanol: London forces and hydrogen bonding.

Test yourself (page 69)

42 Non-polar methane molecules cannot break into the hydrogen-bonded network of water molecules. The strength of the hydrogen bonds keeps the methane molecules out. Ammonia can hydrogen bond with itself and also with water molecules so ammonia can dissolve in water. 43 Iodine molecules are non-polar and are held together by weak London forces. Cyclohexane molecules are the same. London forces between iodine molecules and cyclohexane molecules are about the same strength as the forces in the separate chemicals. So iodine molecules can separate and mix with cyclohexane. Non-polar iodine molecules cannot break into the hydrogen- bonded network of molecules in water. 44 The −OH group in methanol can hydrogen bond with water molecules. So methanol molecules can break into the hydrogen-bonded structure of water by forming new hydrogen bonds between methanol and water molecules. In decan-1-ol the carbon chain is non-polar and is a much larger part of the molecule than the O−H group. Therefore relatively little hydrogen bonding is possible with water and this is insufficient to cause solubility. 45 Very soluble: caesium fluoride, lithium chloride Soluble: magnesium chloride, magnesium sulfate, potassium iodide Slightly soluble: calcium hydroxide, calcium sulfate, lithium fluoride Insoluble: barium sulfate

Test yourself (page 70)

46 Carbon dioxide is a non-polar covalent molecule. The London forces between molecules are weak and easily overcome so carbon dioxide is easily turned to gas. Silicon dioxide is a giant covalent lattice (see Figure 2.48). This lattice is held together by strong covalent bonds between the atoms. These bonds are difficult to break so the melting point of silicon dioxide is very high.

47 A planar representation makes it easier to see the repeating SiO2 units.

Test yourself (page 74)

48 a) Graphite from the pencil is rubbed onto the zip. This acts as a lubricant and the zip moves more freely. b) Oil might leave greasy patches on the garment/cloth to which the zip is attached. Graphite would not cause the same mess. c) Machinery can get very hot with the constant contact of moving parts. Graphite has a very high melting point and does not burn until very high temperatures. Oil, on the other hand, may evaporate at high temperatures or even catch fire. 49 a) Possible answers are oxygen and ozone; red, white and black phosphorus or monoclinic and rhombic sulfur. b) A composite combines two or more materials to produce a material which has the desirable properties of its constituents. For example, reinforced concrete has steel rods in the concrete. This combines the strength and flexibility of the steel with the hardness of the concrete. (Wattle-and-daub combines the flexibility and long support from twigs, branches or straw with the hard filling of dried clay or mud.) 50 a) All the outer shell electrons of the carbon atoms in diamond are held in strong covalent bonds. It has no delocalised electrons which might allow it to conduct electricity. However, the rigidity of the strong covalent bonds in diamond means that thermal energy in the vibrating atoms can travel rapidly through the giant structure. b) The delocalised electrons can move throughout the sheets of graphene, but in fullerenes the electrons can only move about in one molecule and cannot transfer to another.

Test yourself (page 75)

51 Because they are not fixed (‘localised’) to a particular atom but can move freely throughout the metal structure. 52 a) 3 b) 3 c) 12 53 a) Most metals have a close-packed structure in which the atoms are as close together as possible. b) The structures of metals consist of a giant lattice of positive ions in a ‘sea’ of delocalised electrons. When the metal is connected to a battery, the delocalised electrons are attracted to the positive terminal and flow through the metal as an electric current.

Test yourself (page 76)

54 Possible illustrations are: a) Aluminium foil in cooking b) Copper wires for electrical appliances c) Silver and gold jewellery d) Steel cables in cranes 55 a) i) Transition metals ii) Metals in Groups 1, 2 and 3 b) i) Transition metals ii) Group 1 metals

Activity: Identifying and explaining the trends in ionic radii (page 46)

1 Ionic radii increase with increase in atomic number down the group. From one element to the next, the elements have an additional filled shell of electrons. 2 Yes, because down every group in the periodic table, each element has one more filled shell of electrons than the element above it. 3 a) 1s22s22p6 b) 1s22s22p6 c) 1s22s22p6 d) 1s22s22p6 4 a) ‘Isoelectronic’ means having the same electron structure. b) The ionic radii decrease from N3− to Al3+.

c) From N3− to Al3+, the nuclear charge is increasing from 7 protons to 13 protons but the electron structure of all the ions is identical. The increasing nuclear charge attracts the electrons more strongly, pulling them closer, and the ionic radii decrease.

Activity: Interpreting electronegativity values (page 58)

1 He and Ne are noble gases that do not form compounds with other elements. They do not form covalent bonds. 2 The two non-metals with near equal electronegativities are carbon and sulfur. The covalent bonding in carbon disulfide is effectively non-polar. (Potassium and rubidium would form an alloy with metallic bonding.) 3 Rb and F 4 Values of electronegativity increase from left to right across a period. 5 a) Electrons in main shells: Li 2, 1 and F 2, 7. b) The nuclear charge increases from 3+ to 9+ while the number of shielding electrons stays the same. So electrons in the outer, bonding shell of F are attracted much more strongly by the nucleus than the electrons in the bonding shell of Li. 6 Electronegativity values decrease down a group. 7 a) Electrons in main shells: F 2, 7 and Cl 2, 8, 7. b) The increase in nuclear charge from 9+ in F to 17+ in Cl is effectively cancelled out by the corresponding increase in the number of shielding electrons. The electrons in the bonding shell for Cl are further from the same effective nuclear charge of 7+ than the bonding electrons in F so the bonding electrons in Cl atoms are less strongly attracted by the nucleus.

Activity: Intermolecular forces and the properties of alkanes (page 63)

1 Unbranched alkanes: C1 to C4 are gases; C5 to C10 (and up to C17) are liquids; and after C18 they are solids. 2 After butane the approximate increase in boiling temperature is 20–30 °C for each carbon added to the chain up to dodecane. 3 The boiling temperature of dodecane is 216 °C. 4 London forces act between non-polar alkane molecules. 5 The two key features are the number of electrons and the area of contact between molecules. 6 Branching lowers the boiling points of isomeric alkanes. 7 Branched isomers are more compact and have a smaller area of contact between molecules.

8 Similarity: the trend is the same, the more carbon atoms the higher the melting or boiling temperature. Difference: the melting temperatures are lower but the main difference is that the melting temperatures of alkanes with an odd number of carbon atoms fall on a lower trend curve than alkanes with an even number of carbon atoms. 9 This can be explained by suggesting that alkanes with an odd number of carbon atoms pack together less well in the solid state and so are easier to break apart. The alkanes with an even number of carbon atoms have molecules that pack together better so the structure is very regularly organised and requires more energy to break it apart. 10 Polythene can be regarded as a very long chain alkane. Not all the molecules in the polymer are

the same length and the chains may not end in −CH3. The chains are so long that London forces operate over a very large surface area. Also the chains are tangled up together.

Activity: Choosing metals for different uses (page 76)

1 a) Copper is a better conductor of electricity than aluminium, but aluminium has a much lower density than copper so pylons in the National Grid can be much further apart. b) Bridges require a large amount of very strong material. Iron is very cheap relative to other metals and, although its tensile strength is marginally less than other metals, it forms a very strong alloy with small proportions of carbon and other metals as steel. c) Zinc protects the iron in steel underneath even if the surface is scratched and the iron is exposed. d) Nowadays, silver is too costly. e) The tensile strength of aluminium is low. Its alloy with titanium is much stronger and the alloy’s density is still relatively low. f) Copper is a much better conductor of heat than iron. 2 a) Students should plot a graph of density against melting temperature. This should have clear axes with scales and labels and points plotted accurately. b) No, there does not appear to be a relationship as it is not possible to plot a line or smooth curve through the points. 3 a) Students should plot a graph of electrical conductivity (reciprocal of electrical resistivity) against thermal conductivity. b) Yes, there appears to be a relationship as it is possible to draw a smooth curve through, or close to, all the points on the graph.

Activity: Structure, bonding and physical properties (page 77)

1 Completed Table 2.11: The four types of solid structure and some properties.

2 Ions in a lattice are held in position by the strong electrostatic forces and are unable to move about. When the compound is molten or in solution in water, the ions can move towards the electrodes. 3 Group 1 metals have lower melting points from Li (454 K) down to Cs (302 K). These metals have only one electron per atom in their delocalised cloud whereas transition metals have many more s electrons and d electrons. Therefore the metallic bonding in Group 1 metals is much weaker than in transition metals. 4 In covalent compounds all the electrons are used in bonding. There are neither delocalised electrons as in metals, nor charged particles as in ionic compounds so covalent compounds do not conduct electricity. Some covalent molecular compounds such as hydrogen chloride or ammonia react with water to form ions. Their solutions can therefore conduct electricity as the ions formed can move towards the electrodes. 5 Graphite conducts electricity as it contains delocalised electrons. 6 *The actual identity of the substances is not expected, only the type of bonding and structure.

Exam practice questions (pages 79-82)

1 a) Bonding between atoms in a metal is the result of electrostatic attractions between positive metal ions in a lattice [1] and delocalised electrons in the outer shell of the metal atoms. [1] In sodium, this results from ions with a charge of 1+ and one delocalised electron in each atom. In magnesium, the bonding results from ions with a charge of 2+ and two delocalised electrons in each atom. The bonding between atoms is therefore weaker in Na [1] and its melting temperature is much lower.

b) P4 and S8 are relatively simple molecules with weak forces of attraction. However, S8 is larger

and heavier than P4 [1] and it requires higher temperatures before its molecules have sufficient energy to move away from fixed positions [1] in the solid lattice. c) i) Aluminium – giant metallic structure [1]; bonding involves the attraction of positive ions for delocalised electrons. [1] ii) Silicon – giant covalent structure [1]; bonding involves the attraction of positive nuclei for shared electrons [1] in covalent bonds. [1] iii) Chlorine – simple molecular structure [1]; bonding involves the attraction of positive nuclei for shared electrons [1] in covalent bonds between atoms plus weak intermolecular forces. [1] (Any 6 points) 2 a) i) Bonding in calcium involves the electrostatic attraction of positive ions [1] in a giant lattice for the delocalised electrons [1] in the outermost shell of the calcium atoms. [1] ii) Calcium contains delocalised electrons. [1] When calcium is connected to a battery, these delocalised electrons are attracted to the positive terminal [1] of the battery, forming an electric current. b)

[1] for both correct charges, 2 × [1] for each electron structure, [1] for two electrons of Ca transferred to O. c) CaO conducts electricity when molten (liquid). [1] Ca2+ [1] ions are attracted to the negative electrode from which they take electrons and form Ca metal. [1]

2– O [1] ions are attracted to the positive electrode where they give up electrons and form O2 gas. [1] The giving of electrons to the positive electrode and taking of electrons from the negative electrode makes an electric current. [1]

3 a) Hydrogen chloride has covalent bonding. [1] The hydrogen and chlorine atoms share a pair of electrons [1] in a covalent bond. Dot-and-cross diagram. [1] Sodium chloride has ionic bonding. [1] Sodium ions, Na+ and chloride ions Cl– are held in a giant lattice [1] by the attraction between oppositely charged ions. [1] Copper has metallic bonding. [1] This involves the attraction of positive copper ions [1] in a giant lattice for delocalised electrons. [1] b) Sodium chloride does not conduct when solid but will conduct when liquid (molten) or dissolved in water. [1] Under these conditions the ions are no longer held in a lattice and can move through the sodium chloride to the electrodes. [1] In copper, delocalised electrons move through the metallic lattice. [1] c) In copper, the metallic lattice structure means layers of identical atoms move over each other without repulsion. [1] In sodium chloride, the ionic structure means that if one layer of ions is moved on impact the distance of only one ion diameter with respect to another layer [1], then like charges will be adjacent and will repel [1], causing the layers to separate. 4 a)

[1] [1]

+ – b) i) NH3(g) + H2O(l) NH4 (aq) + OH (aq)

[1] for products,⇌ [1] for state symbols ii) A dative covalent bond is a bond in which two atoms share a pair of electrons [1], both of the electrons being contributed by one atom. [1] iii)

[1]

5 a) (trigonal) pyramidal [1] 107° [1] (based on tetrahedral electron arrangement) [2]

b) (square) pyramidal [1] 90° [1] (based on octahedral electron arrangement)

[2]

c) pyramidal [1] 107° [1]

[2]

d) tetrahedral [1] 109.5° [1]

[2]

6 a)

[2] b)

[1] Pyramidal [1]; bond angle 107° [1]

c) There are four pairs of electrons around the phosphorus atom [1] (one lone pair and three shared pairs). These four pairs repel each other as far as possible, taking up tetrahedral

positions around the phosphorus atom [1] and leading to a pyramidal shape for atoms in PCl3. The angle is reduced from 109.5° to 107° by the extra repulsion of the bonding pairs by the lone pair. [1]

d) The lone pair of electrons on the phosphorus atom in PCl3 [1] donates this pair in a dative

covalent bond [1] to the boron atom in BCl3, giving the boron atom a full second shell of electrons. [1] e) Both P and B now have 4 bonding pairs in their outer shell [1] so both have tetrahedral arrangement [1] leading to a bond angle of 109.5°. [1] 7 a) i) London forces. [1] ii) London forces, hydrogen bonding. [1] b) Propane and ethanol are molecules of about the same size and molar mass so the contribution from London forces is about the same for both molecules. [1] Hydrogen bonds are a much stronger type of intermolecular force than London forces. [1] c) Glycerol, with three −OH groups, has much more extensive hydrogen bonding between its molecules than ethanol. All three −OH groups can take part in hydrogen bonding. [1] This extra hydrogen bonding makes it difficult for the molecules to move over each other in the liquid, which is very viscous. [1] It is also harder to separate the glycerol molecules from each other so the boiling temperature of glycerol is higher than that of ethanol. [1]

8 CF4 is tetrahedral [1] and therefore non-polar. [1]

SF4 is distorted tetrahedral (often called ‘see-saw’) [1] with four bonding pairs and a lone pair, so is polar. [1]

XeF4 is square planar [1] and therefore non-polar. [1] 9 a) F is more electronegative than Cl so F−Cl is polar [1] but in a fluorine molecule both atoms are the same so the molecule is not polar. [1]

b) There is a lone pair of electrons on the S atom in SO2 so the molecule is not linear. Hence it is polar because the S=O bonds are polar. [1]

CO2, with just two double bonds, is linear and the polarities of the two bonds cancel. [1]

c) There is a lone pair of electrons on the N in NCl3 so the molecule is pyramidal. With three polar N−Cl bonds the molecule is overall polar. [1]

With no lone pairs on BCl3 the molecule is planar and symmetrical. Hence it is not polar despite the polarity of the B−Cl bonds. [1]

10 a) Hydrogen bonding [1]

A lone pair on the oxygen atom [1] is attracted to the very electron-deficient hydrogen atom [1] on another molecule. b) i) Water is the only one of the compounds affected by hydrogen bonding. Its boiling temperature is much higher than expected. [1] The trend in boiling temperatures for the other hydrides in the group is as expected for a series of compounds with similar structures but increasing numbers of electrons per molecule. The molecules become more polarisable down the group and so London forces increase. [1] ii) Molecules in ice are held together by hydrogen bonding. The molecules form a giant lattice structure in which each oxygen atom is bonded to two hydrogen atoms by covalent bonds and two others by hydrogen bonds. This gives rise to an open lattice. [1] As ice melts the hydrogen-bonded structure collapses bringing the water molecules closer together. So the melt water is denser than the ice. [1] iii) Non-polar pentane molecules [1] cannot break into the hydrogen-bonded network of molecules in water. [1] iv) The intermolecular forces in methoxymethane are London forces and attractions between permanent dipoles. [1] The hydrogen bonding between ethanol molecules is much stronger. [1] 11 a) Allotropes are different forms of the same element [1] in the same state. [1] b) Yes, because they are another solid form of carbon. [1] c) i) Any two from: graphite fibres will conduct electricity; graphite fibres act as lubricants maintaining smooth contact; graphite fibres are flexible and will not break as the contacts open and close. [2] ii) Diamonds do not conduct electricity. [1] Diamonds would be too expensive. [1] Diamonds would scratch any surface with which they made contact. [1]

12 a)

[2] b) i) Trigonal planar [1] ii) Pyramidal [1] c) Any two of: urea provides higher percentage of nitrogen; urea would not result in nitrate contamination of the soil; urea would not cause eutrophication like nitrate; urea is not as

soluble as NH4NO3, so it is not washed out of the soil so readily. [2] 13 a)

[2] b) Allow 106°–108°. [1] The lone pair will repel more strongly than the shared pairs of electrons [1] around the N atoms. So, the H−N−H bond angle will be slightly less than 109.5° [1], similar to that in ammonia.

c) N2H4(g) + O2(g) → N2(g) + 2H2O(g) Formulae of reactants and products [1]; correctly balanced with state symbols [1]

(Allow 2H2O(l). Allow equations in which the products are H2O and N2O or NO or NO2.) d) The molar mass of hydrazine = 32.0 g mol–1

∴ When 1 mol N2H4 burns completely, the enthalpy change of combustion of hydrazine = −(18.3 kJ g−1 × 32.0 g mol−1) = −585.6 kJ mol–1 = −586 kJ mol−1 (3 s.f.) [1] for 586, [1] for sign and units. 14 a) [1]

b) H–C–H is 109.5° [1]; tetrahedral electron arrangement all bonding pairs. [1] C–N–H is 107° [1]; tetrahedral electron arrangement three bonding pairs and one lone pair. [1] Lone pair repels more strongly than bonding pairs so reduces the bond angle. [1]

c) [1]

Lone pair and dotted hydrogen bond [1]; 180° [1]

+ − d) i) CH3NH3 Cl [1] ii) 109.5° [1]; tetrahedral electron arrangement all bonding pairs. [1]

15 ICl3 has five electron pairs so its structure is based on the trigonal bipyramidal electron

arrangement [1] like PF5.

But there are three bonding pairs and two lone pairs, so there are three possible arrangements: A Both lone pairs vertical B One lone pair vertical one C Both lone pairs horizontal horizontal

[1] [1] [1]

Repulsions Repulsions Repulsions lp–lp 1 × 180° lp–lp 1 × 90° lp–lp 1 × 120° lp–bp 6 × 90° lp–bp 2 × 120° & 3 × 90° lp–bp 4 × 90° & 2 × 120° bp–bp 3 × 120° [1] & 1 × 180° bp–bp 2 × 90° & 1 × 180° bp–bp 1 × 120° & 2 × 90° [1] [1]

The shape with the minimum overall repulsions is (C) where the atoms form a ‘T-shaped’ molecule. [1] Although the lone pairs are not as far apart as in shape (A), the planar molecule, the lone-pair bonding pair repulsions are less in (C) than (A) and this is favoured. [1]

16 This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answers taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Points to make, and illustrate, in an answer: • Group 7 elements are non-polar molecules. • Down the group the number of electrons in the molecule increases. • Therefore the London forces between molecules increase. • Group 1 elements are metals. • Down the group, the number of delocalised electrons per atom remains the same but the size of the atoms increases. 17 a) Polar molecules are attracted to the charged rod. The tiny dipoles orientate themselves so that their opposing charge is pointing towards the charged rod. [1] Non-polar molecules have no dipoles and are not attracted by the charged rod. [1] b) Polar molecules that are deflected towards the charged rod: water and bromoethane. Hexane and tetrachloromethane are non-polar molecules overall so they are not deflected 4 × (answer + prediction) for [1] each; total [4] c) Polar molecules always orientate themselves so that the region of the molecules with the opposite charge to the charged rod is closer to the rod. [1] This means that attractive forces are always larger than repulsive forces between the rod and the molecules. [1] 18 Substance a) Bonding b) Identity A Covalent molecular [1] 1-Bromobutane [1] B Metallic lattice [1] Manganese [1] C Ionic lattice [1] Sodium bromide [1] D Giant covalent [1] Silicon dioxide [1] E Covalent molecular [1] Hydrogen bromide [1] F Ionic lattice [1] Aluminium oxide [1] G Mobile metal atoms in liquid [1] Mercury [1]

19 a) Both metals have ions with a charge of 1+ and one delocalised electron in each atom. [1] But sodium is a smaller ion than potassium [1] so the attraction for the delocalised electrons is stronger [1] so more energy is required to separate the ions. [1] b) Oxide ion has greater charge than chloride [1], so force of attraction between ions is greater [1] so more energy is required to separate the ions. [1]

c) Weak forces between chlorine molecules are easily overcome [1] so little energy is needed to boil chlorine. [1] The covalent bonds between chlorine atoms are strong [1] so a high temperature (or UV light) is required to provide the required energy. [1] d) The bonding in pure aluminium chloride is largely covalent [1] because Al3+ polarises chloride

ions. [1] An AlCl3 molecule contains an Al atom with only 6 electrons in its outer energy level. [1] This vacant orbital can accept a pair of electrons from a chlorine atom forming a dative covalent bond. [1]

20 a)

Calculation of molar mass. [1] Axes [1]; points correctly plotted. [1] Good use of scale. [1]

b) The main intermolecular forces between the molecules of H2S and H2Se are permanent dipole–dipole attractions. [1]

–1 c) Extrapolating the graph gives a value of about 18.5 kJ mol for the ∆Hvaporisation of water from intermolecular forces other than hydrogen bonding. [1] d) So the contribution of hydrogen bonding is about 22 kJ mol–1. [1] 21 This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answers taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Points to make, and illustrate, in an answer: • the C–Cl bonds are polar • the C–C single bond in 1,2-dichloroethane can rotate so the C–Cl bonds will not necessarily be opposite

• therefore the molecule has an overall dipole • the C=C bond in 1,2-dichloroethene cannot rotate • therefore the C–Cl bonds in the E-isomer will be opposite so there is no overall dipole • and the C–Cl bonds in the Z-isomer will not be opposite so there is an overall dipole

22 a) i) 2Na(s) + H2(g) → 2NaH(s) [1] for equation; [1] for symbols ii)

[1] for each ion. b) Hydrogen is discharged at the positive electrode. [1]

– − 2H → H2 + 2e [1]

− − c) H + H2O → H2 + OH [1] The hydride ion is acting as a proton acceptor or a base. [1]

2+ + d) The smaller size and larger charge makes Mg more polarising than Na [1], so MgH2 has even

more covalent character than NaH. [1] Heating will decompose MgH2 into its elements [1] or treatment with water:

MgH2 + 2H2O → Mg(OH)2 + 2H2 [1] 23 a) i) Despite the polar COOH group, the large non-polar benzene ring [1] limits the possible interactions with water molecules and the ability of the molecule to break into the hydrogen bonded network of water molecules. [1] ii) [1]

The COOH groups cannot hydrogen bond to the non-polar solvent, [1] but they can form hydrogen bonds to another COOH. [1] The dimer then disperses throughout the non- polar solvent. [1]

b) i) C6H5COOH + NaOH → C6H5COONa + H2O [1]

−1 ii) Mr = 122.0 g mol [1] 2.90 g 2.90 g acid = = 0.02377 mol [1] 122.0 g mol−1 0.02377 mol Volume of 0.500 mol dm−3 NaOH required = 0.500 mol dm−3 = 0.0475 dm3 or 47.5 cm3 [1] iii) The salt formed is ionic [1] Hydration of the ions provides sufficient energy to overcome the lattice energy of the ionic compound. [1]

24 a) Ammonia and water contain hydrogen bonding [1], but the electronegativity of oxygen is greater than nitrogen [1] so the hydrogen bonding in water is stronger than that in ammonia. [1] Methanol contains only one O–H bond per molecule and water contains two O–H bonds. [1] Although there are two lone pairs on each oxygen [1], fewer hydrogen bonds can be formed. [1] b) Hydrogen bonding alone would suggest that the acid should have the higher boiling temperature, so extra interactions must be present. [1]

The lone pair on the N can be delocalised [1] (towards the more electronegative oxygen). This leads to an ionic structure [1] which will exert stronger electrostatic forces on similar species. [1] 25 Tin fluoride: Sn F 61.0 = 0.514 39.0 = 2.05 118.7 19.0

Empirical formula = SnF4 [1] Tin iodide: Sn I 19.0 = 0.160 81.0 = 0.638 118.7 127

Empirical formula = SnI4 [1]

Electronegativity values Sn = 1.8 F = 4.0 difference = 2.2 probably ionic [1] Electronegativity values Sn = 1.8 I = 2.5 difference = 0.7 probably covalent [1]

Melting temperature of SnF4 is quite high so likely to be ionic. [1]

Melting temperature of SnI4 is lower so likely to be molecular (tetrahedral shape so non-polar molecule). [1] Sn4+ is very polarising. [1]

− F is small and not easily polarised [1] so SnF4 is mainly ionic.

− I is large and is easily polarised to give a covalent compound SnI4. [1]

Test yourself (page 83)

1 Oxidised Reduced a) magnesium water (steam) b) hydrogen copper (ii) oxide c) aluminium iron (iii) oxide d) carbon carbon dioxide

Test yourself (page 84)

2 a) H2O(g) + Mg(s) → H2(g) + MgO(s)

b) CuO(s) + H2(g) → Cu(s) + H2O(l)

c) 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

d) CO2(g) + C(s) → 2CO(g)

Test yourself (page 86)

3 a) 2Na → 2Na+ + 2e–

– – Cl2 + 2e → 2Cl b) 2Zn → 2Zn2+ + 4e–

– 2– O2 + 4e → 2O c) Ca → Ca2+ + 2e–

– – Br2 + 2e → 2Br 4 2Ag+(aq) + 2e– → 2Ag(s) Zn(s) → Zn2+(aq) + 2e– 2Ag+(aq) + Zn(s) → 2Ag(s) + Zn2+(aq)

Test yourself (page 88)

5 a) +3 b) –3 c) +5 d) –3 6 a) Oxidised b) Reduced c) Oxidised d) Reduced e) Oxidised

Test yourself (page 89)

7 a) Li2O

b) CO and CO2 c) In order of the formulae: +2, +1, +4, +3, +5

d) NH3 and PH3

e) Only fluorine is more electronegative than oxygen. Oxygen is more electronegative than all other elements except fluorine.

Test yourself (page 90)

8 a) SnO

b) SnO2

c) Fe(NO3)3

c) K2SO4

Test yourself (page 91)

9 a) Carbon(+2) disproportionates to carbon(0) and carbon(+4). b) Manganese(+6) disproportionates to manganese(+4) and manganese(+7). c) Chlorine(0) disproportionates to chlorine(−1) and chlorine(+1).

Test yourself (page 92)

+ – 10 a) H2O2(aq) + 2H (aq) + 2e → 2H2O(l) Fe2+(aq) → Fe3+(aq) + e–

+ 2+ 3+ H2O2(aq) + 2H (aq) + 2 Fe (aq) → 2Fe (aq) + 2H2O(l)

2– 2– + – b) SO3 (aq) + H2O(l) → SO4 (aq) + 2H (aq) + 2e

– – Cl2(aq) + 2e → 2Cl (aq)

2– 2– + – SO3 (aq) + H2O(l) + Cl2(aq) → SO4 (aq) + 2H (aq) + 2Cl (aq)

– + – – c) IO (aq) + 2H (aq) + 2e → I (aq) + H2O(l)

– – + – IO (aq) + 2H2O(l) → IO3 (aq) + 4H (aq) + 4e

– – – 3IO (aq) → 2I (aq) + IO3 (s)

Test yourself (page 93)

10 a) 2Fe(s) + 3Br2(l) → 2FeBr3(s) 2 Fe oxidised from 0 to +3 (up 6). Six Br reduced from 0 to −1 (down 6).

b) 2F2(g) + 2H2O(l) → 4HF(g) + O2(g) Four F reduced from 0 to −1 (down 4). Two O oxidised from −2 to 0 (up 4).

– + – c) IO3 (aq) + 6H (aq) + 5I (aq) → 3I2(s) + 3H2O(l) One I in iodate ions reduced from +5 to 0 (down 5). Five I in iodide ions oxidised from −1 to 0 (up 5)

2− – 2− d) 2S2O3 (aq) + I2(aq) → 2I (aq) + S4O6 (aq) Two I reduced from 0 to −1 (down 2). Four S oxidised from +2 to an average of +2.5 (up 2).

Activity: Preparing a sample of an oxide of nitrogen (page 94)

1 a) Pb +2, N +5, O –2 b) Pb +2, O –2 c) N +4, O –2 d) N +4, O –2 e) O 0 2 a) Neither b) Reduction c) Oxidation d) Neither

3 a) 2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)

b) 2NO2(g) → N2O4(g)

4 1 mol Pb(NO3)2 gives 2 mol NO2 which can condense to form N2O4. 20 Theoretical yield = × 92 g = 5.6 g 331 5 Possible explanations: not all the lead nitrate decomposes, some of the product escapes through the side arm of the test tube without condensing.

Exam practice questions (pages 95-6)

+ − 1 a) 2H (aq) + 2e → H2(g) [1] b) Fe3+(aq) + e− → Fe2+(aq) [1]

+ − c) H2O2(aq) + 2H (aq) + 2e → 2H2O(l) [1] 2 a) Mg(s) → Mg2+(aq) + 2e− [1] b) Sn2+(aq) → Sn4+(aq) + 2e− [1]

− − c) 2I (aq) → I2(aq) + 2e [1]

3+ 2+ 2+ 4+ 3 a) 2Fe (aq) + Sn (aq) [1] → 2Fe (aq) + Sn (aq) [1]

+ 2+ b) Mg(s) + 2H (aq) [1] → Mg (aq) + H2(g) [1]

+ − c) H2O2(aq) + 2H (aq) + 2I (aq) [1] → 2H2O(l) + I2(aq) [1] 4 Oxidation numbers: −1, +1, +3, +5, +7 (3 marks for all 5, lose one for each mistake) 5 Oxidation numbers: 0, −3, −2, +5, +3, −1, +3 (4 marks for all 7, lose one for each mistake) 6 a) Co2+(aq) → Co3+(aq) + e− [1] (oxidation) [1]

+ − b) SO2(aq) + 6H (aq) + 6e → H2S(g) + 2H2O(l) (reduction) [1]

− − c) 4OH (aq) → O2(g) + 2H2O(l) + 4e [1] (oxidation) [1]

+ − d) H2(g) → 2H (aq) + 2e [1] (oxidation) [1] 7 a) Oxygen disproportionates [1] from −1 to −2 and 0. [1] b) Chlorine disproportionates [1] from 0 to −1 and +1. [1]

c) Manganese disproportionates [1] from +6 to +7 and +4. [1] 8 a) 2Fe2+(aq) → 2Fe3+(aq) + 2e− [1] (oxidised)

− − Br2(aq) + 2e → 2Br (aq) [1] (reduced) [1]

− − b) 2I (aq) → I2(aq) + 2e [1] (oxidised)

− − Cl2(aq) + 2e → 2Cl (aq) [1] (reduced) [1] c) Zn(s) → Zn2+(aq) ) + 2e− (oxidised) [1] 2V3+(aq) + 2e− → 2V2+(aq) (reduced) [1]

9 a) 3CuO(s) + 2NH3(g) [1] → N2(g) + 3H2O(l) + 3Cu(s) [1]

b) 8KI(s) + 5H2SO4(l) [1] → 4K2SO4(s) + 4I2(s) + H2S(g) + 4H2O(l) [1]

c) NaIO3(aq) + 5NaI(aq) + 3H2SO4(aq) [1] → 3I2(aq) + 3H2O(l) + 3Na2SO4(aq) [1]

d) Cu(s) + 4HNO3(aq) [1] → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) [1] 10 a) Br oxidised (–1 to 0) [1] S reduced (+6 to +4) [1]

2HBr(g) + H2SO4(l) → Br2(l) + SO2(g) + 2H2O(l) [1] b) Fe oxidised (+2 to +3) [1] Mn reduced (+7 to +2) [1]

− + 2+ 2+ 3+ MnO4 (aq) + 8H (aq) + 5Fe (aq) → Mn (aq) + 5Fe (aq) + 4H2O(l) [1] c) Cr oxidised (+3 to +6) [1]

O in H2O2 reduced (–1 to –2) [1]

3+ − 2− 2Cr (aq) + 10OH (aq) + 3H2O2(aq) → 2CrO4 (aq) + 8H2O(l) [1] d) Cr reduced (+6 to +3) [1] S oxidised (+4 to +6) [1]

2− + 3+ 2− Cr2O7 (aq) + 2H (aq) + 3SO2(aq) → 2Cr (aq) + H2O(l) + 3SO4 (aq) [1]

e) N in N2H4 oxidised (−2 to 0) [1]

N in N2O4 reduced (+4 to 0) [1]

2N2H4(g) + N2O4(g) → 3N2(g) + 4H2O(g) [1] 11 Definitions of the terms: ‘oxidation’ (gain of oxygen, loss of hydrogen, loss of electrons, oxidation number becoming more positive) and ‘reduction’ (loss of oxygen, gain of hydrogen, gain of electrons, oxidation number becoming less positive) [4] a) Up to 4 marks for any key points such as:

H2S → S: sulfur oxidised from −2 to 0 state. [1] Older definition of loss of hydrogen applies. [1] Half-equation to show loss of electrons. [1]

SO2 → S: sulfur reduced from +4 to 0 state. [1] Definition of loss of oxygen applies. [1] Half-equation to show gain of electrons. [1]

b) Up to 4 marks for any key points such as:

− H2 → H : hydrogen reduced from 0 to −1 state. [1] This gives hydrogen in an unusual oxidation state. It is acting as an oxidising agent, not a reducing agent. [1] Half-equation to show gain of electrons. [1] Na → Na+: sodium oxidised from 0 to +1 state. [1] Older definition of gain of hydrogen does not apply. [1] Half-equation to show loss of electrons. [1] c) Up to 4 marks for any key points such as:

H2O2 → H2O: oxygen reduced from −1 to the −2 state. [1] Oxygen has an unusual oxidation state in peroxides. Because atoms of the element are bonded to each other [1] Definition of reduction in terms of loss of oxygen suggests that hydrogen is reduced when it does not change oxidation state [1] Half-equation to show gain of electrons. [1]

− I → I2: iodide oxidised from −1 to 0 state. [1] Older definitions in terms of oxygen and hydrogen do not apply. [1] Half-equation to show loss of electrons. [1] d) Up to 4 marks for any key points such as: This is an example of disproportionation [1]

H2O2 → H2O: as in (c) [1]

H2O2 → O2: oxygen oxidised from −1 to the 0 state. [1] Definition of oxidation in terms of loss of hydrogen could be applied. [1] Half-equation to show loss of electrons. [1] 12 This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. a) Up to 4 marks for key points such as: Sulfite: S is +4. [1] Four outer electrons used for bonding to a more electronegative atom. [1] One lone pair not involved in bonding. [1] Sulfate: S is +6. [1] All six outer electrons used for bonding. [1] b) Indicative content: Thiosulfate: • Applying the normal rules to the thiosulfate ion, without reference to structure, gives an average value of +2. • However, many chemists now regard the thiosulfate ion as analogous to the sulfate ion with all six outer electrons of sulfur involved in bonding and assign the central S atom to the +6 state and the other S atom to the −2 state. Tetrathionate: • The average value for the oxidation state of S in tetrathionate is +2.5. • However, the presence of an S–S bond suggests that the two S atoms involved should be assigned oxidation state −1 by analogy with the –O–O– situation in peroxides.

• In which case the oxidation states of the other two S atoms are +6 corresponding the use of 6 outer electrons in bonding as in the sulfate ion. • The oxidation number rules are, to some extent arbitrary, and work so long as they are applied consistently. c) Indicative content:

2− + • The equation for the reaction is: S2O3 (aq) + 2H (aq) → SO2(aq) + S(s) + H2O(l) • This can be regarded as disproportionation … • … as S in the +2 state → S in the +4 and 0 states. • Alternatively this is seen as a redox reaction in which …. • … the sulfur atom in the +6 state is reduced to the +4 state … • … as it oxidises the other sulfur atom from the −2 to the 0 state.

Test yourself (page 97)

1 a) Thermal decomposition and redox: chlorine reduced from +5 to −1; oxygen oxidised from −2 to 0. b) Thermal decomposition and redox: nitrogen reduced from +5 to +4; oxygen oxidised from −2 to 0. c) Thermal decomposition: no changes of oxidation state. d) Thermal decomposition and redox: chromium reduced from +6 to +3; nitrogen oxidised from −3 to 0.

Test yourself (page 99)

+ − 2 a) Hydrogen ions, H (aq) and nitrate ions NO3 (aq)

+ 2− b) Hydrogen ions, H (aq) and sulfate ions, SO4 (aq)

3 a) Zn + H2SO4 → ZnSO4 + H2

b) CaO + 2HNO3 → Ca(NO3)2 + H2O

c) Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

+ 2− 2+ 2− 4 a) Zn + 2H + SO4 → Zn + SO4 + H2

+ 2+ Zn + 2H → Zn + H2

2+ 2− + − 2+ − b) Ca O + 2H + 2NO3 → Ca + 2NO3 + H2O

2− + O + 2H → H2O

+ 2− + − + − c) 2Na + CO3 + 2H + 2Cl → 2Na + 2Cl + CO2 + H2O

2− + CO3 + 2H → CO2 + H2O

Test yourself (page 100)

2+ 2− 5 a) Barium sulfate: Ba (aq) + SO4 (aq) → BaSO4(s) b) No precipitate

2+ 2− c) Calcium carbonate: Ca (aq) + CO3 (aq) → CaCO3(s)

2+ − d) Lead(ii) chloride: Pb (aq) + 2Cl (aq) → PbCl2(s)

2+ − e) Copper(ii) hydroxide: Cu (aq) + 2OH (aq) → Cu(OH)2(s 6 a) Precipitation b) Thermal decomposition c) Redox d) Acid–alkali

Test yourself (page 101)

7 a) [Ne]2s1 b) [Ar]2s1 8 The charge on the nucleus increases and the number of inner full shells also increases. The shielding effect of the inner electrons means that the effective nuclear charge attracting the outer electron is 1+. Down the group the outer electrons get further and further away from the same effective nuclear charge and so they are held less strongly and the ionisation energies decrease. 9 The loss of the outer shell makes each ion smaller than its corresponding atom.

Test yourself (page 103)

11 a) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)

– + OH (aq) + H (aq) → H2O(l)

b) Na2CO3(aq) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l) + CO2(g)

2– + CO3 (aq) + 2H (aq) → H2O(l) + CO2(g)

12 a) Li2CO3(s) → Li2O(s) + CO2(g)

b) 4LiNO3(s) → 2Li2O(s) + 4NO2(g) + O2(g)

Test yourself (page 104)

13 a) Redox b) Thermal decomposition and redox c) Redox d) Thermal decomposition

Test yourself (page 106)

14 Mg: 1s22s22p63s2 Mg2+: 1s22s22p6 Ca: 1s22s22p63s23p64s2 Ca2+: 1s22s22p63s23p6 Sr: 1s22s22p63s23p63d104s24p65s2 Sr2+: 1s22s22p63s23p63d104s24p6 15 Consider Period 3. Na+ and Mg2+ have the same electron configuration. The charge on the nucleus of a magnesium ion is greater than the charge on the nucleus of a sodium ion, so the electrons in the magnesium ion are held more strongly. 16 The ionic radii increase as the number of full shells in the M2+ ions increases down the group from Be2+ (1s2) to, for example, Sr2+ (1s22s22p63s23p63d104s24p6).

Test yourself (page 107)

17 Oxidation state: −1

18 a) 2Sr(s) + O2(g) → 2SrO(s)

b) Ba(s) + O2(g) → BaO2(s)

c) Sr(s) + Cl2(g) → SrCl2(s)

d) Ba(s) + 2H2O(g) → Ba(OH)2(s) + H2(g)

Test yourself (page 108)

19 a) MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)

b) CaO(s) + H2O(l) → Ca(OH)2(s)

c) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l

Test yourself (page 110)

20 a) One possibility is to have a test tube containing solid metal carbonate with bung and delivery tube dipping into a second tube containing limewater. b) On heating, a few bubbles flow through the limewater as the air expands. Bubbling continues with magnesium carbonate and the limewater turns cloudy as carbon dioxide is given off. Barium carbonate does not decompose. Calcium carbonate only decomposes on very strong heating with a Bunsen flame and signs of decomposition are likely to be limited if the solid is heated in a test tube.

21 a) MgCO3(s) → MgO(s) + CO2(g)

b) MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(g)

c) 2Ca(NO3)2(s) → 2CaO(s) + 4NO2(g) + O2(g)

d) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l

22 2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g) Nitrogen reduced from state +5 to state +4. Oxygen oxidised from state −2 to 0. 23 A calcium ion has a larger charge (2+) and a smaller radius than a potassium (K+) ion. This means that the charge density of a calcium ion is greater than that of a potassium ion. So a calcium ion has a greater polarising power and hence a greater tendency to distort the distribution of bonding electrons in a carbonate ion.

Test yourself (page 112)

24 a) Gas: small molecules with few electrons so weaker London forces than, for example, chlorine molecules. b) Solid: larger molecule with more electrons than an iodine molecule so stronger London forces. 25 a) Cl: 1s22s22p63s23p5 b) Cl–: 1s22s22p63s23p6 c) Br: 1s22s22p63s23p63d104s24p5 d) Br–: 1s22s22p63s23p63d104s24p6 26 a) The number of full inner shells increases down the group b) The atoms gain an electron to form negative ions. The extra electron slightly increases the radius of the outer shell. 27 a) b) c)

Test yourself (page 113)

28 Non-polar molecules, such as the halogen molecules, mix freely with the molecules of non-polar solvents such as hexane. Easy mixing is possible where the intermolecular (London) forces between solute molecules, between solvent molecules and between solute and solvent molecules are all about the same strength. The intermolecular forces in water are much stronger hydrogen bonds. Halogen molecules cannot take part in hydrogen bonding so it acts as a barrier which keeps out the halogen molecules.

29 a) Mg(s) + Br2(l) → MgBr2(s) oxidation numbers: Mg from 0 to +2; Br from 0 to −1

b) 2Fe(s) + 3Cl2(g) → 2FeCl3(s) oxidation numbers: Fe from 0 to +3; Cl from 0 to −1

c) Fe(s) + I2(s) → FeI2(s) oxidation numbers: Fe from 0 to +2; I from 0 to −1 30 a) Chlorine gas is highly toxic. Any excess gas leaving the apparatus must not be released into the laboratory. b) Water vapour mixed with the chorine reacts with iron(iii) chloride as it forms, hydrolysing the product. c) Wool has a larger surface area than lumps so the reaction is faster. d) The product sublimes. It condenses to a solid and deposits in the receiver. 3) The drying agent prevents water vapour in the air getting into the receiver and reacting with the product.

Test yourself (page 114)

31 The reactions with fluorine and chlorine can be explosive; the reaction with bromine vapour goes smoothly with a steady flame; the reaction with iodine is not complete and has to be carried out by heating under pressure. 32 Formula of the product HF. The reaction is highly exothermic.

33 SiCl4 is molecular. The molecule has the shape of a regular tetrahedron and so, despite the polar bonds, it is overall non-polar. The Si–Cl bonds are strong but the forces between the molecules are weak: too weak to hold the molecules in the solid state at room temperature but strong enough to prevent them evaporating.

34 a) 2P(s) + 3Cl2(g) → 2PCl3(s)

2P(s) + 5Cl2(g) → 2PCl5(s) P in oxidation state zero oxidised to oxidation states +3 or +5.

– – b) Cl2(g) + 2I (aq) → 2Cl (aq) + I2(s) Cl in oxidation state zero reduced to oxidation state –1.

− – 35 Br2(aq) + 2e (aq) → 2Br (aq) 2Fe2+(aq) → 2Fe3+(aq) + 2e−(aq)

2+ – 3+ Br2(aq) + 2Fe (aq) → 2Br (aq) + 2Fe (aq)

Test yourself (page 116)

36 The apparatus can consist of a small flask fitted with a bung holding a tap funnel and delivery tube. Sulfuric acid from the tap funnel drips onto solid sodium chloride in the flask. The gas produced by the reaction does not need drying. HCl is denser than air so it can be collected from a delivery tube dipping vertically downwards to the bottom of a test tube.

37 2NaBr(s) + 3H2SO4(l) → 2NaHSO4(s) + Br2(l) + SO2(g) + 2H2O(l)

Test yourself (page 117)

38 a) Ag+(aq) + I–(aq) → AgI(s) b) Ag+(aq) + Br–(aq) → AgBr(s) 39 a) The colourless solution turns to the orange-brown colour of bromine. b) The orange colour of the bromine disappears. The colourless solution of potassium iodide

− turns to yellow-brown as iodine is displaced and remains in solution in excess as I3 ions. As the reaction continues, and more of the iodide ions are oxidised, grey specks of insoluble iodine appear.

40 Strength as reducing agents: iodide > bromide > chloride. Iodine is the least reactive element so has the least tendency to turn into iodide ions. So iodide ions are the ones with the greatest tendency to turn back into molecules. Chlorine is the most reactive element so has the greatest tendency to turn into chloride ions. So chloride ions are the ones with the least tendency to turn back into molecules. 41 Fluorine is sufficiently electronegative for there to be hydrogen bonding between HF molecules. Hydrogen bonding is a stronger type of intermolecular forces than the intermolecular forces between the other hydrides which are London and dipole–dipole forces.

+ + 42 a) NH3 + H → NH4 HBr → H+ + Br− b) The two reactants are gases with weak intermolecular forces. The product is an ionic compound with strong electrostatic forces between the ammonium ions and bromide.

Test yourself (page 118)

– – – 43 a) 3I2 + 6OH → IO3 + 5I + 3H2O Iodine starts at oxidation state 0. Five iodine atoms going from 0 to −1 oxidise 1 iodine atom from 0 to +5. b) Iodine is both reduced from 0 to −1 and oxidised from 0 to +5.

Activity: Water treatment (pages 118-9)

+ – 1 Cl2(aq) + H2O(l) → HClO(aq) + H (aq) + Cl (aq) 2 Liquid chlorine is highly corrosive and the gas is very toxic. Transporting, storing and using the cylinders is hazardous. 3 Chloric (i) acid is a weak acid which means that it has very little tendency to ionise. Chlorate(i) ions react with water. They take hydrogen ions from water to form chloric(i) acid. 4 As shown by the graph, the concentration of HClO falls as the pH rises. If the pH rises above 7.8 the concentration of HClO is too low to kill all bacteria. Above pH 7.8 the water can be irritating to the eyes of swimmers. 5 If too acid, the water can be irritating to the eyes of swimmers. Also if the pool water becomes too acid it can corrode the pipes through which the circulating water flows. It can also start to etch concrete surfaces.

6 a) NH3(aq) + HClO(aq) → NH2Cl(aq) + H2O(l)

b) NH2Cl(aq) + NHCl2(aq) → N2(g) + 3HCl(aq)

c) NH2Cl(aq) + 2HClO(aq) → NCl3(g) + 2H2O(l)

7 The swimming pool smell of chlorine is NCl3, which only forms if the water is over-chlorinated with the chemical that forms HClO.

Exam practice questions (pages 120-2)

1 a) Atomic radius increases down the group [1] because of the increasing number of inner full shells of electrons. [1] The more full shells the larger the atom. [1] b) First ionisation energies decrease down the group [1] because the shielding effect of inner shells means that in all the atoms the effective nuclear charge for the outer electrons is the same [1] but down the group the outer electrons get further away and so are held less strongly. [1] c) Thermal stability of carbonates increases down the group. [1] The greater the polarising power of the metal ion the more unstable the carbonate. [1] Down the group the radii of the 2+ ions increase so their polarising power decreases. [1]

– – – 2 a) Cl2(aq) + 2OH (aq) → ClO (aq) + Cl (aq) + H2O(l) [1] Chlorine (oxidation state 0) is both oxidised to the +1 state and reduced to the −1 state. [1]

– – – b) 3ClO (aq) → ClO3 (aq) + 2Cl (aq) [2]

– – – c) 4ClO3 (s) → 3ClO4 (s) + Cl (s) [1] Chlorine in the +5 state is oxidised to the +7 state [1] while being reduced to the −1 state. [1] 3 a) X is sodium nitrate. Yellow flame indicates sodium. [1] Sodium hydroxide is soluble. [1] The gas is oxygen. [1] This could be from a nitrate decomposing to the nitrite. [1] b) Y is barium chloride. Green flame suggests barium. [1] The white precipitate with silver nitrate indicates chloride ions. [1] c) Z is potassium bromide. Mauve flame suggests potassium. [1] Potassium and bromide ions do not affect the pH of solutions. [1] The cream precipitate with silver nitrate indicates bromide ions. Silver bromide dissolves in concentrated ammonia. [1] Chlorine oxidises colourless bromide ions to orange bromine. [1]

4 a) Expect radium to burn brightly in oxygen forming a white oxide (or peroxide) [1] RaO/RaO2 [1]

2Ra(s) + O2(g) → 2RaO(s) or Ra(s) + O2(g) → RaO2(s) [1] b) Rapid reaction of radium with water producing a colourless gas, hydrogen [1] and an alkaline solution of the hydroxide. [1]

Ra(s) + 2H2O(l) → Ra(OH)2(aq) + H2(g) [1] c) Expect radium oxide to react exothermically with water and to produce a strongly alkaline

solution [1] of the hydroxide Ra(OH)2(aq). [1]

RaO(s) + H2O(l) → Ra(OH)2(aq) [1]

d) Expect radium hydroxide to neutralise dilute hydrochloric acid to produce a colourless

solution [1] of the chloride, RaCl2. [1]

Ra(OH)2(aq) + 2HCl(aq) → RaCl2(aq) + 2H2O(l) [1] e) Expect radium sulfate to be insoluble [1] because the trend is for the sulfates to become less soluble down the group and barium sulfate is insoluble. [1] f) Expect radium nitrate to need strong heating before it decomposes. Likely to behave similarly to barium or lithium and decompose to give the oxide, nitrogen dioxide and oxygen. [1] Nitrogen dioxide is a brown gas. Oxygen relights a glowing splint. [1]

2Ra(NO3)2 → 2RaO(s) + 4NO2(g)+O2(g) [1] 5 a) This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answer taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Points to make in an answer: • Solubility of hydroxides increases down the group. • Solubility of nitrates decreases down the group. • Solubility of carbonates decreases down the group. • The polyatomic nitrate and carbonate ions are relatively large … • … compared to the hydroxide ion, which consists of two relatively small atoms. • On the basis of these examples the generalisation is true. b) Fluorine is an element in the second period. Its atoms and ions are relatively small [1]. In line with the generalisation this means that solubility is expected to increase down the group from magnesium fluoride to barium fluoride. [1] 6 a) Expect astatine to be a solid at room temperature. [1] Iodine is a solid and astatine has larger molecules with more electrons so that it is more polarisable [1] giving rise to strong London forces between the molecules. [1] b) Expect chorine to oxidise astatide ions to astatine molecules [1] just as chlorine oxidises iodide ions. Oxidising power of the halogens decreases down the group. [1] c) Little or no hydrogen astatide expected to form. [1] The astatide ion is the strongest reducing agent of the halide ions [1] and can be expected to reduce concentration sulfuric acid to sulfur and hydrogen sulfide. [1] d) Silver astatide might be an even brighter yellow than silver iodide. [1] Trend in the solubility of AgX in concentrated ammonia solution decreases down the group so At will not dissolve. [1]

7 a) Adding aqueous chlorine displaces orange bromine from a solution of a bromide. [1] On adding aqueous chlorine to a solution of an iodide the mixture turns to a dark yellow-brown colour. [1] Adding a little of an organic solvent such as hexane to the mixtures helps to distinguish the results. Bromine is orange when dissolved in hexane. [1] Iodine is violet. [1]

− – Cl2(aq) + 2Br (aq) → 2Cl (aq) + Br2(aq) [1]

− – Cl2(aq) + 2I (aq) → 2Cl (aq) + I2(aq) [1] b) Silver nitrate gives a cream precipitate with a solution of a bromide and a yellow precipitate with a solution of an iodide. [1] Adding ammonia solution helps to distinguish the precipitates. [1] Silver bromide dissolves in concentrated aqueous ammonia solution. [1] Silver iodide does not dissolve. [1] Ag+(aq) + Br–(aq) → AgBr(s) [1]

Ag+(aq) + I–(aq) → AgI(s) [1] c) Adding concentrated sulfuric acid to the solid salts gives distinctive results. A bromide gives

off orange bromine vapours as the bromide ions reduce H2SO4 to SO2. [1] Some hydrogen

bromide gas forms too. Sulfur dioxide turns acid dichromate(VI) paper from orange to green. [1] Hydrogen bromide fumes in moist air. [1]

2NaBr(s) + 3H2SO4 [1] → 3NaHSO4 + Br2(l) + SO2(g) + 2H2O(l) [1] With an iodide the products are iodine vapour, hydrogen sulfide and sulfur. [1] Very little hydrogen iodide forms. The hydrogen sulfide can be detected by its foul smell. [1] 8 I− > Br− > Cl− This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answer taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Possible examples (one or the other or both): • Halide ions in solution reducing halogen molecules • Reactions of concentrated sulfuric acid with potassium halides. Key components of the answer:

• Description of procedure (to allow comparison of Cl–, Br– and I–) • Observations for tests • Equations for reactions • Interpretation in terms of reducing power to establish trend.

9 a) i) CaF2(s) + H2SO4(l) → CaSO4(s) + HF(g) [1] ii) Fluorine is so electronegative that the electrons in the H–F bond are drawn strongly towards the F atom. [1] This means that there is hydrogen bonding in H–F unlike the other hydrogen halides. [1] Hydrogen bonds are relatively strong intermolecular forces compared to the London forces and dipole–dipole attractions in the other hydrogen halides. [1] b) i) The reactions between chlorine and alkanes are substitution reactions. [1] The products are halogenoalkanes and HCl. [1] Halogenoalkanes are important chemical intermediates so they are manufactured on a large scale producing large amounts of HCl as a by-product. [1] ii) NaOH is made by the electrolysis of sodium chloride solution [1] which produces hydrogen at one electrode and chlorine at the other. [1] The two gases are easily purified (largely by removing water vapour). [1] iii) The reaction of hydrogen with chlorine is highly exothermic and can be explosive. [1] iv) Any suitable example [1] such as the manufacture of fine chemicals such as drugs, or in sensitive chemical analysis. c) Bromide ions are stronger reducing agents than chloride ions [1]; they reduce sulfuric acid to sulfur dioxide and are oxidised to bromine as they do so. [1] 10 a) The boiling temperature of beryllium chloride is very much lower than that of chlorides of the

other metals in the group. [1] BeCl2 vaporises much more easily than the other chlorides. [1] b) The ionic radius of beryllium (1s2) is much smaller than that of magnesium (1s22s22p6). [1] This means the beryllium ion has a very high polarising power. [1] This means that the bonding in beryllium chloride is substantially covalent. [1] It vaporises relatively easily to give

BeCl2 molecules. [1] c) There are only two electrons in the outer shell of a Be atom. [1] The atom forms two covalent bonds with chlorine atoms. [1] When it does so, there are no lone pairs in the outer shell of the Be atom, just two bonding pairs, so these arrange themselves as far apart as possible with a bond angle of 180°. [1] d) The beryllium chloride molecules align so that each chlorine atom forms a covalent bond with one beryllium atom and a dative covalent bond with another Be atom [1]. Thus each Be atom is linked to chlorine atoms by two covalent bonds and two dative covalent bonds. [1] There are no lone pairs on the Be atoms and so the four bonds are arranged tetrahedrally around the Be atom. [1]

[1]

11 By inspection, if x = 3 [1], one iodine atom is reduced from oxidation state +3 to 0 [1] while three iodide ion are oxidised to iodine [1]. There is no change to the oxidation state of chlorine.

− − ICl3 + 3I → 4I2 + 3Cl [1]

The reaction of iodine with chlorine: I2(s) + 3Cl2 → 2ICl3 [1] Iodine is oxidised. Chlorine is reduced. [1] 12 a) The grey precipitate is iodine displaced by chlorine. [1] The orange-brown fumes are bromine displaced by chlorine. [1] The mixture contains potassium bromide and potassium iodide. [1]

− − b) Cl2(aq) + 2Br (aq) → 2Cl (aq) + Br2(g) [1]

− − Cl2(aq) + 2I (aq) → 2Cl (aq) + I2(s) [1] c) The initial precipitate is a mixture of AgBr and AgI [1]. Treatment with concentrated ammonia dissolves the AgBr to leave only AgI. [1] Mass of AgI precipitated = 0.176 g

Amount of AgI precipitated = 0.176 g ÷ 234.8 g mol−1 = 0.000750 mol [1]

Mass of KI in the 0.214 g sample = 0.000750 mol × 166 g mol−1 = 0.124 g [1]

Percentage of KI in the sample = (0.124 g ÷ 0.214 g) × 100% = 57.9% [1] The rest of the sample is KBr, so its percentage in the sample = 42.1% [1] (The same calculation can be worked through on the basis of the mass of AgBr.)

Test yourself (page 125)

1 a) 0.50 mol b) 0.05 mol c) 1.00 mol d) 0.10 mol e) 0.25 mol 2 a) 12.69 g b) 17.75 g c) 36.0 g d) 0.535 g e) 12.0 g 3 a) 2 mol b) 1 mol c) 4 mol 4 a) 3.01 × 1023 b) 24.1 × 1023 = 2.41 × 1024 c) 54.2 × 1023 = 5.42 × 1024

Test yourself (page 127)

1 a) CH4 b) VCl4 c) Na2SO4

6 a) H2SO4 b) C2H6O

Test yourself (page 128)

7 a) Tb = 77 K

b) Tb = 272.5 K

c) Tm = 459 K

d) Tm = 1813 K

Test yourself (page 129) PV 100 000 Pa×× 200 10−63 m 8 Amount of gas = = = 0.00808 mol RT 8.31Jmol−−11 K× 298 K

Mass of gas = 0.356 g

0.356 g − Molar mass = = 44.1 g mol 1 0.00808 mol

PV 100 000 Pa×× 65 10−63 m 9 Amount of vapour = = = 0.00209 mol RT 8.31Jmol−−11 K× 374 K

Mass of sample = 0.163 g

0.163 g − Molar mass = = 78.0 g mol 1 0.00209 mol

Test yourself (page 131)

10 a) 10 mol 48 cm3 31− b) 24 000 cm mol = 2.0 × 10−3 mol 3 dm3 3 c) 24 dm = 0.125 mol 11 a) 48 000 cm3 b) 4.8 cm3 c) 3000 cm3 12 14.0 g 13 2.02 g 14 0.33 kg

Test yourself (page 134)

3 3 15 175 cm O2 is required, forming 100 cm CO2

3 16 0.01 mol Zn gives 240 cm H2

Test yourself (page 135)

17 a) 0.050 mol dm−3 b) 0.40 mol dm−3 c) 0.625 mol dm−3 50

18 a) 1000 × 2.0 mol = 0.1 mol H2SO4 ⇒ 9.81 g H2SO4 100

−3 b) 1000 × 0.01 mol = 10 mol KMnO4 ⇒ 0.158 g KMnO4 250

c) 1000 × 0.2 mol = 0.05 mol Na2CO3 ⇒ 5.3 g Na2CO3

Test yourself (page 136)

19 a) Useful in diagnosis to see if a patient has diabetes. Correct monitoring and treatment depends on accurate values. b) Useful to the police to see if they should prosecute for drink driving. Incorrect analysis could lead to the wrong decision about whether or not to prosecute. c) Useful to a mining company to assess whether an ore is worth mining. Investment in new mines is very costly.

d) Useful to those monitoring air pollution in cities to see if there is a risk to the health of the population. People with breathing problems refer to forecasts and websites to decide whether or not to go out on a particular day. 20 Sodium hydroxide pellets quickly pick up moisture from the air (they are deliquescent). This means that they cannot be weighed accurately. 21 Hydrated sodium carbonate gradually loses its water of crystallisation. A sample of the solid does not have a definite composition. Anhydrous sodium carbonate can be heated strongly to drive off any water. It is a stable compound which does not decompose on heating.

Test yourself (page 139)

22 a) Measure 50.0 cm3 of the acid solution into the flask, make up to the mark and mix well. b) Measure 10.0 cm3 of the NaOH(aq) into the flask, make up to the mark and mix well. 23 a) 0.00100 mol dm–3 b) 0.400 mol dm–3

Test yourself (page 142)

24 Methyl orange solution is red at, or below, pH 3.2; it is yellow at, or above, pH 4.2. At pH 3.7 the solution is about half-way through the colour change, so it contains roughly equal amounts of the red and yellow forms and so it looks orange. 25 0.108 mol dm–3 26 0.222 mol dm–3 27 The data shows that 1 mol of the acid reacts with 2 mol of NaOH.

H3PO3(aq) + 2NaOH(aq) → Na2HPO3(aq) + 2H2O(l)

Test yourself (page 142)

28 a) Pipette: • random – not aligning the bottom of the meniscus with the graduation • systematic – manufacturing tolerance such that the volume is not exactly as stated, not working at the temperature for which the glassware is calibrated, not allowing the pipette to drain under gravity.

b) Burette: • random – errors in taking readings from the bottom of the meniscus and the graduation scale, judging the end-point from the colour change • systematic – manufacturing tolerance such that the volume is not exactly as stated, not working at the temperature for which the glassware is calibrated. c) Making up a standard solution in a standard flask: • random – errors when dissolving the weighed solid and transferring all the solute to the flask after dissolving in a beaker; errors when making up the volume of solution to the graduation scale • systematic – manufacturing tolerance such that the volume is not exactly as stated, not working at the temperature for which the glassware is calibrated.

Test yourself (page 147)

29 1 mol Fe2O3 → 2 mol Fe

159.6 g Fe2O3 → 111.6 g Fe

1000 kg Fe2O3 → × 1000 kg Fe = 699 kg Fe Theoretical yield = 699 kg

Percentage yield = = 90%

30 1 mol CaO → 1 mol Ca(OH)2

56.1 g CaO → 74.1 g Ca(OH)2

500 kg CaO → × 500 kg Ca(OH)2 = 660 kg Ca(OH)2 Theoretical yield = 660 kg

Percentage yield = × 100 = 93.9%

Test yourself (page 148)

31 a) Atom economy = 100% There is only one product which makes up 100% of the molar mass of all the product. 2× 46 g mol−1 (92+ 88) g mol−1 b) Atom economy = × 100 = 51.1%

118. 7 g mol−1 (118. 7+ 56) g mol−1 c) Atom economy = × 100 = 68%

Test yourself (page 149)

32 a) Ionic precipitation to form silver iodide. Ag+(aq) + I–(aq) → AgI(s) b) Acid–base reaction forming magnesium chloride, carbon dioxide and water.

MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l) c) Ionic precipitation to form barium sulfate.

2+ 2− Ba (aq) + SO4 (aq) → BaSO4(s) d) Thermal decomposition and redox to form potassium nitrite.

2KNO3(s) → 2KNO2(s) + O2(g) e) Acid–base to form ammonium chloride.

HCl(g) + NH3(g) → NH4Cl(s) f) Redox to form bromine and potassium chloride solution.

− − Cl2(aq) + 2Br (aq) → 2Cl (aq) + Br2(aq) g) Acid–base to form ammonia gas, aqueous sodium chloride and water.

NH4Cl(s) + NaOH(aq) → NH3(g) + NaCl(aq) + H2O(l)

Activity: Finding the formula of red copper oxide (page 127)

1 Wear eye protection; make sure the combustion tube is filled with natural gas before lighting the excess natural gas; carry out the experiment behind a Perspex safety screen. (Mixtures of natural gas and oxygen in the air are explosive.) 2 Repeat the heating and weighing of the reduced product until two weighings are the same. 3 and 4

Experiment Mass of red Mass of Mass of Amount of Amount of number copper copper/g oxygen/g copper/mol oxygen/mol oxide/g 1 1.43 1.27 0.16 0.020 0.010 2 2.14 1.90 0.24 0.030 0.015 3 2.86 2.54 0.32 0.040 0.020 4 3.55 3.27 0.28 0.051 0.018 5 4.29 3.81 0.48 0.060 0.030

5 The graph should have clear axes with scales and labels, points plotted accurately and a clear thin straight line of best fit.

6 The point from experiment 4 7 a) 2.0 b) 2.0 mol

c) Cu2O 8 The student carried out the experiment five times with different quantities. Plotting a graph was a way of averaging the results and checking their consistency. The graph helps to identify the result that is out of line – and that should, ideally, be repeated. Four of the five experiments give a consistent value of 2.0 for the number of moles of copper combined with one mole of oxygen. The only variable in the experiments was the mass of red copper oxide used. 9 red copper oxide + methane → copper + carbon dioxide + water

4Cu2O(s) +CH4(g) → 8Cu(s) + CO2(g) + 2H2O(g)

Activity: Measuring the molar volume of three gases (page 130)

1 Step A is to check the syringe does not leak. This is particularly important in steps C and D. 2 At the end of step C the plunger has been pulled out but no gas has entered the syringe, so the syringe contains a vacuum. The pressure of the air outside the syringe would immediately force the plunger back into the syringe if it were not held in place with the nail. 3 The syringe is surrounded by a fluid, air. In many experiments the effect of buoyancy is so small that it does not affect the results, but here the masses of gases being weighed are so small that the buoyancy effect is significant because the mass of the air displaced by 50 cm3 of apparatus is similar to the masses of gas being measured. This means that the total volume of syringe, bung and nail must be the same for all measurements. 4 Carbon dioxide can be generated by the action of dilute hydrochloric acid on marble chips. The gas can be dried with a suitable drying agent such as anhydrous calcium chloride or concentrated sulfuric acid. The gas must be dry otherwise it is mixed with water vapour. Water molecules have a different molar mass from that of the gases being investigated so replacing some of the gas molecules in the syringe with water molecules affects the results.

5 Mass of 50 cm3 carbon dioxide = 0.089 g Molar mass of carbon dioxide = 44 g mol−1 Molar volume of carbon dioxide = (50 ÷ 0.089) cm3 g−1 × 44 g mol−1 = 24 700 cm3 mol−1 = 25 000 cm3 mol−1 to 2 s.f. Mass of 50 cm3 methane = 0.034 g Molar mass of methane = 16 g mol−1 Molar volume of methane = (50 ÷ 0.034) cm3 g−1 × 16 g mol−1 = 23 500 cm3 mol−1 = 24 000 cm3 mol−1 to 2 s.f. Mass of 50 cm3 butane = 0.109 g Molar mass of butane = 58 g mol−1 Molar volume of butane = (50 ÷ 0.109) cm3 g−1 × 58 g mol−1 = 26 600 cm3 mol−1 = 27 000 cm3 mol−1 to 2 s.f. 6 The masses of gas are small and even a three-place balance only gives values to 2 s.f. 7 The main source of uncertainty is the measurement of the masses of the gases. The mass readings are relatively large but it is the small differences in the values that are used in the calculation. Measuring the mass of a relatively large piece of apparatus means that the readings are easily affected by how the apparatus is placed on the scale. 8 Using a larger syringe would increase the mass of the gas samples to be weighed. Using a 100 cm3 syringe might give more consistent results.

Core practical 1 (page 133)

1 Students were able to add the magnesium in the tube and fit the bung before the metal mixed with the acid. This meant that no gas was escaped. 2 The graph is very close to straight line which passes through the origin. 3 Molar mass of magnesium is 24.3 g mol−1. Mass of ribbon = 0.200 g ÷ 25.0 cm = 0.0080 g cm−1 Length of ribbon containing 1.0 × 10−3 mol magnesium = 0.0243 g ÷ 0.0080 g cm−1 = 3.04 cm 4 Volume = 24.5 cm3

5 Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

6 1 mol magnesium reacts with excess acid to give 1 mol hydrogen. So 1.0 × 10−3 mol magnesium reacts to give 1.0 × 10−3 mol gas. 7 Molar volume of hydrogen = 24.5 cm3 ÷ 1.0 × 10−3 mol = 24 500 cm3 mol−1 8 The equation PV = nRT can be used to calculate the volume V of n moles of an ideal gas, assuming that the temperature and pressure of the gas are known. 9 The measurements assume that: • the magnesium ribbon is not corroded and that its mass per unit length is uniform • the ribbon has to be cut accurately into measured lengths • the gas collected is dry (free of water vapour) • the syringe plunger is moving freely so that the pressure inside the syringe is atmospheric pressure.

Core practical 2 (page 137)

1 Molar mass of KHP = 204.2 g mol−1 Mass needed for 1 dm3 of a 0.1 mol dm−3 solution = 20.42 g Mass needed to make 250 cm3 of a 0.1 mol dm−3 solution = 5.105 g 2 The molar mass of oxalic acid is lower than that of KHP. The mass of oxalic acid needed to make up 250 cm3 of a 0.1 mol dm−3 solution = 1.575 g. Measurement error for oxalic acid gives rise to a higher percentage error than for KHP. 3 a) The glass rod prevents drips of solution running down the side of the beaker and so ensures that all the KHP in solution is transferred to the graduated flask. b) The beaker is washed out with water several times and then the glass rod and funnel are rinsed with all the washings being carefully added to the graduated flask. 4 There are slight volume changes when solids dissolve in water. Dissolving all the solid in most of the solvent and mixing before making up to the mark allows for this. Adding the last few cm3 of water with a dropper makes it less likely that too much water is added by accident. The flask is calibrated on the assumption that the graduation is tangential to the meniscus when it contains the specified volume of solution. 5 Mass of solid added to the beaker = 5.361 g Amount of KHP in 0.250 dm3 solution = 5.361 g ÷ 204.2 g mol−1 Concentration of the standard KHP solution = 0.105 mol dm−3

Core practical 3 (page 145)

1 A Volumetric glassware must be clean and not contaminated with chemicals from previous titrations. Water drains in a smooth film from grease-free glassware but forms into droplets if it is not clean. B The solution to be measured from the burette must not be diluted with water remaining after washing. Rinsing with a little of the solution to be measured gets rid of the water. Opening the tip fills the space between the tap and the tip of the burette. C Similarly the pipette must be rinsed with the solution to be measured. D The indicator is added to detect the end-point. Methyl orange is yellow in alkali and red in acid. It is orange at the end-point. E The rough titration identifies quickly a small range of volumes in which the correct value lies. This allows most of the acid to be added quickly during accurate titrations. Then the end- point can be approached drop-by-drop. F Reading to the nearest 0.05 cm3 (half a scale division) helps to achieve titres consistent to 0.10 cm3. 2 A pipette is graduated on the assumption that it is allowed to drain for about 20 seconds and that the tip of the pipette is held against the side of the flask for several seconds after movement of the meniscus has ceased. 3 Molar mass of sodium carbonate = 106.0 g mol−1 Amount sodium carbonate in 0.5 dm3 solution = 2.920 g ÷ 106.0 g mol−1 Concentration of the standard solution = 0.0551 mol dm−3 4 Burette readings are taken from the bottom of the meniscus, which shows up more clearly if light is reflected on the meniscus with a white piece of paper or card. 5 Pipette volume = 25.0 cm3 25. 0 Amount of sodium carbonate in the flask = 1000 dm3 × 0.0551 mol dm−3 Equation for the titration reaction:

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

1 mol Na2CO3 reacts with 2 mol HCl Hence amount of HCl added from the burette 25. 0 = 2 × 1000 dm3 × 0.0551 mol dm−3 22. 4 Mean titre = 22.40 cm3 = 1000 dm3

25. 0 Concentration of the hydrochloric acid = 2 × 22. 4 × 0.0551 mol dm−3 = 0.123 mol dm−3 6 Percentage uncertainty in the concentration of the sodium carbonate solution 0. 001 = 0. 055 × 100% = 1.8% 0. 05 Percentage uncertainty in the pipette volume = 25. 0 × 100% = 0.2% 0. 15 Percentage uncertainty in the burette volume = 22. 4 × 100% = 0.7% Total percentage uncertainty = 1.8% + 0.2% + 0.7% = 2.7% Uncertainty in the calculated value = 0.027 × 0.123 mol dm−3 = ± 0.003 mol dm−3

Core practical 7 (part 1) (page 150)

1 Dip the loop in a clean metal wire into concentrated hydrochloric acid and then use it to pick up a few crystals of the salt. Hold the crystals near the bottom of a blue Bunsen flame, near the edge of the flame. Use cobalt blue glass if necessary to screen out a sodium flame. 2 Check that the wire is clean by holding it in the flame before using it in a test. If not clean, repeatedly dip the wire into concentrated hydrochloric acid and then hold it in the hottest part of a Bunsen flame until it no longer colours the flame. 3 Test for oxygen: ignite a splint, blow it out, then lower it into the gas while it is still glowing to see if it relights. Test for acid gas: moisten a length of indicator paper and lower it into the gas and note any colour changes. Test for hydrogen sulfide (not reported in the table): moisten a piece of lead ethanoate paper and lower it into the gas; the paper darkens as lead sulfide forms if hydrogen sulfide is present. 4 X and Y are both potassium compounds. 5 X decomposes on heating giving off oxygen. The product from decomposing X resembles Y. Traces of a purple vapour on heating both compounds suggests the presence of iodine. X could be potassium iodate which decomposes to potassium iodide on heating.

2KIO3(l) → 2KI(l) + 3O2(g)

6 After heating, the solids react in the same way with concentrated sulfuric acid. The gas which has a bad egg smell is hydrogen sulfide. The reduction of sulfuric acid to hydrogen sulfide is characteristic of an iodide. Other gases detected are hydrogen iodide (the fuming gas) and iodine vapour which is purple. 7 Potassium compounds with halogens are generally colourless and soluble in water. The observations are what is seen when an oxidising agent reacts with an iodide. In acid solution potassium iodate(v) oxidises potassium iodide to iodine.

− − + IO3 (aq) + 5I (aq) + 6H (aq) → 3I2(aq) + 3H2O(l) 8 Possible further tests include: • Dissolve Y in water. Acidify with dilute nitric acid and then add silver nitrate solution. A yellow precipitate forms. The precipitate is silver iodide. • Dissolve Y in water. Add aqueous chlorine and then shake with a little cyclohexane. The aqueous solution turns dark brown. A violet colour appears in the organic layer. Chlorine oxidises iodide ions to iodine. • Heat a second small sample of X strongly until decomposed. Cool. Dissolve in water and show that the residue behaves like Y with either silver nitrate solution or aqueous chlorine.

Exam practice questions (pages 152-4)

1 a) Cu2S(s) + 2O2(g) → 2CuO(s) + SO2(g) All numbers correct [1]

b) 2FeS(s) + 3O2(g) + 2SiO2(s) → 2FeSiO3(s) + 2SO2(g) All five numbers correct [3], four numbers correct [2], three numbers correct [1]

c) 4Fe(NO3)3(s) → 2Fe2O3(s) + 12NO2(g) + 3O2(g) All four numbers correct [3], three numbers correct [2], two numbers correct [1]

−1 2 a) Molar mass of O2 = 32.0 g mol [1]

4.0 g O2 = 0.125 mol [1] 0.125 mol contains 0.125 × 6.02 × 1023 molecules = 0.75 × 1023 = 7.5 × 1022 molecules [1]

−1 −1 −1 b) Molar mass of K2O = (2 × 39.1) g mol + 16.0 g mol = 94.2 g mol [1]

9.4 g K2O = 0.0998 mol [1]

23 23 0.0998 mol K2O contains 0.0998 × 6.02 × 10 × 3 = 1.80 × 10 ions [1]

−3 3+ −3 − 3 a) 0.112 × 10 g Fe and 12.40 × 10 g NO3 [1] b) 0.00201 × 10−3 mol Fe3+ = 2.01 × 10−6 mol Fe3+ [1]

−3 − −4 − 0.200 × 10 mol NO3 = 2.00 × 10 mol NO3 [1]

c) 12.1 × 1017 = 1.21 × 1018 Fe3+ ions [1]

20 − 1.20 × 10 NO3 ions [1] 4 a) In 100 g of X there are 3.57 mol C, 2.36 mol H, 1.19 mol N and 2.37 mol O. [2] C : H : N : O is 3 : 2 : 1 : 2. [1]

Empirical formula is C3H2NO2 [1] b) Sum of relative atomic masses in the empirical formula = 84 [1]

Molecular formula is C6H4N2O4. [1] 5 a) Precipitation [1] b) Redox [1] c) Thermal decomposition and redox [1] d) Acid–base (neutralisation) [1] 6 a) Ionic precipitation [1] to form silver iodide. [1] Ag+(aq) + I–(aq) → AgI(s) [1] b) Acid–base reaction [1] forming magnesium chloride, carbon dioxide and water. [1]

MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l) [1] c) Ionic precipitation [1] to form barium sulfate. [1]

2+ 2− Ba (aq) + SO4 (aq) → BaSO4(s) [1] d) Thermal decomposition [1] to give zinc oxide and carbon dioxide. [1]

ZnCO3(s) → ZnO(s) + CO2(g) [1] e) Redox reaction (displacement) [1] to form copper metal and zinc sulfate solution. [1]

Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq) [1] 6.0 7 a) Concentration = mol dm−3 = 6.0 × 10−3 mol dm−3 [1] 1000 b) Relative formula mass of cholesterol = 386 [1] 6 × 386 concentration = g dm−3 = 2.3 g dm−3 [1] 1000 c) Mass = 0.023 g [1] 8 a) i) Volume needed = 50 cm3 [1] ii) Removing drops of the mixture on a glass rod [1] then testing these drops with universal indicator paper [1].

b) i) Ammonium iron(II) sulfate contains equal amounts of moles of (NH4)2SO4 and FeSO4 [1] 25 Amount of ammonium sulfate prepared = × 2 = 5 × 10−2 mol [1] 1000

−2 Amount of FeSO4 required = 5 × 10 mol

−2 −1 Mass of FeSO4 required = 5 × 10 mol × 152 g mol = 7.6 g [1]

ii) Amount of ammonium iron(II) sulfate produced = 5 × 10−2 mol [1]

Molar mass of (NH4)2SO4.FeSO4.6H2O = M((NH4)2SO4) + M(FeSO4) + 6M(H2O) = (132 + 152 + 108) g mol−1 = 392 g mol−1 [1] Maximum possible yield = 392 g mol−1 × 5 × 10−2 mol = 19.6 g [1] Thus a 50% yield produced 9.8 g [1] 9 a) In 100 g of Z there are 4.54 mol C, 9.1 mol H, [1] 2.28 mol O. [1]

C : H : O = 2 : 4 : 1 [1] Empirical formula is C2H4O. [1]

pV 95 000 Pa × 0.0001 m3 b) n = [1] = [1] = 0.00306 mol [1] RT 8.31 J mol−1 K−1 ×373 K Molar mass = 0.270 g ÷ 0.00306 mol = 88 g mol−1 [1]

Molecular formula is C4H8O2 [1]

10 a) Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) All symbols and formulae correct. [1] Correctly balanced. [1] All state symbols correct. [1] b) Diagram showing Ca reacting with HCl(aq) in a conical flask [1] with a delivery tube to an inverted measuring cylinder over water. [1] c) Filter the final solution to remove excess calcium. [1] Collect the filtrate in an evaporating basin and heat it until crystals start to form at the edges of the solution. [1] Set the solution aside to cool and crystallise. [1] Dry the crystals using absorbent paper. [1] d) The Ca is in excess, so the yield is limited by the amount of HCl(aq). [1] Amount of HCl(aq) used = 2.5 × 10−2 mol [1] 2 mol of HCl produce 1 mol of calcium chloride.

1 −2 So, maximum amount of calcium chloride produced = 2 × 2.5 × 10 mol [1] Maximum mass of product = (0.5 × 2.5 × 10−2) mol × 219 g mol−1 = 2.74 g [1] e) Some calcium chloride is lost during filtration. [1] More calcium chloride is lost because the crystals are separated from a saturated solution of calcium chloride. [1] Yet more calcium chloride is lost when the crystals are removed from the evaporating basin and dried. [1] (Any 2 of 3 points.)

11 a) M2CO3(s) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l) [1] b) Amount of HCl = 0.0236 dm3 × 0.150 mol dm–3 = 0.003 54 mol [1]

c) Amount of M2CO3 in the sample = 0.5 × 0.003 54 mol = 0.001 77 mol [1]

0.245 g –1 d) Molar mass of M2CO3 = [1] = 138 g mol 0.001 77 mol

Relative formula mass of M2CO3 = 138 [1] e) Relative atomic mass of M = 0.5 × (138 − 60) = 39 [1] f) M is potassium. [1]

12 a) 2NH3 + NaOCl → N2H4 + NaCl + H2O [1] Molar mass of the desired product = 32 g mol−1 [1] Total molar mass of all products = 108.5 g mol−1 [1] Atom economy = (32 ÷ 108.5) × 100% ≈ 30% [1]

b) i) C2H6 + Br2 → C2H5Br + HBr [1] Molar mass of the desired product = 108.9 g mol−1 [1] Total molar mass of all products = 189.8 g mol−1 [1] Atom economy = (108.9 ÷ 189.8) × 100% ≈ 57% [1]

ii) C2H4 + HBr → C2H5Br [1] The two reactants add together to give a single product. All the reactant atoms end up as product atoms. [1] By inspection, the atom economy is 100%. [1] c) This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answer taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Points to make in the answer: • Yields are calculated by considering only one reactant and one product. • A reaction can have a high percentage yield but also make a lot of waste product. • Processes should be designed so that the maximum amount of all the starting materials ends up in the product. • This kind of reaction has a low atom economy. • This means that there is a minimum of waste to get rid of. • This reduces the cost of disposing of the waste.

−1 −4 13 a) The amount of BaSO4 precipitate = 0.141 g ÷ 233.4 g mol = 6.04 × 10 mol [1] All the sulfate came from the sodium sulfate. [1]

The amount of Na2SO4 in the sample was 0.000604 mol [1] Mass of sodium sulfate in the sample = 0.000604 mol × 142.0 g mol−1 = 0.0858 g [1]

Percentage of Na2SO4 in the sample = (0.0858 g ÷ 0.250 g) × 100% = 34.3% [1]

b) Iron forms iron(II) chloride when it reacts with hydrochloric acid.

Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) [1] Amount of hydrogen formed = (191 cm3 ÷ 24 000 cm3 mol−1) [1] From the equation, 1 mol Fe reacts to give 1 mol hydrogen gas. [1] Mass of iron in the sample = (191 cm3 ÷ 24 000 cm3 mol−1) × 55.8 g mol−1 = 0.444(4) g [1] Percentage of iron in the sample = (0.444 g ÷ 0.500 g) × 100% = 88.8% [1]

14 (COOH)2 + 2NaOH → (COONa)2 + 2H2O(l) [1] Amount of NaOH needed to neutralise 25.0 cm3 of the acid solution = 0.0156 dm3 × 0.160 mol dm–3 [1] Amount of the acid in 25.0 cm3 of the solution = 0.5 × 0.0156 dm3 × 0.160 mol dm–3 [1]

0.5 ×0.0156 dm3 ×0.160 mol dm−3 Concentration of the acid = = 0.0499 mol dm–3 [1] 0.025 dm3

Concentration of the original solution = 1.576 g = 6.30 g dm–3 [1] 0.250 dm3

6.30 g dm−3 Molar mass of the acid = = 126.2 g mol–1 [1] 0.0499 mol dm−3

–1 Molar mass of (COOH)2.nH2O = 90 + (n × 18) g mol [1] n = 2 [1] 15 The two accurate titres are: 20.65 cm3 and 20.55 cm3 [1] Mean titre = 20.60 cm3 [1] Amount of acid needed to neutralise the carbonate in 20.0 cm3 of the solution = 0.0206 dm3 × 0.10 mol dm–3 [1]

Na2CO3(aq) + 2HCl(aq) → 2NaCl + CO2(g) + H2O(l) [1] Amount of sodium carbonate in 20.0 cm3 of the solution = 0.5 × 0.0206 dm3 × 0.10 mol dm–3 [1] Amount of sodium carbonate in original sample 250 = × 0.5 × 0.0206 dm3 × 0.10 mol dm–3 [1] 20 Molar mass of hydrated sodium carbonate = 286 g mol–1 [1]

Mass of sample if it had been pure Na2CO3.10H2O

250 = 286 g mol–1 × × 0.5 × 0.0206 dm3 × 0.10 mol dm–3 20 = 3.682 g [1] Actual mass of sample = 2.696 g Loss in mass = 0.986 g [1] Loss in mass is 26.8% [1] 16 a) Calcium carbonate is insoluble in water. [1] It reacts too slowly for a direct titration. [1] An excess of acid is needed to ensure that all the calcium carbonate in the shell reacts completely in a reasonable time.

b) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) [1] Concentration of the acid in the 250 cm3 graduated flask = 0.0972 mol dm–3 [1] Amount of excess HCl = 0.0243 mol [1] c) Amount of HCl added to the sample of egg shell = 0.0480 mol [1] Amount of HCl that reacted with the egg shell = 0.0237 mol [1]

d) CaCO3(s) + 2HCl → CaCl2(aq) + CO2(g) + H2O(l) [1]

Amount of CaCO3 in the sample = 0.0118 mol [1]

–1 Molar mass of CaCO3 = 100 g mol . [1] Mass of calcium carbonate in the sample = 1.18 g [1] Percentage of calcium carbonate in the egg shell = 78.1% [1] e) A back titration is an procedure used when the reaction between the standard solution and the substance to be analysed is slow. [1] The procedure is to add a measured excess of the standard solution, allow time for the reaction to finish [1] and then to use a titration with a second standard solution to measure how much of the first standard solution remains unused. [1]

Test yourself (page 156)

1 The branch of chemistry concerned with the study of compounds containing C−H bonds. 2 Carbon atoms can form chains, branched chains and rings of varying size. Carbon atoms form C−C and C−H bonds which are relatively inert and unreactive. 3 a) The single C−C bond is stronger than all the other listed single bonds between atoms of the same non-metal except the H−H bond. b) The relatively high strength of the C−C bond will increase the number of carbon compounds.

Test yourself (page 158)

4 a) Because they all have the same functional group which gives them their characteristic properties. b) Because as the hydrocarbon chain increases in size, the number of electrons increase and the London forces increase. 5 a) −OH, alcohols b) −CHO, aldehydes c) −Hal, halogenoalkanes d) >C=C< and −Hal, alkenes and halogenoalkenes

6 H2N–CH2–COOH

Test yourself (page 160)

7 a) 51.62% b) CH3O c) C2H6O2 8 a) 27.3% b) 0.46 g c) 11.1%

d) 0.038 g e) CH f) C2H6

9 a) C2H5 b) C4H10 10 The empirical formula is the simplest whole number ratio of each type of atom. In the examples

in Figure 6.1.8, ethanoic acid in vinegar has molecular formula C2H4O2 (twice its empirical

formula); lactic acid in milk has molecular formula C3H6O3 (three times its empirical formula);

glucose has molecular formula C6H12O6 (six times its empirical formula).

Test yourself (page 164)

11 a) Octane b) 2,3,4-Trimethylhexane c) 3-Methylhexane

d) 2,2-Dimethylbutane

12 a) b)

13 Structural formula Skeletal formula

Test yourself (page 165)

14 a) 2-Methylpropene (the number 2 is optional as the methyl group cannot be anywhere other than on the middle carbon) b) Hex-2-ene c) 2,3-Dimethylpent-2-ene

15 a) b)

16 a)

Test yourself (page 166)

17 a)

b) Chain isomerism

Test yourself (page 167)

18 a)

b) 2-Methylpentane and 3-methylpentane are position isomers. 2,2-Dimethylbutane and 2,3-dimethylbutane are position isomers 19 a)

Test yourself (page 169)

20 a) Elimination b) Hydrolysis c) Addition and reduction (redox) d) Substitution e) Elimination

21 a) CH3CH2CH2CH3 butane

b) CH3CH2CHClCH2Cl 1,2-dichlorobutane

c) CH3CH2CH2CHBrCH3 2-bromobutane

– – 22 a) CH3CHBrCH3 + OH → CH2=CHCH3 + H2O + Br

b) CH3CH2CH2CH2OH → CH3CH2CH=CH2 + H2O 23

24 a) C2H6 + Br2 → C2H5Br + HBr

b) CH3CH2CH2Br + NaOH → CH3CH2CH2OH + NaBr

c) CH3COOCH2CH3 + H2O → CH3COOH + CH3CH2OH

Test yourself (page 171)

25 a) Br—Br → Br• + •Br b) H—Br → H+ + Br− c)

26 a) b)

27 a) Nucleophile b) Electrophile c) Free radical d) Nucleophile

Activity: The alkanes – an important series of organic compounds (page 162)

1 Empirical formula: ethane CH3

Molecular formula: propane C3H8

Structural formulae: methane CH4; butane CH3CH2CH2CH3 Displayed formula: ethane

2 a) Hexane b) Octane

3 C3H8 and C8H18 4 a) 2n + 2 b) 2n + 2

c) CnH2n + 2 5 a) b) There is only one carbon atom in methane and a skeletal formula showing only a single line

would represent CH3CH3, i.e. ethane. 6 a)

b) There are three structures:

Activity: Investigating the mechanism of a hydrolysis reaction (page 172)

1 The ester is split apart by reaction with water. This is hydrolysis. 2 The nucleus of an oxygen-18 atom has two more neutrons than the nucleus of an oxygen-16 atom.

3 They have the same number of protons in the nucleus so they have the same number of electrons. As a result the atoms of the isotopes have the same electron configuration and hence

the same chemical properties. 4 The isotopes can be separated and detected in a mass spectrometer. 5 The central carbon in the ester has three bonds: a C−C bond, a C=O bond and a C−O bond. It is the C−O bond which breaks. 6 The oxygen-18 atoms would have ended up in the ethanol molecules instead of in the ethanoic acid molecules.

Exam practice questions (pages 175-6)

1 a) For 1 mark each, any two of: • ability to catenate/form long chains and rings • the unreactive nature of C−C and C−H bonds • ability to form four bonds • ability to form C−C, C=C and C≡C bonds • ability to form isomers. [2] b) i) Compounds with the same molecular formula [1] but a different structural formula. [1]

ii) CnH2n + 2 [1] Allow any alkane from C6 to C10, e.g. C6H14, C10H22. [1] c) [1]

CnH2n [1]

2 a) The empirical formula of X is C5H10O [1] and its molecular formula is also C5H10O. [1] The skeletal formula of X is [1] A functional group is a group of atoms [1] which gives an organic compound its characteristic properties. [1] The functional groups in X are the alkene group [1] (accept carbon−carbon double bond) and the alcohol group [1] (also allow alkyl). b) i) Pentan-1-ol [1]

ii) CH3CH2CH=CHCH2OH(g) + H2(g) → CH3(CH2)3CH2OH(g) [2] iii) Pentan-2-ol and pentan-3-ol [2]

iv) CH3OCH2CH2CH2CH3 or CH3CH2OCH2CH2CH3 [1] 3 a) i) A compound containing carbon and hydrogen only. [1]

ii) C12H26 [1]

iii) C6H13 [1] b) i) Highly reactive atom or group of atoms [1] with an unpaired electron. [1]

ii) Cl—Cl → Cl• + •Cl [1] iii) Homolytic [1] iv) 5 [1]

v) CH3(CH2)8CH2Cl [1] 1-chlorodecane [1] (or 2-, 3-, 4- and 5-chlorodecane) 4 a) Elimination [1] b) Substitution [1] c) Addition [1] d) Oxidation [1] e) Substitution, hydrolysis [1] f) Polymerisation, addition [1] 5 a) i) An electrophile is a reactive molecule or ion that seeks out and reacts with electrons in molecules. [1] Examples include, H+ ions, Hδ+ atoms in molecules such as H–Br [1] ii)

(3)

The H−Br bond is polar [1] with the hydrogen atom at the δ+ end of the dipole. [1] b) i) A nucleophile is a molecule or ion with a lone pair of electrons which seeks out and forms new bonds with δ+ atoms in molecules. [1] Examples include water molecules, hydroxide ions, cyanide ions, ammonia molecules. [1] ii) [3]

The C−Br bond is polar [1] with the carbon atom at the δ+ end of the dipole. [1] 6 a) Indicative content: Homolytic bond breaking is favoured with non-polar reactant [1] In the gas phase or in non-polar solvents. [1] Absorption of UV light can bring about homolytic fission. [1] Heterolytic bond breaking is favoured if the organic reactant has polar bonds [1] The reagents are ionic or highly polar. [1] This type of bond breaking is also favoured by polar solvents. [1] b) Reagents with a similar classification tend to react in similar ways with compounds of a particular type. [1] The characteristic reactions of functional groups can be related to the types of reagents with which they react. [1] This helps to explain and predict the reactions which are likely to take place. [1] This makes it possible to select reagents for particular purposes. [1]

408 7 a) Number of moles of carbon dioxide in 408 cm3 = = 0.0170 mol [1] 24 000 Mass of carbon in this amount of carbon dioxide = 0.0170 × 12.0 = 0.204 g [1] 2.0 Mass of hydrogen in 0.308 g water = 0.308 × = 0.0342 g [1] 18.0 Mass of oxygen in original compound = (0.292 – 0.204 – 0.0342) g = 0.0538 g [1]

Ratio of masses C : H : O = 0.204 : 0.0342 : 0.0538 0.204 0.0342 0.0538 Ratio of moles C : H : O = : : [1] 12.0 1.0 16.0 = 0.017 : 0.0342 : 0.00336 = 5 : 10 : 1

Empirical formula = C5H10O [1] b) If the compound were a (saturated) alcohol it would have 12 hydrogens. [1] If cyclic, could be a cyclic ether such as [1]

If unsaturated, could be an enol such as [1]

Or an aldehyde such as [1]

Or a ketone such as [1]

8 a) [1] for arrow, [1] for structure of ion

[1] × 2 for arrows

c) [1] × 2 for arrows

d) The nucleophile is ethanol itself. [1] There is a lone pair of electrons on the oxygen atom of

+ the −OH group. So this oxygen attacks the carbon atom of the C−OH2 in an ethanol molecule that has gained a hydrogen ion from the acid. [1]

e) i) (CH3)3COH [1] ii) Physical: ethoxyethane has lower boiling temperature [1] because no hydrogen bonding occurs between molecules Or using IR, ethoxyethane has no O–H alcohols peak. [1] Chemical: alcohol reacts with Na [1] to give effervescence; ethoxyethane does not. [1]

Test yourself (page 178)

1 a) CH4, methane; C2H6, ethane; C3H8, propane; C4H10, butane.

b) CH2 2 In a straight-chain alkane with n carbon atoms, each carbon atom is bonded to two hydrogen atoms except the carbon atoms at the end of the chain, each of which has an extra hydrogen atom. So the total of H atoms is 2n + 2.

3 Liquid: C5H12

Solid: C18H38

Test yourself (page 178)

4 Compounds in a series of organic compounds have similar chemical properties because they have the same functional group, but there is a gradation in physical properties along the series. Compounds of elements in a group have similar chemical and physical properties, again with some changes down the group.

5 a) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

b) CO2, H2O, carbon (soot) and carbon monoxide c) If the boiler is badly maintained, there may be insufficient air for the complete combustion of methane. This could produce poisonous carbon monoxide which could enter the room. 6 Gas cylinders need a good pressure of gas inside to keep the gas flowing smoothly. In a very cold wintry spell, butane (boiling temperature −0.5 °C) would liquefy, unlike propane (boiling temperature −42 °C), and this would significantly reduce the gas pressure.

7 a) Propagation: CH3• + Cl2 → CH3Cl + Cl•

Termination: CH3• + Cl• → CH3Cl The number of radicals present is small so that a methyl radical is much more likely to collide with a chlorine molecule in a propagation step.

b) CH3Cl + Cl2 → CH2Cl2 + HCl Stage 1 Initiation: Cl Cl → Cl• + Cl•

Stage 2 Propagation: Cl• + H–CH2Cl → HCl + •CH2Cl

•CH2Cl + Cl–Cl → CH2Cl2 + Cl• Stage 3 Termination: any two radicals combine including

Cl• + Cl• → Cl2

•CH2Cl + Cl• → CH3Cl

•CH2Cl + •CH2Cl → ClCH2CH2Cl, etc.

c) CH2Cl2 + Cl2 → CHCl3 + HCl, trichloromethane

CHCl3 + Cl2 → CCl4 + HCl, tetrachloromethane d) Dichloromethane can be obtained from the mixture by fractional distillation. 8 a) Bromine does not react with hexane in the dark. But sunlight can cause the homolytic fission of bromine molecules to bromine atoms (free radicals) and this starts the reaction. The products are bromohexane and hydrogen bromide, so the orange colour of bromine fades.

b) Br2(l) + C6H14(l) → HBr(g) + C6H13Br(l) c) The acid fumes are fumes of hydrogen bromide coming off the solution.

Test yourself (page 185)

9 a) It provides a large proportion of our energy needs and is the source of most of our organic chemicals, particularly plastics. b) Because the reserves of crude oil are not limitless and it isnon-renewable. 10 a) Ethanol and ETBE have oxygen in their molecules. This helps to facilitate the combustion of any petrol containing them. b) Cracking and reforming. 11 a) Cracking increases our supplies of the gasoline (petrol) fraction and produces short-chain alkenes that are used to manufacture polymers (plastics). b) High temperatures (c. 500 °C) with zeolite (silicate) catalysts. 12 a)

c) CH3CH2CH2CH2CH2CH3 → + H2

d) CH3CH2CH2CH2CH2CH2—CH2—CH3 → + 4H2

e) C16H34 → C8H18 + 2C3H6 + C2H4

Test yourself (page 188)

13 a) C4H10 + 6½O2 → 4CO2 + 5H2O b) C5H12 + 5½O2 → 5CO + 6H2O

c) C6H14 + 3½O2 → 6C + 7H2O d) C2H5SH + 4½O2 → 2CO2 + SO2 + 3H2O 14 The blue, roaring, hot flame observed when the air hole is fully open becomes quieter, less steady and eventually luminous when the hole is fully closed. The flame with the air hole closed is luminous because of the tiny particles of carbon in the flame. If this flame is used for heating, carbon is deposited on the object being heated.

15 Sulfur impurities in fossil fuels burn to form sulfur dioxide. This reacts with water to form

sulfurous acid, H2SO3. Oxidation of the sulfur dioxide also occurs to form sulfur trioxide; this reacts with water to form sulfuric acid. Nitrogen react with oxygen to form NO either at high temperature in internal combustion engines or when lightning provides sufficient activation energy. The NO reacts with oxygen to

form NO2 which reacts with water and more oxygen to form nitric acid.

Test yourself (page 189)

16 Sustainable development means planning to live within the means of the environment in order that the Earth’s natural resources are not destroyed, but will be available for future generations. In terms of fuels, this means conserving fossil fuels by reducing their consumption not only because they contain valuable non-renewable resources, but also because their combustion

increases the atmospheric concentration of CO2, leading to an increased greenhouse effect and climatic changes throughout the world. One of the means of reducing our use of fossil fuels is by using alternative fuels – biofuels, fuel cells and nuclear fuels. 17 Various answers are possible including: • reduce travel (especially air travel) • reduce the transport of goods (particularly food) over large distances • promote the purchase and use of local food and other goods • encourage motorists to use smaller vehicles (if necessary by heavy taxes on large vehicles) • explore the use of alternative fuels (biofuels, fuel cells, nuclear fuels)

18 Biofuels remove CO2 from the atmosphere during the growth of the plants from which they are

obtained. This same CO2 is returned to the atmosphere when the biofuels burn. So, the balance

of CO2 is unchanged – neutral. However, this takes no account of the energy used in the production of the fuels. This energy is likely to lead to the release of additional carbon dioxide. 19 In Brazil the waste biomass from sugar cane can be the fuel used to distil the ethanol.

Test yourself (page 193)

20 Both carbon atoms in ethene are surrounded by three centres of negative charge (two single bonds and one double bond). These three centres of negative charge will repel one another as far as possible so they will occupy trigonal positions around the carbon atom, resulting in bond angles of 120°.

21 Because rotation could only happen if the double bond is broken, and the energy of molecules at normal temperatures is insufficient for this. 22 IPent-2-ene only 23 a)

b) 1,1-dibromoethene is a position isomer of the other two, which are cis and trans isomers.

Test yourself (page 196)

24 a) i) CH3CH2CH3; propane; platinum or palladium catalyst at room temperature, or nickel catalyst at 150 °C.

ii) CH3CHClCH2Cl; 1,2-dichloropropane; room temperature

b) i) CH3CHClCH2CH3; 2-chlorobutane; room temperature;

ii) CH3CH(OH)CH2CH3; butan-2-ol; acid catalyst such as phosphoric acid adsorbed on silica and at high pressure and high temperature. 25 The purple colour will disappear to leave a colourless solution.

CH3CH=CH2 + [O] + H2O → CH3CH(OH)CH2(OH) Product is propane-1,2-diol 26 a) Addition of hydrogen using a catalyst. b) Saturated organic compounds have only single bonds between the atoms in their molecules. Unsaturated organic compounds contain one or more double or triple bonds between atoms in their molecules. Vegetable oils contain unsaturated compounds with C=C double bonds in their molecules. These can be converted to solid, saturated compounds in products such as low-fat spreads by the addition of hydrogen. c) Saturated fats, such as cream, cause arteriosclerosis in which the linings of the arteries become hardened and furred. Unsaturated cis fats are much less likely to cause this. Trans fats, although unsaturated, are now known to be as harmful as saturated fats.

Test yourself (page 201)

27 a)

Product name is 2,3-dibromobutane. b) Reaction:

Same carbocation intermediate formed in both cases, but in (a) only bromide ions can attack, whereas in the reaction with bromine water the water molecules are more likely to attack than bromide. 28 a) i)

ii) is a secondary carbocation and is the more stable as there are two inductive effects stabilising the C+ rather than one for the other primary carbocation. iii) Major product: 2-chlorobutane

b) i) Primary Tertiary

ii) is a tertiary carbocation and is the more stable as there are three inductive effects stabilising the C+ rather than one for the other primary carbocation. iii) Major product: 2-bromo-2-methylpropane

+ 29 a) The more stable carbocation formed, CH3CH CH2Br, can react with chloride ions to give the major product 1-bromo-2-chloropropane. The less stable carbocation leads to the minor product 2-bromo-1-chloropropane.

b) Brδ+ in bromine molecules can act as an electrophile leading to the overall addition of bromine. Chloride ions, like all negative ions, will be repelled by the π cloud of the double bond so cannot act as electrophiles.

Test yourself (page 205)

30 a) 1,1,2,2-Tetrafluoroethene

c) Poly(tetrafluoroethene) d) Non-stick coating for frying pans (low coefficient of friction); waterproof lining for jackets (water repellent); cover for electrical wires (electrical insulator). 31 a) High temperature, high pressure, with initiators b) Because it has such useful properties as low density, flexibility, easily moulded, transparent, good insulator, resistance to water, acids and alkalis. c) They are not biodegradable. 32 a) A small amount of initiator is used in the first propagation step but further propagation steps require no more initiator. The initiator is bonded at the end of the polymer chain so is used up and is therefore not a catalyst. b)

Activity: Renaming cis–trans as E/Z isomers (page 192)

1 .

4 a) From that of the highest priority, the order is: Br, Cl, C in CH3, H.

b) From that of the highest priority, the order is CH3O−, HOCH2−, CH3CH2− and CH3−. 5 a) Cl− then H−

b) Br− then CH3−. 6

7 a)

8 a) trans-1,2-dichlorothene and E-1,2-dichlorothene: trans and E match b) trans-1-bromo-1,2-dichlorothene and Z-1-bromo-1,2-dichlorothene: Cl atoms are trans, but Cl on the left and Br on the right have priority and they are Z.

Activity: A more sustainable future for polymers (page 202)

1 Various answers are possible including: • reduce travel (especially air travel) • reduce the transport of goods (particularly food) over large distances • promote the purchase and use of local food and other goods • encourage motorists to use smaller vehicles (if necessary by heavy taxes on large vehicles) • explore the use of alternative fuels (biofuels, fuel cells, nuclear fuels) 2 Setting up increasingly higher targets for the reduction of landfill waste. In some cases, the Government will penalise local councils who do not meet their targets by reducing their revenues. 3 Landfill sites are being used up. In addition, huge quantities of materials that could be recycled continue to be thrown into landfill sites. 4 Plastic material may be dirty, may be black, there may be one type of plastic inside a different type of plastic. 5 a) Once the process is running, gaseous hydrocarbon produced by pyrolysis can be used to heat the shredded polymer.

b) Pyrolysis of PVC will also produce the toxic gas HCl. 6 a) Both ‘Yes’ and ‘No’ are possible answers, depending on the quality of the argument for the choice. For example: Yes – incineration can produce energy for heating or power generation. No – incineration may release toxic gases or greenhouse gases. b) Society has a communal and general dislike and distaste for waste. Unsightly waste gases have sometimes been released from incinerators in the past. Waste gases released from incinerators may be toxic and noxious, causing concerns about health and safety, but high temperature treatment in modern incinerators should now prevent this. c) If incineration is complete, any carbon in the waste will be converted to carbon dioxide which would exacerbate the greenhouse effect and increase global warming. d) Acidic gases can be removed by passing the flue gases through material such as powdered calcium carbonate (limestone) or powdered calcium oxide obtained by heating limestone. 7 Although completely biodegradable plastic bags are available, they are still more expensive to produce than traditional plastic bags. Their use will end when public opinion changes so a majority of the population is willing to pay more or government policy dictates that a change is made.

Exam practice questions (pages 206-8)

1 a) i) 150 °C [1] at normal atmospheric pressure. [1] ii) Saturated compounds have only single bonds between the atoms in their molecules. [1] Unsaturated compounds have at least one double or triple bond between atoms in their molecules. [1] Alkenes contain a C=C double bond. [1] iii) Foods containing cis unsaturated fats are healthier than foods containing saturated fats [1] or trans unsaturated fats [1], which are major causes of arteriosclerosis (hardening of the arteries). [1]

b) i) Dilute sulfuric acid [1] and potassium manganate(VII) solution. [1] ii) Ethane-1,2-diol [1] iii) Antifreeze for windscreen washer fluid in winter. [1] c) i) Relatively low temperature and pressure [1] and special catalysts [1] ii) [1]

d) Any four of: light, flexible, relatively strong, transparent, resistant to water, easily drawn into sheets, easily coloured. ([2] for four properties, [1] for two or three properties) 2 a) Indicative content: The crude oil is heated to about 400 °C and then flows into the fractionating column. The column is hotter at the bottom and cooler at the top. Rising vapour condenses when it reaches a tray with liquid at a temperature just below its boiling temperature. A series of horizontal trays up the column Allows hydrocarbons with small molecules (low boiling temperatures) to rise higher up the column while larger molecules stay lower down. Fractions containing molecules of roughly the same mass are drawn off the column at various heights. b) i) Cracking is the conversion of heavier fractions from crude oil [1] containing long-chain alkanes [1] into smaller molecules in a mixture of alkanes and alkenes. [1]

ii) C12H26(g) → 2C2H4(g) + C8H18(g) [1]; octane [1] iii) To speed up the cracking process [1]; to lower the temperature at which cracking occurs and save on energy costs. [1]

c) CH3(CH2)5CH3(g) → + H2(g) [2]

d) i) [1]

ii) Oxygen within their molecules facilitates the combustion [1] of the petrol to which they are added.

3 a) i) C5H12(l) + 8O2(g) → 5CO2 (g) + 6H2O(l) [1]

ii) C6H14(l) + 6½O2(g) → 6CO (g) + 7H2O(l) [1]

iii) C7H16(l) + 4O2(g) → 7C(s) + 8H2O(l) [1] b) Pentene would burn with a more smoky flame [1] because it contains a higher percentage of carbon than pentane. [1] 4 a) i) Electrophilic addition is an addition reaction in which one molecule combines with another to form a single product [1] and in which one of the reacting molecules acts as an electrophile [1], attacking a negatively charged region [1] in the other reactant. ii) 2-Bromopropane [1]

iii) [06_02_A_07]

Arrow from ethene to H [1] Polarity of H–Br and arrow to Br [1] Structure of carbocation [1] Arrow from bromide to C+ [1]

+ iv) When H adds to CH3CH=CH2, two carbocations are produced [1],

+ + + CH3C HCH3 and CH3CH2C H2 [1]. CH3C HCH3 is the more stable because of the inductive effect of two alkyl groups. [1] So it persists longer and reacts with more Br− ions to form the major product. b) 2-Chloro-1-iodopropane. [1] Iodine is δ+ because chlorine is more electronegative [1], so iodine bonds first to the end carbon to form the more stable carbocation. [1]

c) CH3CH=CH2 + Br2 + H2O → CH3CH(OH)CH2Br + HBr [1] 5 a) The change in structure of a molecule from one isomer to another. [1] b) [1]

c) [1]

d) [1]

2,2,3-Trimethylbutane [1] e) Heptane has larger molecules than hexane so more electrons. [1] So it has stronger London forces and its molecules require more energy before they have sufficient energy to escape from the attractions of their neighbours. [1] f) About 91 °C [1] (i.e. halfway between the boiling temperature of 2,3-dimethylpentane and heptane) because 2-methylhexane has about half the branching in its structure compared with 2,3-dimethylpentane but the same relative mass as both 2,3-dimethylpentane and heptane. [1]

6 a) Percentage of hydrogen = 14.3 [1] 85.7 14.3 C: = 7.14; H: = 14.3 [1] 12.0 1.0

Simplest ratio 1 : 2 so empirical formula is CH2. [1]

Mr = 84.0 so molecular formula is C6H12. [1] b)

[1] 2,3-Dichloro-2,3-dimethylbutane [1] 7 a) i) A compound produced by the addition reaction [1] of large numbers [1] of monomers/small molecules. [1] ii) It consists of a mixture of polymer molecules with different chain lengths and different branched sections [1] which do not melt at the same temperature. [1] iii)

[1] iv) Any two of: • sorting polymers prior to recycling using infrared and other techniques • converting polymer waste into simpler compounds to use a feedstock for cracking • incineration of polymer waste to release energy • removal of toxic/harmful gases formed during incineration before release into the atmosphere. [2] b) i) Removal/elimination of hydrogen from a compound. [1]

ii) CH2=CHCH=CH2 [1]

iii) CH3CH=CHCH3 → CH2=CHCH=CH2 + H2 [2]

(Allow [1] for C4H8 → C4H6 + H2) 8 a) The oil does not mix with water [1] so shaking is needed to transfer the bromine into the organic layer so it can react there with the oil. [1]

b) The lower aqueous layer would be very pale yellow/colourless. [1] The organic layer would be yellow/orange as it contains unreacted bromine molecules. [1] c) The volumes of oil measured using a measuring cylinder will vary. [1] Either small amounts of oil could be weighed out or added from a burette [1] or alternatively a larger amount of oil could be added to a volumetric flask and a solution prepared which could be used for several experiments. [1] d) Either the concentration of the bromine water could be reduced [1] or the concentration of the oil solution could be increased. [1] 9 a) Bonds broken C–H (412) and Br–Br (193) [1] 412 + 193 = 605 kJ mol−1 [1] Bonds formed C–Br (276) and H–Br (366) [1] 276 + 366 = 642 kJ mol−1 [1] ∆H = +605 − 642 = −37 kJ mol−1 [1] (Extra (5 × 412) + 348 = 2408 may be included on both sides.) b)

[1] × 2 for equations Propagation [1] 10 a)

[2] b) The excess of propane makes it more likely that chlorine radicals will collide with propane molecules[1] rather than with chloropropane molecules which would lead to further substitution. [1] c) Radical on central C is more stable than radical on terminal C [1] (same as carbocations; similarly due to inductive effects). This would predict that there would be more 2-chloropropane than 1-chloropropane at lower temperatures. [1] d) There are six H atoms on end C atoms and two H on the central C [1] so expected mole ratio statistically is 3 : 1 for 1-chloropropane to 2-chloropropane [1], assuming that substitution of any H is equally likely. [1] e) At higher temperatures, sufficient activation energy is available; reactions are more rapid and stability of intermediates is less important. [1]

11 1000 dm3 of petrol = 1000 × 1000 cm3 = 1.00 × 106 cm3 [1] Mass of petrol = 0.8 g cm−3 × 1.00 × 106 cm3 = 8.00 × 105 g [1] Molar mass of octane = 114 g mol−1 [1] 8.00 × 105 g Moles of octane = = 7.018 × 103 mol [1] 114 g mol−1

On combustion, 1 mole of octane produces 8 moles of CO2 = 8 × 44.0 g CO2 [1]

3 3 6 7.018 × 10 moles of octane produce 8 × 44.0 × 7.018 × 10 g CO2 = 2.47 × 10 g CO2 [1]

Test yourself (page 210)

1 a) CH3CH2CH2I, primary

b) (CH3)2CClCH2CH3, tertiary

c) CH3CH2CHBrCH2CH3, secondary

d) BrCH2CHClCH3, primary bromo, secondary chloro 2 a) 1,3-Dichlorobutane b) 1,1,2-Trichloro-1,2,2-trifluoroethane c) 2-Bromo-4-methylpentane

3 Only CCl4 is non-polar. 4 The boiling temperature rises as the halogen atom gets larger and has more electrons. In this series the molecules become more polarisable and so the intermolecular forces increase. 5 Branched molecules are more compact and so the area over which intermolecular forces can operate is smaller. So the attraction between the molecules is less making it easier to separate the molecules. So the tertiary compound boils at the lowest temperature. 6 Despite the polar bond, the overall polarity of the molecules is not high so they cannot interact strongly with polar water molecules. Also, they cannot form hydrogen bonds with water.

Test yourself (page 211)

7 Chloromethane, 2-bromobutane, 1,2-dibromopropane, chloroethane, 1-chloropropane

8 a) CH3CH2CH2CH3 + Br2 → CH3CH2CH2CH2Br + HBr UV light is needed. 2-Bromobutane will also be formed together with a mixture of compounds where further substitution has occurred. An excess of butane will reduce further substitution.

b) H3CH2CH=CH2 + HBr → CH3CH2CH2CH2Br 1-Bromobutane is the minor product; more 2-bromobutane will be formed via the more stable secondary carbocation.

c) H3CH2CH=CH2 + HBr → CH3CH2CH2CH2Br 9

Test yourself (page 214)

10 Nucleophile: ion or molecule with a lone pair of electrons that attacks positive centres in this case a water molecule. Substitution means replacement of a group or atom; in this case the reaction replaces the halogen atom with an −OH group. Hydrolysis means splitting apart with water. Hydrolysis of a halogenoalkane by water produces an alcohol and a hydrogen halide.

+ – 11 CH3CH2CHBrCH3(l) + H2O(l) → CH3CH2CHOHCH3(l) + H (aq) + Br (aq) Ag+(aq) + Br–(aq) → AgBr(s) 12 a) Silver ions react with halide ions not with covalently bonded halogen atoms. b) The alkali used to hydrolyse the halogenoalkane must be neutralised, otherwise a precipitate of silver oxide forms on adding silver nitrate. Nitric acid is suitable because nitrate ions do not interfere with the test.

– – c) CH3CH2CH2CH2Br(l) + OH (aq) → CH3CH2CH2CH2OH(l) + Br (aq)

+ – H (aq) + OH (aq) → H2O(l) Ag+(aq) + Br–(aq) → AgBr(s) 13 Fluorine is highly electronegative so the C–F bond is very polar and nucleophiles are attracted to the highly δ+ carbon. But the C–F bond is also very strong and is not broken by attack of boiling water or hot alkali.

Test yourself (page 216)

14 a) CH3CH2CH2Br + KCN → CH3CH2CH2CN + KBr b) 1-Bromopropane; heat under reflux in ethanol c) Propan-1-ol 15 a) Nucleophile b) Base

16 CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br

17 (CH3CH2)2NH Yes, this compound, diethylamine, has a nitrogen with a lone pair which can act as a nucleophile

with excess bromoethane to form (CH3CH2)3N triethylamine.

+ − This can again react as a nucleophile to form the salt (CH3CH2)4N Br

Test yourself (page 218)

18 a) Propan-2-ol, CH3CH(OH)CH3

b) Propene, CH2=CHCH3

19 Formation of but-1-ene

Formation of but-2-ene

20 Nucleophilic substitution (Figure 6.3.7) is a relatively simple mechanism; one bond forms as one breaks. Elimination (Figure 6.3.10) is more complex and involves movement of three pairs of electrons (three curly arrows). This will have a higher activation energy and therefore require higher temperature than substitution.

Test yourself (page 220)

21 a) CH3CH2CH2OH, primary b) CH3CH(OH)CH3, secondary

c) (CH3)2C(OH)CH2CH3, tertiary d) (CH3)2CHCH(OH)CH3, secondary

22 a) pentan-1-ol b) pentan-2-ol c) 2-methylbutan-2-ol

23 a) 3-Methylpentan-2-ol, secondary b) Cyclohexanol, secondary

c) 2-Methylpropane-1,2-diol (one tertiary OH and one primary OH)

Test yourself (page 221)

24 For example: the lower the boiling temperature the much less volatile the compound. Ethanol (molar mass 46.0 g mol–1) boils at 78 °C, so is more volatile than propane (molar mass 44.0 g mol– 1), which boils at −42 °C. 25 a)

b) Hydrogen bonds are stronger than London forces between non-polar molecules. There is hydrogen bonding between −OH groups in methanol, which is a liquid, unlike an alkane with a comparable molar mass such as ethane. Methanol molecules can break into the hydrogen- bonded network in water so that the two liquids mix. Ethane gas molecules cannot interact so strongly with water molecules. Hydrogen bonding in water excludes non-polar molecules such as ethane. 26 Propane-1,2,3-triol with three OH groups is very viscous because of the large amount of hydrogen bonding between molecules. The air molecules cannot easily pass through the array of linked molecules. Propan-1-ol has much less hydrogen bonding and so there is little restriction of the movement of molecules through it.

Test yourself (page 222)

27 a) C3H7OH(l) + 4½O2(g) → 3CO2(g) + 4H2O(l)

b) C4H9OH(l) + 2O2(g) → 4C(s) + 5H2O(l)

28 CH3CH2CH(OH)CH3 + HBr → CH3CH2CHBrCH3 + H2O 29 Hydrogen chloride fumes in moist air; it turns blue litmus red; and it forms a white smoke with the ammonia gas given off from concentrated ammonia solution.

30 3CH3CH2CH2CH2OH + PI3 → 3CH3CH2CH2CH2I + H3PO3 31 Concentrated sulfuric acid rapidly oxidises iodide ions to iodine (see Section 4.10).

Test yourself (page 228)

2– 3+ 32 Chromium is reduced from the +6 state in Cr2O7 (aq) to the +3 state in Cr (aq).

33 a) Butanal b) Butanoic acid

c) Butanone d) No reaction

+ − 34 a) CH3CH2CH2OH → CH3CH2CHO + 2H + 2e

CH3CH2CH2OH + [O] → CH3CH2CHO + H2O

+ − b) CH3CH2CH2CH2OH + H2O → CH3CH2CH2COOH + 4H + 4e

CH3CH2CH2CH2OH + 2[O] → CH3CH2CH2COOH + H2O c)

35 a) CH3CH(OH)CH3 → CH3CH=CH2 + H2O elimination

Activity: The preparation of 1-bromobutane from butan-1-ol (page 217)

1 a) Possible hazards: sulfuric acid is corrosive; the bromine which forms during the reaction is corrosive and toxic; butan-1-ol is a flammable liquid; heating a liquid in a small flask can lead to bumping; pressure can build up in a tap funnel when gases are evolved. b) Wear eye protection. Carry out the heating under reflux in a fume cupboard. Heat with an electrical heater instead of a flame. Release the pressure in the tap funnel by inverting it and opening the tap from time to time.

2 NaBr(s) + H2SO4(l) → NaHSO4(s) + HBr(g) 3 Concentrated sulfuric acid rapidly oxidises bromide ions or hydrogen bromide to bromine (see Section 4.10). 4 Possible explanations: the reaction is slow and more time is needed for it to go to completion; some of the butan-1-ol forms by-products. 5 The reflux condenser prevents the loss of volatile liquids from the reaction mixture during heating. 6 1-Bromobutane, unchanged butan-1-ol and hydrogen bromide are volatile; they distil over. Ionic salts such as NaBr are involatile and remain in the flask. 7 1-Bromobutane does not mix with water. 8 Hydrochloric acid is an acid which can transfer a hydrogen ion to the −OH group in the alcohol,

+ turning it into −OH2 . In this ionic form, the compound is more soluble in water. 9 Aqueous sodium hydrogencarbonate is a weak alkali which neutralises acids without hydrolysing the product. Sodium hydroxide is a strong alkali which tends to hydrolyse halogenoalkanes. 10 Fractional distillation separates liquids according to their boiling temperatures. Each fraction contains the compounds that distil over in a narrow range of boiling temperatures. The boiling temperature of the product is 102 °C.

11 1 mol butan-1-ol forms 1 mol 1-bromobutane. − Amount of butan-1-ol = 0 .. 81 g cm 33 × 7 5 cm = 0.0821 mol −1 74. 0 g mol Theoretical yield of 1-bromobutane = 0.0821 mol × 136.9 g mol–1 = 11.2 g 6.8 g Percentage yield = × 100% = 61% 11. 2 g 12 There is a variety of explanations: some unchanged butan-1-ol remains in this preparation even with an excess of sodium bromide and concentrated sulfuric acid; some of the butan-1-ol is converted to by-products; some of the product is lost during the process as the chemicals are mixed, heated, distilled, transferred from one container to another, washed, dried and redistilled.

Core practical 4: Investigation of the rates of hydrolysis of halogenoalkanes (pages 212-3)

1 Ethanol is a solvent that allows the halogenoalkanes to mix with an aqueous solution of silver nitrate. 2 Other variables such as the temperature of the halogenoalkane solution and of the silver nitrate solution, the volumes of ethanol, silver nitrate solution and halogenoalkane were all kept constant. 3 If the silver nitrate solution was added to the three tubes one after the other, then there was a slight difference in start time for the three reactions. 4 The precipitate in Tube 2 was silver bromide and the precipitate in Tube 3 was silver iodide. No silver chloride precipitate appeared in Tube 1. In halogenoalkanes the halogen atoms are bound covalently to carbon atoms. Precipitates cannot form until reaction with water (the nucleophile) forms halide ions which can then give precipitates with silver ions. 5 Rate of hydrolysis: RI > RBr >> RCl. 6 Chlorine is the most electronegative and iodine the least electronegative, so the C–Cl bond is the most polar and the C–I bond the least polar. These relative polarities might predict that chloroalkanes would be the most reactive. So bond polarity is not the key factor that determines the rates of reaction. The reaction rates do, however, correlate with the strength of the bonds. The C–I bond is the longest and the weakest (as measured by the mean bond enthalpy). The C–Cl bond is the shortest and the strongest. This suggests that the C–I bond breaks more readily than the C–Cl bond. 7 To ensure the comparison is as fair as possible: different people could add the samples of silver nitrate solution simultaneously so the start times were more similar; the temperatures could

have been controlled by using a common water bath; the volume of each solution could have been measured more carefully than by using drops from a pipette. 8 It is not possible to draw a firm conclusion based on the information available at AS Level, but it is plausible that a change in mechanism accounts for the difference in rate. The following uses ideas from the second half of the A Level course and would not be expected in an AS answer. The study of the rates of these substitution reactions suggests that primary and tertiary halogenoalkanes react with nucleophiles by different mechanisms. The rate of reaction of a primary halogenoalkane, such as 1-bromobutane, with hydroxide ions, is found to depend on the concentrations of both reactants. This suggests that both halogenoalkane molecules and hydroxide ions are involved in the critical step of the reaction. To account for this, chemists have proposed a mechanism in which the C–Br bond breaks at the same time as the nucleophile, OH−, forms a new bond with the carbon atom.

This mechanism is sometimes called SN2 – substitution, nucleophilic, bimolecular. Bimolecular means that two species, the halogenoalkane molecule and the hydroxide ion, are involved in the critical step of the reaction.

Mechanism of nucleophilic substitution of a primary halogenoalkanes, SN2. When it comes to the hydrolysis of tertiary halogenoalkanes, such as 2-bromo-2-methylpropane, it turns out that the concentration of the halogenoalkane affects the rate, but that the concentration of hydroxide ions does not affect the rate. So the mechanism has to account for the fact that the hydroxide ions are not involved in the critical step that controls the rate of reaction. In this case, the suggested mechanism shows that the C–Br bond breaks first to produce an intermediate, a tertiary carbocation. Only after this has happened does the nucleophile, OH−, form a new bond with carbon in a fast second step.

The critical first step involves only one species, so this mechanism is sometimes called SN1, – substitution, nucleophilic, unimolecular.

TIP Your examinations will only test the mechanism involving primary halogenoalkanes.

Mechanism of nucleophilic substitution of a tertiary halogenoalkane.

Core practical 6: Chlorination of 2-methylpropan-2-ol with concentrated hydrochloric acid (page 223)

2 2-Methylpropan-2-ol is flammable – keep away from naked flames Concentrated hydrochloric acid is corrosive – wear gloves and eye protection 3 The substitution reactions of tertiary alcohols are faster than the reactions of primary alcohols. 4 Purification steps: • adding calcium chloride and running off aqueous layer: separates the bulk of the excess hydrochloric acid • shaking with sodium hydrogencarbonate: neutralises any HCl dissolved in the product • adding anhydrous sodium sulfate: dries the product • distillation: separates the product from organic impurities including any unchanged alcohol which boils at a higher temperature. 5 A strong alkali such as sodium hydroxide tends to hydrolyse the product back to the alcohol. 6 Release the pressure through the tap while holding the stopper of the inverted funnel firmly in place with the palm of the hand. 7 The upper organic layer. 8 To filter out the sodium sulfate drying agent. 9 Fractional distillation separates mixture of liquids with different boiling temperature. Here the fraction boiling in the range close to the boiling temperature of the product is collected. 10 a) The mass of 2-methylpropan-2-ol used is 7.86 g. 7. 86 g This is = 0.106 mol 74. 0 g mol−1 This forms 0.106 mol of 2-chloro-2-methylpropane (Mr = 92.5) Theoretical yield = 9.83 g 0. 75 b) Percentage yield = 100 × = 7.73% 9. 83

Core practical 5: The oxidation of ethanol (page 228)

1 The products have distinct smell. One product is acidic while the other is not. The product from Part 2 can be further oxidised by copper(ii) ions but the product from Part 1 cannot. The oxidation product from Part 1 is ethanoic acid. The oxidation product from Part 2 is ethanal.

2 Na2CO3 + 2CH3COOH → 2CH3COONa + H2O + CO2 3 Copper(i) oxide; ethanoic acid.

+ − 4 a) CH3CH2OH + H2O → CH3COOH + 4H + 4e + − b) CH3CH2OH → CH3CHO + 2H + 2e

5 a) CH3CHO < CH3CH2OH < CH3COOH (1) The strongest intermolecular forces in ethanal are dipole-dipole attractions. These are weaker than hydrogen bonding in ethanol and ethanoic acid. Ethanoic acid has O–H and C=O, so hydrogen bonding between its molecules will be stronger than in ethanol. b) In preparation of ethanal, the low boiling temperature of ethanal allows it to evaporate easily from the warm solution. Using the condenser set for distillation allows the ethanal to be distilled away from the oxidising agent and prevents further oxidation. To oxidise the ethanal to ethanoic acid, the aldehyde must not be allowed to escape. The reflux condenser prevents its escape and allows further oxidation to the acid. 6 The pipette will have concentrated sulfuric acid on its outside and must not simply be placed on the bench. Instead, it should be placed in a beaker prior to careful washing. 7 The flame must be removed or extinguished while the apparatus is re-arranged for distillation because flammable vapour will escape from the pear-shaped flask. 8 Clamps should be fixed to the joints of flasks, i.e. at the neck of the pear-shaped flask, at the lower end joint of the condenser. Clamps should not be fixed in the middle of the condenser where the glass jacket is at its weakest.

Exam practice questions (pages 230-2)

1 a) Heat W under reflux [1] with an excess of a solution of potassium dichromate(VI) [1] in dilute sulfuric acid. [1]

b) CH3CH2CH2CHO [1] Use the same reagent but distil off the aldehyde from the reaction mixture as it forms. [1] c) Heat W [1] with a mixture of red phosphorus and iodine. [1] 2 a) Halogenoalkanes do not mix with aqueous reagents. [1] Halogenoalkanes do dissolve in ethanol and the solution then mixes with aqueous reagents. [1] b) Ethanol is highly flammable so it is much safer to heat it in hot water than with a flame. [1]

c) The halogenoalkane hydrolyses producing halide ions – in this instance bromide ions which precipitate as AgBr. [1] AgBr is cream coloured. It is soluble in concentrated ammonia [1], unlike AgI, which is a brighter yellow. d) The compound is a bromoalkane. [1]

3 a) (CH3)2C=CH2 [1]

b) CH3CH2CH2OH [1]

c) CH3CH2CH2NH2 [1]

d) HOOCCH2CH2COOH [1] e)

4 a) From bromoethane: Step 1: KCN [1] reflux in ethanol [1]

CH3CH2Br + KCN → CH3CH2CN + KBr [1]

Step 2: H2 [1] with Ni catalyst to 150 °C [1]

CH3CH2CN + 2H2 → CH3CH2CH2NH2 [1] From 1-bromopropane:

Step 1: excess [1] NH3 [1] in ethanol [1]

CH3CH2CH2Br + 2NH3 → CH3CH2CH2NH2 + NH4Br [1] b) Disadvantage of two-step route: KCN toxic. Disadvantage of one-step route: further reaction may occur because the product amine can act as a nucleophile. a)

b) Step 1: sodium hydroxide [1]; warm, aqueous [1]; nucleophilic substitution [1]

Step 2: acidified potassium dichromate(VI) [1]; reflux [1]; oxidation [1] 6 a) Reaction 1: KOH [1] aqueous; warm [1] Reaction 2: KBr [1] and conc. sulfuric acid [1]

Reaction 3: H3PO4 [1]; hot [1]

Reaction 4: steam [1]; H3PO4 catalyst [1] Reaction 5: KOH [1]; hot ethanolic [1] Reaction 6: conc. HBr [1]

b) Either: In Reaction 4 [1], the major product would be propan-2-ol rather than propan-1-ol. [1] Or: In Reaction 6 [1], the major product would be 2-bromopropane rather than 1-bromopropane. [1] Because, in both cases, the reaction would proceed via the more stable secondary carbocation. [1]

7 a) Transition state formed with CH3CH2CH2CH2Br is:

[1]

Intermediate carbocation formed with (CH3)3CBr is:

[1]

− For tertiary compound, three CH3 groups hinder attack by OH /steric hindrance inhibits formation of intermediate (trigonal bipyramidal). [1] Tertiary carbocation is more stable then primary. [1] b) Rate would increase [1] because the C–I bond is weaker than the C–Br bond. [1] c) Alternative products:

CH3–CH2–CH=CH2 [1] [1]

8 a) Hydrolysis [1] produces 3-methylbutan-2-ol, (CH3)2CHCHOHCH3 [1] b) Elimination [1] of HBr produces two alkenes:

2-methylbut-2-ene [1], (CH3)2C=CHCH3 [1] and

3-methylbut-1-ene [1], (CH3)2CHCH=CH2 [1]

c) Nucleophilic substitution [1] produces 2-amino-3-methylbutane [1], (CH3)2CHCHNH2CH3 [1]

d) CH3CH(CH3)CH(CH3)CN [1], 2,3-dimethylbutanenitrile [1]

9 The mark for each improvement must be linked with the correct improvement. Thermometer bulb should be level with condenser side arm and not in the liquid. [1] The temperature of the vapour as it condenses gives the boiling temperature of the liquid distilling. [1] The water should flow in at the bottom of the condenser and flow out at the top. [1] This ensures all the condenser surface is cooled and improves the efficiency of the condenser. [1] The ethanal should be collected in a conical flask (preferably cooled in ice) rather than in an open beaker. [1] (See Figure 6.3.16 for suitable apparatus.) The boiling point of ethanal is only just above room temperature and it will easily evaporate from an open beaker. [1] 10 Indicative content • Separating funnel needed. • Sodium hydrogen carbonate added to remove acid. • Mixture shaken and tap opened to release any pressure; lower aqueous layer removed. • Water added to remove any sodium hydrogen carbonate; after shaking, lower aqueous layer removed. • Damp cyclohexene run into conical flask and lumps of anhydrous calcium chloride added. • When liquid is clear, dry cyclohexene decanted off or calcium chloride filtered off. 11 a) 1 mol propan-2-ol forms 1 mol propanone. [1] Separated by distillation [1] 11 cm33× 0.78 g cm− Amount of propan-2-ol = = 0.143 mol [1] 60.0 g mol−1

Mass of pure propanone = 0.80 × 0.143 mol × 58.0 g mol–1 = 6.64 g [1] b) 1 mol butan-1-ol forms 1 mol 1-iodobutane [1] 9.25 cm33× 0.81 g cm− Amount of butan-1-ol = [1] 74.0 g mol−1 = 0.101 mol [1] Mass of pure 1-iodobutane = 0.85 × 0.101 mol × 183.9 g mol–1 = 15.8 g [1]

12 a) Q is CH3CH2CH2CH2CH2OCH3 [1] b) i) Nucleophilic substitution [1]

ii) Q is an ether for which the strongest intermolecular forces are dipole–dipole interactions. [1] Alcohols such as pentan-1-ol have hydrogen bonding, which is stronger. [1]

c) i) R is CH3CH2CH2CH2CH2OCH2CH3 [1] ii) P is a strong base [1] so instead of acting as a nucleophile, it can remove a proton from bromoethane [1] in an elimination reaction. [1] d) i) [1] ii) A nucleophilic substitution [1] reaction occurs when the oxide ion on one end of the chain acts as a nucleophile and attacks the δ+ carbon attached to the bromine at the other end. [1] iii) The ring in S has 6 atoms so the internal bond angle is similar to the tetrahedral angle of 109.5°. [1] If 2-bromopentan-1-ol were used instead, the ring would contain only 3 atoms and be unstably strained. [1] iv) S is a cyclic ether so will only have dipole–dipole interactions. [1] Pentan-1-ol with hydrogen bonding will have the higher boiling temperature. [1]

Test yourself (page 236)

1 a) i) Peak S ii) Relative mass is 58.

+ + + b) P: CH3 , Q: C2H5 , R: C3H7

+ c) CH3 is as relatively unstable ion with the charge on a primary carbon atom. Generally fragmentation favours ions with the charge on secondary and tertiary carbon atoms. d) For example:

+ + CH3CH2CH2CH3 → CH3CH2CH2 + CH3

+ + or CH3CH2CH2CH3 → CH3CH2CH2 + CH3

Test yourself (page 237)

2 In mass spectrometry the sequence is: vaporisation, ionisation, fragmentation, acceleration, deflection according to m/z, detection. 3 It is important to have a high vacuum inside a mass spectrometer so that unstable ions and fragments can exist and so that the ions can travel freely through electric and magnetic fields without bumping into other molecules.

+ 4 m/z = 31 corresponds to CH2OH which must come from propan-1-ol by loss of an ethyl group.

+ Loss of a methyl group from propan-2-ol can happen in two ways to leave CH3CHOH , which will produce a major peak at m/z = 45.

Test yourself (page 240)

5 O−H, C−O and C=O bonds are polar and make a big change to the polarity of the molecule as they vibrate. C−C bonds are not polar and only change the polarity of molecules when they vibrate in combination with polar functional groups. 6 a) A:O-H B:C-H C:C-H D:C=O E:O-H F:C-H G:C=O b) Ethanol: spectrum a Ethanal: spectrum b Ethanoic acid: spectrum c c) The broadening of the peaks is due to hydrogen bonding. Hydrogen bonding is more extensive in ethanoic acid than in ethanol due to the presence of C=O as well as O−H. 7 The sample of propanal may be contaminated by unreacted propan-1-ol and/or by water impurity.

8 Boiling temperatures vary with pressure. During fractional distillation the desired product is likely to distil off over a narrow range of temperature. An infrared spectrum provides a distinctive fingerprint for a compound. Any peaks not found in a reference spectrum signify the presence of impurities.

Core practical 7 (part 2) (page 241)

1 R and U could be hexan-2-ol and hexanal as only these two could be oxidised by acidified potassium dichromate(vi) solution. 2 The test tubes should be warmed in a warm water bath. The compounds under test are flammable and could ignite if the test tubes were heated directly by a Bunsen burner. 3 Fehling’s test uses a mixture of a solution of copper(ii) sulfate and a solution containing 2,3-dihydroxybutanedioate ions in strong alkali. This mixture is warmed with the compound under test. Care needs to be taken, for example eye protection and gloves should be worn, as Fehling’s test solution is strongly alkaline. Also the test tube should be heated in a water bath to prevent the risk of the organic compounds catching fire. 4 When Fehling’s solution reacts with an aldehyde, the blue solution forms an orange-brown precipitate of copper(i) oxide. Hence U is the aldehyde, hexanal, and therefore R is hexan-2-ol. 5 Sodium hydrogencarbonate is used to test for acids, which react to liberate carbon dioxide; an effervescence will be seen. 6 Q is therefore hexanoic acid.

CH3CH2CH2CH2CH2COOH + NaHCO3

→ CH3CH2CH2CH2CH2COONa + H2O + CO2 7 The bromine water would be decolorised. 8 P is hex-1-ene

CH3CH2CH2CH2CH=CH2 + Br2 + H2O

→ CH3CH2CH2CH2CH(OH)CH2Br + HBr 9 A cream precipitate would slowly form. 10 S is hexane and T is 2-bromohexane.

CH3CH2CH2CH2CHBrCH3 + H2O → CH3CH2CH2CH2CH(OH)CH3 + HBr

AgNO3 + HBr → AgBr + HNO3 11 a) Hex-1-ene: C=C stretch absorption between 1669 and 1645 cm−1 Hexan-2-ol: broad alcohols O–H stretch absorption between 3750 and 3200 cm−1 Hexanal: aldehydes C=O stretch absorption between 1740 and 1720 cm−1 Hexanoic acid: carboxylic acids O–H stretch absorption between 3300 and 2500 cm−1 OR carboxylic acids C=O stretch absorption between 1725 and 1700 cm−1.

b) The fingerprint regions in the spectra of these two compounds could be compared with the fingerprint regions of known samples of hexane and 2-bromohexane. Exact agreement would confirm the identity of each compound.

Exam practice questions (pages 243-6)

+ + + + + 1 a) Peak 1 is C2H3 ; peak 2 is C2H5 ; peak 3 is CH2OH ; peak 4 is C2H5O ; peak 5 is C2H5OH . ([3] less [1] for each error)

– + – b) CH3CH2OH + e [1] → CH3CH2OH + 2e [1]

+ + c) i) CH3CH2OH [1] → CH3 + CH2OH [1]

ii) The CH3 fragment is an uncharged radical. [1] A mass spectrometer only detects ions. [1] 2 Indicative content: • There is no broad −OH peak as expected for a carboxylic acid, which suggests that the product of oxidation is an aldehyde or a ketone and not a carboxylic acid. • The different C=O absorptions for aldehydes and ketones are too close so which is formed cannot be decided from this spectrum.

• The oxidation was carried out with a limited amount of potassium dichromate(VI) in acid conditions. • If the aldehyde, the product was distilled off as it formed to avoid further oxidation to the acid.

• Possible primary alcohols are CH3CH2CH2CH2OH (butan-1-ol) or (CH3)2CHCH2OH (2-methylpropan-1-ol).

• If a ketone, the possible secondary alcohol is CH3CH2CH(OH)CH3 (butan-2-ol).

35 37 3 a) i) There are two forms of chloroethane, CH3CH2 Cl (relative mass 64) and CH3CH2 Cl (relative mass 66). [1] ii) The chlorine-35 isotope is three times as abundant as the chlorine-37 isotope. [1]

b) i) Mr = 97 uses the relative atomic mass of 35.5 for chlorine, which is the weighted average of the isotopes. [1] There is no individual Cl atom with this mass. [1]

ii) In a molecular ion of C2H2Cl2 with two chlorine atoms the possibilities are: most likely two atoms of chorine-35 (96) [1]; next, one of chlorine-35 and one of chlorine-37 (98) [1]; and least likely two atoms of chlorine-37 (100). [1] The adjacent peaks differ by two mass units and the expected ratio is 9 : 6 : 1. iii) These two peaks are from fragments formed by the loss of one chlorine atom leaving a single chlorine atom in the fragment. [1] So the ratio of abundances corresponds to the ratio of abundances of the isotopes. [1]

4 a) Mr C5H12 = 72.1483

Mr C4H8O = 72.1054

Mr C3H4O2 = 72.0625 ([1] for all three)

So the molecular formula of the compound is C3H4O2 [1] b) Effervescence confirms the presence of acid group COOH [1] so the structure is

H2C=CH–COOH. [1] 5 The absorptions in their IR spectra at about 1720 cm−1 are due to C=O [1] so A and B are carbonyl compounds. Apart from the molecular ion at m/z = 58, the mass spectrum of A had major peaks at m/z = 43

+ + + and 15, i.e. loss of 15 or CH3 leaving C2H3O or CH3C=O [1] and m/z = 15 is CH3 . So A is

propanone, CH3COCH3. [1] Apart from the molecular ion at m/z = 58, the mass spectrum of B had a major peak at m/z = 29,

+ + i.e. loss of 29 or CH3CH2 leaving CHO [1] or loss of CHO leaving CH3CH2 . So B is propanal

CH3CH2CHO. [1] The IR spectrum of C has absorptions at 1645 cm−1 (corresponding to a C=C bond) [1] and at 3300 cm−1 (corresponding to an O–H alcohol bond). [1]

Possible structures could in theory be H2C=CH–CH2OH or HC(OH)=CH–CH3 or H2C=C(OH)–CH3 Apart from the molecular ion at m/z = 58, the mass spectrum of C had major peaks at m/z = 57 and 31.

To form a fragment of m/z = 31 requires loss of 27 or H2C=CH– [1], so C is prop-2-en-1-ol,

H2C=CH–CH2OH. [1] 6 a) A: hexan-3-one [1] B: hex-1-ene [1] C: hexan-1-ol [1] D: 1-chlorohexane [1] b) i) Warm, aqueous [1] NaOH [1]; nucleophilic substitution [1]

ii) Potassium dichromate(VI) in acid [1] Peak at 3750–3200 cm−1 [1] due to O–H (alcohol) should have disappeared if reaction is complete. [1] 7 C: 54.5% H: 9.1% O: 36.4% [1] 54.5 9.1 36.4 [1] 12.0 1.0 16.0 = 4.54 = 9.1 = 2.28

Empirical formula = C2H4O [1]

Mr = 88, so molecular formula is C4H8O2. [1] IR peak at 3408 cm–1 is due to O–H (alcohol) [1] and peak at 1709 cm−1 is due to C=O. [1]

+ + MS peak at m/z = 43 due to CH3CO [1] (cannot be C3H7 because X is not a carboxylic acid).

Possible structures = CH3COCH2CH2OH or CH3COCH(OH)CH3 [1] Oxidation formed compound Y which gave an IR peak at 3087 cm−1. This is due to O–H (acid). [1]

So X must be a primary alcohol, i.e. CH3COCH2CH2OH. [1]

Y is CH3COCH2COOH. [1]

Test yourself (page 248)

1 Exothermic changes b), c), d) Endothermic changes a), e)

2 a) C(s) + O2(g) → CO2(g) ∆H = −394 kJ b)

3 a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

−1 −1 −1 b) M(CH4) = 12.0 g mol + (4 × 1.0) g mol = 16.0 g mol

c) 0.200 g CH4 produces 11.0 kJ 16. 0

16.0 g CH4 produces 0. 200 × 11.0 kJ = 880 kJ d)

Test yourself (page 252)

4 Energy transferred to water = 200 g × 4.18 J g−1 K−1 × 10.0 K = 8360 J = 8.36 kJ Molar mass of butane = (4 × 12.0) g mol−1 + (10 × 1.0) g mol−1 = 58.0 g mol−1 Hence 0.29 g butane transfers 8.36 kJ on burning. 58 1 mol of butane (58.0 g) transfers 0. 29 × 8.36 kJ = 1672 kJ So, the molar enthalpy change of combustion of butane = −1700 kJ mol−1 (2 s.f.)

5 a) HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)

b) Energy transferred to solution = 50 g × 4.18 J g−1 K−1 × 6.5 K = 1365 J = 1.358 kJ 25 3 −3 Moles of HNO3 used = 1000 dm × 1.0 mol dm = 0.025 mol Enthalpy change of neutralisation per mole of water formed 1. 358 kJ − = 0. 025 mol = −54 kJ mol−1 (2 s.f.)

6 a) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

b) CuSO4(aq) is the limiting reagent. Energy transferred to solution = 25 g × 4.18 J g−1 K−1 × 9.5 K = 992.8 J 25 3 −3 Amount of CuSO4 used = 1000 dm × 0.2 mol dm = 0.005 mol Enthalpy change for the molar amounts in the equation 992. 8 J − = 0. 005 mol = −198 560 J mol−1 = −200 kJ mol−1 (2 s.f.)

Test yourself (page 257)

7 A balanced equation for the standard enthalpy change of formation of carbon dioxide is:

C(graphite) + O2(g) → CO2(g) This equation also shows the standard enthalpy change of combustion of carbon (graphite).

8 a) 2Al(s) + 1½O2(g) → Al2O3(s)

b) ½H2(g) + ½Cl2(g) → HCl(g)

c) 3C(graphite) + 4H2(g) → C3H8(g) 9 The temperature rise is the same (6.5 °C). The energy given out from double the amounts of acid and alkali is twice as much but this heats up twice the volume of water.

Test yourself (page 260)

10 The Hess cycle for the formation of methanol with its combustion and the combustion of its constituent elements is shown below.

From the Hess cycle:

∆H2 = ∆fH [CH3OH(l)] + ∆H3

∆cH [C(graphite)] + (2 × ∆cH [H2(g)])

= ∆fH [CH3OH(l)] + ∆cH [CH3OH(l)]

∆fH [CH3OH(l)] = ∆cH [C(graphite)] + (2 × ∆cH [H2(g)]) − ∆cH CH3OH(l)]

–1 –1 –1 ∆fH [CH3OH(l)] = −394 kJ mol + (2 × −286 kJ mol ) − (−726 kJ mol ) = −394 kJ mol–1 − 572 kJ mol–1 + 726 kJ mol–1 = −240 kJ mol–1 11 Standard enthalpies of combustion enable chemists to calculate the standard enthalpies of formation for many substances which cannot be measured directly.

Test yourself (page 261)

12 T12C(graphite) + 11H2(g) + 5 O2(g) → C12H22O11(s) 13 Because standard enthalpy changes should relate to substances in their normal (stable) states at the specified temperature 298 K (25 °C) and 1 atmosphere pressure. Under these conditions, water is a liquid and not a vapour (steam).

Test yourself (page 262)

From the energy cycle, ∆rH = ∆H2 − ∆H1

∆rH = {+∆fH [N2(g)] + 2∆fH [H2O(l)]}

– {∆fH [N2H4(l)] + ∆fH [O2(g)]}

–1 –1 ∆rH = {+(0) + (2 × −286 kJ mol )} – {(+51 kJ mol ) + (0)} = − 572 kJ mol–1 − 51 kJ mol–1 = − 62 kJ mol–1

Test yourself (page 267)

15 a) 2 H−H + O=O → 2 H−O−H b) i) + (2 × 436 kJ) + 498 kJ = 1370 kJ ii) + (4 × 464 kJ) = 1856 kJ iii) Energy released = 1856 − 1370 = 486 kJ 16 a) H−H + Cl−Cl → 2 H−Cl Bonds broken Bonds formed 1 H−H 2 H−Cl 1 Cl−Cl

–1 –1 –1 −1 ∆rH = +436 kJ mol + 243 kJ mol − (2 × 432 kJ mol ) = −185 kJ mol

b) The energy level diagram should have labelled axes; the reactants, H2(g) + Cl2(g) above the product, 2HCl(g); with an arrow from the reactants to the product and labelled ∆H = −185 kJ mol−1. 17 a) The average of the successive O−H bond enthalpies in water 498+ 428 = 2 kJ mol–1 = 463 kJ mol−1 b) This average is very close to the mean bond enthalpy of the OH bond, 464 kJ mol−1. The mean bond energy is averaged over a range of molecules, not just water. 18 a)

Bond Mean bond Bond length/nm enthalpy/kJ mol−1

C−C 347 0.154

C=C 612 0.134

C≡C 838 0.120

18 b) Bond strength: C≡C > C=C > C−C Bond length: C≡C < C=C < C−C As the strength of the bond increases, the length of the bond decreases.

Bonds broken Bonds formed 1 C=C 2 C−H 1 H−H 1 C−C

Enthalpy change = +612 kJ mol−1 + 436 kJ mol−1 − (2 × 413 kJ mol−1) − 347 kJ mol−1 = +1048 kJ mol−1 − 1173 kJ mol−1 = −125 kJ mol−1 20 a) Enthalpies of formation b) Enthalpies of formation are calculated or measured for the specific substance involved whereas mean bond enthalpies are average values for the bond concerned in all substances containing the bond.

21 a) H2(g) + Cl2(g) → 2HCl(g) b) The Cl−Cl bond is weaker than the H−H bond, so the Cl−Cl bond will probably break first. c) The F−F bond is much weaker than the Cl−Cl bond. The H−F bond is stronger than the H−Cl bond. So, less energy is needed to break the F−F bond than the Cl−Cl bond, and more energy is given out as an H−F bond forms compared with an H−Cl bond. The fluorine/hydrogen reaction is much more exothermic than the chlorine/hydrogen reaction.

Activity: Measuring and evaluating the enthalpy change for the reaction of zinc with copper(ii) sulfate solution (pages 253- 254)

1 The graph should have: • axes with scales and labels • points plotted accurately • a clean, smooth curve of best fit including the extrapolation. 2 a) 36.4 ± 0.2 °C b) ∆T = (36.4 ± 0.2) − 24.1 = 12.3 ± 0.2 °C 3 Energy given out = 50 g × 4.18 J g−1 K−1 × 12.3 K = 2571 J 50 4 a) Amount of copper(ii) sulfate = 1000 dm3 × 0.25 mol dm−3 = 0.0125 mol b) From the equation, the same amount of zinc reacts = 0.0125 mol

5 0.0125 mol of Zn and CuSO4 give out 2571 J. 2571

⇒ 1 mol of Zn and CuSO4 give out 0. 0125 J = 205 680 J

−1 ∴ ∆rH = −206 kJ mol 6

Measurement Value Uncertainty Percentage uncertainty

Concentration of 0.25 mol dm−3 ± 0.01 0. 01 copper(ii) sulfate ×=100 4%

0. 25 solution

Volume of copper(ii) 50 cm3 ± 1 1 sulfate solution ×=100 2%

Temperature rise, 12.3 ± 0.2 for the extrapolated 03. ×=100 2. 4% ∆T, estimated from value and ± 0.1 for the 12. 3 the difference in initial temperature two readings = ± 0.3

7 Total percentage uncertainty = 4 + 2 + 2.4 = 8.4% 8 Total uncertainty in the calculated 84. −1 −1 ∆rH = 100 × 206 kJ mol ≈ 17 kJ mol

−1 9 Enthalpy change for the reaction, ∆rH = −206 ± 17 kJ mol 10 The main sources of error are: • loss of energy used to heat the polystyrene cup and the air • estimation of the maximum temperature of the reaction mixture and the temperature rise • measurement of the copper(ii) sulfate solution using a 100 cm3 measuring cylinder • concentration of the copper(ii) sulfate solution given to only 2 s.f. 11 Errors could be reduced by: • preparing copper(ii) sulfate solution with a concentration accurate to 3 s.f. • measuring the volumes of copper(ii) sulfate solution with greater accuracy • using a calorimeter with greater insulation. The accuracy of the result would be improved by repeating the experiment several times and taking an average value.

Core practical 8 (pages 262-263)

1 The amount of water (2.5 mol) is 100 times the amount of MgSO4 (0.025 mol). 2 Both reactions have to end up with the same conditions in order to apply Hess’s Law. In 0.025 mol of the hydrated salt there are 7 × 0.025 mol = 7 × 0.025 mol × 18 g mol−1 = 3.15 g water. So when the experiment is carried out with the hydrated salt the beaker contains 41.85 g + 3.15 g = 45.0 g water, which is the same as it was for the anhydrous salt. 3 a) Mass of anhydrous salt = 3.01 g Temperature change = +11.3 K b) Energy given out by the reaction = 45.0 g × 4.18 J g−1 K−1 × 11.3 K = 2130 J = 2.13 kJ

−1 Amount of MgSO4(s) added = 3.01 g ÷ 120 g mol = 0.0250 mol Enthalpy change for the exothermic reaction,

∆H2 = −2.13 kJ ÷ 0.0250 mol = −85.0 kJ mol−1 4 a) Mass of hydrated salt = 6.16 g Temperature change = −1.4 K b) Energy taken in by the reaction = 45.0 g × 4.18 J g−1 K−1 × 1.4 K = 263 J = 0.26 kJ

−1 Amount of MgSO4.7H2O(s) added = 6.16 g ÷ 246 g mol = 0.0250 mol Enthalpy change for the exothermic reaction,

∆H3 = +0.26 kJ ÷ 0.0250 mol = + 10.4 kJ mol−1

5 According to Hess’s law: ∆H2 = ∆H2 + ∆H3.

So ∆H1 = ∆H2 − ∆H3.

−1 −1 6 ∆H1 = (−85.0 kJ mol ) − (+ 10.4 kJ mol ) = −95.4 kJ mol−1 = −95 kJ mol−1 (2 s.f.) The temperature changes are small. The uncertainty in each temperature reading is ±0.1 K. A total uncertainty of 0.2 K in a 1.4 K change gives rise to a percentage uncertainty that is greater than 10%. The values for the temperature changes can only be calculated to 2 s.f.

7 A Hess’s Law cycle shows that:

ΔrH = sum of ΔfH [products] − sum of ΔfH [reactants]

ΔrH = ΔfH [MgSO4.7H2O(s)] – {ΔfH [MgSO4(s)] + 7 × ΔfH [H2O(l)]} = (−3389 kJ mol−1) – [−1285 kJ mol−1 + (7 × −286 kJ mol−1)] = −3389 kJ mol−1 + 3287 kJ mol−1 = −102 kJ mol−1 The value calculated from standard enthalpy changes of formation lies within the range of uncertainty for the experimental value. 8 A total uncertainty of 0.2 K in a 1.4 K change gives rise to a percentage uncertainty that is greater than 10%. This is a much greater percentage uncertainty than an uncertainty of 0.01 g in a mass of 3.01 g that gives rise to a percentage uncertainty that is less than 0.5%. Using a three-place balance would not significantly reduce the accuracy of the results. 9 One way of improving the procedure would be to aim to reduce energy transfers to and from the surroundings while the solids were dissolving. This could be done by adding a lid to the calorimeter and putting the expanded polystyrene cup inside a second, similar cup to increase the insulation. A second approach is to estimate the rate of energy transfers by recording and plotting a series of temperature measurements for a few minutes before adding a solid until all the solid has dissolved. Each curve can then be extrapolated back to the time when the solid was first added to estimate the maximum, or minimum, temperature to use in the calculations (see the Activity in Section 8.4).

Exam practice questions (pages 269-271)

1 a) Insulated [1] polystyrene beaker/container [1], 0–50 °C thermometer. [1]

b) i) Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s) [1] ii) Energy transferred = 50 g × 4.18 J g−1 K−1 × 5.0 K [1] = 1045 J [1] iii) Any two of: • the density of the solution = 1.0 g cm−3 • the specific heat capacity of the solution = 4.18 J g−1 K−1 • the mass of magnesium added is ignored. [2]

50 3 −3 iv) Amount of CuSO4 used = dm × 0.040 mol dm = 0.0020 mol [1] 1000 1045 J Energy transferred for 1 mol CuSO4 = [1] 0.0020 mol = 522 000 J mol−1 Enthalpy change = −520 kJ mol−1 (2 s.f.) [1]

2 a) Energy needed = 500 g × 4.18 J g−1 K−1 × 80 K = 167 kJ [1] b) Energy from burning butane = 2876 kJ mol−1 Amount of butane burnt to heat the water = 167 kJ ÷ 2876 kJ mol−1 = 0.0581 mol. [1] c) Volume of butane needed = 0.0581 mol × 24 dm3 mol−1 = 1.4 dm3 (2 s.f.) [1] d) The calculation assumes that combustion of butane to carbon dioxide and water is complete [1], and that all the energy from the flame heats the water and not the container or the surroundings. [1]

3 a) CH3COONa(aq) [1] + CO2(g) + H2O(l) [1]

b) To avoid any spillage/loss of reactants due to the effervescence of CO2(g). [1] c) Energy change = 50 g × 4.18 J g−1 K−1 × 8.0 K [1] = 1672 J [1] 50 d) Amount of ethanoic acid = dm3 × 1.0 mol dm−3 = 0.050 mol [1] 1000 e) 0.050 mol ethanoic acid absorbs 1672 J 1672 J Therefore 1 mol ethanoic acid absorbs = 33 400 J mol−1 [1] 0.050 mol Standard enthalpy change of reaction = +33 000 J mol−1 or + 33 kJ mol−1 [2] ([1] for 33, [1] for sign and units) f) Two significant figures because the data for the volume of acid, the concentration of acid and the temperature rise are given to 2 significant figures. [1] 4 a) Upper level [1]; lower levels with labelled axes. [2]

b) Enthalpy change = +44 kJ mol−1 [2] ([1] for sign, [1] for the value)

5 a) The standard enthalpy change of combustion of a substance, ∆cH , is the enthalpy change when one mole of the substance [1] burns completely in oxygen [1] under standard conditions. [1]

b) CH3CH2CH3(l) + 5O2(g) → 3CO2(g) + 4H2O(l) [2] (combustion products [1]; balancing [1])

c) The standard enthalpy change of formation of a compound, ∆fH , is the enthalpy change when one mole of the compound [1] forms from its elements under standard conditions [1] with the elements and the compound in their standard (stable) states. [1]

d) 3C(s, graphite) + 3H2(g)+ ½O2(g) → CH3CH2CHO(l) [2] (elements in correct states [1], balancing [1])

6 a) CaSO4.2H2O(s) → CaSO4(s) + 2H2O(g) [2] ([1] for balanced equation, [1] for state symbols)

b) Endothermic [1], because bonds must be broken between CaSO4 and H2O molecules [1] and no new bonds are formed. c) Because it is difficult to determine just how much energy is required to decompose a known mass of the hydrate. The salt has to be heated to a high temperature at which it is impossible to measure the energy changes. [1]

d) ∆rH = ∆fH [CaSO4(s)] + 2∆fH [H2O(l)] − ∆fH [CaSO4.2H2O(s)] [2] = (−1434 kJ mol−1) + (2 × −286 kJ mol−1) − (−2023 kJ mol−1) [1] = +17 kJ mol−1 [2] ([1] for 17, [1] for sign and units) 7 a) Reaction 1

∆rH = ∆fH [CO2(g)] − ∆fH [SnO2(s)] [1] = (−394 kJ mol−1) − (−581 kJ mol−1) = +187 kJ mol−1 [2] ([1] for 187, [1] for sign and units) Reaction 2

∆rH = 2∆fH [CO(g)] − ∆fH [SnO2(s)] [1] = (2 × −110 kJ mol−1) − (−581 kJ mol−1) = −220 + 581 = +361 kJ mol−1 [2] ([1] for 361, [1] for sign and units) b) Both reactions are endothermic requiring an input of energy using fuels. The energy input and therefore fuel used in Reaction 1 is less than Reaction 2 [1], so on just this basis Reaction 1 is more economic. [1]

8 a) i) Box top right: 2CO2(g) + 3H2O(l) [1]

Box below: 2C(s) + 3H2(g) + 3.5O2(g) [1] ii) The standard enthalpy change of formation of a compound is the enthalpy change when one mole of the compound [1] forms from its elements [1] under standard conditions with the elements and the compound in their standard (stable) states. [1]

−1 iii) ∆H1 = ∆fH [C2H5OH(l)] = −277 kJ mol [1]

∆H2 = 2∆fH [CO2(g)] + 3∆fH [H2O(l)] [1] = (2 × −394 kJ mol−1) + (3 × −286 kJ mol−1) = −1646 kJ mol−1 [1]

∆cH [ethanol] = −∆H1 + ∆H2 [1] = −(−277 kJ mol−1) + (−1646 kJ mol−1) = −1369 kJ mol−1 [2] ([1] for 1369, [1] for sign and units) b) i) Energy transferred = 150 g × 4.18 J g−1 K−1 × 14.8 K [1] = 9279.6 J [1] ii) Amount of ethanol burnt = 0.898 g ÷ 46 g mol−1 = 0.019 52 mol [1] Energy transferred per mole of ethanol burnt = 9279.6 J ÷ 0.019 52 mol = 475 390 J mol−1 = 475 kJ mol−1 (3 s.f.)

−1 ∆cH[ethanol] = −475 kJ mol [2] iii) The experimental method is very crude and greatly underestimates the enthalpy change for the combustion of ethanol. Most of the energy from the flame is not transferred to the water but heats up the rest of the apparatus and the surrounding air. [1] Enthalpy changes of formation in tables of data are derived from enthalpy changes measured by much more sophisticated apparatus such as a bomb calorimeter. [1]

9 a) [4]

([1] for each molecule) b) Bonds broken Bonds formed 3 C−C 8 C=O 10 C−H 10 H−O [1]

1 6 2 O=O [1]

× −1 × −1 1 × −1 c) Enthalpy change = +(3 347 kJ mol ) + (10 413 kJ mol ) + (6 2 498 kJ mol ) [1] − (8 × 805 kJ mol−1) − (10 × 464) kJ mol−1 [1] = +8408 kJ mol−1 − 11 080 kJ mol−1 = −2672 kJ mol−1 [2] ([1] for 2672, [1] for sign and units) d) The bond enthalpies values are averages for a range of molecules and so are not exactly the right values for all the molecules in this reaction. [1] Bond enthalpy values apply to molecules in the gas phase. Water is a liquid at 298 K and the energy change for water condensing from a gas to a liquid alters the value at 298 K. [1]

10 a) Reaction 1

Bonds broken Bonds formed 1 C−H 1 C−Br 1 Br−Br 1 H−Br [1] Enthalpy change = + 413 kJ mol−1 + 193 kJ mol−1 − 290 kJ mol−1 − 366 kJ mol−1 [1] = −50 kJ mol−1 [1] Reaction 2

Bonds broken Bonds formed 1 C−C 2 C−Br 1 Br−Br [1] Enthalpy change = +347 kJ mol−1 + 193 kJ mol−1 −(2 × 290) kJ mol−1 [1] = −40 kJ mol−1 [1] b) Any two good points in an argument [2] such as: • The enthalpy change for Reaction 1 is more negative than for Reaction 2, which might explain why it is the one that is observed to happen. • However the difference is small. As explained in Section 2.6 of Student Book 1, the enthalpy change for a reaction is not a reliable guide to the preferred direction of change. Other factors are involved. • Sometimes it is the reaction that goes faster that is preferred even when an alternative is more exothermic. In this instance, the energy required to break a C−H bond in Reaction 1 is greater than the energy required to break a C−C bond in Reaction 2. Easier breakage of the C−C bond in Reaction 2 could suggest that this reaction is likely to happen more easily. c) Mean bond enthalpies do not always tally with the bond strengths in specific compounds. [1] Mean bond enthalpies are calculated assuming substances are gases whereas bromine in

both reactions and C2H5Br in Reaction 1 are liquids. [1]

11 a) i) C6H12(l) + H2(g) → C6H14(l) [1] A metal catalyst is used, usually one of nickel, palladium or platinum. [1]

ii)

Cycle [1] Balanced combustion products [1]; adding in 9.5 O2 twice. [1]

iii) ∆H1 = ∆cH [C6H12(l)] + ∆cH [H2(g)] = −4003 kJ mol−1 + (−286 kJ mol−1) = −4289 kJ mol−1 [1]

−1 ∆H2 = ∆cH [C6H14(l)] = −4163 kJ mol [1]

−1 −1 −1 ∆rH = ∆H1 − ∆H2 = −4289 kJ mol −(−4163 kJ mol ) = −126 kJ mol [1] b) For all the alkanes the reaction is essentially the same. Each alkene has a C=C double bond. One of the two C−C bonds breaks and an H−H bond breaks. Two new C−H bond form. [1] The only difference in each case is the number of unreactive carbon and hydrogen atoms in the alkene. [1] The molecules are all very similar and so the bond enthalpies have almost the same value. [1] 12 a) i) This is important in many forms of transport but especially in aviation because the fuel has to be lifted. [1] ii) Energy given out per gram from burning methanol: 726 kJ mol–1 ÷ 32 g mol–1 = 22.7 kJ g–1 [1] Energy given out per gram from burning octane: 5470 kJ mol–1 ÷ 114 g mol–1 = 47.5 kJ g–1 [1] b) i) This is important wherever the space for storing the fuel is limited, for example on a motorcycle. ii) Energy given out per cm3 from burning methanol = (726 kJ mol–1 × 0.793 g cm–3) ÷ 32 g mol–1 = 18.0 kJ cm–3 [1] Energy given out per cm3 from burning octane = (5470 kJ mol–1 × 0.703 g cm–3) ÷ 114 g mol–1 = 33.7 kJ cm–3 [1] c) Indicative content: • Octane provides more energy per gram or per cm3. • All the atoms in the fuel react with oxygen when it burns, unlike methanol which includes oxygen atoms in its molecules. • For aviation there is a clear benefit to using the hydrocarbon fuel. • However, octane is obtained from a non-renewable resource

• Octane requires much more oxygen for complete combustion (over 8 times as much per mole) so a much larger volume of air has to be supplied to burn the fuel. • Methanol has the potential to become a more sustainable fuel because it can be made from a wide range of renewable resources, including many wastes.

Test yourself (page 273)

1 a) Keep the moisture level as low as possible, coat the iron with paint or plastic to keep out air and water. b) Turn off the toaster to let the bread cool. c) Keep the milk cool in a refrigerator. 2 a) Warm the dough. b) Blow air onto the fuel to keep up the oxygen concentration. c) Warm the object being repaired. d) Use a catalyst in a catalytic converter.

Test yourself (page 274)

3 a) Rate = 4.8 cm3 s–1 b) Rate = 0.0002 mol s–1 (given that the volume of 1 mol gas is 24 000 cm3)

c) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Rate of removal of Mg(s) = rate of formation of MgCl2(aq) = 0.0002 mol s–1 Rate of removal of HCl(aq) = 0.0004 mol s–1

Test yourself (page 277)

4 a) In the Haber process a mixture of nitrogen and hydrogen gases flows through a reactor containing solid granules of iron. There are two phases in the reactor: a gas phase containing the reactants and a solid phase made up of the catalyst. This is a heterogeneous system. The catalyst is in a separate phase from the reactants. b) The advantage of a heterogeneous catalyst is that it can be held in a reaction vessel as the reactants flow in and the products flow out. There no difficulty in separating the products from the catalyst

Test yourself (page 281)

5 At a higher temperature the molecules move faster so there are more collisions per second, and more of the collisions have an energy greater than the activation energy.

Test yourself (page 282)

6 a) The bottom of the dip between the two peaks of the curve indicates the formation of an intermediate that is unstable relative to both reactants and products.

b) There are transition states at the tops of the two peaks. The height of the left-hand peak is the activation energy barrier from the formation of the intermediate from the reactants. The height of the right-hand peak represents the activation energy barrier that has to be surmounted as the intermediate changes to the products. 7 a) This reaction tends to go but it has a high activation energy. At room temperature the molecules do not collide with enough energy to break the bonds in the reactants and start the reaction. A flame or spark heats the gas mixture and increases the energy of the colliding methane and oxygen molecules. b) The reaction is exothermic, so once it starts it keeps the gas mixture hot enough to continue reacting. 8 Catalysts often form an intermediate by combining in some way with the reactants and this may only happen with one specific reactant for a particular catalyst. Catalysts can be designed to interact with molecules with a particular shape, size and group of reactive atoms. In the case of enzymes and some solid catalysts, specificity arises from the shape of the active site.

Activity: Investigation of the effect of concentration on the rate of a reaction (page 275)

1 The investigation aims to find out how the time taken for the cross to be obscured varies with the concentration of the thiosulfate ions. All other variables that might affect the results must be kept constant. The volume of the solution of sodium thiosulfate solution must be the same each time so that the depth of solution in the flask is constant. This is to ensure that the cross is obscured each time when the same quantity of sulfur has precipitated. Other variables to be kept constant are the volume and concentration of the added hydrochloric acid and the temperature. 2 Mix 40 cm3 of 0.15 mol dm3 sodium thiosulfate solution with 10 cm3 water. 3 The reaction produces sulfur dioxide, some of which can escape from the solution as a gas. Sulfur dioxide gas can cause breathing problems, especially for asthmatics. Sulfur dioxide is an acidic oxide that reacts with alkaline sodium carbonate solution to form sodium sulfite, which is ionic and stays in solution. 4 The build-up of a translucent layer of sulfur on the bottom of the flask shortens the time taken for a precipitate in the next experiment to obscure the cross. 5 Rate = 0.018 s−1 6 Suitable graph with scales, labelled axes and accurately plotted points. 7 The rate of reaction is directly proportional to the concentration of thiosulfate ions.

Activity: Investigating the effect of surface area on the rate of a reaction (page 276)

1 a) The experiment is designed to measure the mass of carbon dioxide given off by measuring the loss in mass as carbon dioxide escapes from the flask. The total mass of the apparatus and other chemicals has to be recorded – at first with the marble in a suitable holder beside the flask, then once the reaction has started. Everything must be on the balance throughout so that the only change in mass is that caused by the escape of carbon dioxide gas. b) The reaction of marble with acid produces gas bubbles which may produce a spray above the solution in the flask. The cotton wool plug ensures that only gas escapes from the flask and not liquid droplets. 7 Graph with appropriate scales, labelled axes and accurately plotted points joined by a smooth curve.

Exam practice questions (page 284)

1 a) Apparatus described in words or a diagram to include: flask, bung and delivery tube [1] leading to a gas syringe or to a graduated cylinder inverted over water. [1] Any appropriate technique for adding the catalyst after the bung is in place, such as standing a small sample tube containing the catalyst in the flask until the bung is in place [1], and then shaking to tip over the tube and mix the catalyst with the solution. [1]

b) 2H2O2(aq) → 2H2O(l) + O2(g) [1] c) Appropriately labelled axes with scale [1]; accurate plot with smooth curve. [1] d) The reaction is fastest at the start so that is when the graph is steepest. [1] As the hydrogen peroxide concentration falls the rate decreases [1] and the graph becomes less steep finally levelling off when the reaction stops. [1] e) i) Graph steeper at the start [1] but levelling off to the same final volume of gas collected at room temperature. [1] ii) Initial rate unchanged [1] but final volume of gas is 80 cm3. [1] iii) Slower initial rate [1] but the final volume of gas is unchanged at 40 cm3. [1] iv) Slower initial rate [1] and the final volume of gas is 20 cm3. [1] 2 a) Only collisions with energy greater than the activation energy lead to reaction. [1] The activation energy is typically well to the right of the Maxwell–Boltzmann distribution, so only a minority of colliding molecules have enough energy to react. [1] b) By raising the concentration or increasing the pressure so that reactant molecules are closer together. [1]

c) Raising the temperature shifts the Maxwell–Boltzmann distribution to the right. [1] The value of the activation energy does not change [1] but the area under the curve of the distribution representing the number of molecules with enough energy to react can increase markedly for a small rise in temperature. [1] 3 Graph similar to Figure 9.16(b) in the Student’s Book, but differing in detail: • curve with two peaks showing reactants, intermediate and products [1] • products at a higher energy level than the reactants [1] • activation energy for the change from reactants to intermediate greater than the activation energy for the change of the intermediate to products. [1] 4 a) Maxwell–Boltzmann distribution sketched with the right shape and with labelled axes as in Figure 9.15 in the Student’s Book. [2] b) Two activation energies drawn in to the right-hand end of the distribution [1] with the activation energy for the reaction with the catalyst being lower than that for the uncatalysed reaction as in Figure 9.15. [1] c) Area under the distribution to the right of the activation energy indicates the number of molecules with enough energy to react. [1] This area is greater when the activation energy is lower in the presence of a catalyst [1] so more collisions can lead to reaction. [1] 5 a) Up to 4 marks for key points such as the following: Energy of collisions not great enough to break bonds at room temperature. [1] A spark raises the temperature of the gas mixture. [1] At a higher temperature more molecular collisions have enough energy for reaction. [1] Platinum acts as a catalyst for the reaction [1] by providing a reaction pathway with a lower activation energy [1] such that collisions at room temperature have enough energy for reaction. The reaction is highly exothermic so it gets hotter and hotter once it starts so that the reaction gets very fast. [1] b) Up to 4 marks for key points such as the following: Double bond in oxygen molecules and triple bond in nitrogen molecules are strong. [1] Nitrogen is a relatively inert gas. [1] At room temperature molecular collisions do not have enough energy for reaction. [1] In the cylinder of an engine the gas mixture is hot and under pressure. [1] Under these conditions there is enough energy in the molecular collisions to overcome the activation energy. [1] c) Up to 4 marks for key points such as the following: Lumps of coal or sacks of flour are combustible but they burn slowly. [1] In a mine or mill some of the coal or flour becomes finely divided so that it has a high surface area making it potentially much more reactive. [1] Activity in mine or mill can raise dust unless care is taken to damp it down. [1] The mixture of air and finely divided fuel can lead to a very rapid reaction if ignited. [1]

6 Indicative content: • an outline of collision theory and … • … the concept of activation energy • the Maxwell–Boltzmann distribution and the values of activation energies in relation to the energies of the bulk of the molecules in a gas mixture • reasons why activation energies vary depending on the chemical nature of the reactants • the effects of temperature, concentration, catalysts and … • … the explanation of these effects in terms of collision theory, activation energy and the Maxwell–Boltzmann distribution • the state of the reactants (solid, liquid, gas) and the factors affecting rates at the interface between phases.

7 Indicative content: • Models are the basis for explanations for observed patterns/generalisations. • For example: the gas laws summarise patterns in observations; or the kinetic theory explains the patterns; or observations of the effect of different factors on the rates of reaction are explained in terms of collision theory. • Models are based on a set of assumptions which may too simple and not fully reflect reality. • This is illustrated by the assumptions underlying kinetic theory which apply to ideal gases but not exactly to real gases. • Good models can explain a wide range of phenomena. • For example: kinetic theory can account for the differing effects of pressure, temperature and volume changes on a gas; or collision theory can account for the effects of changes in temperature, concentration surface area and catalysts on the rates of reactions. • Models can be used to make predictions that can be tested. • Ideally models make possible quantitative predictions that can be tested this applies to both these models (though this has not been covered in this chapter).

Test yourself (page 287)

1 a) CaCO3(s) + CO2(aq) + H2O(l) → Ca(HCO3)2(aq)

b) Ca(HCO3)2(aq) → CaCO3(s) + CO2(aq) + H2O(l) 2 a) Raising the temperature makes the ice melt again. b) Adding alkali turns the litmus blue again. c) Allowing the white solid to cool and then adding water re-forms the blue compound. 3 In a sealed bag the water in the washing evaporates until the air in the bag is saturated with water vapour. The liquid and gaseous water are in equilibrium with equal rates of evaporation and condensation. In the open the moisture is blown away and so evaporation can continue until the clothes are dry.

Test yourself (page 289)

4 a) At 0 °C b) At 100 °C (if the pressure is normal atmospheric pressure) c) When the solution is saturated 5 The diagram shows particles close together and regularly arranged in a solid, crystal lattice. Arrows and labelling indicate that some of these particles are leaving the solid state and passing into solution where they mix with solvent molecules. In the diagram dissolved solute particles are spread out between solvent molecules. More arrows indicate some solute particles returning to the solid crystal. When the solution is saturated solute particles return to the solid from solution as fast as other particles dissolve.

Test yourself (page 290)

6 Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq)

Disturbance ⇌ How does the The result equilibrium mixture respond?

Increasing the It moves to the right More Ag(s) precipitates. The concentration of solution appears more yellow Fe2+(aq)

Lowering the It moves to the left Some Ag(s) reacts. The solution concentration of Ag+(aq) is less yellow

2– + 2– 7 2CrO4 (aq) + 2H (aq) Cr2O7 (aq) + H2O(l)

+ a) Adding acid: increases⇌ the concentration of H (aq), this shifts equilibrium to the right, turns from yellow to orange.

b) Adding alkali: lowers the concentration of H+(aq), this shifts equilibrium to the left, turns from orange to yellow.

8 CaCO3(s) CaO(s) + CO2(g)

Allowing the⇌ carbon dioxide to escape stops the back reaction so the system cannot reach equilibrium. The forward reaction continues until all the limestone decomposes.

Test yourself (page 292)

9 Temperature: the forward reaction producing ammonia is exothermic so, according to Le Chatelier’s principle, the proportion of ammonia at equilibrium is increased by lowering the temperature. However, the lower the temperature the slower the reaction. So a compromise temperature is chosen that allows a significant proportion of nitrogen and hydrogen to react to form ammonia while the gases are flowing through the catalyst. Pressure: the equation shows that 4 mol of gas molecules react to form 2 mol of ammonia. So, according to Le Chatelier’s principle, the proportion of ammonia at equilibrium is increased by raising the pressure. However, the higher the operating pressure the higher the capital cost of building a chemical plant to withstand the pressure and the higher the running costs. In practice, a compromise pressure is chosen.

10 a) CO(g) + 2H2(g) CH3OH(g)

b) Spreading the catalyst⇌ over the surface of pellets increases the surface area exposed to the reacting gases and so speed up the reaction. If the catalyst is expensive, then coating it onto an inert support makes it possible to fill the reactor with a large area of active catalyst relatively cheaply. c) Le Chatelier’s principle suggests that the conversion of the reactants to methanol by this exothermic reaction is favoured by high pressure and a low temperature. The limits to the pressure arise from the capital and running costs of a plant operating at high pressures. The limit to the low temperature arise from the need for the reaction to be fast enough. The catalyst is inactive and the reaction too slow if the temperature is too low.

Test yourself (page 294)

11 a) H2(g) + I2(g) 2HI(g)

2 [HI (g)]⇌ Kc = [H22 (g)][I (g)]

b) 2NO(g) + O2(g) 2NO2(g)

2 [NO2 (g)]⇌ Kc = 2 [NO(g)] [O2 (g)]

c) N2(g) + 3H2(g) 2NH3(g)

2 [NH3 (g)]⇌ Kc = 3 [N22 (g)][H (g)]

Test yourself (page 294)

4 [H2 (g)] Kc = 4 12 a) [H2 O(g)] [H S(g)] K = 2 c [H (g)] b) 2 [Fe3+ (aq)] Kc = ++2 c) [Ag (aq)][Fe (aq)]

13 a) H2(g) + I2(g) 2HI(g)

b) H2O(g) + CO(g)⇌ H2(g) + CO2(g) 3+ 2+ 2+ c) 2Cr (aq) + Fe(s)⇌ 2Cr (aq) + Fe (aq) ⇌ Test yourself (page 295)

14 Pure water contains (1000 g ÷ 18 g mol-1) = 55.6 mol dm-3 of water molecules. So in dilute solutions, even up to 1 mol dm-3 of reactants, the change in the concentration of water molecules is negligible, even if the water molecules are involved in the reaction.

Activity: An equilibrium involving iodine and chlorine (page 291)

1 Chlorine is a toxic gas. Iodine and iodine monochloride are both harmful. The demonstration must be carried out in a fume cupboard. Precautions must be taken to avoid skin contact with the iodine at the start and the iodine monochloride at the end.

2 I2(l) + Cl2(g) → 2ICl(s)

3 ICl(l) + Cl2(g) ICl3(s)

4 When the chlorine⇌ supply is stopped and the U-tube is stoppered. A dynamic equilibrium can only exist in a closed system when the forward and reverse reactions take place at equal rates. 5 Only the forward reaction is possible on first adding chlorine to the brown liquid so the trichloride forms faster than it decomposes. At room temperature and pressure most of the ICl is converted to the yellow solid so long as there is excess chlorine to maintain the rate of the forward reaction. 6 Pouring out the dense chlorine gas lowers its concentration in the tube so the forward reaction with ICl slows while the back reaction continues at the same rate. Once all the chlorine has poured from the tube, the ICl3 disappears.

Activity: The manufacture of ethanol (page 292)

1 Ethene is made by cracking hydrocarbons. The hydrocarbons can be ethane from natural gas or one of the fractions from distilling crude oil. 2 The equation shows that 1 mol ethene reacts with 1 mol steam. Le Chatelier’s principle suggests that an excess of steam would move the position of equilibrium to the right and increase the percentage of ethene converted to ethanol. However, too much steam dilutes the catalyst and can wash it off the support. To avoid this, less steam is used than is needed to convert all the ethene to ethanol. 3 The forward reaction is exothermic. Le Chatelier’s principle suggests that the temperature should be as low as possible to shift the equilibrium to the right. However, the lower the temperature the slower the reaction. The choice of 500 K is the compromise between equilibrium and rate factors that allows 5% of the ethene to react with steam in the short time it is in the presence of the catalyst. 4 The equation shows that 2 mol of gaseous reactants form 1 mol of gaseous product in the reactor. Le Chatelier’s principle shows that the higher the pressure the further the equilibrium position favours the product side of the reversible reaction. However, the pressure must not be too high because ethene polymerises to polythene under pressure. Polymerisation wastes ethene and damages the reactor by forming plastic where it is not wanted. 5 Cooling the gases leaving the reactor condenses ethanol with water. Unchanged ethene does not condense and can be recycled and fed back into the reactor with more steam. 6 Fractional distillation can be used to concentrate ethanol. There is a limit to this. It turns out to be impossible to obtain a distillate that contains more than 95% ethanol. This is because a mixture of 95% ethanol and 5% water distils without changing composition. Other methods have to be used to obtain pure ethanol.

Exam practice questions (pages 296–7)

1 a) CO2(g) + aq CO2(aq) [1]

b) At equilibrium,⇌ CO2 molecules are constantly leaving the solution and entering the gas phase. They are also leaving the gas phase and passing into solution. [1] The two processes happen at the same rate and so overall cancel each other out. [1] c) Reducing the pressure on the gas slows the rate of dissolution but does not affect the rate that gas molecules leave the solution and enter the gas phase. [1] The net effect is that overall gas bubbles out of solution. [1] d) Adding an alkali neutralises the hydrogen ions and removes them from the equilibrium. [1] As a result the equilibrium shifts to the right and more carbon dioxide dissolves. [1]

2 a) In the lungs the oxygen concentration is relatively high. [1] This pushes this equilibrium over to the oxygenated Hb side. [1] b) Around muscle cells the oxygen concentration is relatively low. [1] This means that the equilibrium shifts over to the deoxygenated Hb side. [1] c) The carbon dioxide concentration is higher where cells are respiring and releasing carbon dioxide into the blood. [1] This helps the Hb to release oxygen to the respiring cells that need it. [1] 3 a) i) The reaction makes no difference the amount of gas in moles [1], so this change is not affected by raising the pressure. [1] ii) The reaction is slightly exothermic. Raising the temperature tends to reduce the formation of HI. [1] b) i) There is a slight decrease in volume when salt dissolves in water so raising the pressure tends to favour the solution side of the equilibrium. [1] ii) It is an endothermic process. [1] Raising the temperature tends to increase the solubility of the salt. [1] c) i) The change from graphite to diamond leads to a reduction of volume. [1] This is favoured by high pressure. [1] ii) The formation of diamond from graphite is endothermic [1] so it is favoured by high temperatures. [1]

4 a) I2(s) + aq I2(aq)

This equilibrium⇌ is well over to the left-hand side. [1] However in KI(aq) the iodide ions react − with iodine to form I3 ions. [1] This removes free aqueous iodine molecules from the above equilibrium, which favours the forward reaction, and so more iodine dissolves. [1] b) The dark yellow-brown colour is caused by the presence of the tri-iodide ions. [1]

– – I2(aq) + I (aq) I3 (aq)

Thiosulfate ions⇌ react with iodine molecules, removing them from the above equilibrium. [1] 2− 2− – 2S2O3 (aq) + I2(aq) S4O6 (aq) + 2I (aq)

The equilibrium shifts⇌ to the left so the concentration of tri-iodide ions falls. [1] In dilute – solution the tri-iodide ion is pale yellow. [1] If sufficient thiosulfate is added, the I3 (aq) concentration is so low that the solution becomes colourless. [1]

5 a) 2SO2(g) +O2(g) 2SO3(g) [1]

b) Adding an excess⇌ of oxygen gas [1] to the reaction mixture tends to push the equilibrium to

the right and favour the conversion of SO3 to SO2. [1] The reaction from left to right reduces the amount of gas in moles, [1] so this change is favoured by raising the pressure. [1] The reaction is highly exothermic. [1] Lowering the temperature tends to increase the

proportion of SO2 converted to SO3 at equilibrium. [1]

c) There is a limited flow rate of gases through the reactor. [1] Adding more air to give an excess of oxygen means that less of the gas mixture flowing through the reactor in a given time is

sulfur dioxide. [1] This increases the percentage of SO2 converted to SO3 but reduces the

amount of SO3 formed in a given time. [1] d) Indicative content: • The chosen temperature is a compromise. • It must be as low as possible … • … but high enough for the catalyst to be active and… • … for the product to form quickly enough.

• The percentage conversion of SO2 to SO3 in this process is sufficiently high at atmospheric pressure … • … so that the cost of operating the plant at a higher pressure is not justified.

6 a) H2(g) + Cl2(g) 2HCl(g) [1]

[HCl(g)]⇌2 Kc = numerator [1], denominator [1] [H2 (g)][Cl2 (g)]

b) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) [1]

4 6 [NO(g)] [H2O(g)]⇌ Kc = 4 5 numerator [1], denominator [1] [NH3 (g)] [O2 (g)]

c) NH4HS(s) NH3(g) + H2S(g) [1]

Kc = [NH3(g)][H⇌ 2S(g)] numerator [1], no denominator [1]

7 a) CH4(g) + H2O(g) 3H2(g) + CO(g) [1]

The reaction is endothermic.⇌ [1] The amount of gas in moles increases from left to right. [1] Formation of products is favoured by: high temperatures [1], low pressures [1] and an excess of steam. [1]

b) i) CO(g) + H2O(g) H2(g) + CO2(g) [1]

[H22 (g)][CO (g)] Kc = ⇌ numerator [1], denominator [1] [H2 (g)][CO(g)]

ii) At the higher temperature the value of Kc is smaller. This means that as the temperature rises the proportion of products at equilibrium decreases. [1] Le Chatelier predicts that raising the temperature shifts the equilibrium of an exothermic reaction to the left. [1] The reaction is exothermic. [1] iii) The temperature has to be high enough for the catalyst to be active [1] and the reaction to proceed fast enough. [1] c) Zeolites are crystalline materials with channels through which small molecules (hydrogen) and pass freely [1] while larger molecules (carbon dioxide) are trapped. [1]

d) Carbon dioxide is a greenhouse gas [1] and so processes are designed to minimise its release into the atmosphere. [1] The carbon dioxide captured can be useful. [1] (For example: in systems to extinguish fires; in the solid state as a coolant in mobile refrigeration systems; to control pH levels in paper making and water treatment; and as a gas under pressure to increase the recovery of oil from oil wells.) e) Hydrogen is used on a large scale to manufacture ammonia and methanol. It is used in oil refining (for example to remove sulfur). The gas is also been used on a small scale in fuel cells. (Any two for [2])

Test yourself (page 299)

1 a) This is true. At equilibrium there is no apparent change taking place and the concentrations of reactants and products are constant. b) This is not true. Some reversible reactions go almost to completion before they reach an equilibrium state. For other reactions, the equilibrium mixture consist largely of reactants. c) This is true. Chemical equilibrium is dynamic at a molecular scale. Forward and back reactions proceed at the same rate so that overall there is no change. 2 a) The solid and liquid states of a pure chemical at the melting temperature, such as ice and water at 273 K. b) Any salt in equilibrium with its saturated solution in water. c) The distribution of a solute (such as iodine) between solutions in aqueous (such as aqueous potassium iodide) and a non-aqueous solvent. d) Any reversible reaction at equilibrium, such as the reaction of nitrogen with hydrogen to form ammonia.

3 a) Figure 11.2: At first there are only H2 and I2 molecules. As they collide and react, HI molecules

form. As the concentrations of H2 and I2 fall, the rate of the forward reaction decreases. As the concentration of HI rises there is a growing number of collisions between these molecules so the rate of the backward reaction increases. Equilibrium is reached when the rates of the forward and backward reactions are equal.

b) Figure 11.3: At first there are only HI molecules. As they collide and react, H2 and I2 molecules form. As the concentration of HI molecules falls, the rate of the forward reaction decreases.

As the concentrations of H2 and I2 molecules rise there is a growing number of collisions between these molecules so the rate of the backward reaction increases. Again, equilibrium is reached when the rates of the forward and backward reactions are equal.

Test yourself (page 301) [NO(g)]2 4 a) Kc = [N (g)][O (g)] 22 [NO(g)] Kc b) = 1/2 1/2 [N22 (g)] [O (g)]

2 [NH3 (g)] c) Kc = [N (g)][H (g)]3 22 [N (g)][H (g)]3 d) Kc = 22 [NH (g)]2 3 5 A system is homogeneous if all the substances involved are in the same phase.

Test yourself (page 4)

−−33 6 Kc = [PCl 32 (g)][Cl (g)] = 0.30 mol dm× 0.12 mol dm −3 [PCl5 (g)] 0.72 mol dm

−1 3 Kc = 0.05 mol dm (mol dm−32 ) 7 a) Units: = dm3 mol−1 (mol dm−−32 ) (mol dm3 ) b) Square root of the value for the equation in the question:

3 1.5 −0.5 Kc = 1.26 × 10 dm mol c) The reciprocal of the value for the equation in the question:

−7 −3 Kc = 6.25 × 10 mol dm 8 The system is not at equilibrium

3 −1 Substituting in the expression forKc gives a value of 3750 dm mol . This is much greater than

the given value of Kc. There is a tendency for the equilibrium to shift to the left to increase the

concentration of NO2.

Test yourself (page 305)

9 In a heterogeneous system the substances involved are in more than one phase. 10 Before the bottles was opened it contained a gas phase consisting of a small volume of carbon dioxide under pressure in equilibrium with an aqueous phase. The aqueous phase is a solution of carbon dioxide in water.

2 –6 11 a) Kc = [H2S(g)][NH3(g)]; units: mol dm [Sn2+ (aq)] b) Kc = [Pb 2 + (aq)] ; units: none 2 –3 c) Kc = [HCl(aq)] ; units: mol dm [BiCl3 (aq)] 12 In an undisturbed cave where water is slowly dripping from stalactites. The stalactites consist of solid calcium carbonate. The water is a dilute solution containing dissolved carbon dioxide and calcium hydrogencarbonate. 13 Pure water contains (1000 g ÷ 18.0 g mol−1) = 55.6 mol dm−3 of water molecules. So in dilute solutions, even up to 1 mol dm−3 of reactants, the change in the concentration of water molecules is negligible, even if the water molecules are involved in the reaction.

Test yourself (page 306)

14 a) Under these conditions, the forward reaction tends to go essentially to completion. b) Under these conditions, the backward reaction tends to go essentially to completion. c) Under these conditions, the equilibrium mixture contains significant amounts of reactants and products.

© Andrew Hunt, Graham Curtis, Graham Hill 2019 11 Equilibrium ΙΙ Answers 15 Correspondingly small. The value of the equilibrium constant of the reverse reaction is the reciprocal of the value for the forward reaction.

Test yourself (page 307) 15.6 4.4 16 a) Mole fraction nitrogen = = 0.78; mole fraction oxygen = = 0.22 20 20 b) Partial pressure of nitrogen = 0.78 × 1 atm = 0.78 atm Partial pressure of oxygen = 0.22 × 1 atm = 0.22 atm 17 a) Amount of propane = 22 g ÷ 44.0 g mol−1 = 0.50 mol Amount of 2-methylpropane = 11 g ÷ 58.0 g mol−1 = 0.19 mol Total amount of gas = 0.69 mol 0.50 Mole fraction of propane = = 0.72 0.69 Mole fraction of 2-methylpropane = 1 − 0.72 = 0.28 b) Partial pressure of propane = 0.72 × 1.5 atm = 1.1 atm The total pressure is the sum of the partial pressures. Partial pressure of 2-methylpropane = 1.5 atm – 1.1 atm = 0.4 atm

Test yourself (page 309)

2 26 ()pSO ()pp× ( ) 3 −1 N22 HO Kp = 2 ; units: atm Kp = ; units: atm 18 a) (pp )× b) 43 SO22 O (pp )× () NH32 O c) Kp = pCO2 ; units: atm ()p 2 19 Kp = NO2 p NO24 3.8 pNO2 = × 1.20 atm = 0.383 atm 8.1+ 3.8 pN2O4 = (1.20 − 0.383) atm = 0.817 atm (0.383 atm)2 Kp = = 0.18 atm 0.817 atm

pp× 20 Kp = CH24 H 2 p CH26 Equilibrium amounts are:

Amount C2H6 = (5.0 − 1.8) mol = 3.2 mol;

Amount C2H4 = 1.8 mol = amount H2 3.2 pC2H6 = × 1.80 atm = 0.847 atm 6.8

1.8 pC2H4 = pH2 = × 1.80 atm = 0.476 atm 6.8

(0.476 atm)2 Kp = = 0.27 atm 0.847 atm

21 Adding alkali neutralises H+(aq) ions. So the concentration of aqueous hydrogen ions in solution falls. As predicted by Le Chatelier’s principle, the equilibrium shifts in the direction that forms more H+(aq) ions. Dichromate(vi) ions change to chromate(vi) ions. The solution turns from orange to yellow.

22 a) At equilibrium: Kc = [ethyl ethanoate][water] [ethanol][acid] Adding more ethanol raises [ethanol] in the mixture.

Temporarily, Kc > [ethyl ethanoate][water] [ethanol][acid] So some of the added ethanol reacts with some acid to form more ethyl ethanoate and water until the concentrations of ester and water have risen enough, and the concentrations of alcohol and water fallen enough until once again:

Kc = [ethyl ethanoate][water] [ethanol][acid] b) The answer to part (a) shows that the equilibrium shifts to the right producing more product. As predicted by Le Chatelier’s principle, this tends to counteract the effect of adding extra ethanol.

Test yourself (page 312)

23 a) In the equation: 3 mol reactants 2 mol product. Raising the pressure increases the

proportion of the product in the equilibrium⇌ mixture. b) In the equation: 2 mol reactants 4 mol products. Raising the pressure decreases the

proportion of the products in the⇌ equilibrium mixture. c) In the equation: 2 mol reactants 2 mol product. Raising the pressure has no effect on the

composition of the equilibrium mixture.⇌ 24 In the mixture: 2 2 pNO (1.2 atm) 2 = = 0.6 atm > Kp (= 0.11 atm) p 2.4 atm NO24

The system is not at equilibrium. The partial pressure of N2O4 tends to increase. 25 a) The new equilibrium mixture contains more methanol and less carbon monoxide. b) The new equilibrium mixture contains more methanol and less hydrogen and carbon monoxide. c) The new equilibrium mixture contains less methanol and more carbon monoxide. With the total pressure the same, the addition of argon with its partial pressure, reduces the partial pressures of the other gases. (The presence of argon in reactant mixtures is a practical problem in processes such as the manufacture of ammonia where argon enters the system

© Andrew Hunt, Graham Curtis, Graham Hill 2019 11 Equilibrium ΙΙ Answers with air used to supply oxygen. Argon builds up in the process as unreacted gases are constantly separated from the product and recycled.)

Test yourself (page 313)

26 As the temperature rises the value of Kp gets smaller. This means that as the temperature rises there is more sulfur dioxide and oxygen at equilibrium and less sulfur trioxide. This is as predicted by the Le Chatelier’s principle because the equilibrium shifts in the direction that is endothermic as the temperature rises.

27 a) Raising the temperature favours the decomposition of the pale yellow gas N2O4 and the

formation of the dark brown NO2. b) In Figure 11.8 the gas mixture is darkest when in hot water and palest when in iced water. This as predicted in (a).

c) The decomposition of N2O4 gas must be endothermic (ΔH is positive), since the equilibrium shifts to the right as the temperature rises. 28 Less HI(g) is present in the equilibrium mixture as the temperature rises. The reaction of hydrogen with iodine is exothermic. The reverse reaction, the decomposition of HI(g), is endothermic.

Activity: Testing the equilibrium law (page 299)

1 H2(g) + I2(g) 2HI(g)

3 −2 2 a) In tube 1,⇌ the amount of the initial iodine in each dm that has reacted is 1.26 × 10 mol.

1 mol I2 reacts with 1 mol H2. So the expected amount of hydrogen left at equilibrium is (2.40 − 1.26) × 10−2 mol dm−3 = 1.14 × 10−2 mol dm−3. b) In tube 2, the amount of the initial iodine in each dm3 that has reacted is 1.48 × 10−2 mol.

1 mol I2 produces 2 mol HI. So the expected amount of HI at equilibrium is (2 × 1.48) × 10−2 mol dm−3 = 2.96 × 10−2 mol dm−3 3 In tubes 5 and 6, the tubes contained only HI at the start. When HI decomposes it produces equimolar amounts of hydrogen and iodine.

4 a) and b) 2 [HI (g)]eqm [HI (g)] eqm

[H2 (g)] eqm [I 2 (g)] eqm [H2 (g)] eqm [I 2 (g)] eqm

The expression that corresponds to Kc does indeed give a value that is almost constant.

5 The mean value for Kc at 731 K is 46.7. Note that the units cancel in this example.

Activity: Analysing the result of an experiment to measure Kc (page 303)

1 The acid was added as a catalyst. It was not used up in the reaction. 2 Molar mass of HCl = 36.5 g mol−1. Mass of 0.010 mol HCl = 0.365 g Mass of water in the added hydrochloric acid = 5.17 g − 0.365 g = 4.81 g 3 Mass of water added at the start = 0.99 g + 4.81 g = 5.80 g Amount of water added at the start = 5.80 g ÷ 18.0 g mol−1 = 0.322 mol

4 Both HCl and CH3COOH react 1 : 1 with NaOH. Total amount of acid at equilibrium from the titration 39.2 = dm3 × 1.00 mol dm–3 = 0.0392 mol 1000 Taking away the amount of hydrochloric acid added at the start gives (0.0392 mol − 0.010 mol) = 0.0292 mol ethanoic acid 5 The equation shows that: • the amount of ethanol formed from the ester is the same as the amount of ethanoic acid formed • the amount of ester left at equilibrium = 0.0413 mol − 0.0292 mol = 0.0121 mol • Th amount water left at equilibrium = 0.322 mol − 0.0292 mol = 0.293 mol

[CH3 COOC 25 H (l)][H 2 O(l)] (0.0121 /VV )(0.293 / ) 6 Kc = = = 4.2 [CH3 COOH(l)][C25 H OH(l)] (0.0292/VV )(0.0292/ ) 3 7 In the expression for Kc, V is the volume of the mixture in dm . Here the volume terms cancel because there are two concentration terms on the top line and two on the bottom line. This also

means that Kc for this equilibrium has no units. 8 Initial amount of ethyl ethanoate = 4.51 g ÷ 88.1 g mol−1 = 0.0512 mol As in the first experiment, the initial mass of water from the added HCl(aq) = 4.81 g Initial amount of water = 4.81 g ÷ 18.0 g mol−1 = 0.267 mol Total amount of acid at equilibrium from the titration 41.30 = dm3 × 1.00 mol dm−3 = 0.0413 mol 1000 Taking away the amount of hydrochloric acid added at the start gives (0.0413 mol – 0.010 mol) = 0.0313 mol ethanoic acid at equilibrium.

VV Kc = (0.0199 / )(0.236 / ) = 4.7 (0.0313/VV )(0.0313/ )

9 The two values for Kc are reasonably close considering that the results were obtained by students with school equipment. The masses and volumes that were measured out at the start to make the mixtures were all small. In particular an error in measuring the 5.0 cm3 of acid would have a significant effect on the overall result because it would not only affect the value of the initial mass of water, but would also change the titration result and so affect all the other quantities

substituted in the expression for Kc.

Exam practice questions (pages 315-7)

[Sn4++ (aq)][Fe 22 (aq)] 1 a) i) Kc = [2] [Sn2++ (aq)][Fe 32 (aq)]

–3 3 ii) Kc has no units [1]; the concentration terms cancel – (mol dm ) appears on the top and

the bottom of the expression for Kc. [1]

–10 b) i) The reciprocal of the value given in the question [1]: Kc = 1 × 10 [1]

5 ii) The square root of the value given in the question [1]: Kc = 1 × 10 [1] (0.070 mol dm−32 ) 2 a) Kc = [1] (0.010 mol dm−−33 )× (0.010 mol dm ) = 49 [1]; no units [1] b) i) The quantity of iodine that has reacted = 0.0030 mol dm–3 This reacts with 0.0030 mol dm–3 hydrogen and produces 0.0060 mol dm–3 HI. So the new concentrations are:

–3 –3 [H2(g)] = 0.017 mol dm [1] and [HI(g)] = 0.076 mol dm [1].

ii) Substituting in the expression for Kc gives: (0.076 mol dm−32 ) = 48.5 [1] (0.017 mol dm−−33 )× (0.0070 mol dm )

This is very close to the value for Kc calculated in (a). The system is again at equilibrium. [1] c) Doubling the hydrogen concentration causes a shift in the position of equilibrium to the right giving more HI and less iodine [1]; some, but not all, of the added hydrogen reacts. [1]

© Andrew Hunt, Graham Curtis, Graham Hill 2019 11 Equilibrium ΙΙ Answers 3 a) At equilibrium the forward and backward reactions happen at the same rate so that overall there is no change. [1] Nitrogen and hydrogen react to form ammonia as fast as ammonia decomposes back to nitrogen and hydrogen. [1]

b) i) Mole fractions: NH3, 0.137 [1]; H2, 0.655 [1]; N2, 0.209 [1]

ii) Partial pressures: NH3, 1.37 atm; H2, 6.55 atm; N2, 2.09 atm [2] ()p 2 NH3 iii) Kp = [1] pp×()3 NH22

= 3.20 × 10–3 [1]; atm–2 [1] 4 a) Chemical equilibrium is dynamic [1]; on a molecular scale there is constant change with opposing reactions cancelling each other out. [1] However, the overall macroscopic effect is that nothing appears to be changing when the system is at equilibrium. [1]

b) Kc does not vary with concentration. [1] However, it is true that adding more of a reactant to a system at equilibrium means that the mixture is no longer at equilibrium [1]; some of the added chemical reacts and more products form until the system is again at equilibrium. [1]

c) Adding a catalyst does not affect the value of Kc. [1] At a constant temperature, the yield of product at equilibrium is not changed by adding a catalyst [1]; however, many processes are so slow that they never reach equilibrium in the absence of a catalyst, so in that sense an added catalyst increases the yield by enabling the reaction mixture to reach equilibrium. [1] d) This is true if the reaction to make products is endothermic. [1] If the reaction is exothermic, then the equilibrium yield of product falls as the temperature rises. [1] However, as with catalysts, raising the temperature can increase the yield of a product by allowing a reaction mixture to reach equilibrium that reacts too slowly at room temperature. [1] e) A reaction is feasible in the thermodynamic sense if the total entropy change for the process is positive. [1] Adding a catalyst speeds up the rate of reaction but cannot change the feasibility of a reaction at a particular temperature. [1] Only if the reaction is not going because it is kinetically inert can a catalyst enable a reaction to happen at a particular temperature. [1] 17 5 a) Concentration in aqueous layer = × 0.50 mol dm–3 [1] 10 = 0.85 mol dm–3 [1] 6.0 b) Concentration in organic solvent = × 0.01 mol dm–3 [1] 10 = 0.006 mol dm–3 [1] 0.85 mol dm−3 c) Kc = = 140 [1]; no units [1] 0.006 mol dm−3

[PCl32 (g)][Cl (g)] 6 a) Kc = [1] [PCl5 (g)]

1 3 –1 b) Kc = = 125 [1] dm mol [1] 8× 10−−33 mol dm

c) [PCl3(g)] = [Cl2(g)] = x [1] x2 = 8 × 10–3 mol dm–3 [1] 5× 10−−23 mol dm x = 2 × 10–2 mol dm–3 [1]

d) i) Some, but not all, of the added PCl5 decomposes [1] and increases the concentrations of

PCl3 and chlorine. [1] ii) Increasing the pressure causes the equilibrium to shift in the direction that gives fewer

moles of gas [1]; so the concentration of PCl5 increases while the concentrations of PCl3 and chlorine decrease. [1] iii) Raising the temperature causes the equilibrium to shift in the direction that is

endothermic [1]; so the concentration of PCl5 decreases while the concentrations of PCl3 and chlorine increase. [1]

e) i) Changing concentrations does not affect the value of Kc. [1]

ii) Changing the pressure does not affect the value of Kc [1]

iii) Kc increases [1] For an endothermic reaction the value of Kc gets larger as the temperature rises [1] pp×()3 CO H2 7 a) Kp = [2] pp× CH42 H O

b) i) Kp does not change [1]

ii) Kp increases [1]

iii) Kp does not change [1] c) i) Concentration of products decreases and the concentration of reactants increases. [1] ii) Concentration of products increases and the concentration of reactants decreases. [1] iii) Composition of the equilibrium mixture does not change. [1]

8 a) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) [2]

b) In theory the yield⇌ of NO at equilibrium is high if the temperature is low [1] and the pressure low [1]; adding excess oxygen also favours the conversion of ammonia to NO. [1] c) There is an excess of oxygen (even though only a fifth of the air is oxygen). [1] The temperature has to be high because reaction only happens at a practicable rate if the catalyst is hot. [1] Raising the pressure compresses the gases and this increases the mass of material flowing through the reactor and hence the mass of product produced in a given time from a given piece of plant. [1] (The chosen pressure also depends on the price of platinum. Working at higher pressures increases the wear on the catalyst and shortens the life of the catalyst gauzes.) d) The gases flow through the solid catalyst gauzes and so the catalyst is used continuously and does not have to be recovered from the reaction mixture. [1]

© Andrew Hunt, Graham Curtis, Graham Hill 2019 11 Equilibrium ΙΙ Answers e) The gases flow through the catalyst gauzes so the mixture of product and unchanged reactants is only in contact with the catalyst for a short time [1]; as soon as the gases flow away from the catalyst reaction stops – the mixture has time to approach equilibrium but not reach equilibrium. [1] f) The hot gases flow though a heat exchanger where they turn water to hot steam [1]; that can be used in other processes or to generate electricity. [1] 9 a) COCl2(g) CO(g) + Cl2(g)

Initial amount/mol 1 ⇌ 0 0

Amount at equilibrium/mol (1 − α) α α

Equilibrium partial pressure/Pa (1− α)P α P α P

(1+ α) (1+ α) (1+ α)

Terms in second row of table [1]; total amounts in moles [1]; mole fractions [1]; partial pressures. [1] α 2P Kp = (working [1], answer [1]) (1+−αα )(1 )

b) i) α = 1.4 × 10–3 (working [1], answer [1]) ii) α = 1.0 × 10–3 (working [1], answer [1]) 10 a) i) Equal volumes of gases contain equal amounts, in moles, of gas. So the mole ratio is

1 : 1. [1] According to the equation 1 mol SO2 reacts with 0.5 mol O2, so the oxygen is in excess. [1] ii) According to the equilibrium law (and Le Chatelier’s principle), adding an excess of a reactant tends to increase the conversion of other reactants to products. [1] Excess oxygen favours the formation of sulfur trioxide. [1] iii) The oxygen comes from the air and is mixed with nitrogen. [1] Adding more air would dilute the sulfur dioxide so much that too little of the gas would be flowing into the plant to produce product at a reasonable rate. [1] b) 700 K is hot enough for the catalyst to be effective and for the reaction to be fast enough. [1]

The temperature is not higher because the reaction is exothermic and so the value of Kp falls as the temperature rises. [1] The equilibrium yield of product falls as the temperature rises. [1] c) Raising the pressure would increase the yield of product at equilibrium because 3 mol reactants combine to give 2 mol product. [1] Increasing the pressure would also increase the mass of chemicals flowing through the plant. However a high-pressure plant costs more to build and raises the running costs because of the energy needed to run the pumps. [1]

Since the conversion to SO3 is close to 100% it is generally not worth operating at a higher pressure. [1]

© Andrew Hunt, Graham Curtis, Graham Hill 2019 11 Equilibrium ΙΙ Answers d) The reaction is exothermic so at each stage the gas mixture tends to heat up. [1] The gas mixture is cooled between catalyst beds to ensure that the equilibrium yield of product is as high as possible. [1] (The energy is used to raise steam, which can be used in other processes or used to generate electricity.) e) Removing the product from the gas stream when the sulfur dioxide concentration is low helps to shift the equilibrium to the right [1] to ensure that almost all the sulfur dioxide is converted to sulfur trioxide. [1] f) Sulfur dioxide is an acidic gas which causes acid rain. It is not acceptable to release any traces of sulfur dioxide into the air. [1] There is a cost to removing the last traces of sulfur dioxide from the remaining gases (which are mainly nitrogen and excess oxygen). [1]

11 a) 2HI(g) H2(g) + I2(g)

Initial amounts:⇌ HI(g) = 0 mol, H2(g) = 2 mol and [I2(g)] = 1 mol Let the equilibrium amount of HI = 2x Then the equilibrium concentrations: 2x (2− x ) (1− x ) [HI(g)] = , [H2(g)] = and [I2(g)] = [1] 1 mol dm−3 1 mol dm−3 1 mol dm−3 (2−−xx )(1 ) Kc = = 0.020 [1] 4x2 Hence: 0.92x2 – 3x + 2 = 0.020 [1] Solving shows that x = 0.935 [1] Hence the equilibrium concentrations are:

[I2(g)] = 0.065 mol, [H2(g)] = 1.06 mol and [HI(g)] = 1.87 mol [2]

b) C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)

Initial amounts/mol: 3.0 1.0 ⇌ 0 3.0 [1] Equilibrium amounts: (3.0 – x) (1.0 – x) x (3.0 + x) [1] xx(3.0+ ) Kc = = 4.0 [1] (3.0−−xx )(1.0 ) because the volume terms cancel. Hence: 3x2 – 19x + 12 = 0 [1] x = 0.71 [1] Amount of ethyl ethanoate at equilibrium = 0.71 mol [1]

– – c) I3 (aq) I2(aq) + I (aq)

Initial amounts/mol: 0 ⇌ 0.01 0.01 [1] x (0.01− x )2 (0.01− x )2 Equilibrium concentrations/mol dm–3: [1] 1 1 1 − 2 (0.01x ) –3 Kc = = 1.5 × 10 [1] x x2 – 0.0215x + 0.0001 = 0 [1]

– –3 – –3 [I3 (aq)] = 0.0068 mol dm and [I (aq)] = 0.0032 mol dm [1]

d) CO(g) + Cl2(g) COCl2(g) Initial amounts/mol: 0.20 0.10 0 [1] (0.20− x ) (0.10− x ) x Equilibrium concentrations/mol dm–3: [1] 3 3 3 3x Kc = = 0.41 [1] (0.20−−xx )(0.10 ) 0.41x2 – 3.123x + 0.0082 = 0 [1] x = 0.0262 [1]

–3 The equilibrium concentration is [COCl2(g)] = 0.026 mol dm [1]

e) 2NO2(g) N2O4(g)

Initial amounts/mol: 0 ⇌ 1 Equilibrium amounts/mol: 2x (1 – x) [1] Total amount at equilibrium/mol =(1 + x) 2x (1− x ) Equilibrium partial pressures/atm: × 1 × 1 [1] (1+ x ) (1+ x )

(1−+xx )(1 )2 Kp = = 7.1 [1] (1+ xx )4 2 29.4x2 –1 = 0 x = 0.184 [1] 1− 0.184 So partial pressure of N2O4 = × 1 = 0.69 atm [1] 1+ 0.184 2× 0.184 and partial pressure of NO2 = × 1 = 0.31 atm [1] 1+ 0.184

Test yourself (page 319)

1 Heating sodium chloride with concentrated sulfuric acid produces hydrogen chloride which dissolves in water to make hydrochloric acid. Concentrated nitric acid can be distilled from a mixture of potassium nitrate and concentrated sulfuric acid. 2 Many common acids are oxoacids; these form when the acidic oxides of non-metals dissolve in

water. Examples include H2SO3, H2SO4, HNO2, HNO3 and H3PO4.

3 Examples include HF, HBr, HI and H2S. 4 a) The formula of any acid contains hydrogen. See answers to Questions 2 and 3 for examples. b) There are many examples in organic chemistry including hydrocarbons and alcohols.

Test yourself (page 320)

+ − 5 Neutralisation is summed up by the equation: H (aq) + OH (aq) → H2O(l) 6 Two suggestions: • Dip an identical pair of electrodes into the two solutions, apply the same voltage across the electrodes and measure the current which flows – the current is much larger with the strong acid. • Add identical strips of magnesium to the two solutions and observe the rate of reaction, which is faster with the solution of the strong acid.

+ 2+ 7 a) Zn(s) + 2H (aq) → Zn (aq) + H2(g)

2− + b) CO3 (aq) + 2H (aq) → CO2(g) + H2O(l)

+ 2+ c) CaO(s) + 2H (aq) → Ca (aq) + H2O(l)

− + d) OH (aq) + H (aq) → H2O(l)

Test yourself (page 321)

8 a) Dative covalent bond. b)

The ion has a positive charge because the water molecule uses one lone-pair to form a dative bond with a proton, H+. So in the ion the number of protons is one more than the number of electrons. c) trigonal pyramidal

Test yourself (page 322)

10 a) Hydrogen chloride and sodium hydrogensulfate:

NaCl(s) + H2SO4(l) → NaHSO4(s) + HCl(g)

b) Protons transfer from the acid H2SO4 molecules to chloride ions. The proton acceptor (the base) is the chloride ion. c) HCl is a gas, so it escapes from the reaction mixture as it forms. This means that the backward reaction does not start so the forward reaction continues essentially to completion. 11 a) The oxide ion is a base accepting protons from dilute hydrochloric acid, forming water molecules. b) Hydrogen ions in dilute sulfuric acid combine with ammonia molecules, forming ammonium ions. c) Ammonium ions are proton donors giving protons to hydroxide ions, forming ammonia molecules and water molecules. d) Hydrogen ions in dilute hydrochloric acid combine with the carbonate ions. The carbonic acid formed decomposes to carbon dioxide and water. 12 a) Dative covalent bond. b)

Test yourself (page 323)

13 Nitrate ion, ethanoate ion, hydrogensulfate ion, carbonate ion. 14 Hydroxide ion, water molecule, ammonium ion, hydrogencarbonate ion, carbonic acid, hydrogensulfate ion. 15 a) If an acid is strong it means that it tends to ionise in water to form hydrated hydrogen ions and the conjugate base. In other words, the conjugate base has a weak hold on protons in competition with water molecules. b) If a base is strong it tends to accept protons and turn into its conjugate acid. The conjugate acid has a correspondingly limited tendency to give up the protons and ionise.

Test yourself (page 324)

16 a) pH = 1.00 b) pH = 2.00 c) pH = 3.00 17 pH = 1.09 18 a) 5.01 × 10−4 mol dm−3 b) 3.98 × 10−6 mol dm−3

c) 1.99 × 10−7 mol dm−3 d) 3.16 × 10−11 mol dm−3

Test yourself (page 326)

19 a) Ionisation is endothermic. b) Extent of ionisation increases as the temperature rises and the hydrogen ion concentration rises as the temperature rises. The number value of the pH falls. c) No, because [H+(aq)] always equals [OH−(aq)]. 20 a) pH = 14.00 b) pH = 12.30

− −3 –3 c) [OH ] in 0.0010 mol dm Ba(OH2) = 2 × 0.0010 mol dm × 14 2− 6 So [H+(aq)] = 1.0 10 mol dm = 5.0 × 10−12 mol dm–3 2× 0.0010 mol dm−3 pH = 11.30

Test yourself (page 327)

21 A solution with pH = 3 could, for example, be a dilute solution of a strong acid or a concentrated solution of a weak acid. To determine the strength of an acid it is necessary to know the concentration of the acid as well as the pH. 22 Ethanoic acid in solution is only partially ionised into hydrogen ions and ethanoate ions. However, on adding alkali the hydrogen ions are neutralised and turned to water. As a result the equilibrium shifts to the right to form more hydrogen ions – which in turn are neutralised. This continues until all the ethanoic acid has turned into sodium ethanoate. 1.0 mol ethanoic acid reacts with 1.0 mol NaOH, just as 1.0 mol HCl reacts with 1.0 mol NaOH.

Test yourself (page 328)

23 a) F−

− b) HCO2 c) ClO−

Test yourself (page 329) + 2 [H][CN]+− [H ] [HCN] + − [HCN] 24 Ka = ; but [H ] = [CN ], so Ka =

+ 2 [H ] = Ka × [HCN] = 4.9 × 10−10 mol dm−3 × 0.010 mol dm−3 = 4.9 × 10−12 mol2 dm−6 [H+] = 2.21 × 10−6 mol dm−3 So pH = 5.66

+ 2 25[H ] = Ka × [CH3COOH] = 1.7 × 10−5 mol dm−3 × 0.050 mol dm−3 = 8.5 × 10−7 mol2 dm−6 [H+] = 9.2 × 10−4 mol dm−3 So pH = 3.04 +− [H ][HCO2 ] + − 26Ka = and [H ] = [HCO2 ] [HCOOH] pH = 2.55 so [H+] = 2.82 × 10−3 mol dm−3 (2.82× 10−−3 mol dm32 ) Ka = 0.050 mol dm−3 = 1.59 × 10–4 mol dm−3

−5 −3 27 Using the same method as for Question 22, Ka = 1.40 × 10 mol dm .

Test yourself (page 331)

28 pKa = 3.8

−5 −3 29 Ka = 6.3 × 10 mol dm 30 First example: pH = ½(4.9 + 2) = 3.45

Second example: pKa = (4.86 – 1) = 3.86;

−4 −3 Ka = 1.38 × 10 mol dm

Test yourself (page 334)

31 pH = 2.88

+ [CH3 COOH] 32 H ] = Ka × − [CH COO ] 3 3 15 cm = 1.7 × 10−5 mol dm−3 × 10 cm3 = 2.55 × 10−5 mol dm−3 pH = 4.59 33 The ethanoate ion is the conjugate base of a weak acid so it is a relatively strong base. Ethanoate ions in solution take protons from water to form ethanoic acid molecules and hydroxide ions. 34 In a titration the equivalence point is determined by the concentrations of the acid and the alkali and not by their strengths. 35 The ammonium ion is the conjugate acid of a weak base so it is a relatively strong acid. Ammonium ions in solution dissociate to form hydrogen ions and ammonia molecules.

− + 36 CH3COOH(aq) + NH3(aq) H3COO NH4 (aq)

⇌ Test yourself (page 338)

37 a) HIn is yellow. In− is red. b) In a more acid solution the hydrogen ion concentration is higher. This tends to shift this equilibrium to the left, giving more of the HIn, yellow form:

HIn(aq) H+(aq) + In–(aq)

In a less ⇌acid, or neutral, solution the hydrogen ion concentration is lower. This tends to shift the equilibrium to the right, giving more of the In–(aq) red form. 38 a) The indicator would change colour before the titration reached the equivalence point. b) The indicator would not change colour until the titration has passed the equivalence point. c) Strong acid/strong base titration: methyl orange, methyl red or bromothymol blue. Weak acid/strong base titration: phenolphthalein. Strong acid/weak base titration: methyl orange or methyl red. d) There is only a slight, and rather gradual change of pH at the end-point of the titration. There needs to be a steeply rising part of the graph spanning at least two pH units to give a sharp colour change with an indicator. 39 The theory showing that an indicator typically changes colour over two pH units assumes that the two colours are equally intense so that the human eye is equally sensitive to both of them. This is not always the case.

Test yourself (page 339)

40 a) Strong acid/strong base: pH = 7 b) Weak acid/strong base: pH above 7 c) Strong acid/weak base: pH below 7 d) Weak acid/strong base: pH above 7

Test yourself (page 340)

41 In the apparatus shown, some of the energy from the reaction heats up the container and the surroundings and this is not allowed for in the calculation. 42 HCl is a strong acid and fully ionised before mixing with the alkali. Ethanoic acid is only very slightly ionised in solution. The base has to break the O−H bonds in the weak acid.

−1 43 HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l) ∆nH = –57.6 kJ mol

−1 HCl(aq) + NH3(aq) → NH4Cl(aq) + H2O(l) ∆nH = –53.4 kJ mol

−1 CH3COOH(aq) + NH3(aq) → CH3COONH4(aq) ∆nH = –50.4 kJ mol

Test yourself (page 341)

44 In a solution of a weak acid there are plenty of acid molecules, but very few of the ions that are the conjugate base. So, the solution can neutralise added alkali, but it can do nothing to respond to added acid. A salt of the acid must be added to provide the conjugate base so that the equilibrium can move in both directions

Test yourself (page 343)

+ [acid] −3 –3 45 a) Strong acid/strong base: [H ] = Ka × = Ka = 6.3 × 10 mol dm [salt] So pH = 6.2

b) 2.2 g of C6H5COOH is 0.1 mol

7.2 g of C6H5COONa is 0.05 mol

+ [acid] [H ] = Ka × [salt] 0.1 = 6.3 × 10−5 mol dm–3 × 0.05 = 1.26 × 10−4 mol dm–3 So pH = 3.9 c) pH = 3.9 because the ratio is the same. 46 [H+] = 4.0 × 10−6 mol dm–3 [acid] [H+ ] 4.0× 10−−63 mol dm = = = 0.24 × −−53 [salt] []Ka 1.7 10 mol dm

Activity: The effect of dilution on the degree of dissociation of a weak acid (page 331)

1 An accurate, ten-fold dilution is required. This can be done by using a pipette to measure 25.0 cm3 of 0.10 g dm−3 solution of acid into a 250 cm3 graduated flask and adding distilled water up to the graduation mark and mixing well. 2 Distilled water contains dissolved carbon dioxide. This acidic oxide affects the pH of water. Boiling removes dissolved gases because the solubility of gases falls as the temperature rises. 3 The properties of the electrode in the probe of a pH meter are not constant and can vary when the electrode is stored and when it is used. So the probe of a pH meter has to be calibrated before use. 4 The pH electrode is sensitive to, and responds to, the presence of hydrogen ions. If dipped into the most concentrated solution first, it would be difficult to ensure that the electrode was rinsed sufficiently with water to remove traces of the more concentrated solution before dipping it into a more dilute solution. 5 Reading down from the top of the column, the entries are: 4.0, 3.0 and 2.0. 6 a) The pH values of HCl(aq) are calculated assuming that the acid is fully ionised. At each concentration the hydrogen ion concentration in the ethanoic acid solution is lower than in the hydrochloric acid solution. This shows that the ethanoic acid is only partially dissociated into ions. b) The pH values for ethanoic acid get closer to the values for HCl(aq) as the acids get more dilute. This shows that the degree of dissociation of the weak acid increases as it is diluted.

+ − c) CH3COOH(aq) H (aq) + CH3COO (aq) +− [H (aq)][CH32 CH COO (aq)] −5 −3 Ka = ⇌ = 1.7 × 10 mol dm [CH32 CH COOH(aq)] Diluting the ethanoic acid lowers the concentration of ethanoic acid molecules, hydrogen ions

and ethanoate ions. There are two concentration terms on the top of the expression for Ka and only one on the bottom. After each dilution, the equilibrium has to shift to the right in

−5 −3 order for the expression for Ka to continue to equal 1.7 × 10 mol dm at equilibrium.

Activity: Titration of a strong acid with a strong base (page 332)

1 pH = −log 10−1 = 1 2 At the equivalence point, the solution just contains a neutral salt with an anion which is a very weak base and a cation which is not an acid. With hydrochloric acid and sodium hydroxide, for example, the solution at the equivalence point is a neutral solution of sodium chloride. 3 a) pH = 3.7 b) pH = 10.3 4 pH = 12.2 5 The equivalence point on the titration curve is at 7. There is a very steep rise of pH at the end-point. Adding just two drops of the alkali raises the pH by over 6 pH units. 6 An accurate titrations depends on a very sharp change of pH at the equivalence point. This means that there is a sudden colour change if an indicator is used, with just one drop of added alkali being enough to reach the end-point (Section 12.6).

Activity: Blood buffers (page 334)

+ − 1 CO2(aq) + H2O(l) H (aq) + HCO3 (aq)

2 Breathing faster and⇌ more deeply speeds up the removal of carbon dioxide from blood in the lungs. This lowers the concentration of carbonic acid in the blood making it more alkaline, and so raises the blood pH. 3 a) During exercise the cells in muscles produce more carbon dioxide. In solution in blood this raises the hydrogen ion concentration and lowers the pH. Also, in vigorous exercise anaerobic respiration produces lactic acid, which tends to acidify the blood further. b) People breathe faster and more deeply when running because the raised levels of carbon dioxide in the blood stimulate the centres in the nervous system that control the lungs. Increasing the rate of breathing helps to remove of the extra carbon dioxide in the blood. [H+− (aq)][HCO (aq)] 4 a) Ka = 3 [CO (aq)] 2

Ka2× [CO (aq)] − [HCO3 (aq)]

b) i) [H+(aq)] =

(4.5 × 10−−73 mol dm ) ×× (1.25 10 −− 33 mol dm )

= 2.5 × 10−−23 mol dm = 2.25 × 10−8 mol dm−3 ii) pH = 7.65 c) The pH of the blood is a little outside the normal range – it is slightly more alkaline than it should be in a healthy person. This could be the result of excessive respiration or due to some illness. 5 The mixture of carbon dioxide, water and hydrogencarbonate ions is in equilibrium (see answer to Question 1). The equilibrium includes hydrogen ions. If acid is added to the system this increases the hydrogen ion concentration. In response the equilibrium shifts to the left to remove most of the added hydrogen ions. Conversely, if some of the hydrogen ions are neutralised by added alkali the equilibrium shifts to the right. For these changes to happen, there must be significant amounts present in solution of both carbon dioxide and water on one side of the equation and hydrogencarbonate ions on the other. 6 Respiration in all the cells in the body continuously produces carbon dioxide. If this is not steadily removed in the lungs, it builds up in the blood and overwhelms the blood buffers. 7 After breathing thick smoke, the lungs may be damaged so that breathing is less effective. As a result, carbon dioxide may not be removed from the blood as efficiently as normal. A build-up of carbon dioxide in the blood makes it more acidic. A buffered, intravenous drip can help a patient to recover.

Core practical 9: Finding the Ka value for a weak acid (page 344)

1 The glass electrode in the probe of a pH meter responds to the concentration of hydrogen ions without adding or removing ions from the solution. So using a pH meter does not disturb the equilibrium system being studied. 2 a) A buffer solution consists of a weak acid mixed in solution with one of its salts. Significant amounts of both acid and salt must be present, as is the case during the titration for titres in the range 5–18 cm3. The buffering action is shown by the fact that after adding 5 cm3 alkali the pH was about 2.5, and the pH did not rise above 3.5 until about 18 cm3 of alkali had been added. This was a change of 1 pH unit while adding 13 cm3 alkali. This contrasts with the change of 5 pH units on adding a couple of drops of alkali at the end-point.

+ – b) CH2ClCOOH(aq) H (aq) + CH2ClCOO (aq)

3 a) Halfway to the equivalence⇌ point, the added alkali converts half of the weak acid to its salt. At this point:

– [CH2ClCOOH(aq)] = [CH2ClCOO (aq)]

+ K × [CH ClCOOH(aq)] So: [H (aq)] = a2 = Ka − [CH2 ClOO (aq)]

+ Hence, at this point: [H (aq)] = Ka From the graph, the end-point of the titration was reached when the titre was 24.0 cm3. Also from the graph, halfway to the end-point (at 12.0 cm3) pH = 2.9.

–2.9 –3 –3 –3 b) Ka = 10 mol dm = 1.3 × 10 mol dm This is based on the equilibrium mixture consisting of a mixture of a weak acid and one of its salts, where the weak acid is only slightly ionised, while the salt is fully ionised. The assumption is that all the negative ions come from the salt present and all the un-ionised molecules from the acid. 4 The graph of the pH readings shows the end-point and so it is possible to find the pH halfway to this point without knowing the exact concentration of the acid or the alkali.

Exam practice questions (pages 346–8)

+ – 1 a) NH3(aq) + H2O(l) NH4 (aq) + OH (aq) [1]

– Pair 1: acid – H2O;⇌ base – OH [1] + Pair 2: acid – NH4 ; base – NH3(aq) [1] b) Sodium chloride is the salt of a strong acid. [1] The chloride ion is a very weak base and does not take protons from water molecules so a solution of sodium chloride is neutral. [1] Sodium ethanoate is a salt of a weak acid. [1] The ethanoate ion is a strong enough base to take protons from water molecules to produce an alkaline solution with OH– ions. [1] c) The pH of a solution of a weak acid depends on both the concentration of the acid and its degree of ionisation. [1] The acid dissociation constant is an equilibrium constant [1] that applies at all equilibrium mixtures at a given temperature. [1] 2 a) HX(aq) H+(aq) + X–(aq) [1] [H+− (aq)][X (aq)] b) K = ⇌ (1 a [HX(aq)]

c) i) As the temperature rises the equilibrium shifts in the direction that is endothermic. So

the acid ionises more as the temperature rises and this is because the value of Ka increases. [1] ii) Since the acid ionises more as the temperature rises, the hydrogen ion concentrations rises and the pH decreases. [1]

iii) Ka does not change. Ka is a constant and only varies if the temperature changes. [1]

+ – 3 a) In water: H2O(l) H (aq) + OH (aq) [1]

+ – Kw = [H (aq)][OH⇌(aq)] [1] b) i) [H+(aq)] = 1 × 10–7 mol dm–3 [1] Also, in pure water [H+(aq)] = [OH–(aq)] So: [OH–(aq)] = 1 × 10–7 mol dm–3 [1]

–14 2 –6 Hence Kw = 1 × 10 mol dm

ii) The value of Kw becomes more positive as the temperature rises, which means that at a higher temperature the equilibrium shifts to the right, increasing the concentration of hydrogen and hydroxide ions. [1] This shows that the ionisation of water is endothermic. [1] c) i) [H+(aq)] × 0.30 = 1 × 10–14 mol2 dm–6 [H+(aq)] = 3.33 × 10–14 mol dm–3 [1] pH = 13.48 [1] ii) After dilution [OH–(aq)] = 0.030 mol dm–3 [1] [H+(aq)] = 33.33 × 10–14 mol dm–3 [1] pH = 12.48 [1] ii) Amount of NaOH added to the mixture 25 = dm3 × 0.30 mol dm–3 = 0.0075 mol [1] 1000 Amount of HCl added to the mixture 75 = dm3 × 0.20 mol dm–3 = 0.0150 mol [1] 1000 Volume of the solution = 100 cm3 [H+(aq)] = 7.5 × 10–2 mol dm–3 [1] pH = 1.12 [1] 4 a) A strong acid ionises completely in aqueous solution. [1] An example is hydrochloric acid: HCl(aq) → H+(aq) + Cl–(aq) [1] A weak acid only ionises to a slight extent in solution. [1]

+ – An example is ethanoic acid: CH3COOH(aq) H (aq) + CH3COO (aq) [1]

b) i) 2 [1] ⇌ ii) [OH–(aq)] = 0.01 mol dm–3 1× 10−−14 mol 2 dm 6 [H+(aq)] = = 1 × 10⁻12 mol dm–3 [1] 0.01 mol dm−3 pH = 12 [1]

c) i) [H+ (aq)] = (1.5 ×× 10 −−53 0.020) mol dm [1]

pH = 3.26 [1] ii) Line of sketch graph rises at first from pH = 3.3; partly levels off through buffer region [1]; then rises to give a steeply rising curve when the titre is 25 cm3 [1]; halfway up the steeply rising portion is above pH 7 [1]; the curve then levels off at about pH 12. [1] iii) Thymolphthalein [1]; full colour change must take place within the steeply rising part of the curve. [1] 5 a) This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answer taking

into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Points to make in the answer: • a buffer solution evens out the large swings in pH on adding small amounts of acid or alkali which can happen without a buffer • a typical buffer mixture consists of a solution of a weak acid and one of its salts, for example a mixture of ethanoic acid and sodium ethanoate • there must be plenty of both the acid and its salt in the buffer mixture • adding a small amount of strong acid temporarily increases the concentration of H+(aq) so the equilibrium shifts to the left to counteract the change • adding a small amount of strong alkali temporarily decreases the concentration of H+(aq) so the equilibrium shifts to the right to counteract the change • the pH of blood, for example, is closely controlled by buffers within the narrow range 7.35 to 7.45; or chemists use buffers when they want to investigate chemical reactions at a fixed pH. 0.080 mol dm−3 b) i) [H+(aq)] = 1.7 × 10–5 mol dm–3 × [1] 0.040 mol dm−3 = 3.40 × 10–5 mol dm–3 pH = 4.47 [1] 0.06 mol dm−3 ii) [H+(aq)] = 1.7 × 10–5 mol dm–3 × [1] 0.06 mol dm−3 = 1.7 × 10–5 mol dm–3 pH = 4.77 [1] 1× 10−−14 mol 2 dm 6 iii) [H+(aq)] = 0.02 mol dm−3 = 5.0 × 10–12 mol dm–3 [1] pH = 11.30 [1] iv) The change of pH is very much less when adding the alkali to a buffer solution than when adding it to water. [1] 6 a) i) HClO(aq) H+(aq) + ClO–(aq) [1]

+− [H (aq)][ClO⇌ (aq)] ii) Ka = [1] [HClO(aq)]

b) [H+(aq)] = 10–7.43 [1] = 3.7 × 10–8 mol dm–3 [1] (3.7 × 10−−83 mol dm )(0.050 mol dm − 3 ) Ka = [1] 0.050 mol dm−3 = 3.7 × 10–8 mol dm–3 [1] 7 a) i) 10 cm3 of the acid reacts with 10 cm3 of 0.01 mol dm–3 solution of sodium hydroxide. [1] The acid is monobasic. So the acid concentration is 0.01 mol dm–3. [1]

ii) At the start pH = 3.5 [1] iii) Acid is weak. For a 0.01 mol dm–3 monobasic strong acid, pH = 2 [1]

b) pKa = 2pH + log cA = (7 − 2) = 5 [1]

–5 –3 Ka = 1 × 10 mol dm [1] c) i) The pH did not rise fast between titres of 2.5 cm–3 and 7.5 cm–3. [1] In this region the flask contains the partly neutralised weak acid and the salt formed by neutralising the acid. [1] This series of mixtures of the acid and its conjugate base are buffer solutions. [1]

ii) Halfway to the end-point, [acid] = [salt] and pH = pKa. [1]

From the graph pKa = 5.0 [1]

–5 –3 Hence Ka = 1 × 10 mol dm d) i) The equivalence point is at pH = 8.5, which is halfway up the near vertical part of the curve. [1] ii) At the equivalence point the flask contains a solution of the sodium salt of a weak acid. The anion in the salt is the conjugate base of the weak acid. [1] If an acid is weak its conjugate base is relatively strong and so gives an alkaline solution in water. [1]

8 a) i) [H+(aq)] = 1.5×× 10−−53 mol dm 0.1 mol dm − 3 = 1.23 × 10–3 mol dm–3 [1] pH = 2.91 [1]

ii) Trichlorethanoic acid: [H+(aq)] = 2.3×× 10−−13 mol dm 0.1 mol dm − 3 = 0.15 mol dm–3 [1] pH = 0.82 [1] b) i) Fluoroethanoic acid > chloroethanoic acid > iodoethanoic acid > ethanoic acid [1] ii) Substituting an electronegative atom [1] on the carbon atom next to the −COOH group makes the acid stronger. [1] The more electronegative the atom, the greater the effect. [1] A possible explanation is that the electronegative atom pulls electrons away from the carboxylic acid group so that it holds onto to its proton less strongly. [1] c) i) Trichloroethanoic acid > dichloroethanoic acid > chloroethanoic acid > ethanoic acid [1] ii) This is consistent with (b) (ii). Increasing the number of electronegative atoms [1] pulling electrons away from the carboxylic acid group [1] increases the strength of the acid. d) i) 2-chlorobutanoic acid > 3-chlorobutanoic acid > 4-chlorobutanoic acid > butanoic acid [1] ii) Substituting electronegative chlorine atoms into the hydrocarbon chain increases the acid strength of butanoic acid. [1] However, the effect falls away as the chlorine atom gets further from the carboxylic acid group. [1] The effect is most marked when the chlorine atom is on the carbon atom next to the —COOH group. [1] 9 a) Use an insulated container: polystyrene cup or vacuum flask. [1] b) Indicative content: • In the mixtures with 0 cm3 to 67 cm3 NaOH(aq), the acid was in excess. • The reaction is exothermic.

• The more alkali there is in these mixtures, the more the energy is given out and the greater the temperature rise. • In the mixtures with 67 cm3 to 100 cm3 NaOH(aq), the alkali was in excess. • The more acid there is in these mixtures, the more energy is given out and the greater the temperature rise. • So there are two rising lines meeting at the point corresponding to exact neutralisation. c) i) 67 cm3 of 2.0 mol dm–3 NaOH(aq) reacts with 33 cm3 of the 2.0 mol dm–3 acid. [1]

ii) 2 mol NaOH reacts with 1 mol of the acid. [1] The acid is dibasic, H2X. [1] d) i) 16 K [1] ii) Energy given out = 100 g × 16 K × 4.18 J g–1 K–1 [1] = 6690 J = 6.69 kJ [1] iii) Amount of acid in the mixture = 0.033 dm3 × 2 mol dm–3 = 0.066 mol [1] Energy change per mole of the acid (exothermic) = 6.69 kJ ÷ 0.066 mol = 102 kJ mol–1 [1]

–1 ∆nH = −102 kJ mol

–1 iv) For the neutralisation of a monobasic acid ∆H ≈ –57.5 kJ mol . [1] The value for H2X is much less than twice this value. This suggests a weak acid. [1]

Test yourself (page 350)

1 a) b)

2 Metals have only 1, 2 or 3 electrons in their outer shell. It is much easier to lose these few electrons and obtain a more stable electron structure like that of the next noble gas than it is to gain 7, 6 or 5 electrons. By losing these electrons, metals form positive ions. Most non-metals have 5, 6 or 7 electrons in their outer shell. In forming ions, it is easier for them to gain 3, 2 or 1 electrons respectively and form a stable ion rather than lose 5, 6 or 7. By gaining electrons, the non-metals form negative ions. 3 The ions in magnesium oxide each carry two unit charges, whereas those in sodium fluoride carry only one unit charge. So, the ionic forces of attraction in magnesium oxide are greater than those in sodium fluoride, and this results in a higher melting point. 4 a) Ions are the charge carriers in an electrolyte when it conducts electricity. The electrolyte can only conduct if the ions are free to move towards the electrodes, which they can do in the liquid state but not in the solid state. b) At the cathode: Mg2+(g) + 2e–(g) → Mg(s)

– – At the anode: 2Cl (g) → Cl2(g) + 2e (g) 5 The electron configuration of a chloride ion is: 2,8,8 so the ion has one more full shell of electrons compared to a sodium ion with the configuration 2,8. 6 Electrostatic forces are inversely proportional to the square of the distance between the ions. There are many pairs of oppositely charged ions that are next together in the lattice and so much closer together than any of the pairs of ions with like charges.

Test yourself (page 351)

7 The standard enthalpy change of formation of a substance is the enthalpy change when one mole of it forms from its elements under standard conditions. So, the standard enthalpy change of formation of an element is the enthalpy change when one mole of the element forms from the element under standard conditions.

This involves no change, so ∆fH [element] = 0. 8 Standard enthalpy changes of formation are based on the formation of compounds from elements in their most stable states. Graphite is the more stable from of carbon. The change from diamond to graphite is exothermic.

9 Ca(s) + I2(s) → CaI2(s)

Test yourself (page 354)

10 a) Mg(s) + Br2(l) → MgBr2(s)

2+ – b) Mg (g) + 2Br (g) → MgBr2(s) c) Mg+(g) → Mg2+(g) + e– d) Mg(s) → Mg(g) e) Br(g) + e– → Br–(g) 11

12 Because water is a liquid at the temperature and pressure prescribed for standard conditions. 13 The gain of the first electron produces a Cl– ion with a more stable electron structure of 2,8,8. This additional stability of the Cl– ion is accompanied by a loss of energy and the process is exothermic. Adding a second electron to the negative Cl– ion requires energy to overcome the repulsion between the negative electron and the Cl– ion. This process is therefore endothermic. .14 Because energy is always required to remove an electron from an atom or a positive ion. .15 The first ionisation energy involves the removal of an electron from a neutral atom. This is always easier than removing the second electron from an ion with one positive charge for the second ionisation. So, the second ionisation energy is more endothermic than the first. .16 Because more energy is required to separate Li+ and F– ions in lithium fluoride than to separate the larger Na+ and Cl– ions in sodium chloride.

Test yourself (page 356)

17 a) Lattice energy is the energy change when one mole of an ionic solid forms from free gaseous ions. As the oppositely charged ions come close together in the solid, strong ionic bonds are formed and energy is given out from the system to the surroundings – the change is exothermic. So the lattice energy is negative. b) It is impossible to produce a gaseous mixture of oppositely charged ions in order to measure the lattice energy directly. 18 Hess’s law states that the overall enthalpy change of reduction is the same whether reactants are converted to products directly or indirectly via a series of intermediates. In a Born-Haber cycle, elements are converted to the solid ionic product directly as the enthalpy change of formation,

and also indirectly via a series of intermediates involving atomisations, the formation of ions and the lattice energy.

19 a) ∆H 1 is the standard enthalpy change of formation of MgCl2(s).

∆H 2 is the standard enthalpy change of atomisation of magnesium.

∆H 3 is the standard first ionisation energy of magnesium.

∆H 4 is the standard second ionisation energy of magnesium.

∆H 5 is twice the standard enthalpy change of atomisation of chlorine.

∆H 6 is twice the standard first electron affinity of chlorine.

∆H 7 is the lattice energy of MgCl2(s). a) The lattice energy of magnesium chloride,

–1 –1 –1 ∆H 7 = – (–698 kJ mol ) – (+244 kJ mol ) – (+1451 kJ mol ) – (+738 kJ mol–1) – (+148 kJ mol–1) + (–641 kJ mol–1) = –2524 kJ mol–1

Test yourself (page 358)

20 a) N3–, O2– and F– have the same electron structure (2,8). As the nuclear charge increases from N3– to F–, the electrons are attracted more strongly and pulled closer to the nucleus resulting in decreasing radii. b) i) Mg2+ and Li+ are roughly the same size, but Mg2+ has twice the positive charge and therefore twice the charge density. So, it exerts a greater attraction (i.e. greater polarising power) on the outermost electrons of anions. ii) Li+ is much smaller than K+, so it has a much larger charge density. This exerts greater attraction (i.e. greater polarising power) on the outermost electrons of anions. iii) Al3+ is much smaller than Na+ and it has three times the charge. Therefore, Al3+ has a very much greater charge density and polarising power on the outermost electrons of anions. iv) N3– and F– have the same electron structure. N3– with the larger ionic radius also has the smaller nuclear charge. So, the nucleus of the N3– ion has a weaker attraction for its outermost electrons which are therefore more polarisable than those in F–. 21 a) I– ions are larger than F– ions. They cannot get as close to Li+ ions as F– ions. So, the lattice energy of LiI is less exothermic than that of LiF. b) LiF c) I– ions, with more electrons and being much larger than F– ions, are more polarisable and therefore LiI has a greater degree of covalency in its bonding.

22 Smaller ions with the same charge have a greater attraction for oppositely charged ions and therefore result in more exothermic lattice energies, the order of the compounds to match the values given are MgO, MgS, BaO, BaS.

Test yourself (page 362)

23 Polar water molecules hydrate both the ions in sodium chloride. These are exothermic processes and the energy given out matches the energy needed to remove the ions from the crystal lattice. Hexane is non-polar and so its molecules are not attracted to ions. In hexane the ions remain firmly held to the lattice. 24 a) As ionic charge increases, the enthalpy change of hydration becomes more exothermic.

+ –1 b) ΔhydH [Na (g)] = −444 kJ mol

2+ –1 ΔhydH [Mg (g)] = −2003 kJ mol

3+ –1 ΔhydH [Al (g)] = −2537 kJ mol 25 a) As ionic radius increases, the enthalpy change of hydration becomes less exothermic. b) An appropriate series of ions, showing their ionic radii from Table 13.1.2 and their enthalpy change of hydration from the data sheet, is Li+, Na+ and K+.

Test yourself (page 363)

–1 26 a) ΔsolH [LiF(s)] = –11 kJ mol

–1 ΔsolH [LiI(s)] = –96 kJ mol b) The F– and I– ions carry the same charge, but the I– ion is much larger. This means that the lattice energy and the hydration enthalpy of lithium iodide are less exothermic than those of lithium fluoride. c) Chemists often use ΔH values to predict the direction of change. The more exothermic the value of ΔH, the more likely it is that the change will occur. But, ΔH is not always a reliable guide (see Chapter 13.2). In this case, however, LiF and LiI are very similar compounds and the significantly more

exothermic ΔsolH of LiI is consistent with its greater solubility. 27 The ions, Mg2+ and O2–, in magnesium oxide each carry twice the charge of those in sodium fluoride, Na+ and F–. Q ×Q Using the equation F ∝ 1 2 d2 and assuming that the Na+ and Mg2+ ions are roughly the same size and the F– and O2– ions are also about the same size, the overall electrostatic forces of attraction in MgO is approximately four times greater in MgO than in NaF. This results in a lattice energy roughly four times larger in MgO than NaF.

Activity: The stability of ionic compounds (page 359)

1 Applying Hess’s law to Figure 13.1.12,

–1 –1 –1 ∆fH [MgCl(s)] = +148 kJ mol + 738 kJ mol + 122 kJ mol – 349 kJ mol–1 – 753 kJ mol–1 = –94 kJ mol–1

2 MgCl(s) is more stable than its elements Mg(s) and Cl2(g). 3 Applying Hess’s law to Figure 13.1.13,

–1 –1 ∆rH = –(2 × –94) kJ mol + (–641 + 0) kJ mol = +188 kJ mol–1 – 641 kJ mol–1 = −453 kJ mol–1

4 MgCl(s) is unstable relative to MgCl2(s) and Mg(s). Therefore, MgCl2(s) forms in preference to MgCl(s) in any reaction between magnesium and chlorine.

5 a) MgCl3(s) is massively unstable relative to its elements, Mg(s) and Cl2(g). It is also massively

unstable relative to MgCl2 and Cl2. MgCl2(s) always forms in preference to MgCl3(s) in any reaction between magnesium and chlorine.

b) In assessing the value of ∆fH [MgCl3(s)], there are a number of endothermic changes in forming Mg3+ and Cl– ions. The most endothermic of these changes is the third ionisation energy of magnesium, because it involves not only the removal of an electron from an Mg2+ ion with two positive charges, but also the removal of an electron from the filled second quantum shell so that the effective nuclear charge holding this third electron is 10+ rather than 2+.

3+ – 6 a) Mg (g) + 3Cl (g) → MgCl3(s) b) The combination of gaseous Cl– ions with gaseous Mg3+ ions is more exothermic because the electrostatic forces between Mg3+(g) and Cl–(g) are greater than those between Mg2+(g) and Cl–(g).

Exam practice questions (pages 365–6)

1 a) i) Calcium changes from 0 to the + 2 state: oxidation. [1] Fluorine changes from 0 to the –1 state: reduction. [1] ii)

[2] (one mark for each ion – no circles needed) b) i) Properties such as: relatively high melting points, conduct electricity when molten but not when solid. (2 marks for any two)

ii) Giant lattice of oppositely charged ions. [1] Strong attraction between oppositely charged ions accounts for the high melting points. [1] Ions can only acts as charge carriers and conduct electricity if the ions are free to move. [1] 2 a) The first electron affinity of an element is the energy change when one mole of gaseous atoms [1] gain electrons to form one mole of gaseous ions, each with a single negative charge. [1] b) i) The second electron affinity involves the addition of a negative electron to an ion with one negative charge. [1] This requires energy to overcome the repulsion. [1] So the process is endothermic. ii) N(g) + e– → N–(g) (1 N2–(g) + e– → N3–(g) [1] c) i) Polarisation of an ion is the distortion of the electron cloud [1] around the ion by a nearby positive charge. [1] ii) The large charge density in the small Mg2+ ion. [1] The large, highly polarisable I– ion. [1] 3 a) i) A is Ag(g) [1]; B is Ag+(g) [1]; C is 2F(g) [1] ii) D is the second ionisation energy of silver. [1] E is the 2 × electron affinity of fluorine. [1]

F is the lattice energy of silver(II) fluoride [1] b) Lattice energy is the energy change when one mole of an ionic solid [1] forms from free gaseous ions. [1]

–1 –1 c) ∆fH [AgF2(s)] = + 289 kJ mol + 732 kJ mol + 2070 kJ mol–1 + (2 × +79 kJ mol–1) + (2 × –348 kJ mol–1) + (–2650) kJ mol–1 [1] = –97 kJ mol–1 [1] for 97, [1] for sign and units d)

[3]

–1 –1 ∆fH [AgF(s)] = +289 kJ mol + 732 kJ mol +79 kJ mol–1 – 348 kJ mol–1 – 955 kJ mol–1 [1] = –203 kJ mol–1 [1] for 203, [1] for sign and units

–1 –1 e) i) ∆rH = –203 kJ mol – (–97 kJ mol ) = –106 kJ mol–1 [1] for 106, [1] for sign

ii) AgF2 has an exothermic ∆fH so is stable with respect to its elements [1], but is

unstable with respect to AgF(s) so tends to decompose to form AgF(s) and ½F2(g). [1] 4 a)

–1 –1 –1 –1 –1 b) ∆H lattice[RbI(s)] = – (–295 kJ mol ) – 403 kJ mol – 107 kJ mol – 81 kJ mol – 334 kJ mol = – 630 kJ mol–1 [1] for 630, [1] for sign and units c) The Li+ ion is much smaller than the Rb+ ion [1] and can get closer to I– ions in the solid ionic crystal. [1] This results in a more exothermic process when LiI forms. [1] d) This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answer taking into account both the key points made (up to 4 marks) and the logic and coherence of the discussion (up to 2 marks). Points to make in the answer: • Li+ ions are smaller that Rb+ ions because they have fewer inner full shells. • This means that Li+ ions have a higher charge density than Rb+ ions. • The Li+ ions therefore cause greater polarisation of the I– ions. • This distorts the negative charge cloud around the I– ions, pulling the electrons towards the Li+ ions. • This gives rise to a degree of electron sharing (covalent bonding) … • … which is not accounted for by the ionic model.

5 a)

–1 –1 –1 b) ∆solH[CaCl2(s)] = –(–2258 kJ mol ) + (–1657 kJ mol ) + (2 × –340 kJ mol ) [1] = +2258 kJ mol–1 – 1657 kJ mol–1 – 680 kJ mol–1 = –79 kJ mol–1 [1] for 79, [1] for sign and units c) The Ca2+ ion has a higher charge and a larger radius than Li+. The higher charge tends to make its electron density greater than Li+ [1] but its larger radius tends to make its electron density smaller than Li+. [1] d) Water molecules are polar with a δ+ charge between the H atoms and a δ– charge on the O atom. [1] The δ+ charge attracts anions and the δ– attracts cations. [1] So the hydration enthalpies of both anions and cations are exothermic/negative. 6 a) i) B: 1s22s22p1 [1] ii) The first electron is shielded from the 5+ nuclear charge by the two electrons in the first shell and partially by the two 2s electrons. It is easiest to remove. [1] The next two electrons are shielded by the two 2s electrons so that they are removed against an effective nuclear charge of 3+. [1] There are no electrons shielding the last two electrons, which are closes to the nucleus and have to be removed from the full 5+ nuclear charge, and so there is a big jump in the ionisation energies once electrons are removed from the inner shell. [1] b) The following presented as a calculation, energy cycle or energy level diagram [1]: • energy needed to atomise two B atoms = 2 × +590 kJ mol–1 = +1180 kJ mol–1 [1] • energy needed to ionise two B(g) atoms = 2 × (800 + 2400 + 3700) kJ mol–1 = 13 800 kJ mol–1 [1] • energy needed to atomise three O atoms = 3 × +250 kJ mol–1 = +750 kJ mol–1 [1] • energy needed to turn three O(g) atoms into ions = 3 × (–140 + 790) kJ mol–1 = 1950 kJ mol–1 [1] • lattice energy = –1270 kJ mol–1 – (1180 + 13 800 + 750 + 1950) kJ mol–1 = –18 950 kJ mol–1 [1]

c) Boron is a first row element and so its atom and ion are relatively small. The ion has a 3+ charge. So the polarising power of the ion is relatively large. [1] This suggests significant deviation from pure ionic bonding in its compounds. [1] So probably poor agreement between the calculated value and a theoretical value based on the ionic model.

d) i) B2O3(s) + 3H2O(l) → H3BO3(aq) [1] ii) Formation of an acidic oxide is characteristic of the covalent, molecular oxides of non- metal elements. [1] This property of boron oxide is consistent with the idea that the B3+ ion means that there is extensive covalent bonding in the oxide. [1] e) The data shows that the chloride is likely to be a gas a room temperature. [1] This is characteristic of covalent, molecular chlorides of non-metals. [1] Such chlorides are hydrolysed by water to give hydrogen chloride and an oxoacid. [1]

Test yourself (page 368)

1 a) Feasible, fast b) Not feasible c) Feasible, slow d) Not fast e) Feasible, fast

Test yourself (page 372)

2 a) Mercury is a liquid so its atoms are free to move and so are more disordered than the atoms in a copper crystal. b) Under standard conditions ammonia is a gas while water is a liquid. Molecules in a gas have much more freedom of movement and are more disordered than the molecules in a liquid. c) Argon consists of single atoms. Propane has molecules made of several atoms. Larger molecules with more atoms have more ways to rotate and vibrate increasing the number of ways that the energy of the molecules can be distributed.

Test yourself (page 373)

3 a) Bromine gas. b) Liquid water. c) Ammonia gas which can rotate and vibrate in more complex ways than hydrogen fluoride. d) Ethane gas which can rotate and vibrate in more complex ways than methane. e) Aqueous sodium chloride. f) Ethene gas rather than the solid polymer. g) Pentene gas. The ring in cyclopentane restricts the freedom of the molecules to rotate about single bonds so it is more ordered

Test yourself (page 374)

4 a) Increase b) Increase c) Decrease d) Increase e) Increase. f) Increase

5 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆S = +185 J mol–1 K–1. The reaction leads to an increase in the total number of moles of gas.

Test yourself (page 375)

6 a) The reactants include a gas but the single product is a solid. The system becomes more ordered. −1 + 602 000 J mol –1 –1 b) ΔS surroundings = = +2020 J mol K 298 K –1 –1 –1 –1 ΔS total = – 217 J mol K + 2020 J mol K = +1803 J mol–1 K–1 The reaction is so exothermic that overall the total entropy change is positive.

–1 –1 –1 –1 7 a) ΔSsystem = (203 J mol K + 384 J mol K ) – (189.2 J mol–1 K–1 + 99.7 J mol–1 K–1) = +298 J mol–1 K–1 Two solids react to produce a solid and a gas. There is a higher degree of disorder in the products. The entropy of the system increases as expected. b) ΔrH = (–1552 kJ mol–1) – (–1573 kJ mol–1) = + 21 kJ mol–1 = + 21 000 J mol–1

–1 –1 c) ΔS surroundings = –70.5 J mol K –1 –1 d) ΔS total = +228 J mol K This endothermic reaction is feasible under standard conditions because of the positive entropy change of the system. The formation of ammonia gas from two solids is responsible for the size of the positive entropy change of the system.

Test yourself (page 376)

− −1 8 a) ΔSsurroundings = 28 100 J mol 298 K = – 94.3 J mol–1 K–1 ΔSsystem = 113 J mol–1 K–1 + 146 J mol–1 K–1 – 151 J mol–1 K–1 = +108 J mol–1 K–1 ΔStotal = 108 J mol–1 K–1 – 94 J mol–1 K–1 = +14 J mol–1 K–1 b) The positive entropy change of the system is not unexpected when ions leave a regular crystal lattice and pass into solution. Here the positive entropy change of the system is large enough to outweigh the negative entropy change of the surroundings, and so the reaction is feasible.

Test yourself (page 379)

9 No. If the entropy change is negative the free energy change becomes more positive as the temperature rises.

10 a) Two solids react to produce a solid and a gas. There is a higher degree of disorder in the products. b) –TΔS = – (298 K × 0.558 kJ mol–1 K–1) = –166 kJ mol–1 Under standard conditions the value of –TΔS is not sufficiently negative to outweigh the positive value for the enthalpy change of reaction. c) The reaction becomes feasible when the –TΔS term is more negative than +468 kJ mol–1. 468 kJ mol−1 T = = 839 K 0.558 kJ mol−1 K−1 Test yourself (page 380) ΔG − 11 a) ln K = RT (+163 000 J mol−1) = − 8.31 J K−1 mol−1 ×298 K = –65.8 K = e–65.8 = 2.6 × 10–29 b) The positive value of ΔG suggests that the reaction is not feasible at 298 K. This is confirmed

by the very small value of K which shows that that the equilibrium is well over to the left hand side. At room temperature, ozone tends to decompose into oxygen.

Test yourself (page 382)

12 Taking Table 13.2.7 row by row: Reactants: hydrogen chloride. Products: hydrogen and chlorine. Reactants: methane and oxygen. Products: carbon dioxide and steam.

Reactants: PCl5. Products: PCl3 and chlorine. Reactants: ammonia and hydrogen chloride. Product: ammonium chloride. 13

Activity: The thermal stability of Group 2 carbonates (page 382)

1 A solid decomposes to give a solid and a gas. Gases have higher molar entropies than solids. 2 a) i) ∆G = +117 kJ mol–1 – (298 K × 0.175 kJ mol–1 K–1) = +64.9 kJ mol–1

ii) ∆G = +268 kJ mol–1 – (298 K × 0.172 kJ mol–1 K–1) = +216.7 kJ mol–1 b) The compounds are stable at room temperature. The free energy changes for both decomposition reactions are positive and so the reactions are not feasible. 3 The reaction becomes feasible when ΔH –TΔS < 0. This is when TΔS > ΔH

For MgCO3 this happens when T = 117 kJ mol–1 ÷ 0.175 kJ mol–1K–1 = 669 K

For BaCO3 this happens when T = 268 kJ mol–1 ÷ 0.172 kJ mol–1 K–1 = 1558 K 4 Down the group the carbonates become more stable relative to decomposition into the oxide and carbon dioxide. 5 Exothermic: the two lattice energies. Endothermic: enthalpy changes for breaking bonds, and the enthalpy change for the decomposition of the metal carbonate. 6 An Mg2+ ion is smaller than a Ba2+ ion. So in MgO the 2+ and 2– ions are closer together and the forces of attraction between them are stronger.

2– 3– 7 The O and CO2 ions both carry a 2– charge but the oxide ion is smaller. With a smaller anion the ions in the oxide pack closer together so the charges are closer and the attraction between them greater. So more energy is given out per mole when the ions of the oxide come together to form a crystal lattice. 8 The key energy quantities in Figure 13.2.20 are: • the energy needed to break up the carbonate ion into an oxide ion and carbon dioxide • the energy given out as the 2+ and 2– charges get closer together when the larger carbonate ions break up into smaller oxide ions and carbon dioxide (this is the difference between the less negative lattice enthalpy of the carbonate and the more negative lattice enthalpy of the oxide). The key factor in explaining the trend in stability is the energy released as the ions get closer

together. This energy is the difference between the lattice enthalpies of the carbonate (MCO3) and the oxide (MO). The difference is greater when the metal ion is small than when it is large. The lattice enthalpy for an ionic compound is inversely proportional to the distance between the ions. Down Group 2, lattice enthalpies decrease in magnitude (become less negative) from one compound to the next as the ionic radii of the metal ions increase. For a large anion, such as the carbonate ion, the lattice enthalpy decreases (becomes less negative) only slowly as the metal ion radius increases down the group from Mg2+ to Ba2+. For the smaller oxide ion, however, the magnitude of the lattice enthalpy decreases more quickly as the cation radius increases. As a result, the difference between the lattice enthalpies gets smaller

down the group, so less energy is released as the lattice contracts when the compounds decompose and the carbonates become more stable. 9 The higher the charge on a positive ion and the smaller its size, the greater is polarising power. So a smaller magnesium ion has a greater polarising power than a larger barium ion. So the thermal stability of the Group 2 metal carbonates increases down the group as the polarising power of the metal ions decreases.

Exam practice questions (pages 385–6)

1 a) At 0 K water is solid and the atoms are highly ordered. [1] The crystals are not disrupted by molecular movement. [1]

b) The rises in entropy at T1 and T2 correspond to the changes of state: melting and vaporising. [1] There is a greater increase in disorder when molecules separate from a liquid and turn to gas than when molecules melt [1] because of the much greater freedom of movement of molecules in the gas phase. [1] c) i) Water and steam are in equilibrium [1] at 373 K and 1 atmosphere pressure so there is no tendency for the change to go in either direction. [1] ii) ΔG = ΔH – TΔS [1] ∆H Since ΔG = 0 kJ mol–1, ΔS = [1] T 41.1 kJ mol−1 ΔS = [1] 373 K = 0.110 kJ mol–1 K–1 = 110 J mol–1 K–1 [1]

2 a) NH4NO3(s) → N2O(g) + 2H2O(g) [1] Solid decomposing to two gases. Increased disorder. ΔS positive. [1] b) KCl(s) + aq → KCl(aq) [1] Ordered crystals breaking up to produce hydrated ions in solution. Increased disorder. ΔS positive. [1]

c) N2(g) + 2O2(g) → 2NO2(g) [1] Three moles of gas combining to give two moles of gas. Less disorder. ΔS negative. [1]

d) 2H2O2(aq) → 2H2O(l) + O2(g) [1] Two moles of reactant producing three moles of product, one of which is a gas. Increased disorder. ΔS positive. [1]

e) NH3(g) + HI(g) → NH4I(s) [1] Two moles of gas combining to give one mole of solid. Less disorder. ΔS negative. [1]

3 a) Al2O3(s) + 3C(s) → 2Al(s) + 3CO(g) [1]

b) Total enthalpy change of formation of products = 3 × (–111 kJ mol–1) = –333 kJ mol–1 [1] Total enthalpy change of formation of reactants = –1669 kJ mol–1 [1] Overall enthalpy change = –333 kJ mol–1 – (–1669 kJ mol–1) [1] = +1336 kJ mol–1 c) Total entropy of reactants = 50.9 J mol–1 K–1 + (3 × 5.7 J mol–1 K–1) = 68 J mol–1 K–1 [1] Total entropy of products = (2 × 28.3 J mol–1 K–1 + (3 × 198 J mol–1 K–1) = 651 J mol–1 K–1 [1] Overall entropy change = 651 J mol–1 K–1 – 68 J mol–1 K–1 [1] = 583 J mol–1 K–1 d) i) ∆G = ∆H – T∆S [1] ∆G = +1336 kJ mol–1 – (298 K × 0.583 kJ mol–1 K–1) [1] = +1162 kJ mol–1 ii) The reaction is not feasible at 298 K because the free energy change is positive. [1] e) The reaction becomes feasible when ΔH – TΔS < 0. [1] This is when TΔS > ΔH. This happens when T = 1336 kJ mol–1 ÷ 0.583 kJ mol–1 K–1 = 2292 K [1] f) Reactions between solids are slow. The reaction rate has to be very high for it to proceed at a measurable rate. [1]

–1 –1 –1 –1 –1 –1 –1 –1 4 a) (6 × 205 J mol K ) + S (C6H12O6) – (6 × 214 J mol K + 6 × 70 J mol K ) = –256 J mol K [1]

–1 –1 S (C6H12O6) = +218 J mol K [1] b) ∆G = +2879 kJ mol–1 – (298 K × – 0.256 kJ mol–1 K–1) [1] = +2955 kJ mol–1 [1] c) i) ∆G is large and positive at 298 K so the reaction is not feasible under standard conditions. [1] The entropy change is negative and so –TΔS is positive at all temperatures. [1] Changing the temperature cannot make the free energy change for the reaction negative. [1] ii) A plant leaf is not a closed system. It is the input of energy from the Sun which makes photosynthesis possible. [1] 5 a) ΔG = ΔH – TΔS [1] ΔG = –519 kJ mol–1 – (298 K × 0.082 kJ mol–1 K–1) [1] = –543 kJ mol–1

b) i)

Labelled axes and appropriate scales [1]; accurate plot giving straight line. [1]

ii) The gradient of the graph is –ΔS. [1] Since the graph is a straight line this shows that –ΔS does not vary with temperature. [1] c) i) ΔG is negative across the whole temperature range [1], so the reaction is feasible at all these temperatures. [1] ii) At 298 K the activation energy of the reaction is too high (the bonds in the molecules are too strong) [1] and the energy of collisions between molecules too low for the reaction to happen at a measurable rate. [1] d) In the reaction forming CO and steam, 2.5 mol gases react to form 3 mol of gas. [1] This increases disorder so that the entropy change is substantially positive. [1] In the reaction forming soot, 2 mol gases react to form an ordered solid and 2 mol gas. [1] Overall there is little change in entropy. [1] 6 a) Bromine is a liquid under standard conditions while iodine is a solid. For comparable substances, liquids generally have higher standard entropy values than solids.[1] This is because there is greater randomness in the distribution of molecules and energy quanta in a liquid compared to a solid. [1]

b) At the boiling point the liquid and vapour are in equilibrium and so ΔStotal = 0 [1]

ΔStotal = ΔSsystem + ΔSsurroundings ∆H ∆H 0 = ΔSsystem – , hence ΔSsystem = [1] T T 26 000 J mol−1 = = 84.4 J K–1 [1] for 84.4, [1] for units 308 K c) i) Highly endothermic change: ΔH +ve [1] 1 mol liquid produces 1.5 mol gas, so ΔS positive. [1] Reaction is the reverse of hydrogen burning and so not expected to be feasible, so ΔG positive. [1]

ii) Highly exothermic change: ΔH negative [1]

1 mol liquid fuel reacts with oxygen to produce more moles of gas (CO2 and H2O), so ΔS positive. [1] ΔG must be negative [1] d) i) ΔG = ΔH – TΔS [1] ΔG = +132 kJ mol–1 – (298 K × 0.233 kJ mol–1 K–1) [1] = +62.5 kJ mol–1 ∆G +62 500 J mol−1 ii) ln K = – = − [1] for numerator, [1] for denominator RT 8.31 J mol−−11 K× 298 K = – 25.2 K = e–25.2 = 1.1 × 10–11 [1] iii) The positive value of ΔG suggests that the reaction is not feasible at 298 K. [1] This is

confirmed by the very small value of K, which shows that that the equilibrium is well over to the left-hand side. [1] iv) The reaction becomes feasible when TΔS > ΔH [1] This happens when T = 132 kJ mol–1 ÷ 0.233 kJ mol–1 K–1 = 566 K [1]

Test yourself (page 389)

1 a) Ca(s) + I2(s) → CaI2(s) Oxidation of Ca to Ca2+ – loss of electrons: Ca → Ca2+ + 2e–

– Reduction of I2 to I – gain of electrons:

– – I2 + 2e → 2I b) Oxidation of O2– ions by loss of electrons at the anode:

2– – 2O (l) → O2(g) + 4e Reduction of Al3+ ions by gain of electrons at the cathode: Al3+(l) + 3e– → Al(l)

c) 2Fe(s) + 3Cl2(g) → 2FeCl3(s) Oxidation of Fe to Fe3+ – loss of electrons: 2Fe → 2Fe3+ + 6e–

– Reduction of Cl2 to Cl – gain of electrons:

– – 3Cl2 + 6e → 6Cl 2 a) Fluorine is the most reactive and most electronegative non-metal. Therefore, the oxidation

number of fluorine in OF2 is –1.

As OF2 is a neutral molecule, this means that the oxidation number of oxygen in OF2 is +2.

+ b) Sodium forms Na ions in all its compounds including peroxides such as Na2O2. If the

oxidation number of sodium in Na2O2 is +1, the oxidation number of oxygen in Na2O2 must be

–1, because the sum of the oxidation numbers on the atoms in Na2O2 must be zero.

– 3 The oxidation number of two Br ions increases from –1 to 0 in Br2. The oxidation number of one

S atom in H2SO4 decreases from +6 to +4 in H2SO3.

4 a) K +1, I +5, O –2 b) N +5, O –2

c) H +1, O –1 d) S +6, F –1 e) Na +1, H –1

2– 2– 5 a) SO2, H2SO3, SO3 and S b) SO3, H2SO4, SO4

2– c) H2S, S 6 a)

7 a) Oxidised b) Reduced c) Reduced d) Oxidised

Test yourself (page 392)

8 a) Fe3+(aq) + e– → Fe2+(aq)

– – b) Br2(aq) + 2e → 2Br aq)

+ – c) H2O2(aq) + 2H (aq) + 2e → 2H2O(l) 9 a) Zn(s) → Zn2+(aq) + 2e

– – b) 2I (aq) → I2(aq) + 2e c) Fe2+(aq) → Fe3+(aq) + e–

Test yourself (page 393)

10 The chlorine oxidises iodide ions on the starch-iodide paper to iodine. This reacts with the starch to form a blue/black colour. Only a small amount of chlorine is needed to produce a distinct blue colour with starch, so this is a very sensitive test for chlorine. 11 a) Fe3+(aq) + e– → Fe2+(aq)

– – 2I (aq) → I2(aq) + 2e

3+ – 2+ b) 2Fe (aq) + 2I (aq) → 2Fe (aq) + I2(aq) c) The yellow iron(iii) chloride solution turns brown due to the formation of iodine. d) 1

e) Iodide ions reduce iron(iii) ions to iron(ii) ions so FeI3 cannot exist as a stable compound.

+ – 2+ 12 a) MnO2(s) + 4H (aq) + 2Cl (aq) → Mn (aq) + 2H2O(l) + Cl2(g)

– + 2+ b) Cu(s) + 2NO3 (aq) + 4H (aq) → Cu (aq) + 2NO2(g) + 2H2O(l)

Test yourself (page 395)

– + – 2+ 13 a) MnO4 + 8H + 5e → Mn + 4H2O (Fe2+ → Fe3+ + e–) × 5

– + 2+ 2+ 3+ MnO4 + 8H + 5 Fe → Mn + 5Fe + 4H2O

– + – 2+ b) (MnO4 + 8H + 5e → Mn + 4H2O) × 2

+ – (H2O2 → O2 + 2H + 2e ) × 5

– + 2+ 2MnO4 + 6H + 5H2O2 → 2Mn + 5O2 + 8H2O

2+ – 14 a) 5 mol of Fe ≡ 1 mol MnO4 20.0 cm3 of 0.100 mol dm–3 Fe2+ contains 20.0 2.00 dm–3 × 0.100 mol dm–3 of Fe2+ = mol of Fe2+ 1000 1000 – 2.00 1 So, the amount of MnO4 reacting = × mol 1000 5 3 –3 – 2.00 1 If V cm of 0.0200 mol dm MnO4 contains × mol, then 1000 5 V V 2.00 1 dm3 × 0.0200 mol dm–3 = × 0.0200 mol = × mol 1000 1000 1000 5 ⇒ V = 20.0 cm3

– b) 5 mol of H2O2 ≡ 2 mol MnO4

3 –3 20.0 cm of 0.200 mol dm H2O2 contains

20.0 3 –3 4.00 dm × 0.200 mol dm of H2O2 = mol 1000 1000 – 4.00 2 So, the amount of MnO4 reacting = × mol 1000 5 3 –3 – 4.00 2 If V cm of 0.0200 mol dm MnO4 contains × mol, then V 4.00 2 1000 5 × 0.0200 mol = × mol 1000 1000 5 ⇒ V = 80.0 cm3

Test yourself (page 397)

15 a) 2+ – Mg(s) → Mg (aq) + 2e

2+ – Cu (aq) + 2e → Cu(s) 2+ 2+ Mg(s) + Cu (aq) → Mg (aq) + Cu(s)

2+ Mg is oxidised and Cu is reduced.

b) Cl2(aq) + 2e– → 2Cl–(aq)

2Br–(aq) → Br2(aq) + 2e–

Cl2(aq) + 2Br–(aq) → 2Cl–(aq) + Br2(aq)

– Cl2 is reduced and Br is oxidised.

c) 2Ag+(aq) + 2e– → 2Ag(s)

Cu(s) → Cu2+(aq) + 2e–

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

Ag+ is reduced and Cu(s) is oxidised.

Test yourself (page 398)

16 In the Zn2+(s)|Zn(s) half-cell, zinc atoms are oxidised to zinc ions as they give up electrons to the metal electrode. In the Cu2+(aq)|Cu(s) half-cell, copper(ii) ions are reduced to copper atoms as they take electrons from the metal electrode. 17 Electrons flow out of the zinc electrode which is therefore negative. Electrons flow in the external circuit from the negative electrode towards the positive electrode. The copper electrode is positive because it loses electrons as copper(ii) ions turn into copper atoms.

Test yourself (page 399)

18 a) Mg(s) → Mg2+(aq) + 2e– Zn2+(aq) + 2e– → Zn(s)

2+ 2+ b) Mg(s)|Mg (aq) ⁞⁞ Zn (aq)|Zn(s) Ecell = +1.61 V

Test yourself (page 401)

19 It is difficult to maintain a constant pressure of hydrogen gas and an equilibrium of H+(aq) and

H2(g) on the platinum black electrode. Also, in time the platinum black loses its effectiveness because it is inactivated by impurities. 20 Sodium reacts violently with water and cannot be used as an electrode in aqueous systems. 21 Because it is a good conductor and also very unreactive so it does not interfere with the redox reaction being investigated. 22 a) Sn2+(aq) + 2e– → Sn(s) E = −0.14 V

– – b) Br2(aq) + 2e → 2Br (aq) E = +1.07 V c) Cr3+(aq) + 3e– → Cr(s) E = −1.01 + 0.27 = −0.74 V

Test yourself (page 402)

+ 2+ 23 Pt[H2(g)]|2H (aq) ⁞⁞ Cu (aq)|Cu(s) Ecell = +0.34 V

2+ 2+ Cu(s)|Cu (aq) ⁞⁞ Pb (aq)|Pb(s) Ecell = −0.47 V

+ 2+ Pt[H2(g)]|2H (aq) ⁞⁞ Pb (aq)|Pb(s) Ecell = −0.13 V 24 a) The e.m.f. when a calomel electrode is connected to a standard hydrogen electrode is +0.27 V. The e.m.f. when a standard Cu2+(aq)|Cu(s) electrode is connected to a standard hydrogen electrode is +0.34 V. So, the e.m.f. when a calomel electrode is connected to a standard Cu2+(aq)|Cu(s) electrode is + 0.07 V and the positive electrode is Cu2+(aq)|Cu(s).

– – b) At the calomel electrode: 2Hg(l) + 2Cl (aq) → Hg2Cl2(s) + 2e At the Cu2+(aq)/Cu(s) electrode: Cu2+(aq) + 2e– → Cu(s)

Test yourself (page 403)

2+ 3+ 2+ 25 a) Zn(s)|Zn (aq) ⁞⁞ V (aq), V (aq)|Pt Ecell = +0.50 V Zn(s) + 2V3+(aq) → Zn2+(aq) + 2V2+(aq)

– – b) Pt|2I (aq), I2(aq) ⁞⁞ Br2(aq), 2Br (aq)|Pt Ecell = +0.55 V

– – Br2(aq) + 2I (aq) → 2Br (aq) + I2(aq)

– + 2+ c) Pt|2Cl (aq), Cl2(aq) ⁞⁞ [PbO2(s) + 4H (aq)],[Pb (aq) + 2H2O(l)]|Pt Ecell = +0.10 V

+ – 2+ PbO2(s) + 4H (aq) + 2Cl (aq) → Pb (aq) + 2H2O(l) + Cl2(aq)

Test yourself (page 405)

26 a) Li > K > Ca > Na > Mg b) Zn > Fe > Sn > Pb > Cu

Test yourself (page 406)

– 2– 3+ 27 a) H2O2 > MnO4 > Cr2O7 > Fe

– b) H2O2 > ClO > Cl2 > Br2 > O2

Test yourself (page 407)

+ – 28 a) 2H (aq) + O2(g) + 2e → H2O2(aq) E = +0.68 V

+ – H2O2(aq) + 2H (aq) + 2e → 2H2O(l) E = +1.77 V

Overall: 2H2O2(aq) → 2H2O(l) + O2(g) The more positive half-reaction acts as the oxidising agent. Hydrogen peroxide reacting to give water tends to oxidise hydrogen peroxide to oxygen. So oxygen in the –1 state in hydrogen peroxide disproportionates to oxygen gas (oxidation state 0) and water (oxidation state of oxygen –2). At room temperature the reaction is slow in the absence of a catalyst.

– + – b) NO3 (aq) + 3H (aq) + 3e → HNO2(aq) + H2O(l)] E = +0.94 V

+ – HNO2(aq) + H (aq) + e → NO(g) + H2O(l)] E = +0.99 V The more positive half-reaction acts as the oxidising agent. Nitrous acid reacting to give NO tends to oxidise nitrous acid to nitric acid. So nitrogen in the +3 state in nitrous acid disproportionates to NO gas (nitrogen in oxidation state +2) and nitric acid (oxidation state +5).

Test yourself (page 408)

29 a) Al3+(aq) + 3e– → Al(s) E = –1.66 V

+ – 2H (aq) + 2e → H2(aq) E = 0.00 V The more positive half-reaction tends to go from left to right and the less positive, or more negative, half-reaction from right to left. So in this case hydrogen ions tend to oxidise aluminium atoms to aluminium ions. b) Aluminium metal is coated with a very thin layer of aluminium oxide which protects it from corrosion in air and slows down its reaction with dilute acids.

+ – 30 2H (aq) + 2e → H2(aq) E = 0.00 V Cu2+(aq) + 2e– → Cu(s) E = +0.34 V

– + – 2NO3 (aq) + 4H (aq) + 4e → 2NO2(g) + 2H2O(l)] E = +0.80 V Copper, with its positive electrode potential, does not tend to reduce hydrogen ions to hydrogen. The feasible reaction goes the other way. However, copper is a strong enough reducing agent to reduce nitrate ions to nitrogen dioxide in acid solution.

Test yourself (page 411)

31 a) Pb2+(aq) + 2e– → Pb(s) E = −0.13 V The electrolyte in a lead–acid cell is sulfuric acid so the lead ions do not stay in solution around the negative electrode but precipitate as lead sulfate. The lead metal in the electrode is not dipping into a 1.0 mol dm–3 solution of lead(ii) ions. b) 2.0 V

2+ + 2+ c) Pb(s)|Pb (aq) ⁞⁞ [PbO2(s) + 4H (aq)], [Pb (aq) + 2H2O(l)]Pb(s) Ecell = +2.0 V Pb(s) → Pb2+(aq) + 2e– E = +0.13 V

+ 2+ Therefore: [PbO2(s) + 4H (aq)], [Pb (aq) + 2H2O(l)]|Pb(s) E = +1.87 V d) During recharge: Cathode: Pb2+(aq) + 2e– → Pb(s)

(in PbSO4)

2+ + – Anode: Pb (aq) + 2H2O(l) → PbO2(s)+ 4H (aq)+ 2e

(in PbSO4)

Test yourself (page 413)

– – 32 a) At the negative electrode: H2(g) + 2OH (aq) → 2H2O(l) + 2e

– – At the positive electrode: ½O2(g) + H2O(l) + 2e → 2OH (aq)

b) H2(g) + ½O2(g) → H2O(l)

c) Ecell = (+0.40 V) – (–0.83 V) = 1.23 V

3 33 a) CH3OH(l) + /2O2(g) → CO2(g) + 2H2O(l) b) Methanol advantage: a liquid at room temperature and so easier to store and transport than hydrogen gas, which has to be compressed or liquefied; much less flammable and so not such a fire or explosion hazard; can be manufactured from renewable resources. Methanol disadvantage: it is toxic; its use adds to carbon emissions especially if it is made from synthesis gas produced from fossil fuels.

Core practical 11: A redox titration (page 396)

– – 1 2I (aq) → I2(aq) + 2e

– + – – 2 ClO (aq) + 2H (aq) + 2e → Cl (aq) + H O(l) 2 2I (aq) → I (aq) + 2e 2 ClO (aq) + 2H (aq) + 2I Cl (aq) + H O(l) + overall: → 2 (aq) I (aq) 2

3 I2(aq) + 2e– → 2I–(aq)

2S2O32–(aq) → S4O62–(aq) + 2e–

overall: I2(aq) + 2S2O32–(aq) → 2I–(aq) + S4O62–(aq)

2– – 4 2 moles S2O3 (aq) ≡ 1 mole I2(aq) ≡ 1 mole ClO (aq).

2– 26.60 3 –3 5 Amount of S2O3 (aq) in titration = dm × 0.10 mol dm 1000 = 2.66 × 10–3 mol So amount of NaClO in 10 cm3 diluted bleach = 1.33 × 10–3 mol 6 Amount of NaClO in 100 cm3 of undiluted bleach = 1.33 × 10–1 mol So mass of NaClO in 100 cm3 of undiluted bleach = 1.33 × 10–1 mol × 74.5 g mol–1 = 9.91 g 7 Random errors • Variations arising both from the judgements about the position of the meniscus relative to graduation marks, and from the judgements involved when deciding when the end-point has been reached in the titrations – these are allowed for by repeating the titrations to obtain accurate titres to within 0.10 cm3 and then taking the average. Systematic errors concentration(s) were not standard • Use of grade B glassware with graduations that are not quite correct – this can be corrected by calibrating the glassware before use.

• Use of glassware that is not clean – this can be corrected by making sure that the glassware is clean and thoroughly washed. • Taking graduated flask, pipette and burette readings consistently from the wrong angle – this can be corrected by ensuring that the position of the meniscus (in burette, pipette or graduated flask) is read at eye level. 8 a) Uncertainty % Uncertainty

3 i) Volume of undiluted bleach 1 cm 1.0% ii) Volume of diluted bleach pipetted 0.05 cm3 0.5% iii) Volume of thiosulfate titrated 0.15 cm3 0.6% iv) Volume of thiosulfate titrated 0.15 cm3 0.6% b) i) Total % uncertainty = 3.1% 9 Mass of sodium chlorate(i) in undiluted bleach = 9.9 ± 0.3 g per 100 cm3.

Core practical 10: Investigating some electrochemical cells (page 404)

2 A simple salt bridge consists of a strip of filter paper soaked in potassium nitrate solution. This provide an electrical connection between the solutions. A metal connection cannot be used because this would add unknown electrode potentials where the wire dips into the solutions. In the salt bridge, ions carrying charge can move. The use of potassium ions and nitrate ions minimises contamination of the cells and avoids any danger of precipitates forming where the salt bridge connects with the cells. 3 a) Cu(s) → Cu2+(aq) + 2e– The copper atoms give up electrons to the metal electrode as they react and go into solution. This adds electrons to the electrode and makes it negative.

b) Ag+(aq) + e– → Ag(s) The silver ions take electrons from the metal electrode as they turn into atoms. This makes the electrode positive. 4 Cell B At the negative electrode: Zn(s) → Zn2+(aq) + 2e– At the positive electrode: Cu2+(aq) + 2e– → Cu(s) Cell C At the negative electrode: Zn(s) → Zn2+(aq) + 2e– At the positive electrode: Ag+(aq) + e– → Ag(s) 5 Cell A: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) Cell B: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Cell C: Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

2+ + 6 Cell A: Cu(s)|Cu (aq) 2Ag (aq)|Ag(s) Ecell = (+0.80 V) – (+0.34 V) = +0.46 V

2+ 2+ Cell B: Zn(s)|Zn (aq) ⦙⦙ Cu (aq)|Cu(s) Ecell = (+0.34 V) – (–0.76 V) = +1.10 V 2+ + Cell C: Zn(s)|Zn (aq) ⦙⦙ 2Ag (aq)|Ag(s) Ecell = (+0.80 V) – (–0.76 V) = +1.56 V The experimental values⦙⦙ are close to the values expected from standard electrode potentials. Some discrepancy is expected for the cells involving silver electrodes because the silver nitrate solution was not at the standard concentration of 1.0 mol dm–3. 7 Figure 14.19 shows that zinc metal displaces reddish copper from a solution of copper(ii) sulfate. The copper(ii) sulfate solution gets paler because the blue copper(ii) ions are replaced by colourless zinc ions. This is predicted by the observations with cell B, in which the right-hand electrode is positive so that the reaction goes as shown in the answer to Question 5. Figure 14.20 shows that copper metal displaces shiny crystals of silver from a solution of silver nitrate. The silver nitrate solution turns blue because colourless silver ions are replaced by blue copper(ii) ions. This is predicted by the observations with cell A, in which the right-hand electrode is positive so that the reaction goes as shown in the answers to Questions 3 and 5.

Exam practice questions (pages 415–7)

1 a) i) Nitrogen oxidised from –3 to 0. [1] Chlorine reduced from 0 to –1. [1] ii) Copper oxidised from +1 to +2 [1] and reduced from +1 to 0. [1] iii) Oxygen oxidised from –2 to 0. Nitrogen reduced from +5 to +3. [1]

2– + – 3+ b) i) Cr2O7 (aq) + 14H (aq) + 6e → 2Cr (aq) + 7H2O(l) [1]

2– 2– + – SO3 (aq) + H2O(l) → SO4 (aq) + 2H (aq) + 2e [1]

2– + 2– 3+ 2– ii) Cr2O7 (aq) + 8H (aq) + 3SO3 (aq) → 2Cr (aq) + 4H2O(l) + 3SO4 (aq) [1]

iii) The yellow/orange dichromate solution [1] turns green as chromium(III) ions form. [1] iv) 3 [1]

– – 2 a) i) Cl2(aq) + 2I (aq) → 2Cl (aq) + I2(aq) [1]

2– 2– – ii) 2S2O3 (aq) + I2(aq) → S4O6 (aq) + 2I (aq) [2] b) Starch [1] c) On mixing the swimming pool water with excess potassium iodide the colourless solutions react to give a dark yellow-brown solution. [1] During the titration the colour of the solution becomes paler. Eventually it turns pale yellow. [1] On adding starch the solution turns blue- black. At the end-point the black colour disappears to give a colourless solution. [1] d) Amount of sodium thiosulfate added to reach the end-point = 0.0194 dm3 × 0.0050 mol dm–3 = 0.000 097 mol [1] This reacted with the iodine displaced by the chlorine in 20 cm3 of the water. [1]

2– 3 From the equations 2S2O3 (aq) ≡ Cl2(aq), so the amount of chlorine in the 20 cm water = 0.5 × 0.000 097 mol [1] So the concentration of chlorine in the water 0.5× 0.000 097 mol = = 0.002 42 mol dm–3 [1] 0.020 dm3 3 a) Fe3+(aq) + e– → Fe2+(aq) [1] reduction Cu(s) → Cu2+(aq) + 2e– [1] oxidation [1] b) Fe is reduced from +3 to +2 [1] Cu is oxidised from 0 to +2 [1] c) Cu(s)|Cu2+(aq) Fe3+(aq), Fe2+(aq)|Pt(s) [2]

Cell e.m.f. = Ecell⦙⦙ = +0.77 – 0.34 = +0.43 V [1]

4 a) Using filter paper soaked in saturated KNO3(aq) or a U-tube of agar gel containing saturated

KNO3(aq). [1] b) i) Cu(s) → Cu2+(aq) + 2e– [1] ii) Ag+(aq) + e– → Ag(s) [1] c) 2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq) [1]

d) Ecell = – 0.34 V + 0.80 V [1] = +0.46 V [1] e) At the silver electrode [1] where Ag+ ions gain electrons. Silver changes oxidation number from +1 to zero. [1] f) Any two of the following points for [2]: • temperature was not standard • concentration(s) were not standard • conditions were not standard.

+ – 5 b) i) O2(g) + 4H (aq) + 4e → 2H2O(l) + – 2H2(g) → 4H (aq) + 4e [1]

Overall: O2(g) + 2H2(g) → 2H2O(l) [1]

ii) From the hydrogen half-cell [1] b) Fuel cells differ from storage cells in that: • instead of recharging the cells, the reactants are fed into the cell continuously [1] • the products are drawn off continuously [1] • the voltage in the cell remains constant so long as fuel and oxygen are supplied. [1] c) Fuel cells produce electric power from the chemical free energy of the fuel and oxygen [1] much more efficiently. [1] Fuel cells do not pollute the atmosphere with pollutants from combustion [1] such as oxides of nitrogen and particulates; and they do not pollute rivers with warm water. [1] d) Indicative content • Advantage that a fuel cell is more efficient as a power source than a combustion engine.

• Advantage that there is no local air pollution (no CO2, no particulates, no oxides of nitrogen) because the only product is steam. • Advantage, in the future, that it may become practicable to make the hydrogen using replenishable energy sources. • Disadvantage that hydrogen is currently made from fossil fuels. • Disadvantage there are many technical difficulties associated with the handling, transport and storage of hydrogen for use in vehicles. • Disadvantage that hydrogen fuel cells are more costly than combustion engines and have a shorter life. • Disadvantage that the public may not be willing to accept the use of hydrogen as a fuel because of perceptions of its flammability and the danger of explosions.

– + – 2+ 6 a) MnO4 + 8H + 5e → Mn + 4H2O

2– – C2O4 → 2CO2 + 2e

2– – ⇒ 5 mol C2O4 ≡ 2 mol MnO4 [1]

−3 2– 7.445 g dm –3 Concentration of C2O4 = = 0.055 56 mol dm [1] 134 g mol−1

2– 25.00 3 –3 –3 ⇒ Amount of C2O4 reacting = dm × 0.055 56 mol dm = 0.001 389 mol dm [1] 1000

– 3 2× 0.001 389 Amount of MnO4 in 28.85 cm solution = mol [1] 5

2× 0.001 389 –3 ⇒ Concentration of the KMnO4 solution = = 0.019 26 mol dm [1] 5×0.028 8 5 b) The half-equations for the reaction are:

– + – 2+ 2+ 3+ – MnO4 + 8H + 5e → Mn + 4H2O and (Fe → Fe + e ) × 5

2+ – ⇒ 5 mol Fe ≡ 1 mol MnO4 [1]

– 26.75 3 –3 0.535 Amount of MnO4 reacting = dm × 0.0200 mol dm = mol [1] 1000 1000

0.535 2.675 ⇒ Amount of Fe2+ reacting = × 5 mol = mol [1] 1000 1000 2.675 So the mass of iron in the ore = mol × 55.8 g mol–1 = 0.15 g [1] 1000 0.15 g ⇒ Percentage iron in the iron ore = × 100 = 11.2% [1] 1.34 g

c) The half-equations for the reaction are:

– + – 2+ 2+ 3+ – MnO4 + 8H + 5e → Mn + 4H2O and (Fe → Fe + e ) × 5

2+ – ⇒ 5 mol Fe ≡ 1 mol MnO4

Zinc reduces iron(III) to iron(II): Zn → Zn2+ + 2e– [1] In the first titration:

– 18.00 3 –3 0.360 Amount of MnO4 reacting = dm × 0.0200 mol dm = mol 1000 1000 0.360 ⇒ Amount of Fe2+ in 25.00 cm3 of the solution = × 5 mol 1000 0.360 Concentration of Fe2+ in the solution = × 5 mol = 0.0720 mol dm–3 [1] 1000 × 0.0250

In the second titration, after reduction of iron(III) to iron(II):

– 22.50 3 –3 0.450 Amount of MnO4 reacting = dm × 0.0200 mol dm = mol 1000 1000 0.450 ⇒ Amount of Fe2+ and Fe3+ in 25.00 cm3 of the solution = × 5 mol [1] 1000 0.450 Concentration of Fe2+ and Fe3+ in the solution = × 5 mol = 0.0900 mol dm–3 1000 × 0.0250 Total concentration of Fe2+ and Fe3+ = 0.0900 mol dm–3 [1] Concentration of Fe3+ = (0.0900 – 0.0720) mol dm–3 = 0.0180 mol dm–3 [1]

2– 7 a) 1 mol S2O3 ≡ 1 mol electrons [1]

2– Amount of S2O3 (aq) needed in the titration = 0.0180 dm3 × 0.0760 mol dm–3 = 0.001 37 mol [1] Amount of copper ions in the pipette volume of solution 3.405 g = × 0.1 = 0.001 36 mol [1] 249.5 g mol−1

So there is a change of 1 mol of electrons per mol of copper(II) ions when they oxidise iodide ions. [1]

Copper(II) is reduced to copper(I). [1]

2+ – b) 2Cu (aq) + 4I (aq) → 2CuI(s) + I2(aq)

2– – 2– I2(aq) + 2S2O3 aq) → 2I aq) + S4O6 aq)

2– 2+ 2 mol S2O3 ≡ 1 mol I2 ≡ 2 mol Cu [1]

2– 22.5 3 –3 Amount of S2O3 reacting = dm × 0.14 mol dm 1000 3.15 = mol [1] 1000 3.15 So the amount of Cu2+ reacting = mol [1] 1000 3.15 ⇒ Mass of copper reacting = mol × 63.5 g mol–1 = 0.20 g [1] 1000 0.20 g Percentage of copper in the alloy = × 100 = 72.7% [1] 0.275 g

2+ – 2+ – 8 a) i) Fe(s) + I2(aq) → Fe (aq) + 2I (aq) or Cd(s) + I2(aq) → Cd (aq) + 2I (aq) [1]

because each of these reactions would have a large positive Ecell . [1] ii) Fe(s) + Cd2+(aq) → Fe2+(aq) + Cd(s) [1]

because this produces a very small positive Ecell . [1] b) i) Labels on the diagram should point to • Fe(s) electrode, in Fe2+(aq) solution, [1] 1 mol dm–3 [1]

– –3 • Pt electrode [1], in a solution containing I2(aq) and I (aq), both at 1 mol dm [1] • salt bridge. [1] ii) Arrow from Fe electrode to Pt electrode and labelled electron flow. [1]

iii) Ecell = + 0.54 – (– 0.44) = + 0.98 V [1] for magnitude, [1] for sign to correspond to that of the right-hand electrode iv) There would be a greater tendency for Fe(s) to form Fe2+(aq) [1], which would make the Fe2+|Fe electrode more negative [1], thus increasing the magnitude of the cell e.m.f. [1] c) i) +0.34 V [1] ii) When log[Cu2+(aq)] = 0, [Cu2+(aq)] = 1.0 mol dm–3 [1] and the value of E is the standard electrode potential. [1] iii) There is a linear relationship between the electrode potential and log[Cu2+(aq)]. [1] The equation for a linear equation takes the general form: y = mx + c In this instance this can be written as: E = constant × log [Cu2+(aq)] + E [2]

Test yourself (page 420)

1 a) 1s22s22p63s23p63d14s2 b) 1s22s22p63s23p6 c) 1s22s22p63s23p63d54s2 d) 1s22s22p63s23p63d5 2 a) [Ar] 3d6 b) [Ar] 3d5 c) Fe3+, because it has a half-filled 3d sub-shell in which the charge will be distributed more evenly. d) Cu2+, because all its shells of electrons are completely full.

Test yourself (page 423)

3 They are d-block elements because the last electron added to form their atomic structures enters a d sub-shell. They are not transition metals because neither of them forms a stable ion with a partially filled d sub-shell 4 Zinc atoms have the electronic structure [Ar] 3d10 4s2. When zinc reacts, both the electrons in the 4s sub-shell are lost to form stable Zn2+ ions, in which all the electron shells are completely full. 5 a) E values for the systems M2+(aq)|M(s) become less negative (more positive). b) The metals become less reactive in terms of the tendency for their atoms to form aqueous ions. 6 From titanium to chromium, the positive charge on the nucleus increases from 22 to 24 while electrons are being added to an inner shell of electrons. The increasing nucleus charge attracts the outer 4s electrons more strongly and this causes a decrease in the atomic radius.

Test yourself (page 425)

7 Possible statements are: • most transition metals can exist in more than one oxidation state • elements in the middle of the series have the greatest range of oxidation states • all the transition elements form compounds in the +2 state • the +2 state becomes more stable relative to the +3 state across the series.

2– 2– 8 a) +3: CrCl3, Cr2(SO4)3; +6: Cr2O7 , CrO4

b) +2: MnCl2, MnSO4; +7: KMnO4

c) +2: FeCl2, FeSO4; +3: FeCl3, Fe2(SO4)3

d) +1: CuCl, CuI; +2: CuCl2, CuSO4 (Only one compound is expected for each oxidation state.)

+ 9 a) Oxygen in VO2 has an oxidation state of –2, so the two O atoms contribute –4 to the overall

+ charge. So the oxidation state of V in VO2 must be +5 to give an overall charge of +1.

+ 2+ b) The solution is green because it contains a mixture of yellow VO2 ions and blue VO ions.

+ + – 2+ 2+ – 10 a) i) VO2 + 2H + e → VO + H2O ii) Zn → Zn + 2e b) i) From the data sheet of standard electrode potentials:

2+ + 3+ [VO (aq) + 2H (aq)], [V (aq) + H2O(l)]|Pt E = +0.34 V

– I2(aq), 2I (aq)|Pt E = +0.54 V

+ + 2+ [VO2 (aq) + 2H (aq)], [VO (aq) + H2O(l)]|Pt E = +1.00 V The more negative electrode system, 2I–(aq) → I2(aq), tends to reduce the more positive electrode system, VO2+(aq) → VO2+(aq). However, the reduction cannot go further to vanadium(iii). ii) From the data sheet of standard electrode potentials: V3+(aq), V2+(aq)|Pt E = –0.26 V Sn2+(aq)|Sn(white, s) E = –0.14 V

2+ + 3+ [VO (aq) + 2H (aq)], [V (aq) + H2O(l)]|Pt E = +0.34 V The more negative electrode system, Sn(s) → Sn2+(aq) tends to reduce the more positive electrode system, VO2+(aq) → V3+(aq). However the reduction cannot go further to vanadium(ii). iii) From the data sheet of standard electrode potentials: Zn2+(aq)|Zn(s) E = –0.76 V V3+(aq), V2+(aq)|Pt E = –0.26 V The zinc electrode system is sufficiently negative to reduce vanadium(v) all the way to vanadium(ii). 11 a) i) Cu+(aq) → Cu2+(aq) + e– ii) Cu+(aq) + e– → Cu(s) b) Cu+(aq) → Cu2+(aq) + e– E = –0.15 V Cu+(aq) + e– → Cu(s) E = +0.52 V c) Adding the two half-equations in part (b) gives

+ 2+ 2Cu (aq) → Cu (aq) + Cu(s) Ecell = +0.52 V – 0.15 V = +0.37 V The overall value of +0.37 V for the above reaction indicates that Cu2+(aq) + Cu(s) are more thermodynamically stable than 2Cu+(aq). So, under standard conditions, aqueous Cu+(aq) ions are oxidised and reduced simultaneously (i.e. disproportionate), forming Cu2+(aq) and Cu(s).

Test yourself (page 426)

12 From the data sheet of standard electrode potentials: Zn2+(aq)|Zn(s) E = –0.76 V Cr3+(aq), Cr2+(aq)|Pt E = –0.41 V The more negative Zn2+(aq)|Zn(s) electrode system reduces the more positive Cr3+(aq), Cr2+(aq)|Pt electrode system.

2– + 2– 13 a) 2CrO4 (aq) + 2H (aq) Cr2O7 + H2O(l)

Yellow ⇌ orange Hydroxide ions from the alkali react with hydrogen ions, removing them from the equilibrium system. The equilibrium shifts to the left to compensate, thus converting orange dichromate ions into yellow chromate ions. Adding acid replaces the hydrogen ion and restores the equilibrium to its previous position so that the orange colour returns. b) This is not a redox reaction. Chromium is in the +6 state in both ions.

Test yourself (page 428)

14 a) The electronic configurations are: Zn2+ [Ar] 3d10; Cu+ [Ar] 3d10; Sc3+ [Ar] In Zn2+ and Cu+, the 3d level is full and in Sc3+ the 3d level is empty, so electron transitions within the d level are not possible. b) Red – orange – yellow

Test yourself (page 431)

15 a) 4 b) 6 c) 6 16 a) Hexaamminecobalt(iii) b) Tetrahydroxozincate(ii) c) Tetrahydridoaluminate(iii) d) Hexaaquanickel(ii) 17 a) +2 b) +1 c) +3 d) +2

2– 3– – 18 AgBr(s) + 2S2O3 (aq) → [Ag(S2O3)2] (aq) + Br (aq)

Test yourself (page 435)

19 a) Linear b) Octahedral c) Tetrahedral d) Octahedral

20 Using the lone pair of electrons on the nitrogen atom and one of the two lone pairs of electrons on either of the oxygen atoms, glycine can make two co-ordinate bonds from the same molecule to a cation. 21 a) Because one EDTA4– ion can make six co-ordinate bonds with the same metal ion. b) Octahedral 22 a) b) Octahedral

2– 2+ 2+ 23 a) [Ni(EDTA)] > [Ni(en)3] > [Ni(NH3)6] b) The stability of complex ions depends on the strength and number of co-ordinate bonds from individual ligands. So, EDTA4–, a hexadentate ligand, forms the most stable complex ions with Ni2+, then 1,2-diaminoethane (en), a bidentate ligand, and then ammonia, a monodentate ligand.

4– 24 a) [Fe(CN)6] b) Fe4[Fe(CN)6]3

Test yourself (page 436)

25 Carbon monoxide reacts with haemoglobin. Through the process of ligand exchange, the carbon monoxide takes the place of oxygen by binding irreversibly with the iron(ii) ions. As a result the haemoglobin can no longer carry oxygen to the tissues in the body.

2+ 2+ 26 a) Co(H2O)6 (aq) + 6NH3(aq) → Co(NH3)6 (aq) + 6H2O(l)

2+ – 2– b) Co(NH3)6 (aq) + 4Cl (aq) → CoCl4 (aq) + 6NH3(aq)

2+ – 4– c) Fe(H2O)6 (aq) + 6CN (aq) → Fe(CN)6 (aq) + 6H2O(l)

2+ – 2– 27 a) Co(H2O)6 (aq) + 4Cl (aq) → CoCl4 (aq) + 6H2O(l) pink colourless blue colourless b) When the filter paper soaked in pink cobalt(ii) chloride solution is dried in an oven, all the water is lost and the substance on the filter paper contains blue tetrachlorocobaltate(ii) ions,

2– 2– 2+ CoCl4 . In the presence of water, the blue CoCl4 ions react to form pink Co(H2O)6 ions.

2– 2+ – CoCl4 (aq) + 6H2O(l) → Co(H2O)6 (aq) + 4Cl (aq) blue pink

Test yourself (page 438)

28 a) The solution turns from pink to green. b) The solution turns from green to pink.

4– 2+ 29 EDTA is a hexadentate ligand forming six co-ordinate bonds with Co ions, whereas NH3 and

H2O are only monodentate ligands.

4– 2+ 2+ When EDTA reacts with either Co(H2O)6 or Co(NH3)6 , there is a large increase in entropy.

2– ΔSsystem and ΔStotal are both positive, which means that Co(EDTA) is thermodynamically more

2+ 2+ stable than Co(H2O)6 and Co(NH3)6 .

Test yourself (page 441)

30 a) In ammonia solution, the following equilibrium operates:

+ – NH3(aq) + H2O(l) NH4 (aq) + OH (aq)

– Initially OH ions ⇌react with aqueous copper ions to form a pale blue precipitate of copper(ii) hydroxide:

2+ – Cu (aq) + 2OH (aq) → Cu(OH)2(s) More specifically:

2+ – Cu(H2O)6 (aq) + 2OH (aq) → Cu(OH)2(H2O)4(s) + 2H2O(l) b) The precipitate dissolves on adding more ammonia to form a deep-blue solution containing

2+ [Cu(NH3)4(H2O)2] ions:

Cu(OH)2(H2O)4(s) + 4NH3(aq)

2+ – → [Cu(NH3)4(H2O)2] (aq) + 2OH (aq) + 2H2O(l)] 31 a) The hydrated iron(iii) ion is acidic. The ligand water molecules bound to the highly polarising Fe3+ ion can donate protons to free water molecules:

3+ 2+ + [Fe(H2O)6] [Fe(H2O)5OH] (aq) + H (aq)

On adding alkali,⇌ further protons are removed to give a neutral complex that is not soluble in water:

– [Fe(H2O)5(OH)2] + 2OH (aq) [Fe(H2O)3(OH)3](s) + 2H2O(l)

Iron(iii) hydroxide is not amphoteric⇌ and so does not dissolve in excess alkali. b) Alkali removes protons from the hydrated cobalt(ii) ion to give a neutral complex that precipitates:

2+ – [Co(H2O)6] + 2OH (aq) [Co(H2O)4(OH)2](s) + 2H2O(l)

⇌ Test yourself (page 442)

32 a) SO2(g) + V2O5(s) → SO3(g) + V2O4(s)

V2O4(s) + ½O2(g) → V2O5(s) b) The catalyst is reduced and reoxidised during the reaction but ends up as vanadium(v) oxide ready to act as a catalyst again. Overall the catalyst is not used up.

Test yourself (page 444)

33 a) In a continuous process, the reaction mixture (gas or liquid) can flow through a solid heterogeneous catalyst in a reactor and there is no problem in separating the products from the catalyst. b) In a batch process, the solid catalyst can be suspended in the reaction mixture or agitated with it and then easily recovered by filtration for the next batch. 34 a) In the absence of a catalyst the two reactants are anions that tend to repel each other because they carry the same negative charge. In each step of the catalysed reaction a negative ion reacts with a positive ion. Ions with opposite charges attract each other. b) In the first stage, Fe3+ ions are reduced to Fe2+ as they oxidise iodide ions to iodine: 2Fe3+(aq) + 2e– → 2Fe2+(aq)

– – 2I (aq) → I2(aq) + 2e

2– 2+ 3+ – Then in the second stage, S2O8 ions oxidise Fe ions back to Fe ready to oxidise more I : 2Fe2+(aq) → 2Fe3+(aq) + 2e–

2– – 2– S2O8 (aq) + 2e → 2SO4 (aq)

2– 2+ 3+ 3+ – c) Yes, because S2O8 ions can oxidise Fe ions to Fe , and the Fe ions can then oxidise I ions

2+ 2– to I2, producing Fe ions which can be oxidised by more S2O8 . 35 a) Any two of: • add some Mn2+ ions • heat the solution

– • increase the concentration of MnO4 (aq)

2– • increase the concentration of C2O4 (aq).

b) i) At the start, the mixture is purple due to unreacted but diluted KMnO4. A few bubbles of

CO2 are be produced. Any change is slow. ii) As the reaction gets underway, the formation of Mn2+ ions start to catalyse the reaction so

– that it speeds up. The purple colour fades and disappears as MnO4 ions form almost colourless Mn2+ ions if ethanedioate is in excess. At the same time, an increasing number

of bubbles of CO2 are produced.

Test yourself (page 445)

36 a) CH3OH(l) + CO(g) → CH3COOH(l) b) 100% 37 Questions to consider when hearing about a new scientific claim: • Where were the findings published – was it in a peer-reviewed journal? • Have other scientists been able to replicate the findings? • Where was the research carried out – was it in a university or in a commercial organisation?

• What is the reputation of the scientist, or team of scientists, who did the work?

Activity: Studying the ionisation energies of transition metals (page 421)

1 a) [Ar] 3d104s2 b) [Ar] 3d10 c) [Ar] 3d104s1 d) [Ar] 3d5 e) [Ar] 3d54s1 2 a) Cr+(g) → Cr2+(g) + e– b) Fe2+(g) → Fe3+(g) + e– 3 From Sc to Zn, the nuclear charge is increasing while electrons are being added to an inner shell. The increasing nuclear charge attracts the outer shell electrons more strongly and so the ionisation energies increase from Sc to Zn. However, the electrons added to the inner 3d sub- shell increase the shielding of outer electrons and this restricts the increase in ionisation energies. 4 a) The first ionisation energy of zinc is more endothermic than that of any of the other d-block elements in Period 4. b) The electronic configuration of zinc atoms is more stable than those of the other elements. c) The electronic configuration of zinc atoms is [Ar] 3d104s2. It has a completely full third shell and a filled 4s sub-shell. None of the other elements have such a stable structure. 5 a) The electronic structure of Cu+ ions must be more stable than those of the equivalent ions of its neighbours. b) The electronic configuration of Cu is [Ar] 3d10 4s1 and that of Cu+ is [Ar] 3d10. This latter structure with a completely full third shell is relatively stable. 6 a) The electronic structure of Cr+ ions must be more stable than those of the equivalent ions of its neighbours. b) The electronic configuration of Cr is [Ar] 3d5 4s1 and that of Cr+ is [Ar] 3d5. This latter structure with a half-filled 3d sub-shell and an even distribution of charge in the five 3d orbitals is more stable than the equivalent ions of its neighbours (V+ – [Ar] 3d34s1 and Mn+ – [Ar] 3d54s1). 7 a) Mn and Zn b) Mn2+ ions should have a more stable electronic configuration than the equivalent ions of its neighbours. Its proposed electronic structure is [Ar] 3d5. This structure with a half-filled 3d sub-shell is more stable than those of its neighbours (Cr2+ – [Ar] 3d4 and Fe2+ – [Ar] 3d6). Zn2+ ions should have a more stable electronic configuration than the equivalent ions of its neighbours. Its proposed electronic structure is [Ar] 3d10. This structure with a completely full third shell is much more stable than those of its neighbours (Cu2+ – [Ar] 3d9 and Ar] 3d10).

Activity: Cis-platin – an important chemotherapy drug (page 434)

1 Because a square planar structure for PtCl2(NH3)2 makes possible the existence of cis/trans isomers. This is not possible with a tetrahedral structure. A tetrahedral structure is chiral and gives rise to isomers only if it has four different groups around the central ion. 2 cis/trans isomerism 3 a) +2 b) cis-diamminedichloroplatinum(ii) c)

4 Because it is uncharged, like the molecules in the cell membrane. So, it is able to form intermolecular forces with molecules in the cell membrane of roughly the same strength as those between its own molecules. 5 The chemical/compound which has the actual physiological/medical effect.

+ 6 Because it is now charged and the forces between ions of [Pt(NH3)2(Cl)(H2O)] and between these

+ ions and polar water molecules are much stronger than those between [Pt(NH3)2(Cl)(H2O)] ions

+ and neutral molecules in the cell membrane. So, [Pt(NH3)2(Cl)(H2O)] ions remain inside the cell. 7 Because it cannot replicate its DNA successfully and then divide into two cells. 8 Because the N and O atoms have lone pairs of electrons which can form co-ordinate bonds to the Pt2+ ion. C and H atoms in the bases have no lone pairs 9 Because they are replicating faster. 10 Because it can enter normal cells as well as cancerous cells and bind with DNA. However, its effect is not as great with normal cells because these are usually replicating more slowly than cancerous cells. 11 The trans isomer is not effective as a drug to treat cancer and is more toxic than the cis isomer.

Activity: Catalytic converters (page 443)

1 The catalyst is expensive. Spreading the catalyst thinly over a large area of an inert material means that a little catalyst provides a large surface area for reaction. 2 a) At first the catalyst is cold. Hot gases from the engine heat the catalyst to its optimum working temperature. b) The catalyst can only remove all the pollutants if there is an appropriate mixture of gases that need to be oxidised and gases that need to be reduced.

3 Overall: 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) In the usual type of catalytic converter this happens in two stages. In the first stage oxides of nitrogen decompose:

2NOx(g) → N2(g) + xO2(g) In the second stage carbon monoxide is oxidised:

2CO + O2 → 2CO2 4 The catalyst absorbs the pollutants onto its surface from the stream of exhaust gases. The catalyst then provides a reaction mechanism with an activation energy which is low enough to convert the pollutants quickly into less harmful products. 5 a) The bonding between reactant molecules and the catalyst has to be strong enough to adsorb the molecules and weaken covalent bonds thus lowering the activation energy and allowing the reaction to proceed. b) If the products are strongly held to the catalyst surface they block the active sites so that the catalyst rapidly becomes ineffective. 6 Lead is a catalyst poison. Using leaded petrol quickly renders the catalyst ineffective. 7 Possible ways to reduce local air pollution in cities: • introducing policies that favour vehicles running on batteries of fuel cells • reducing the number of vehicles on the roads by favouring mass/public transport or by encouraging more journeys on foot or by bicycle. 8 Catalytic converters do nothing to decrease carbon dioxide emissions from vehicles so they do not help to solve the problem of climate change.

Core practical 12: The preparation of a transition metal complex (page 438)

1 Particularly hazardous chemicals used include • Concentrated ammonia solution which is corrosive and gives off pungent, choking fumes – it should be handled in a fume cupboard. • 20-volume hydrogen peroxide which irritates the skin and eyes but does not need more than the usual precautions of wearing protective clothing and eye protection. • Concentrated hydrochloric acid which is corrosive and gives off acidic fumes – it should be handled in a fume cupboard.

2+ 2+ 2 [Co(H2O)6] + 6NH3(aq) [Co(NH3)6] + 6H2O(l)

3 A mixture of ammonium⇌ ions and ammonia molecules in solution form a buffer solution. The buffer mixture stabilises the pH of the reaction mixture. 4 a) The oxidised form of a more positive electrode system tends to take in electrons and oxidise the reduced form of a more negative electrode system. So here H2O2(aq) can oxidise [Co(NH3)6]2+(aq) but not [Co(H2O)6]2+(aq).

b) In aqueous solution chromium(iii) ions are powerfully oxidising and have a strong tendency to be reduced to the chromium(ii) state. In the presence of ammonia, the chromium(ii) is relatively easily oxidised to the +3 state.

+ 2+ 3+ 5 H2O2(aq) + 2H (aq) + 2[Co(NH3)6] (aq) 2[Co(NH3)6] (aq) + 2H2O(l)

6 The charcoal is separated from the reaction⇌ mixture by filtration showing that it is present as a separate solid phase. 7 Steps F and G involve recrystallisation of the product. In step F the impure solid is dissolved in water. The solution is filtered hot to remove the charcoal and any other insoluble impurities. On cooling in step G the relatively concentrated product crystallises leaving traces of soluble impurities in solution. 8 Crystals form when the hot solution is cooled in step G. There is an equilibrium between the crystals and ions of the product remaining in the cold solution.

3+ – [Co(NH3)6]Cl3(s) [Co(NH3)6] (aq) + 3Cl (aq)

Adding hydrochloric⇌ acid increases the concentration of chloride ions, thus tending to make this equilibrium shift to the left. This tends to decrease the solubility of the product and increase the amount that crystallises. 9 The cold water rinses away any of the solution from which the product crystallised so that it does not contaminate the crystals as they dry. A little cold water is used to minimise the amount of product that dissolves and gets washed away. Ethanol mixes with water and is volatile. Rinsing with ethanol removes most of the residual water around the crystals, and then it evaporates relatively quickly so that dry crystals are obtained faster.

–1 10 Molar mass of Co(H2O)6Cl3 = 238 g mol Amount of cobalt chloride starting material = 12 g ÷ 238 g mol–1 1 mol hydrated cobalt(ii) chloride → 1 mol product. Molar mass of product = 267 g mol–1 Theoretical yield = 267 g mol–1 × (12 g ÷ 238 g mol–1) = 13.5 g Actual yield expected if the percentage yield is 70% = 10 g

Core practical 15 (part 1): Analysis of an inorganic unknown (page 440)

1 Chromium(iii), copper(ii), iron(ii) and nickel(ii) salts can all be green. Of these, it is iron(ii) salts, such as iron(ii) sulfate, that have the palest green colour. 2 Chromium(iii) ions are green. The dichromate(vi) has been reduced to chromium(iii). Sulfur dioxide is an acidic gas that is a reducing agent.

3 Chromium(iii) ions give a precipitate with aqueous sodium hydroxide but the precipitate re- dissolves on adding excess alkali. The precipitate of iron(ii) hydroxide does not dissolve in either excess aqueous sodium hydroxide or in excess aqueous ammonia. 4 The formation of a browny-red precipitate with alkali suggests that iron(iii) hydroxide has precipitated. Chlorine is an oxidising agent. It oxidises iron(ii) to iron(iii) which is yellow.

2+ 3+ – 2Fe (aq) + Cl2(aq) → Fe (aq) + 2Cl (aq) 5 Test F shows that X is not a halide (chloride, bromide or iodide) because no silver halide precipitate forms. Test G shows that X is a sulfate.

2+ 2– Ba (aq) + SO4 (aq) → BaSO4(s)

6 The salt X is FeSO4. The sulfur is in the +6 oxidation state. On heating, test B shows that sulfur dioxide forms. The sulfur is reduced, so the iron must be oxidised. This accounts for the formation of a reddish-brown solid. This solid is iron(iii) oxide.

2FeSO4 → Fe2O3(s) + SO2(g) + SO3(g) Balancing the equation shows that S + 3O are left over after accounting for the formation of

iron(iii) oxide and sulfur dioxide. Sulfur does form a second oxide, SO3. This is also formed during the reaction.

7 FeSO4 gives off sulfur trioxide and steam on strong heating. Water condenses on cooling the gas mixture. Sulfur trioxide reacts with water to form sulfuric acid.

SO3(g) + H2O(l) → H2SO4(aq)

Exam practice questions (pages 447-9)

1 a) i) 1s22s22p63s23p63d14s2 [1] ii) 1s22s22p63s23p63d104s1 [1] iii) 1s22s22p63s23p63d9 [1] b) In d-block elements, the last electron added enters a d sub-shell [1]; which applies to both Sc and Cu. Transition elements have at least one stable ion [1]; with a partially filled d sub-shell. [1]

c) i) Tetraamminediaquacopper(II) [1] ii) Octahedral [1] d) In the Cu2+ ion, the energy levels of the five d-orbitals are split [1] by ligands. Electrons in the slightly lower 3d level can absorb energy from certain wavelengths in the visible spectrum [1] and jump to the slightly higher 3d level. [1] The light transmitted is therefore richer in certain wavelengths than others in the visible spectrum, so it appears coloured [1].

2 a) i) A transition metal is an element with at least one stable ion [1] with a partially filled d sub-shell. [1] Example: electron configuration of a transition metal ion with the 3d orbitals shown between d1 and d9. [1] ii) An oxidation number is a number assigned to an atom or ion [1] to describe its relative state of oxidation or reduction. [1]

Example: oxidation number of Fe in any iron(II) compound is +2. [1] iii) A complex ion is an ion in which a number of molecules and/or anions (ligands) [1] are bound to a central metal ion by co-ordinate bonds. [1]

2+ Example: Cu(H2O)6 [1] b) Indicative content: • Precipitation reactions – example with equation and colour of precipitate. • Complexing (ligand substitution) – example with equation and colours of complex ions. • Redox reactions – example with equation (half-equations) and colour change. • Catalytic action – example with equation(s) and evidence of catalytic activity. 3 a) +6 [1] b) Yellow [1]; to orange [1]

2− + 2− 2CrO4 (aq) + 2H (aq) → Cr2O7 (aq) + H2O(l) [1] c) Green [1]; Cr3+ [1]

2− + − 3+ d) i) Cr2O7 (aq) + 14H (aq) + 6e → 2Cr (aq) + 7H2O(l) [1] 20 ii) dm3 × 0.100 mol dm−3 = 0.002 mol [1] 1000 iii) 0.002 × 6 = 0.012 mol [1]

iv) 0.012 mol of VOClx gives up 0.012 mol of electrons. Therefore 1 mol of VOClx gives up 1 mol of electrons. So, vanadium changes its oxidation state by +1. [1]

v) Oxidation state of vanadium in VOClx is +4. So the formula of the oxochloride is VOCl2. [1] 4 a) i) Heterogeneous [1] ii) Homogeneous [1] iii) Heterogeneous [1] iv) Heterogeneous [1] b) In the presence of a catalyst, the reaction pathway has an activation energy which is much lower than when there is no catalyst. [1] Tungsten metal adsorbs hydrogen onto the crystal structure as single atoms. [1] So the catalyst breaks the bonds between the atoms in one of the reactants. [1] The pathway with a lower activation energy allows the reaction to proceed much faster. [1]

2+ 5 a) A contains [Cu(H2O)6] [1]

B contains Cu(OH)2(H2O)4 [1] (accept Cu(OH)2)

2− C contains [CuCl4] [1]

2+ 2+ D contains [Cu(NH3)4(H2O)2] [1] (accept Cu(NH3)4 ) b) The yellow/green solution changes colour through green [1] to blue. [1]

When conc. HCl is added to the blue copper(II) sulfate solution (A), the following equilibrium [1] is set up:

2+ − 2− Cu(H2O)6 (aq) + 4Cl (aq) CuCl4 (aq) + 6H2O(l) [2] blue yellow ⇌ In excess HCl, the equilibrium lies to the right so the solution appears yellow/green. [1] As excess water is added, the equilibrium shifts to the left [1]; and the solution becomes blue. (maximum 6 marks)

2+ − c) Cu(H2O)6 (aq) + 2OH (aq) → Cu(H2O)4(OH)2(s) + 2H2O(l) [1]

2+ − (accept Cu (aq) + 2OH (aq) → Cu(OH)2(s)) EDTA4− is a very powerful [1]; hexadentate [1]; ligand. [1]

It forms a chelated complex/chelate [1]; with copper(II) ions and prevents them from reacting with OH− ions. [1] (maximum 5 marks) 6 a) Homogeneous catalysis involves catalysis in which the catalyst and the reactants are in the same phase. [1] The most important feature of transition metal ions in homogeneous catalysis is that they can change their oxidation state. [1]

− 2− 2− b) i) 2I (aq) + S2O8 (aq) → I2(aq) + 2SO4 (aq) [1] for balanced equation, [1] for state symbols ii) Either – the reactants are both negative ions which will repel each other;

2− 2− or – there are bonds to break in the S2O8 to form SO4 . [1]

3+ − 2+ iii) 2Fe (aq) + 2I (aq) → 2Fe (aq) + I2(aq) [1]

2+ 2− 3+ 2− 2Fe (aq) + S2O8 (aq) → 2Fe (aq) + 2SO4 (aq) [1] c) i) Adsorption of gases on the metal surface is so strong that reactions cannot occur [1]; adsorption of gases on the metal surface is so poor that there are insufficient atoms for the reaction to occur at a useful rate. [1] ii) Adsorption is neither too strong nor too weak [1]; reactants can be adsorbed in sufficient numbers, react and then get desorbed from the metal surface at a good rate. [1] d) Pt alloy or Pt or Pt/Rh [1]; the reducing agent is CO/carbon monoxide [1];

2CO(g) + 2NO(g) → 2CO2(g) + N2(g) [1]

− − 7 a) H2NNH2(aq) + 4OH (aq) → N2(g) + 4H2O(l) + 4e [2] b) Its E value is more positive than +0.66 V [1] but less positive than +1.32 V. [1]

c) The two half-equations are:

− − H2NNH2(aq) + 4OH (aq) → N2(g) + 4H2O(l) + 4e and

+ − 2+ − (VO2 (aq) + H2O(l) + e → VO (aq) + 2OH (aq)) × 4 [1]

+ 2+ − Overall: H2NNH2(aq) + 4VO2 (aq) → N2(g) + 4VO (aq) + 4OH (aq) [1] d) i) Disproportionation is the simultaneous oxidation and reduction of atoms or ions of the same element. [1] For example, Cu+ ions can disproportionate to Cu2+ and Cu. [1] ii) Disproportionation requires the oxidation state of an element to both rise to a higher state and fall to a lower state. [1]

2+ 3+ + iii) 2VO (aq) → V (aq) + VO2 (aq) [1]

2+ − 3+ − iv) VO (aq) + H2O(l) + e → V (aq) + 2OH (aq) E = −1.32 V

+ − + − VO2 (aq) + 2OH (aq) → VO2 (aq) + H2O(l) + e E = +0.66 V This disproportionation will not occur under standard conditions in alkaline solution,

+ because VO2 (aq) (the oxidised form of the more positive electrode system) [1] tends to oxidise V3+(aq) (the reduced form of the less positive electrode system) to VO2+(aq). [1]

So vanadium(V) and vanadium(III) tend to react to give vanadium(IV), which is the reverse of disproportionation. [1] 8 a) A complex is a molecule or ion in which ligands [1] are attached to a central metal cation by co-ordinate bonds. [1] A stereoisomer is one of two or more molecules or ions with the same molecular/ionic formula and the same structural formula [1], but different stereo formulae (different positions of the atoms in space). [1] b) i) +2 [1] ii) +3 [1] c) i) 4 [1] ii) 6 [1]

d) i) Diamminedichloronickel(II) [1]

ii) Tetraamminedichlorocobalt(III) [1] e)

2 marks for each diagram: [1] for lone pairs on N; [1] for correct shape. f)

2 marks for each diagram: [1] for lone pairs on N; [1] for correct shape.

Test yourself (page 451)

1 a) Concentrated hydrochloric acid reacts much faster with marble chips than dilute hydrochloric acid. b) Any reaction involving gases such as the manufacture of ammonia from nitrogen and hydrogen. c) Magnesium powder reacts much faster with dilute hydrochloric acid than magnesium ribbon. d) Catalytic converters are only effective in speeding up the reactions which remove pollutants from car exhausts once they are hot. e) A platinum–rhodium alloy catalyses the oxidation of ammonia to nitrogen oxide. 2 a) In a more concentrated solution there are more collisions per second between reactants and so an increased rate of reaction. b) Increasing the pressure forces the molecules closer together and increases the rate of collisions which lead to reaction. c) When a solid reacts with a liquid or gas the reaction takes place on the interface where the reactants meet. The larger the surface, the greater the area open to reaction. d) At a higher temperature the Maxwell–Boltzmann distribution shifts to the right so the proportion of molecules moving fast enough to collide with energies greater than the activation energy increases. This means there are more collisions that are energetic enough to lead to a reaction. e) A catalyst provides an alternative reaction pathway with a lower activation energy. With a lower activation energy there are more collisions with enough energy to lead to a reaction. ( . − . ) −3 3 Rate = 0 55 0 42 mol dm = 0.0087 mol dm–3 s–1 15 s 4 a) 2N2O5(g) → 4NO2(g) + O2(g)

b) 2 mol of NO2 are formed from 1 mol of N2O5, so rate = 7.0 × 10–4 mol dm–3 s–1

Test yourself (page 453)

5 a) Collect the gas in a graduated syringe. b) Remove samples at intervals, stop the reaction by cooling and then titrate against alkali the acid produced by the reaction. c) Measure the conductivity of the solution to follow the increase in the concentration of ions. d) Carry out the reaction in a flask, with a loose plug of cotton wool in its neck, on a balance and record the loss in mass at regular intervals.

Test yourself (page 455)

6 Rate = k[peroxide] 7.4×10−6 mol dm−3 s−1 k = = = 3.7 × 10–4 s–1 0.02 mol dm−3 7 Rate = k[ester][OH–] rate 0.000 69 mol dm−3 s−1 k = [ester][OH− ] = 0.05 mol dm−3 ×0.10 mol dm−3 = 0.138 dm3 mol–1 s–1

Test yourself (page 456)

8 a) The graph is a straight line so the reaction is first order with respect to bromine. b) i) Half-lives are all close to 200 s wherever they are read from the graph. ii) This is consistent with the answer to part (a). The half-life for a first-order process is independent of the starting concentration. 9 For a first-order reaction the gradient of the rate–concentration graph gives the rate constant.

Test yourself (page 457)

– 10 a) Rate = k[C3H7Br][OH ] b) Second order c) Units: dm3 mol–1 s–1

Test yourself (page 458)

11 a) Graph similar to Figure 16.14.

b) 2NH3(g) → N2(g) + 3H2(g) Rate = k

Test yourself (page 460)

12 a) Rate = k[RBr][OH–] −3 −1 rate 1.36 mol dm s b) k = = = 3400 dm3 mol–1 s–1 [RBr][OH− ] 0.02 mol dm−3 × 0.02 mol dm−3 13 a) Rate = k[R´Br] rate −3 −1 b) k = = 40.40 mol dm s = 2020 s–1 [R´Br] 0.02 mol dm−3 14 a) 2H2(g) + 2NO(g) → N2(g) + 2H2O(g)

2 b) Rate = k[H2(g)][NO(g)]

Test yourself (page 461)

15 The true initial rate can be found from the gradient of a tangent to the curve drawn at t = 0. Reference to Figure 16.15 shows that a straight line to represent the average rate until the time of the colour change of a clock reaction would be less steep than a tangent drawn at the very start. This is because the reaction begins to slow down as soon as concentrations of reactants start to fall. The rate is at its initial maximum before the reactants have started to be used up.

Test yourself (page 461)

16 There are several steps to picking up a meal from a canteen counter. The rate at which the queue moves can be slowed down if there is a rate-determining step such as waiting for a toaster to make pieces of toast or for a coffee machine to deliver cups of coffee. Similarly the flow of traffic along a motorway slows down overall if there are lane closures and the traffic has to travel at 50 mph along a coned-off section of the road.

Test yourself (page 464)

17 In the first step a strong covalent bond has to break – this is slow. In the second step two oppositely charged ions, which attract each other, come together to form a bond – this is fast.

18 a) NO2(g) + CO(g) → NO(g) + CO2(g)

2 b) Rate = k[NO2(g)] c) Zero order

Test yourself (page 466)

19 The intermediate has a double bond as in an alkene, and an –OH group as in an alcohol. Hence it is called an enol. 20 Hydrogen ions act as the catalyst for the reaction. They are not used up in the reaction and so do not appear in the balanced equation. However, hydrogen ions are involved in the rate- determining step so that the concentration of hydrogen ions affects the rate of reaction. As a result, the concentration of hydrogen ions appears in the rate equation. 21 In both reactions the rate is determined by the rate of formation of the enol, which depends on the concentrations of propanone and hydrogen ions. Bromine or iodine then react very quickly with the enol intermediate as soon as it is formed.

Test yourself (page 467)

22 Reaction between molecules involve the breaking of covalent bonds. The range of activation energies roughly corresponds to the range of values for covalent bond energies.

Test yourself (page 468)

23 a) These are the units for a first-order reaction. b) As the value of k gets larger, the rate gets faster for a given concentration of the reactant. So the reaction speeds up as the temperature rises c) The 10 degree rise from 298 K to 308 K brings about a 5-fold increase in rate. The 20 degree rise from 278 K to 298 K brings about just over a 25-fold increase in rate.

24 a) Ea/RT becomes smaller in magnitude as T rises. Because of the negative sign in the equation, this means that ln k becomes more positive. So k gets larger as T rises and the rate is faster.

b) The larger the activation energy the larger the value of Ea/RT and so the smaller the magnitude of ln k. Hence k gets smaller and the rate is slower. E 25 ln (4.93 × 10–4) = constant – a . × 8 31E 295 ln (1.40 × 10–3) = constant – a 8.31×305 Subtract and solve for Ea

–1 Ea = 78 kJ mol

Activity: Investigating the effect of concentration on the rate of a reaction (page 454)

1 The orange-brown colour of bromine fades as the reaction proceeds and bromine molecules turn into colourless bromide ions. 2 Any strong acid, such as hydrochloric acid. 3 Adding a large excess of methanoic acid means that the concentration of this reactant stays effectively constant during the reaction. This allows a study of the effect of the changing bromine concentration on the rate. 4 The data points falls on a smooth curve. 5 At 100 s, rate = 2.4 × 10–5 mol dm–3 s–1 At 500 s, rate = 0.6 × 10–5 mol dm–3 s–1 6 The graph is a straight line. 7 The bromine concentration falls as the time increases. 8 The rate of reaction falls as time increases. 9 The rate of reaction is proportional to the concentration of bromine.

Core practical 13b: Investigating the reaction of hydrogen peroxide with iodide ions by a clock reaction (page 462)

1 The concentration of iodide ions is varied by using different proportions of KI(aq) and water. All other concentrations are kept the same by ensuring that they are diluted each time by the same total volume of KI(aq) and water.

2 [I–] volume of KI(aq)

2– 2– – 3 2S2O∝3 (aq) + I2(aq) → S4O6 (aq) + 2I (aq) In each run the same amount of iodine has to form before all the thiosulfate is used up and a slight excess of iodine gives the blue-black colour with starch. The amount of thiosulfate ion is small so the reaction has not gone far when the colour change is seen, hence fixed amount of iodine formed time is a measure of the initial rate. 1 Hence initial rate t

m – n + p 4 The full rate equation∝ is: rate = k[H2O2] [I ] [H ] All the terms on the right-hand side of the equation are constant under the conditions of the experiment apart from [I–]n. Hence rate = constant × [I–]n 5 log (rate) = log (constant × [I–]n) = log (constant) + log [I–]n = log (constant) + n log [I–] A plot of log (1/t) against log (volume of aqueous KI) is equivalent to a plot of log(rate) against log [I–]. The plot is a straight line with gradient n. 6

Volume of 0.01 mol dm–3 KI(aq) log (volume of aqueous KI) log (1/t) /cm3

5.0 0.70 –2.41

10.0 1.00 –2.13

15.0 1.18 –1.99

20.0 1.30 –1.86

25.0 1.40 –1.76

7 The gradient of the graph shows that the reaction is first order with respect to iodide ions. 8 The volume of potassium iodide solution in step B should be kept constant. The volume of hydrogen peroxide used in step A should be varied while keeping the total volume of the mixed solutions the same by varying the volume of water added in step B.

Core practical 13a: Investigating the rate equation for the reaction of iodine with propanone (page 465)

1 The student added the samples to excess alkali to neutralise and the acid catalyst and so stop the reaction. Then the titration could be carried out knowing the time that the reaction stopped.

Neutralising the reaction mixture is also necessary to stop the acid reacting with the thiosulfate solution. 2 The iodine concentration is proportional to the volume of sodium thiosulfate solution added during the titration. The rate of change of iodine is therefore proportional to the slope of the graph in Figure 16.19. At this initial stage of the reaction the graph is a straight line so the rate is a constant. 3 In this experiment the iodine concentration is much lower than the concentrations of propanone and acid. This means that at the start only the iodine concentration is varying significantly. As the iodine concentration goes down the rate of reaction stays constant. The reaction is zero order with respect to iodine. 4 In the second experiment the propanone and acid are in even larger excess compared to the iodine and their concentrations remain constant while the added iodine is all used up. The first experiment shows that the iodine concentration does not affect the rate of reaction and so using the time for the iodine colour to disappear to calculate the initial rate is justified. 5 In the second experiment all the volumes of the solutions are small so it is important to measure them with as accurately as possible. Small errors in the measured volumes could lead to substantial percentage errors in the results. 6 The volume of I2(aq) is not the same each time so instead of using 1/t as the measure of the initial rate, the student used (volume of I2(aq)/t) as a measure of the initial rate. 7 Comparing runs 1 and 4, the iodine concentration is halved while other concentrations remain the same. The rate of reaction stays the same confirming that the reaction is zero order with respects to iodine. 8 Comparing runs 1 and 2, the acid concentration is halved while other concentrations remain the same. The rate of reaction is roughly halved suggesting that the reaction is first order with respects to hydrogen ions. Comparing runs 1 and 3, the propanone concentration is halved while other concentrations remain the same. The rate of reaction is roughly halved suggesting that the reaction is first order with respects to propanone.

1 + 1 0 + 9 Rate = k[CH3COCH3(aq)] [H ] [I2] = k[CH3COCH3(aq)][H (aq)] 10 The rate equation suggests that propanone and hydrogen ions are involved in the rate- determining step but not iodine molecules.

Core practical 14: Finding the activation energy of a reaction (page 469)

1 After mixing the reaction produces iodine. At first the iodine reacts with thiosulfate ions and its turned back into iodide ions. Each time, one the same, small amount of iodine has been formed, all the thiosulfate is used up and a slight excess of iodine gives the blue-black colour with starch.

The amount of thiosulfate ion is small, so the reaction has not gone far when the colour change is seen, hence 1/t is a measure of the initial rate. 2 The solutions in both tubes must be at the same temperature before they are mixed. On mixing, timing can then start without any further adjustment to the temperature. 3

Temperature/°C 30 36 39 45 51

Temperature/K 303 309 312 318 324

Time, t, for the blue 204 138 115 75 55 colour to appear/s

ln (1/t) –5.32 –4.93 –4.75 –4.32 –4.01

1000 /K–1 3.30 3.24 3.21 3.14 3.09 T

4 Multiplying the values of 1/T by a constant factor makes them easier to plot without affecting the gradient of the resulting graph.

5 6 The gradient of the graph = –6.36 × 103 K

−Ea Hence R = –6.36 × 103 K

3 –1 –1 So –Ea = –6.36 × 10 K × 8.31 J K mol

–1 –1 The activation energy Ea = 52.9 kJ mol

Practice questions (pages 471-2)

+ – 1 a) H2O2(aq) + 2H (aq) + 2I (aq) → 2H2O(l) + I2(aq) [1]

– b) Rate = k[H2O2(aq)][I (aq)] [2] c) Second order [1] d) The first step [1], which includes the two species that feature in the rate equation. [1] 2 a) Labelled axes [1]; accurate plot [1]; smooth curve. [1]

b) Three half-lives = 40 × 103 s [2]; half-life of first order reaction independent of initial concentration. [1] c) Method of calculating gradient [1]; values for gradients [2]; units: mol dm–3 s–1. [1] d) Axes [1]; accurate plot with straight line. [1] k = 1.8 × 10–5 s–1 [1] for value, [1] for units 3 a) Axes [1]; plot and curve. [1] Half-life from 20 × 10–3 mol dm–3: 3.6 × 103 s [1] Half-life from 12 × 10–3 mol dm–3: 7.6 × 103 s [1] Lower starting concentration – higher half-life. [1] b) Tangents drawn [1]; gradients calculated [1]; concentration units included. [1] c) Log (rate) and log (concentration) values quoted. [2] Graph correctly plotted with labelled axes. [2] Order 2 derived from line of best fit. [2] 4 a) i) Changing [X] does not affect the rate [1]; zero order. [1] ii) Rate quadruples if [Y] doubles [1]; second order. [1] b) Rate = k[Y]2 [2] (1 × 10−4 mol dm −− 31 s ) c) k = = 1 × 10–2 dm3 mol–1 s–1 [1] for value, [1] for units (1 × 10−−1 mol dm32 )

d) Rate-determining step involves two Y molecules. [1]

Y + Y → Y2 slow [1]

Y2 + X → XY2 fast [1] e) Range of answers possible (1 mark for each point made up to 3). Such as: • Rate equations help to deduce reaction mechanisms; this information can inform methods of chemical synthesis. • Understanding catalytic activity (including enzymes) can lead to the development of more efficient manufacturing processes or guide drug design and development. • Rate measurements are important in controlling chemical changes in laboratory and industrial processes; chemical engineers use rate data when designing chemical plants. 5 a) The concentrations of all the reacting chemicals are known at the start before they have begun to be used up. [1] It is then only necessary to follow the concentrations of one of the reactants long enough to determine the initial rate from a concentration–time graph. [1] b) i) The reaction is second order with respect to hydrogen. [1] ii) The reaction is first order with respect to NO. [1]

2 c) Rate = k[H2] [NO] [1] (1.20 mol dm−−31 s ) d) k = = 4.2 × 105 dm6 mol–2 s–1 (0.012 mol dm−−32 ) (0.002 mol dm3 ) [1] for value, [1] for units

e) Most reactions take place in several steps. Only one of these steps is rate determining. [1] It is the concentrations of the molecules or ions involved in the rate-determining step that appear in the rate equation. [1] f) i) As the temperature rises the value of the rate constant gets larger. [1] ii) The rate of reaction increases as the temperature rises. This happens because as it gets hotter the mean kinetic energy of the molecules increases. [1] This is shown by a shift in the curve of the Maxwell–Boltzmann distribution to the right. The consequence is that there are more molecules with enough energy to react when they collide. [1] The collisions have enough energy to exceed the activation energy for the reaction. [1] The rate increases even if the concentrations of reactants stays the same. This is explained, in the rate equation, by the fact that the value of the rate constant increases. [1] 6 a) The clock procedure measures the initial rate – in this case by finding the time taken for a small, fixed amount of iodine to form. [1] At first the thiosulfate ions immediately turn the iodine back to iodide ions [1]; but once all the thiosulfate is used up the iodine produces a deep blue-black colour with the starch indicator. [1] b) Completion of table as in Figure 16.10 [3]:

Temperature, T/K 288 292.5 299 308 315 Time, t, for the blue 10.0 7.0 5.0 3.5 2.5 colour to appear/s ln (1/t) –2.30 –1.95 –1.61 –1.25 –0.92 1/T /K–1 0.00347 0.00342 0.00334 0.00325 0.00318

Plot of graph of ln (1/t) against 1/T with appropriate scales and accurate linear plot. [3] −E Gradient = –4180 K [1] = a R

–1 Ea = 35 kJ mol [1]

c) The catalyst provides an alternative reaction pathway with a lower activation energy. [1] One

possible alternative route is that when the catalyst is present, the iron(III) ions oxidise iodide

ions to iodine [1], and that peroxodisulfate(VI) ions then oxidise the resulting iron(II) ions

back to iron(III), thus regenerating the catalyst. [1] Perhaps the negative peroxodisulfate(VI)

ions react more readily with positive iron(II) ions than with negative iodide ions. [1]

– – – 7 a) 3ClO (aq) → ClO3 (aq) + 2Cl (aq) [1]

Chlorate(I) (+1) is oxidised to chlorate(V) (+5) and reduced to chloride (–1). [2] b) The rate decreases by a factor of 4. [1]

c) The rate-determining step appears to involve two chlorate(I) ions. [1] Step 1: slow – rate-determining [1]

– – – 2ClO (aq) → ClO2 (aq) + Cl (aq) [1] Step 2: fast [1]

– – – – ClO2 (aq) + ClO (aq) → ClO3 (aq) + Cl (aq) [1]

– d) One possibility is to look for evidence that ClO2 (aq) exists as an intermediate. [1] It might be detected spectroscopically. [1] This might be hard to do since the second step is fast and so the intermediate disappears very quickly. 8 a) Reactions happen when molecules collide. Collisions with an energy greater than the activation energy lead to reaction. [1] Typically the activation energy is to the right-hand end of the Maxwell–Boltzmann distribution, so that only a small proportion of collisions lead to reaction. [1] Raising the temperature shifts the Maxwell–Boltzmann distribution to the right. [1] The value of the activation energy does not change but the area under the curve of the distribution representing the number of molecules with enough energy to react can increase markedly for a small rise in temperature. [1] b) Raising the temperature increases the speed with which molecules in a gas move so it increases their kinetic energy. [1] Raising the temperature increases not only the number of collisions per second, but also the energy of the collisions. [1] So the increase in reaction rate is the result of two factors: collision rate and the energy per collision. [1] c) Plot of graph of ln (1/t) against 1/T with appropriate scales and accurate linear plot. [3] The gradient of the graph = –2.24 × 104 K [2]

4 –1 –1 Ea = 2.24 × 10 K × 8.31 J K mol [1]

–1 –1 = 186 000 J mol = 186 kJ mol [1]

Test yourself (page 474)

1 hexane 2-methylpentane

3-methylpentane 2,3-dimethylbutane

2 2-bromopentane 3-bromopentane

3 One alcohol and one ether selected from the following four alcohols and three ethers. Four alcohols: butan-1-ol butan-2-ol

2-methylpropan-1-ol 2-methylpropan-2-ol

Three ethers: Ethoxyethane 1-methoxypropane 2-methoxypropane

4 In general, the more branched the hydrocarbon the lower its boiling temperature: 2,2- dimethylpropane (10 °C); 2-methylbutane (28 °C); pentane (36 °C). Boiling temperature depends on the strength of the intermolecular forces. The more branched isomer has lower surface area and so weaker London forces. Pentane 2-methylbutane 2,2-dimethylpropane

Test yourself (page 475)

5 a) E-pent-2-ene (trans) Z-pent-2-ene (cis)

b) E-2-bromobut-2-ene (cis) Z-2-bromobut-2-ene (trans)

c) E-1-chloro-2-methylbut-1-ene Z-1-chloro-2-methylbut-1-ene

The cis–trans system breaks down for this pair of isomers 6 a) b)

Test yourself (page 479)

7 Glove, shoe, screw, spiral binding of the notebook, jug, recorder.

8 CH3CHClBr, CH3CH(NH2)COOH 9 Butan-2-ol, pentan-2-ol 10

Test yourself (page 480)

11 a) 3-Chloro-3-methylhexane is a tertiary halogenoalkane. These compounds tend to undergo

nucleophilic substitution by the SN1 mechanism. In the first step the C−Cl bond breaks to give a planar intermediate with a positive charge on the carbon atom that was originally chiral. Iodide ions, at least in theory, can then attack from either side of the intermediate giving both possible isomers of the chiral product. The result is a mixture of the two isomers.

b) Iodide ions are relatively large. If the new bond between carbon and iodine forms quickly after the old carbon–chlorine bond breaks, then it may be that the iodide ion is less likely to attack from the side of the molecule from which the chloride ion departs. If so, the outcome will favour more of one isomer than the other and there will not be equal amounts of the two isomers present. Overall the mixture will rotate the plane of plane-polarised light slightly in one direction.

12 IThe statement that the products of the reactions are stereospecific shows that the reactions must be proceeding by the SN2 mechanism. In this mechanism the molecule turns inside out (inverts). It is not always the case that the (+) isomer of a reactant will give the (−) isomer of the product, though that is what happens here.

Activity: Chirality and living things (page 481)

1 a) Two C=C double bonds, as in alkenes, and the carbonyl group of a ketone. b) A carboxylic acid group and an amine group. 2 a) Isoleucine – the carbon atom at the centre of the tetrahedron with four different groups attached to it. b) Carvone – the carbon atom in the ring at the bottom of the hexagon drawn in Figure 17.1.15. It is much easier to see why this atom is chiral with a molecular model. Do not forget that hydrogen atoms are not shown in a skeletal formula. Also note that the arrangement of atoms round this carbon atom is tetrahedral with four different groups: and, and in the ring on one side and on the other side. c) Thalidomide – the carbon atom in the right-hand ring joined to the nitrogen in the five- membered ring has four different groups arranged tetrahedrally around it – don’t forget the hydrogen atom not shown on this carbon in Figure 17.1.16. 3 There is a chiral centre in alanine because the central carbon atom has four different groups around it and is asymmetric. Glycine has two hydrogen atoms on the central carbon atom and so it has a plane of symmetry and is not chiral. 4 Often one optical isomer is active in biological systems while the other is less active, inactive or even has quite different (and possibly harmful) effects. Wherever possible, therefore, it is better to market a product containing only one isomer. For this reason chemists have developed methods for separating optical isomers from racemic mixtures and methods of chiral synthesis that produce only one of the two isomers. 5 The benefits in treatment of leprosy, HIV-related conditions and some cancers are undeniable, but there are risks of new thalidomide-affected children in countries where controls on the use of the drug may not be efficient. The World Health Organization does not recommend the use of thalidomide in leprosy. See www.who.int/lep/research/thalidomide/en/: Because of its known teratogenic effects, WHO does not recommend the use of thalidomide in leprosy. Experience has shown that it is virtually impossible to develop and implement a fool- proof surveillance mechanism to combat misuse of the drug. Today, a number of thalidomide babies continue to be born each year reflecting regulatory insufficiency and widespread use under inadequate supervision.

Exam practice questions (page 483)

1 a) Two structural isomers with the same functional group [1]; structures of 1-chloropropane and 2-chloropropane. [1]

1-chloropropane 2-chloropropane b) Six isomers in total: three structural isomers, each of which have E/Z forms. • F and H on the same C atom [2]

Z-1-bromo-1-chloro-2-fluoroethene E-1-bromo-1-chloro-2-fluoroethene • F and Cl on the same C atom [2]

Z-2-bromo-1-chloro-1-fluoroethene E-2-bromo-1-chloro-1-fluoroethene • F and Br on the same C atom [2]

Z-1-bromo-2-chloro-1-fluoroethene E-1-bromo-2-chloro-1-fluoroethene c) Structural isomerism (position) and optical isomerism.

1-bromo-2-chloroethane [1] 1-bromo-1-chloroethane [1] 1-Bromo-1-chloroethane exists as a pair of optical isomers:

[1] 2 a) A chiral centre is typically a carbon atom with four different groups around it [1]; in general a molecule is chiral if it is asymmetric. b) The mirror-image forms of a chiral molecule cannot be superimposed on one another [1]; they have opposite effects as they rotate the plane of plane-polarised light. [1]

c) Only CH3CHOHCOOH is chiral and can exist as two optical isomers [1]; the other two compounds both have two hydrogen atoms on the central carbon atom.

d) A racemic mixture contains equal amounts of the two optical isomers [1]; the two isomers rotate the plane of plane-polarised light equally but in opposite directions, so their effects cancel. [1]

e) i) C13H21O3N [1] ii) There is one chiral carbon atom: [1]

3 Stereoisomerism involves molecules with the same molecular formula [1]; and the same structural formula [1]; but different three-dimensional shapes [1]; in which their atoms occupy different positions in space. [1] There are two forms of stereoisomerism. E/Z (cis/trans) isomerism [1]; which occurs principally in alkenes and other compounds with C=C double bonds. [1] In E/Z isomerism, the Z-isomer has atoms, attached to the C atoms of the double bond, with higher atomic number [1] on the same side of the double bond. [1] In the E-isomer, atoms attached to the C atoms of the double bond with higher atomic number are on opposite sides of the double bond. [1] An example of E/Z isomerism is 1,2-dibromoethene. Named structures of the E- and Z-isomers. [2] Optical isomerism [1] involves non-superimposable mirror images. [1] In most cases, this requires the molecules to possess a C atom with four different atoms or groups attached to it. [1] These mirror images are labelled (+) and (−) forms. [1] The (+) form rotates plane-polarised light clockwise [1] and the (−) form rotates plane-polarised light anticlockwise. [1] A named example of optical isomerism [1]; e.g. alanine or bromochlorofluoromethane. Diagram showing structures of the optical isomers of the example suggested as mirror images. [1] (Any 15 of these or other legitimate points)

4 a) There is a chiral centre in C2H5CHBrCH3. This means that the molecule is asymmetric and has distinct mirror-image forms. [1] b) Indicative content: . At 25oC, reaction mechanism is SN2. . Molecule inverts, so optical activity is retained. . At 80oC, the energy is sufficient to break the CBr bond. . So the mechanism is SN1 mechanism in which the first step forms a carbocation. . The carbocation is planar so hydroxide ions can attack from either side of the plane. . Giving rise to a roughly equal mixture of the two optical isomers, i.e. optically inactive racemic mixture.

c) H2C=CHCH2CH3; but-1-ene [1]

CH3CH=CHCH3; but-2-ene [1]; both E- and Z-isomers. [1] 5 a) Electrophilic addition [1]

[4] for the diagram: [1] for each of the three curly arrows and [1] for the structure of the intermediate ion. b) HBr can add across the double bond in two ways via different carbocations. [1] The secondary carbocation is more stable than the primary because of the extra inductive effect [1], so 2-bromopentane is the major product. [1] c) In the mechanism, the planar carbocation [1] can be attacked by the bromide ion from above or below [1] with equal probability [1] so an optically inactive racemic mixture is formed. [1] 6 a)

A B C D [1] [1] [1] [1] b) i) Call the isomers A, B, C & D as shown. A possible method is: Add bromine water to all four [1]: C & D decolourise it [1]; no change with A & B. [1]

Add acidified potassium dichromate(VI) to C & D [1]: D turns it from orange to green [1]; no change with C. [1]

Add acidified potassium dichromate(VI) to A & B. [1]: B turns it from orange to green [1]; no change with A. [1] ii) A – absorption at 1720–1700 cm–1 for a ketone C=O. [1] B – absorption at 1740–1720 cm–1 for an aldehyde C=O. [1] C – absorption at 1669–1645 cm–1 for an alkene C=C but no absorption for an O–H alcohol around 3750–3200 cm–1. [1] D – absorption at 1669–1645 cm–1 for an alkene C=C and also an absorption for an O–H alcohol around 3750–3200 cm–1 . [1] c) and

Four-membered ring is strained as the bond angle is 90° [1] compared with the tetrahedral 109.5°. [1] Three-membered ring is even more strained as the bond angle is only 60°. [1] The four-membered ring is likely to be more stable. [1]

7 The final step gives equal amounts of two optical isomers. [1]

The molecules of R must have a chiral centre: CH3CH2CH(CH3)COOH. [1] The change from P to Q suggests the hydrolysis of the double ethyl ester of a carboxylic acid. [1] So Q must have two carboxylic acid groups. [1]

Q decomposes to R by losing CO2. [1] Acids with two carboxylic acid groups attached to the same carbon atom lose a carboxylic acid group relatively easily. [1] Q: [1]

P: [1]

Test yourself (page 486)

1 a) b) c)

2 a) CH3CH2CH2CH2OH(l) + [O] → CH3CH2CH2CHO(l) + H2O(l) b) The reaction of butan-1-ol with the oxidising agent is vigorous. If the alcohol is added too quickly the mixture can get very hot. There is a danger of loss of volatile organic compounds before all the alcohol has been added if it is added too quickly. c) The conditions are chosen to make the aldehyde and avoiding oxidising the aldehyde further to butanoic acid. The aldehyde distils off as soon as it forms.

Test yourself (page 487)

3 a) b) c)

4 C10H18O

5 Both have the molecular formula C3H6O but one is a ketone and the other an aldehyde.

6 CH3CH2CHOHCH3(l) + [O]→ CH3CH2COCH3(l) + H2O(l) 7 At least three carbon atoms are needed for a ketone so ethanone itself cannot exist. Phenylethanone has a two carbon chain attached to a benzene ring so the carbonyl group is attached to a carbon on each side

Test yourself (page 488)

8 a) Propanone and propanal are liquids at room temperature, boiling at 56 °C and 49 °C respectively. Butane, which has the same molar mass, is a gas boiling at –0.5 °C. Propan-1-ol with a very similar molar mass boils at 97 °C. b) The C=O bond is polar and so aldehyde molecules have a permanent dipole. Molecules with permanent dipoles have stronger intermolecular forces than alkanes, which are non-polar. Hydrogen bonding, such as between the –OH groups of alcohols, is considerably stronger than other intermolecular forces. 9 The oxygen atom in the carbonyl group of propanone can take part in hydrogen bonding with the –OH groups in water molecules. This allows propanone molecules to break into the hydrogen- bonded network of molecules in water and mix freely.

10 In any aldehyde, solubility is a balance between the attraction water molecules have for the electronegative oxygen atom which can take part in hydrogen bonding, and the hydrophobic properties of the hydrocarbon part of the molecule. The much longer hydrocarbon chain in hexanal means that this is more significant.

Test yourself (page 490)

11 Cr is reduced from +6 to +3. 12 a) Mix with excess of an acidic solution of potassium dichromate(vi) and then heat under reflux. b) i) A flask fitted with a reflux condenser ii) Apparatus rearranged for distillation of the product from the reaction flask

+ – c) CH3CH2CH2CHO(l) + H2O(l) → CH3CH2CH2COOH(l) + 2H (aq) + 2e

13 a) HOCH2CH2OH + 2[O] → OHCCHO + 2H2O b) OHCCHO + 2[O] → HOOCCOOH 14 Butanal is reduced to the primary alcohol, butan-1-ol. Butanone is reduced to the secondary alcohol, butan-2-ol. 15 C=O becomes CH−OH (in an aldehyde or a ketone).

16 a) CH3CH2CH(CH3)CHO(l) + 2[H] → CH3CH2CH(CH3)CH2OH(l) b)

Test yourself (page 490)

17 a) The cyanide ion is negatively charged and has a lone pair of electrons. b) Heterolytic bond breaking c) The C−O bond which breaks was formed using one electron from C and one from O. When both electrons are transferred to O it gains one extra electron and so has a single negative charge. d) HCN acts as an acid, donating a proton to the intermediate. 18 a) The product of the reaction with ethanal has four different groups attached to the central carbon atom. It is asymmetric.

The product of the reaction with propanone has only three different groups attached to the central carbon atom. It is not asymmetric.

b) Ethanal forms 2-hydroxypropanenitrile. Propanone forms 2-hydroxy-2-methylpropanenitrile. 19 a)

b) Nucleophiles attack the carbonyl group. Double bonds in alkenes undergo electrophilic addition. Hydride ions are nucleophiles and these negative ions would be repelled by the electron-rich C=C bonds.

Test yourself (page 494)

2+ – 20 Cu (aq) + 2OH (aq) → Cu(OH)2(s)

21 a) CH3CH2CHO(l) + [O] → CH3CH2COOH(l) b) Silver ions in solution (oxidation state +1) are reduced to silver metal (oxidation state 0).

2+ – – 22 a) 2Cu (aq) + 2OH (aq) + 2e → Cu2O(s) + H2O(l)

+ – b) [Ag(NH3)2] (aq) + e → Ag(s) + 2NH3(aq) 23 Two possible sets of answers. EITHER A (1-chlorobutane) is hydrolysed to B (the primary alcohol, butan-1-ol) which is oxidised to C (the aldehyde, butanal). OR A (1-chloro-2-methylpropane) is hydroysed to B (the primary alcohol, 2-methylpropan-1-ol) which is oxidised to C (the aldehyde 2-methylpropanal).

Test yourself (page 494)

24 a) The reagent first oxidises the alcohol to a carbonyl compound, which can then take part in the triiodomethane reaction. b) Sodium chlorate(i) is a solution of chlorine in sodium hydroxide. The chlorate(i) ions oxidise iodide ions to iodine.

25 Pentan-2-one 3-methylbutan-2-one

26 Ethanal

– – – 27 a) i) I2 + 2OH → I + OI + H2O

– – ii) RCOCH3 + 3OI → RCOCI3 + 3OH

– – iii) RCOCI3 + OH → CHI3 + RCOO b) 3 × equation (i) is:

– – – 3I2 + 6OH → 3I + 3OI + 3H2O Adding this to equations (ii) and (iii) and cancelling produces

– – – 3I2 + 6OH + RCOCH3 + 3OI + RCOCI3 + OH

– – – – → 3I + 3OI + 3H2O + RCOCI3 + 3OH + CHI3 + RCOO

– – – 3I2 + 4OH + RCOCH3 → 3I + 3H2O + CHI3 + RCOO

Activity: Chemicals in perfumes (page 489)

1 2 Advantages of synthetic perfume chemicals include: • uniformity – the quality of oils from natural sources (plants or animals) can vary with the conditions for growth and the circumstances in which the material was collected or harvested and then stored • reliability of supply – weather conditions can cause a crop to fail or yield poorly, while political or economic conditions can restrict access to raw materials • stability of price – the cost of synthetic chemicals is more predictable than the cost of chemicals extracted from natural sources. 3 Compounds with an aldehyde group: undecanal, citral. Compounds with a ketone group: hexamethyl tetralin musk, methyl dihydrojasmonate. 4 a)

b) The carbon skeleton of a geraniol molecule is based on two 2-methylbuta-1,3-diene molecules joined together. 5 a) London forces increase with increasing molecular size because the higher the number of electrons in a molecule, the greater the fluctuating dipoles that lead to weak intermolecular forces.

Molecules with fewer than five carbon atoms are generally too volatile for use in perfumes: they are either gases at room temperature or liquids with such low boiling temperatures that they vaporise too easily. Molecules with more than 17 carbon atoms are generally not volatile enough. Intermolecular hydrogen bonding is possible between molecules of an alcohol, but not between aldehyde molecules. Alcohols therefore have higher boiling temperatures than aldehydes of similar molar mass. b) Musks are near the upper limit of molecular size for perfume chemicals. They may act as fixatives by means of hydrogen bonding or dipole–dipole interactions with more volatile components. c) Each –OH group in an alcohol can link to two or three water molecules by hydrogen bonding. As a result, alcohols are more soluble in water than other perfume ingredients such as carbonyl compounds.

Activity: Identifying an unknown carbonyl compound (page 495)

1 The derivative must be pure if the melting temperature is to be accurate. Impurities lower the melting temperature of a substance. 2 The derivative and the impurities dissolve in the hot ethanol. On cooling, the derivative recrystallises because it is present in relatively large amounts. The much smaller quantities of impurities remain in solution. Washing the purified crystals with pure solvent in the filter funnel makes sure that impurities in solution are rinsed away and do not recrystallise as the derivative dries. 3 The solvent must dissolve the derivative and any impurities. The derivative must be soluble in the hot solvent but very much less soluble in the cold solvent. 4 Pure solids melt suddenly at a definite temperature. Impure solids soften and melt over a range of temperatures. 5 Butanone 6 Insoluble impurities need to be removed by filtration of the hot solution in ethanol. In order to prevent crystals forming in the filter paper or funnel during filtration, more ethanol should be used than the minimum needed to dissolve the derivative. A fluted filter paper in a hot short- stem filter funnel should also be used to speed up filtration. As the solution is less concentrated, a lower yield of crystals will be obtained on cooling

Exam practice questions (pages 498–9)

Reactant Reagent Organic product Name Displayed formula CH3CH2CHO Tollens’ reagent Propanoic acid [1]

[1] 2– + CH3CH2CHO Cr2O7 /H [1] Propanoic acid

[1] CH3CH2CHO LiAlH4 Propan-1-ol [1]

[1]

2 a) C10H16O [1] b) i) Carbon–carbon double bond as in an alkene [1]; aldehyde group. [1] ii) Ketone group. [1] c) i) The blue reagent gives an orange-brown precipitate [1] with citral; but not with β-ionone. [1] ii) The reagent gives a bright orange precipitate [1]; both with citral and with β-ionone. [1] iii) The reagent gives a pale yellow precipitate [1] with β-ionone; but not with citral. [1] d) i) ii)

[1] [1]

e) i) Both compounds have E/Z isomers. [1] ii) EITHER OR

[1] [1]

Z-citral [1] Z-β-ionone [1] iii) Neither compound has a chiral centre and so neither has optical isomers. [1]

3 Indicative content: • X must have four separate groups around the central carbon atom • Including an acid group to give carbon dioxide with a carbonate • Oxidation gives a carbonyl compound Y which is a ketone because it does not react with Fehling’s solution: • But Y is not a methyl ketone as doesn't react with iodine and sodium hydroxide. • X is CH3CH2CHOHCOOH. • Y is CH3CH2COCOOH.

4 a) W: empirical formula CH2, molecular formula C4H8 [1]; could be but-1-ene or but-2-ene [1]; because both give 2-bromobutane when they add HBr. X: 2-bromobutane [1]; because it hydrolyses to butan-2-ol. [1] Y: butan-2-ol [1]; a secondary alcohol because it oxidises to a ketone. [1] Z: butanone [1]; a ketone because it does not react with Benedict’s reagent. [1]

b) CH3CH=CHCH3(l) + HBr(g) → CH3CH2CHBrCH3(l) [1]

CH3CH2CHBrCH3(l) + NaOH(aq) → CH3CH2CHOHCH3(l) + NaBr(aq) [1]

CH3CH2CHOHCH3(l) + [O] → CH3CH2COCH3(l) [1] (oxidation) c) W: Electrophilic addition [1] X: Nucleophilic substitution [1] 5 i) a) Tollens’ [1]: pentanal gives a silver mirror [1]; no reaction with pentan-3-one. [1]

b) Acidified K2Cr2O7 [1]: no reaction with pentan-3-one [1]; solution turns from orange to green with pentan-3-ol. [1] c) Iodine and sodium hydroxide [1]: no reaction with pentan-3-ol [1]; yellow precipitate with pentan-2-ol. [1] ii) a) Infrared absorptions: pentanal gives an absorption in infrared in the region 1740–1720 cm–1 [1]; pentan-3-one gives an absorption in infrared in the region 1720–1700 cm–1. [1] b) Infrared absorptions: pentan-3-one gives an absorption in infrared in the region 1720–1700 cm–1 [1]; pentan-3-ol gives an absorption in the region 3750–3200 cm–1. [1] c) Optical activity: pentan-2-ol has a chiral centre so each enantiomer will show optical activity [1]; pentan-3-ol does not have a chiral centre so does not show any optical activity. [1]

+ – 6) a) CH3CH2CHO + H2O → CH3CH2COOH + 2H + 2e [1]

– – – b) CH3CH2CHO + 3OH → CH3CH2COO + 2H2O + 2e [1] c)

[1]

7 a)

Cyanide ion plus lone pair plus arrow [1]; propanal plus arrow. [1] Structure of intermediate ion. [1] Arrow from lone pair on O to H–CN and arrow from H–CN bond to C of CN. [1] b) If the pH is too high, then [H+] is low [1] (and [HCN] falls), so the second stage becomes slow. [1] If the pH is too low, then [CN–] is low [1] and the first stage becomes slow. [1] c) A racemic mixture is formed [1] because attach by the nucleophile can occur from above or below the plane of the carbonyl group. [1]

8 a) i) C6H12O6 [1]

ii) The formula can be rewritten as C6(H2O)6. So glucose contains carbon with hydrogen and oxygen in the same proportions as in water. [1] Hence the name carbohydrate. b) The large number of −OH groups means that there is extensive hydrogen bonding between glucose molecules. [1] These intermolecular forces are strong enough for the relatively small molecules to form a solid at room temperature. [1] Glucose molecules can also form many hydrogen bonds with water molecules [1]. The interaction with water is strong enough for glucose molecules to break away from the solid and dissolve. [1] c) The functional group in open chain glucose is an aldehyde group. Glucose is an aldose. [1] d) i) Glucose gives an orange precipitate when it reacts with 2,4-dintrophenylhydrazine. [1] ii) 2,4-Dinitrophenylhydrazine reacts with the aldehyde group in the open chain form. [1] Only 1% of glucose is in the open chain form in solution. [1] e) i) As an aldehyde, glucose reacts with Tollens’ reagent to give a silver mirror. [1]

ii) Glucose is a reducing agent. It reduces silver(I) ions in Tollens’ solution to metallic silver.

[1] (Glucose similarly reduces copper(II) to copper(I) when it reacts with Fehling’s or Benedict’s reagent.)

f) The oxygen on carbon number 5 [1] uses its lone pair to act as a nucleophile [1] to attack the δ+ carbon of the carbonyl group on carbon number 1. Nucleophilic addition [1] occurs to produce the six-membered ring structure. [1] A racemic [1] mixture is formed of α and β glucose. 9 Equations to describe the mechanism of addition to C=C and C=O. (2 × [2])

Propene with bromine HCN with propanone

Similarities Addition reaction Heterolytic bond breaking

Two-step process (any 2 for [2]) Differences Electrophilic reagent Nucleophilic reagent π δ Attacks electron-rich bond Attacks + end of polar C=O bond Intermediate is a positive ion Intermediate is a negative ion (any 2 for [2]) (any 2 for [2])

10 a) CH3COCH2CH3 + 2[H] → CH3CHOHCH2CH3 [1]

b) i) LiAlH4 effectively contains hydride ions so reacts by nucleophilic addition. [1] C=O bonds are polar and the nucleophile can attack the δ+ carbon. [1] C=C bond in alkenes is non-polar so nucleophiles cannot attack OR The π cloud of electrons in alkenes repels the negative hydride ion. [1]

ii) Hydride ions react with δ+ H atoms [1] in water molecules so LiAlH4 is decomposed by solvents which can donate protons. [1]

c) LiAlH4 is the stronger [1] as it is a better hydride donor.

– – AlH4 ions contain H ion co-ordinately bonded to Al (using vacant orbital in level 3). [1]

– – These hydride ions are less strongly held than the H ion co-ordinately bonded to B in BH4 ions, which use vacant orbital in level 2. [1] 11 a) A molecule of P adds two molecules of HCN so P has two aldehyde groups. [1]

P is OHCCH2CH2CHO [1] rather than CH3CH(CHO)2. [1]

Q is HOCH(CN)CH2CH2CH(CN)OH. [1] Oxidation converts aldehyde groups to acid groups. [1]

R is HOOCCH2CH2COOH. [1] 1.0 g R is 0.00847 mol of the compound. [1] In the titration this reacts with 0.0169 mol sodium hydroxide [1]; so 1.0 mol of the acid reacts with 2.0 mol of the alkali. [1] This confirms that there are two acid groups in R.

b) The ratio of amounts in Z is C : H : O = 5.36 : 7.11 : 1.79 [1], which is 3 : 4 : 1.

The simplest (empirical) formula of Z is C3H4O. [1]

The molar mass of Z is 56 and its molecular formula is also C3H4O. [1] 0.00179 mol Z [1] reacts with 0.00356 mol hydrogen. [1]

So 1 mol Z reacts with 2 H2, showing that there are two double bonds in the molecule. [1] Z reduces Fehling’s solution so one of the double bonds is in an aldehyde group. [1]

Z is CH2=CH–CHO. [1] 12 Step 1: Warm 2-bromobutane with aqueous/alcoholic sodium hydroxide [1] to form butan-2-ol. [1]

Step 2: Heat with acidified potassium dichromate(VI) [1] to form butanone. [1]

Step 3: Add iodine and sodium hydroxide [1] to form propanoate ions [1] (and CHI3). Step 4: Acidify with dilute hydrochloric acid. [1]

Test yourself (page 522)

1 The molar mass (or relative molecular mass) is needed. This is determined by mass spectrometry.

2 One possibility is CH2=CH–C≡C–CH=CH2. In this structure: • the carbon–carbon bond lengths will not all be the same • there will be several isomers of the 1,2-dichloro compound • the double and triple bonds will be very reactive.

3 a) C7H8 b) C7H8 c) C6H5–CH3

Test yourself (page 524)

4 a) Benzene is 152 kJ mol–1 more stable than the Kekulé structure. b) –240 kJ mol–1 for cyclohexa-1,4-diene (independent C=C bonds). Less exothermic than –240 kJ mol–1 for cyclohexa-1,3-diene (some delocalisation of the C=C bonds). 5 a) All the carbon–carbon bonds will have exactly the same length and strength. b) All six carbon–carbon bonds are exactly the same, so all molecules with two chlorine atoms on any pair of adjacent carbon atoms are identical. c) The delocalised electrons stabilise benzene, which does not undergo addition reactions like cycloalkenes. Any addition would disrupt the ring of delocalised electrons and form an energetically less stable product. 6 Because benzene does not have three localised C=C bonds which the name cyclohexa-1,3,5- triene suggests.

Test yourself (page 525)

7 Because the substituents in the benzene ring should be named alphabetically, i.e. bromo-, then chloro-. 8 1,3-Dinitrobenzene

9 a) 1-Methyl-3-nitrobenzene

b) 4-Bromophenol c) Benzene-1,2-dicarboxylic acid d) Benzene-1,2-diol

10 C6H4Cl2 (3 isomers) 1,2-dichlorobenzene 1,3-dichlorobenzene 1,4-dichlorobenzene,

1,2,3-trichlorobenzene 1,2,4-trichlorobenzene 1,3,5-trichlorobenzene

C6H4Cl2 (3 isomers) Think of these as two hydrogens at positions 1,2, 1,3 and 1,4. 1,2,3,4-tetrachlorobenzene 1,2,3,5-tetrachlorobenzene

1,2,4,5-tetrachlorobenzene C6HCl5 (1 isomer) pentachlorobenzene

Test yourself (page 526)

11 There are only 10 carbons in naphthalene so only 10 electrons in p orbitals can be delocalised. Two circles would imply 12 delocalised electrons. The Kekulé-type structure shown (with delocalisation understood) does not imply 12 delocalised electrons. 12 When benzene molecules approach each other, the movement of electrons (in particular the delocalised π electrons) sets up temporary dipoles. The attractions between these temporary dipoles give rise to weak intermolecular forces, London forces.

13 a) The attractions between water molecules are relatively strong because of hydrogen bonding. The hydrogen bonding is so strong that benzene molecules, which are non-polar, cannot interact and mix with water molecules. b) Hexane, cyclohexene, methylbenzene, etc.

14 a) C6H6(l) + 7½O2(g) → 6CO2(g) + 3H2O(l)

b) C7H8(l) + 2O2(g) → 7C(s) + 4H2O(l)

Test yourself (page 528)

15 a) Reflux boiling benzene with bromine and either iron filings or iron(iii) bromide in a fume cupboard. b) The iron reacts with bromine to form iron(iii) bromide. Iron(iii) bromide polarises the bromine molecules as Brδ+–Brδ– and the Brδ+ atom can then react as an electrophile in attacking the delocalised system in benzene.

In the absence of a halogen carrier (Fe or FeBr3) the bromine is not sufficiently polarised to attack the delocalised system in benzene. 16 a) Iron does not react sufficiently with iodine to produce iron(iii) iodide and therefore iodine is not sufficiently polarised to attack the benzene. b) Iδ+–Clδ– c) The ICl is strongly polarised and the Iδ+ can react as an electrophile, attacking the delocalised π electrons in benzene.

Test yourself (page 529)

17 The π electrons in benzene would repel the nitrate ion; there is no suitable electrophile in the absence of concentrated sulfuric acid.

+ – + 18 HNO3 + 2H2SO4 → NO2 + 2HSO4 + H3O 19 a)

b) Because this would be called 1,2-dinitrobenzene.

20 To provide the oxygen that TNT needs to form CO2, N2 and H2O when it explodes. 21 a)

Test yourself (page 531)

22 a)

+ – c) H + AlCl4 → HCl + AlCl3

23 a) Because AlCl3 reacts with water, forming fumes of hydrogen chloride and an acidic solution

containing hydrated aluminium ions. This reaction will inactivate the AlCl3 catalyst. b) i) ii)

c) The C–I bond in an iodoalkane is weaker than the C–Cl bond in the corresponding chloroalkane. Therefore, the C–I bond will break more easily and the reaction will be faster.

Test yourself (page 532)

24 a) Because its large surface area increases the effectiveness of the catalyst. b)

c) Because they react like alkenes with hydrogen. Their reaction with hydrogen is much faster than the reaction of benzene with hydrogen due to the increased stability of the delocalised electron system in benzene. 25 Each carbon is bonded to a chlorine atom and a hydrogen atom. The C-C bonds in the ring cannot rotate so there are isomers with chlorine atoms on adjacent carbon atoms on the same side or on opposite sides of the ring.

Test yourself (page 534)

26 a) Hydrogen bonding between the –OH groups in phenol molecules gives rise to much stronger intermolecular forces than the weak attractions between the temporary dipoles in benzene. b) Hydrogen bonding between phenol molecules and water molecules makes phenol slightly soluble in water. c) The relatively large non-polar benzene ring in phenol limits its solubility in water much more

than the C2H5– group in ethanol. 27 The phenol first melts and then burns with a very smoky flame. 28 Similarity – form hydrogen bonds with water – dissolve in water

– do not react with carbonates to produce CO2. Difference – phenol (unlike ethanol) is acidic enough to form phenoxides with aqueous alkalis. 29 a) A milky white precipitate as phenol separates from the aqueous solution. b) Phenol forms soluble sodium phenoxide in sodium hydroxide solution. When HCl(aq) is

+ – added, H (aq) ions in the acid react with phenoxide ions, C6H5O (aq), in the solution. The

product is phenol, C6H5OH, which is only slightly soluble and this soon precipitates as a white solid.

Test yourself (page 535)

30 a)

2,4,6-trichlorophenol (commonly known as TCP)

b) A lone pair of electrons on the –OH group interacts with the delocalised electrons in the benzene ring, releasing electrons into the ring and increasing the electron density in the ring. This makes electrophilic attack easier so electrophilic substitution takes place under much milder conditions with phenol than with benzene.

Activity: Studying the reaction of benzine with chlorine (page 533)

1 Because benzene is carcinogenic and toxic. Prolonged exposure, through inhalation or contact with the skin, causes serious damage to health. Also, excess toxic chlorine is continuously released from the reaction mixture. 2 a) To ensure that the excess chlorine and any benzene vapour are removed from the vicinity of the reacting mixture and do not pollute the laboratory. b) To avoid using naked flames because benzene is highly flammable. c) To avoid the benzene boiling. Its boiling temperature is 80 °C.

3 a) 2Fe(s) + 3Cl2(g) → 2FeCl3(s) b) 4

5 a) Because they facilitate the reaction by polarising halogen molecules, which can then attack the delocalised electrons in benzene. b) Because they are not used up in the reaction, which takes place at a faster rate in their presence. c) Because it is used up in forming iron(iii) chloride prior to acting as the halogen carrier. 6 a) More effective.

b) Because aluminium ions/atoms in AlCl3 are smaller than iron ions/atoms in FeCl3. The relative positive charge on the aluminium will, therefore, have a greater density than that on the iron

and this will cause greater polarisation of the Cl2 molecules.

Activity: Manufacturing phenol (page 536)

1 a) Fractional distillation followed by reforming. b) Fractional distillation followed by catalytic cracking.

+ + 2 a) CH3CH2C H2 and CH3C HCH3

+ b) CH3C HCH3 is more stable because it is a secondary carbocation and has the electron- donating (inductive) effect from two alkyl groups to stabilise its positive charge. The alternative primary carbocation is attached to one alkyl group only and so is less stabilised.

c) Propylbenzene, C6H5CH2CH2CH3 3 a)

b) Because H+ ions from the acid react with propene to form the carbocation electrophiles. These ions react with benzene much faster than propene itself can. The H+ ions are then regenerated when the substitution reaction is complete.

4 C6H5CH(CH3)2 + O2 → C6H5OH + CH3COCH3

5 C6H6 + CH3CH=CH2 + O2 → C6H5OH + CH3COCH3 1 mol benzene produces 1 mol phenol + 1 mol propanone If the reaction is 100% efficient, 78.0 g benzene → 94.0 g phenol + 58.0 g propanone But, the reaction is only 85% efficient, so: 85 85 78.0 g benzene → 94.0 × g phenol + 58.0 × g propanone 100 100 → 79.9 g phenol + 49.3 g propanone ∴ 79.9 g phenol and 49.3 g propanone are produced from 78.0 g benzene 49.3 ⇒ 1 g phenol and g propanone are produced from 78.0 g benzene 79.9 79.9 So, 1 tonne of phenol is produced from 0.98 tonnes of benzene, and 0.62 tonnes of propanone are formed at the same time.

Exam practice questions (pages 538-540)

1 a) Indicative content: • Benzene is a planar molecule. • A hexagon with equal length CC bonds. • Each carbon atom forms normal covalent (σ) bonds with its two adjacent carbon atoms and also a hydrogen atom. • The fourth electron on each carbon atom in a p orbital (at right angles to the benzene ring); p orbital overlaps sideways • Forming π bonds above and below the plane. • These electrons are delocalised.

b) The delocalised π electron system in benzene is disrupted initially by formation of an intermediate cation [1], but is restored by elimination of H+ and a reaction involving substitution. [1] In ethene, the localised π bond is weaker than the delocalised π system in benzene. [1] The breakage of the π bond in ethene and the formation of two single bonds by an addition reaction occurs easily. [1] So addition is energetically more favourable than substitution [1] (maximum 4 marks) 2 a) An electrophilic substitution is a reaction in which an electron-deficient species, an electrophile [1], replaces an atom or group of atoms. [1] b) Methylbenzene [1]

Iron(III) bromide or aluminium bromide or their chlorides. [1] c)

d) Friedel and Craft [1]

3 a) A is cyclohexane; B is hydrogen; C is aluminium chloride or iron(III) chloride; D is ethylbenzene; E and F are concentrated nitric acid and concentrated sulfuric acid. [6]

+ + b) i) CH3C H2 [1]; NO2 [1] ii) Temperature 50–60 °C. [1]

+ – HNO3 + H2SO4 → H2O + NO2 + HSO4 [1]

[2] iii) Explosives/further reactions to produce dyes. [1] c) i) Temperatures >150 °C, and high pressure. [1] ii) As a solvent [1] d) i)

n [1]

ii) As an insulator or for packaging. [1]

4 a) i) Aluminium chloride or iron(III) chloride [1] ii) Bubble chlorine into a boiling benzene and catalyst mixture [1]; under reflux [1]; in a fume cupboard. [1]

b) i) I is 1-chloro-4-methylbenzene. [1] II is (chloromethyl)benzene. [1] ii) Formation of I is by electrophilic substitution. [1] Formation of II is by radical substitution. [1] c) i) II reacts with sodium hydroxide. [1]

– OH can attack the δ+ carbon of the CH2Cl group in II. [1] The electron cloud of the benzene ring will repel the OH– so no reaction will take place with I. [1]

ii) C6H5CH2OH [1]; nucleophilic substitution. [1] 5 a) Any two from (maximum 2 marks):

• limiting the amount of concentrated HNO3 used • adding the nitrating mixture to the methyl benzoate in small amounts • keeping the temperature below 15 °C. b) Methyl 2-nitrobenzoate [1]; methyl 4-nitrobenzoate. [1] c) To remove any nitric and sulfuric acid. [1] d) The impurities dissolve in the hot ethanol but there is little present [1] so that none crystallises out when the ethanol is cooled down [1]. e) i) Some will remain dissolved in the ethanol even when it is cooled in ice. [1] ii) Taking the minimum volume of ethanol needed to dissolve all the impure solid at the high temperature. [1] Cooling the ethanol solution so that more crystallises. [1] f) Larger crystals form if the solution cools slowly. [1] g) Methyl benzoate is the limiting reactant. Volume of methyl benzoate used = 5 cm3 Mass of methyl benzoate used = 5.5 g [1] 5.5 g Moles of methyl benzoate used = = 0.0404 mol [1] 136.0 g mol−1

From the equation: 1 mol methyl benzoate → 1 mol methyl 3-nitrobenzoate 0.0404 mol methyl benzoate → 0.0404 mol methyl 3-nitrobenzoate Theoretical yield of methyl 3-nitrobenzoate = 0.0404 mol × 181.0 g mol–1 [1] = 7.3 g [1] 3.5 h) ∴Percentage yield = × 100% = 48% (2 s.f.) [1] 7.3 6 a) A is dilute nitric acid. [1]

b) B C

[1] [1] c) i) EITHER – warming [1] with dilute nitric acid [1] OR – conc. nitric acid [1] at room temperature. [1] ii) 2,4-dinitrophenol, 2,5-dinitrophenol, 2,6-dinitrophenol, 3,4-dinitrophenol, 3,5-dinitrophenol. All 5 correct [3]; 3 or 4 correct [2]; 1 or 2 correct [1]. (Subtract [1] for any incorrect isomers to a minimum of 0.)

7 a) i) CH3CH2COCl [1]; catalyst AlCl3 [1]

+ – ii) CH3CH2COCl + AlCl3 → CH3CH2C O + AlCl4 [1]

[1] [1]

b) LiAlH4 [1]; nucleophilic addition. [1] c) Conc. phosphoric or conc. sulfuric acid [1]; dehydration or elimination. [1] d) Step 2: Optical isomerism occurs as product has four groups attached to central C. [1] Racemic mixture of optical isomers formed. [1] H– can attack from either side of planar C=O. [1] Step 3: E/Z isomerism. [1] No rotation about C=C. [1] Two different groups on each end of the C=C. [1] 8 a) i)

[1]

ii)

[1] b)

[1]

c) A lone pair of electrons on each of the –OH groups in L-dopa interacts with the delocalised electrons in the benzene ring [1], releasing electrons into the ring and making it more susceptible to electrophilic attack. [1]

d) i) Primary amines are organic compounds containing the –NH2 group. ii)

9 a)

ΔH1 = 6 ΔatH(carbon) + 6 ΔatH(hydrogen) [1]

ΔH1 = 6(+715) + 6(+218) [1]

–1 ΔH1 = +5598 kJ mol [1] For Kekulé benzene:

ΔH2 involves formation of 3 C–C bonds + 3 C=C bonds + 6 C–H bonds. [1]

–1 ΔH2 = 3(–347) + 3(–612) + 6(–413) kJ mol [1]

–1 ΔH2 = –5355 kJ mol [1]

–1 ΔfH = ΔH1 + ΔH2 = +5598 – 5355 kJ mol = +243 kJ mol–1 [1] b) Experimental value is +82 kJ mol–1. Difference is 161 kJ mol–1. [1] So actual benzene is 161 kJ mol–1 lower in energy than the Kekulé structure. [1] Delocalisation of electron density gives extra stability to benzene. [1] 10 a) Indicative content • Ethanol is much weaker acid than phenol. • Dissociation of alcohols is according to ROH(aq) RO–(aq) + H+(aq).

• For phenol, the negative charge is delocalised into⇌ benzene ring. • Therefore the anion is stabilised and proton loss is easier. • For ethanol, the inductive effect pushes electrons towards the oxygen so its electron density is increased. • Therefore the anion is more likely to bond to H+ to reform the OH bond.

b) Ethanol forms CH3COOCH2CH3; ethyl ethanoate [1]

Phenol forms CH3COOC6H5; phenyl ethanoate [1]

Test yourself (page 542)

1 a) b) c) d) e) 2 a) Secondary b) Alcohol, phenol 3 a) Primary

butylamine 2-aminobutane 2-methylpropylamine 2-amino-2-methylpropane b) Secondary

Tertiary

4 a) Secondary alcohol b) Primary amine c) Tertiary halogenoalkane d) Tertiary amine e) Secondary amine f) Primary alcohol

Test yourself (page 543)

5 a) b)

c) d)

Test yourself (page 545)

6 Methylamine, like ammonia, is capable of hydrogen bonding with itself and with water molecules. The strength of these hydrogen bonds is roughly the same as those between water molecules, so methylamine will mix and dissolve in water in all proportions.

7 a) Ethane, CH3CH3, has non-polar molecules held together by nothing more than relatively weak intermolecular induced dipole attractions. So, its boiling temperature is very low at –89 °C.

Methylamine, CH3NH2, has polar molecules held together by permanent dipole attractions and much stronger hydrogen bonds in addition to weak induced dipole attractions. Therefore, its boiling temperature (–6 °C) is much higher than that of ethane. b) The molecules of methylamine and dimethylamine both have hydrogen atoms bonded to nitrogen atoms that carry a lone pair of electrons. So, they are held together by hydrogen bonds in addition to weaker induced dipole and permanent dipole attractions. On the other hand, molecules of trimethylamine have no hydrogen atom attached to their nitrogen atom, so they cannot hydrogen bond. Therefore, the forces between trimethylamine molecules are weaker than those between dimethylamine molecules and its boiling temperature is lower.

+ – 8 C6H11NH2(l) + H2O(l) C6H11NH3 (aq) + OH (aq)

+ – C6H5NH2(l) + H2O(l) ⇌ C6H5NH3 (aq) + OH (aq) The nitrogen of cyclohexylamine⇌ has the C6H11 cyclic alkyl group attached, so the electron density of the lone pair is increased. The lone pair on the nitrogen in phenylamine is delocalised into the π cloud and is less available to accept protons.

+ – 9 a) CH3CH2CH2NH2(g) + HBr(g) → CH3CH2CH2NH3 Br (s) propylammonium bromide b) Propylammonium bromide is a salt containing positive propylammonium ions and negative bromide ions. c) The reactants are simple molecular substances held together by much weaker intermolecular forces than the strong electrostatic forces between the ions in propylammonium bromide.

+ 2– 10 a) 2CH3NH2(g) + H2SO4(l) → (CH3NH3 )2SO4 methylammonium sulfate

+ 2– b) 2(CH3)2NH(g) + H2SO4(l) → ((CH3)2NH2 )2SO4 dimethylammonium sulfate

Test yourself (page 549)

2+ 11 [Co(H2NCH2CH2NH2)3]

+ – 12 a) CH3Cl + CH3CH2CH2NH2 → [CH3–N H2–CH2CH2CH3] Cl N-methylpropylammonium chloride

b) CH3CH2COCl + CH3CH2NH2 → CH3CH2CONHCH2CH3 + HCl N-ethyl propenamide hydrogen chloride

13 a) CH3Cl + C2H5NH2 → CH3NHC2H5 + HCl

b) CH3NHC2H5 + CH3Cl → (CH3)2NC2H5 + HCl

– (CH3)2NC2H5 + CH3Cl → (CH3)3NC2H5 Cl

Test yourself (page 551)

14 a) Nucleophilic substitution

The lone pair of electrons on the nitrogen atom of ammonia is attracted to the δ+ carbon atom of the C–Cl bond in 1-chloropropane. As a bond forms between the nitrogen and carbon atoms, the chlorine atom takes both electrons as the C–Cl bond breaks. b) A solution of ammonia in water is alkaline containing OH– ions. The OH– ions would compete

with NH3 molecules for reaction with the halogenoalkanes. Reaction with hydroxide ions produces alcohols. 15 Ethylamine can be prepared from: a) bromoethane by reaction with an excess of ammonia in ethanol. A nucleophilic substitution occurs. Disadvantage: further substitution may occur so the product may be contaminated with some secondary or tertiary amines. b) bromomethane by reaction with potassium cyanide in ethanol. A nucleophilic substitution occurs and ethanenitrile is formed.

The nitrile can then be reduced to ethylamine by reacting with LiAlH4 in ether (or by

hydrogenation with Ni/H2). Disadvantages include toxic reagent or the likelihood that two steps will give a lower yield than one. 16 Tin and concentrated HCl

Test yourself (page 553)

17 a) 2-Aminopropanoic acid b) 2-Amino-3-phenylpropanoic acid c) 2-Amino-3-hydroxypropanoic acid 18 20 × 20 = 400 19 There are two hydrogen atoms attached to the central carbon atom in glycine. So, the molecule has a plane of symmetry and cannot have mirror images that are different. 20 Make up solutions of the two forms with the same concentration. These two solutions will rotate the plane of plane-polarised light equally, but in opposite directions.

Test yourself (page 553)

21 The strongest forces between butylamine molecules and between propanoic acid molecules are hydrogen bonds. These bonds are much weaker than the ionic bonds between zwitterions in glycine. This explains why the melting temperatures of butylamine and propanoic acid are so much lower than that of glycine. 22 a) i)

ii)

b) They carry only one charge, in contrast to the two charges on a zwitterion. 23 Because the transfer of H+ ions occurs before the solid crystallises and the oppositely charged sections of the zwitterions are stabilised by attraction in the solid.

Test yourself (page 556)

Hydrolysis involves splitting (‘lysing’) with water. When the peptide bond breaks, –OH is added to the C=O group and –H is added to the N–H group. The –OH and –H come from water.

26 a) Carboxylic acid, –COOH; amine, –NH2; peptide, ; phenyl, C6H5–; ester, b) It contains an ester link, otherwise it would be the dipeptide, asp-phe c) Hot water will hydrolyse the molecule, splitting the peptide bond and the ester bond. d) Hydrolysis by acids in the digestive system will release the amino acid phenylalanine.

Test yourself (page 559)

27 a) Condensation polymerisation b) and 28 a) Carboxylic acid and amine groups b) Condensation polymerisation c) Water d)

29 Similarities: • In both, the monomers are linked by the amide group, .

• In both, the chains are held together by hydrogen bonds. Differences: • In nylon-6,6, there are six carbon atoms between each amide group but in proteins, there is only one carbon atom between each amide group. • In nylon-6,6, there are no side groups attached to the carbon atoms but in proteins, there are side groups attached to the carbon atoms between each amide group. • In nylon the monomers are diacids and diamines, but in proteins there is one acid and one amine on each monomer.

Activity: Paracetamol – an alternative to aspirin (page 548)

2 a) HO–C6H4–NH2 + CH3COCl → HO–C6H4–NHCOCH3 + HCl b) Because the δ+ carbon atom in the –COCl group is attacked by the δ– nitrogen atom of the

–NH2 group in 4-aminophenol with its lone pair of electrons. 3 a) Because the δ– oxygen atom in the –OH group of 4-aminophenol, with two lone pairs of electrons, is also capable of acting as a nucleophile.

b) HO–C6H4–NH2 + CH3COCl → CH3COO–C6H4–NH2 + HCl

4 HO–C6H4–NH2 + (CH3CO)2O → HO–C6H4–NHCOCH3 + CH3COOH

5 a) The replacement of an H atom in an –OH and/or –NH2 group with the ethanoyl group, b)

6 a) Aspirin is a strong pain-relieving drug. b) Poisoning due to deliberate or accidental overdose; a cause of internal bleeding and gastric ulcers. c) Paracetamol is a good pain-relieving drug without the harmful side-effects of aspirin. However, recent studies have found that, if used over long periods, paracetamol can lead to an increased risk of high blood pressure and also gastrointestinal problems.

Activity: Modelling and synthesising polyamides (page 560)

1 a) The monomers used to make Kevlar are: benzene-1,4-dicarboxylic acid and benzene-1,4-diamine . Condensation reactions can occur over and over again between the –COOH groups of

benzene-1,4-dicarboxylic acid and the –NH2 groups in benzene-1,4-diamine to produce a polymer.

b) If the Kevlar molecules are arranged so that their chains are parallel and close together, every N–H group can be hydrogen bonded to a C=O group. With all these hydrogen bonds, the inter-chain bonding is extremely strong. c) The 1,4 positions are opposite each other on the benzene ring and so the chains are more or less straight and the hydrogen bonding is at a maximum. A polymer based on functional groups in the 1,2 or 1,3 positions would have a very irregular shape. This would reduce the possibility of hydrogen bonding. 2 Nylon-6,6 has two sections of six carbon atoms joined by single C–C bonds in every repeat unit. There is free rotation about all these single bonds allowing the molecule to rotate, twist and flex at every carbon atom. Kevlar, on the other hand, has two benzene rings each containing six carbon atoms in every repeat unit. The structure of the benzene ring does not allow any of the carbon atoms to move out of its flat planar arrangement. This makes Kevlar very inflexible compared to nylon-6,6. 3 a)

b) Because the sodium carbonate reacts with the HCl eliminated between each pair of monomers as condensation reactions occur. This pulls the reaction towards completion. c) Cross-linking can occur between the –OH group on one of the monomer units and the –COOH group on the other monomer unit forming ester links. These esterifications will occur if the polymer chains are warmed with a little concentrated sulfuric acid as catalyst. d) The extent of polymerisation can be controlled either by adjusting the concentration of the monomers in an inert organic solvent or through the pH. In acid conditions, the amine groups

+ + will accept H ions, forming –NH3 groups which are unreactive towards the carbon atom of the C=O group. Cross-linking between the chains can be controlled by adjusting the temperature of the

mixture after adding the concentrated H2SO4 catalyst.

Exam practice questions (pages 564-7)

1 a) D b) C c) E d) B e) A f) D [1] for each 2 a) Proton acceptor [1] lone pair which can be donated. [1]

+ 2– b) (CH3CH2CH2CH2NH3 )2 SO4 [1] c) Bases are lone pair donors. [1] Inductive effect of alkyl group increases electron density on N in butylamine so better lone pair donor. [1] Lone pair is delocalised into aromatic ring in phenylamine so poorer lone pair donor. [1]

2+ d) Deep blue solution [1]; acts as a ligand [1]; [Cu(C4H9NH2)4(H2O)2] (aq). [1] 3 a) i) Insertion of to give:

[2] (wrong way around gets [1]) ii) All four chiral centres circled [3]; three chiral centres circled [2]; two chiral centres circled [1].

b) The compound has at least one carbon atom with four different atoms or groups attached to it. [1] Molecules of the compound are asymmetric. [1] The compound has optical isomers. [1] c)

[1] [1]

+ 4 a) The −NH2 group in amino acids can act as a base accepting H ions and the −COOH group can act as an acid, donating H+ ions. [1] Therefore, in solution, amino acids, like glycine, can exist

+ – as zwitterions such as H3 N−CH2−COO (aq). [1] At one particular pH, the glycine particles will form zwitterions with an overall net charge of zero and this pH is the isoelectric point of the amino acid. [1] b) The isoelectric point of the amino acid will be dictated by the relative abilities of the −COOH

+ group to donate H ions and the −NH2 group to accept them. [1] These relative abilities are not equal, so the isoelectric point is not at pH = 7. [1] As the isoelectric point of glycine is at pH = 6; this suggests that the −COOH group loses

+ – H ions to form −COO ions more readily than the −NH2 group accepts them to form

+ −NH3 ions. [1]

However, at pH = 6 (i.e. [H+] > [OH–]), the dissociation of −COOH is suppressed and the

+ acquisition of H by −NH2 is raised so that the two processes are equally likely. [1] At this pH, the zwitterions are most stable and the average net charge on glycine particles is zero. [1]

– c) Glutamic acid has two −COOH groups and only one −NH2 group. This will mean that −COO

+ – anions form more readily than −NH3 cations. [1] In order to suppress the formation of −COO

+ groups and promote the formation of −NH3 groups relative to the glycine situation, the pH must be more acidic. [1] Therefore, the isoelectric point of glutamic acid will be less than pH = 6. [1]

5 a) i) CH3CH2Br + NH3 → CH3CH2NH2 + HBr [1]

Note that the HBr formed will react with NH3 to form the salt NH4Br, so strictly the equation should be:

CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br ii) Nucleophilic substitution [1]

iii) (CH3CH2)2NH and (CH3CH2)3N [2] iv) The ethylamine which forms is a better nucleophile than ammonia [1] and may react with any unreacted bromoethane forming diethylamine. [1]

CH3CH2Br + CH3CH2NH2 → (CH3CH2)2NH + HBr diethylamine The diethylamine can also react as a nucleophile [1] with any excess bromoethane to form triethylamine.

CH3CH2Br + (CH3CH2)2NH → (CH3CH2)3N + HBr triethylamine Note that the HBr formed in each of these reactions will react with the amines in the reaction mixture to form substituted ammonium salts. b) Ethylamine vapour and hydrogen chloride gas will diffuse away from the cotton wool plugs

along the tube [1]. The relative molecular mass of HCl (Mr = 36.5) is marginally less than that

of ethylamine (Mr = 45.0) so it will diffuse slightly faster along the tube. [1] When the vapours meet (approximately midway along the tube, but slightly closer to the ethylamine end [1]), a white smoke will form, eventually settling on the bottom of the tube as a white powder [1]. This is ethylammonium chloride.

+ – CH3CH2NH2(g) + HCl(g) → CH3CH2NH3 Cl (s) [1] 6 a) i) Fishy smell [1] ii) 1,5-Diaminopentane [1]

– + + – iii) Cl H3N−CH2CH2CH2CH2CH2NH3 Cl [1]

+ – iv) NH4 Cl [1]; ammonium chloride [1]

v) NaCl [1]; H2O [1]

b) i) Primary amines have one alkyl or aryl group attached to an NH2 group. [1] Secondary amines have two alkyl or aryl groups attached to an N–H group. [1] Tertiary amines have three alkyl or aryl groups attached to a nitrogen atom. [1] ii) Cadaverine is a primary amine [1]; piperidine is a secondary amine. [1] c) Their infrared spectra will be very similar [1]; because they have exactly the same types of bonds. [1] 7 a)

[1]

b) CH3CH=CHCH3 [1]; but-2-ene [1] c) i)

[1] [1] ii) 1,4-Dihydroxybenzene (4-hydroxyphenol) or pentanedioic acid [1] d) Ester links in the condensation polymer can be hydrolysed by water [1]; and this process will be catalysed if the conditions are acidic or alkaline. [1]

[1]

The C−C and C−H bonds in polymer A are strong and non-polar, so are not degraded by sunlight, water, acids, alkalis or microorganisms. [1] 8 a) i) Mass C : H : N = 61.0 : 15.3 : 23.7 [1] 61.0 15.3 23.7 Moles C : H : N = : : 12.0 1.0 14.0 = 5.08 : 15.30 : 1.69 [1] = 3 : 9 : 1 [1]

Empirical formula = C3H9N ii) Its relative molecular mass (or molar mass). [1]

iii)

[1] for each. b) The alkyl groups provide an inductive effect towards the nitrogen atom. This increases the electron density on the nitrogen atom and also its basic character. [1] Therefore, the predicted order of basic strength from the strongest to the weakest should be: trimethylamine (3 alkyl groups ) > ethylmethylamine and 2-aminopropane (2 alkyl groups) > propylamine (1 alkyl group). [1] c) In water, the amines are protonated by the water:

+ – R2NH + H2O R2NH2 + OH

+ – R3N + H2O ⇌ R3NH + OH (Both equations needed for [1].) Because of⇌ the extra N–H bonds in the protonated secondary amines there is more hydrogen + bonding possible. [1] This stabilises these ions more than the equivalent R3NH . [1] 9 a) i)

ethenol [1] repeat unit [1]

ii)

[1]

iii) Relative mass of one poly(ethenol) unit = 44.0 [1] Relative mass of one poly(ethenyl ethanoate) unit = 86.0 [1] Relative molecular mass of polymer = relative mass of 1800 poly(ethenol) units + relative mass of 200 poly(ethenyl ethanoate) units = (1800 × 44.0) + (200 × 86.0) [1] = 79 200 + 17 200 = 96 400 [1]

iv) Almost every monomer unit contains an −OH group which can hydrogen bond with water. [1] The extensive hydration of the −OH groups by water molecules results in a soluble polymer. [1] b) i) Because there are fewer −OH groups on the polymer which can form hydrogen bonds with water molecules. [1] (Water molecules can, of course, still form hydrogen bonds with the O atoms in the C=O bonds of the ethanoate groups, but the larger size of the ethanoate group relative to the −OH group will reduce its solubility.) ii) Because the −OH groups on the poly(ethenol) molecules will form hydrogen bonds with each other rather than with water molecules. [1] iii) The poly(ethenol) forms a crystalline solid in which extensive hydrogen bonding between −OH groups along the polymer chain holds the molecule firmly together. [1] Few, if any, −OH bonds are available for hydrogen bonding with water and the poly(ethenol) is insoluble. [1] (Allow credit for any sensible suggestions.) 10 a) i) Induced dipole attractions [1] ii) Low density polythene has short side chains attached to the main C−C chain. [1] This means that its molecules cannot pack as closely as those in high density polythene. [1] iii) High density polythene has the higher melting temperature. [1] Molecules in high density polythene can pack closer than those in low density polythene. [1] This results in more and stronger induced dipole attractions. More energy is therefore needed to overcome the stronger forces in high density polythene, [1] so its melting temperature is higher. b) i) A polyamide is a polymer with amide links between monomer units. [1] ii)

[1]

iii) Hydrogen bonds [1] iv) Hydrogen bonds between the H atom of an N−H group on one chain and the N atom on another chain. [1] Hydrogen bonds between the H atom of an N−H group and the O atom of a C=O group on another chain. [1] c) i) In the crystalline regions, the chains of different molecules are packed in a regular, parallel fashion, but in the non-crystalline regions, the molecules are more randomly placed. [1]

ii) In the crystalline regions, more hydrogen bonds can form between the parallel chains making the total intermolecular forces stronger. [1] 11 a) To prevent loss of nitrobenzene if the flask gets hot. [1] b) The reaction is too slow if the temperature is too low. [1] c) To allow unreacted nitrobenzene to evaporate. [1] d) To neutralise the HCl so oily phenylamine forms from the water-soluble salt. [1] To convert the tin compounds into soluble hydroxy complex ions. [1] e) To saturate the aqueous layer and reduce the solubility of phenylamine in the water. [1] f) Although the densities are similar, the salt increases the density of water so the phenylamine floats [1]; it is the upper layer. [1] g) To absorb water. [1] (Other drying agents react with the phenylamine.) h) 4.2 cm3 nitrobenzene have mass 4.2 × 1.20 = 5.04 g [1] 5.04 Amount nitrobenzene = mol = amount phenylamine [1] 123.0 5.04 Max. mass phenylamine expected = × 93.0 = 3.81 g [1] 123.0

+ − + i) C6H5NO2(aq) + 7H (aq) + 6e → C6H5NH3 (aq) + 2H2O(l) [1]

– 2– − Sn(s) + 6Cl (aq) → [SnCl6] (aq) + 4e [1]

+ – + 2– 2C6H5NO2(aq) + 14H (aq) + 3Sn(s) + 18Cl (aq) → 2C6H5NH3 (aq) + 3[SnCl6] (aq) + 4H2O(l) [1] 12 Indicative content: • C=O bond in ethanamide not in ethylamine • Lone pair on N in propylamine accepts protons so is basic. • C=O withdraws electrons from lone pair on N, therefore ethanamide is not basic. • Hydrogen bonding in propylamine, therefore liquid at room temperature • Delocalisation of lone pair from N onto C=O. • Leads to a separation of charge and therefore stronger electrostatic forces in ethanamide.

b) Fragment corresponding to peak at m/z 17 is –OH Possible fragments corresponding to peak at m/z 29 are:

CH CH — or 3 2 Possible fragments corresponding to peak at m/z 31 are: CH O– or –CH OH 3 2 c) Therefore, possible structures for the compound are:

CH CH CH OH or 3 2 2

(not CH O—CH CH or because they would not 3 2 3 produce an –OH fragment.)

Test yourself (pages 575-576)

6 a) Primary alcohol, secondary alcohol and aldehyde. b) Carbonyl group in a ketone and carboxylic acid. 7 Carbonyl (ketone), alkene and carboxylic acid. 8 Dextropropoxyphene contains two phenyl groups, a tertiary amine group and an ester group. • It will rotate the plane of plane-polarised light. • As a result of the phenyl groups it will burn with a very smoky flame/ probably be insoluble in water. • As a result of the amine group it will: – react with acids to form substituted ammonium salts – react as a nucleophile due to the lone pair of electrons on the nitrogen atoms. • As a result of the ester group it will: – react with hot aqueous acids to form carboxylic acid and alcohol groups – react with hot aqueous alkalis to form carboxylate and alcohol groups. 9 a) D b) C c) B d) A e) A f) A

Test yourself (pages 577-578)

10 a) CH CH Br, addition, HBr is an electrophile. 3 2

− b) CH CH CH OH, substitution, KOH (OH ) is a nucleophile. 3 2 2

+ c) C H NO , substitution, NO ions from HNO /H SO act as electrophiles. 6 5 2 2 3 2 4 11 a) A is lithium tetrahydridoaluminate (LiAlH ), B is ethanol, 4 C is ethylamine or ethylammonium chloride. b) D is CH COCl or (CH CO) O; E is CH CH MgI. 3 3 2 3 2

12 a) But-1-ene 2-bromobutane but-2-ene

b) Propanone propan-2-ol

2-chloropropane propene

c) Bromoethane ethene

1,2-dibromoethane ethane-1,2-diol

Test yourself (page 580)

13 a) CH CH CH CN 3 2 2

− + b) CH CH COO Na + CHI 3 2 3 14 a) Magnesium metal is added to a solution of 1-bromopropane dissolved in dry ethoxyethane. b) i) CH CH CH CH OH butan-1-ol 3 2 2 2 ii) CH CH CH COOH butanoic acid 3 2 2

iii) 1-propyl cyclohexan-1-ol

iv) heptan-4-ol

15 Step1: KCN in ethanol. Step 2: Hydrogen/nickel catalyst or LiAlH in ethoxyethane. 4 16

Test yourself (page 582)

17 a) Method (ii) b) Method (iii) c) Method (i) 18 The stopper is in the top of the funnel. Air cannot get in to replace any liquid lost, so none drains out

Test yourself (page 583)

19 When the pressure is reduced, there are fewer air molecules around and the water molecules can diffuse more rapidly to the drying agent. There is also no build-up of water vapour above the damp solid and so fewer water molecules return to the solid from the gas phase.

2+ 20 [Co(H O) ] 2 6

Test yourself (pages 587-588)

21 a) The bulb will record the temperature at which the vapour condenses, which is identical to the temperature at which the liquid turns to vapour, i.e. its boiling temperature. b) The higher atmospheric pressure, the higher the boiling temperature. c) Impurities in the liquid in the flask will raise the temperature above the boiling temperature. 22 a) Ethene 1,2-dibromoethane

1,2-diaminoethane

b) Bromine should be used in excess in step 1 and ammonia in excess in step 2 CH =CH → CH Br–CH Br → H NCH –CH NH 2 2 2 2 2 2 2 2 Assuming 100% yields in both steps: 1 mol of CH =CH produces 1 mol of H NCH –CH NH 2 2 2 2 2 2 But, the yield in step 1 is 60% followed by 40% in step 2. 40 ∴ the overall % yield = 60 × 100 = 24% So, in practice: 0.24 mol of H NCH –CH NH is obtained from 1 mol of CH =CH 2 2 2 2 2 2 ∴ (0.24 × 60.0) g of H NCH –CH N is obtained from 28.0 g of CH =CH 2 2 2 2 2 2 14.4 g of H NCH –CH NH is obtained from 28.0 g of CH =CH 2 2 2 2 2 2 28.0 ∴ 2.0 g of 1,2-diaminoethane is obtained from 14.4 × 2.0 g of ethene = 3.9 g of ethene 23 Any of the following reasons: • The rate of the reaction is so slow that some of the reactant is not converted to the product. • The reaction comes to an equilibrium containing a proportion of unchanged reactants. • Side reactions may occur which use up some of the reactants. • Losses of reactants can occur during refluxing and transferring liquids when reactions in the synthesis are carried out. • Losses of the product occur during processes such as filtration, washing, drying and recrystallisation when the crude product is being purified. • There are further losses in transferring the product from one vessel/container to another during the purification process. 24 a) Benzene nitrobenzene phenylamine

b) C H C H NO C H NH 6 6 6 5 2 6 5 2 Assuming 100% yield at each step: 1 mol of C H → 1 mol of C H NO → 1 mol of C H NH 6 6 6 5 2 6 5 2 So 78.0 g of C H → 123.0 g of C H NO → 93.0 g of C H NH 6 6 6 5 2 6 5 2

In step 1: 123.0 Theoretical yield = 78.0 × 18 g = 28.4 g nitrobenzene 22 So % yield in step 1 = 28.4 × 100% = 77% (2 s.f.) In step 2: 93.0 Theoretical yield = 123.0 × 22 g = 16.6 g phenylamine 12 So % yield in step 2 = 16.6 × 100% = 72% (2 s.f.) 72 c) The overall % yield = 77 × 100 = 55% 25 a) It is important to use the minimum volume of solvent because some of the product will remain dissolved in the solvent (and therefore lost) during recrystallisation. By using hot solvent, the amount of product recovered is maximised. b) To lower the temperature of the solution and increase the amount of product that crystallises out. c) Leave the solid in a warm place (e.g. on a radiator) or put it in a warm oven or evaporate under reduced pressure.

Activity: Converting one functional group to another (page 580) Reagents and conditions for Figure 18.3.16:

1 H /Ni and 150 °C 2 2 Cl and UV 2 3 HHal (HCl or HBr) 4 Heat with NaOH in ethanol 5 Conc. NH in ethanol 2 6 KCN in ethanol 7 H /Ni and 150 °C, or LiAlH in ethoxyethane, then hydrolysis with water 2 4 8 NaOH(aq) 9 Conc. HHal or PCl 5 10 Reflux secondary alcohol with acidified K Cr O 2 2 7 11 LiAlH in ethoxyethane, then hydrolysis with water 4

12 Primary alcohol with acidified K Cr O and distil off product 2 2 7 13 LiAlH in ethoxyethane, then hydrolysis with water 4 14 KCN with HCN 15 KCN with HCN 16 Reflux with acidified K Cr O 2 2 7 17 LiAlH in ethoxyethane, then hydrolysis with water 4 18 Heat with HCl(aq) 19 Mix equimolar amounts of acid and alcohol, then heat with a little conc. H SO 2 4 20 Reflux with aqueous HCl or H SO (or reflux with NaOH to give salt of acid) 2 4 21 PCl at room temperature 5 22 H O at room temperature 2 23 Alcohol at room temperature 24 Conc. ammonia 25 Magnesium in ethoxyethane 26 Forms alcohols: with methanal to primary; with aldehydes to secondary; with ketones to tertiary. 27 With carbon dioxide, then hydrolysis with water.

1 a) Ethene chloroethane ethylamine

b) Divide the ethanol into two equal portions. Ethanol ethanoic acid ethyl ethanoate

(one portion) c) Propanoic acid propanoyl chloride propenamide

2 a) Ethene chloroethane ethanol

ethanoic acid

b) Propan-2-ol 2-chloropropane propene

propane

Activity: Preparation and purification of an organic liquid (page 585)

1 To prevent escape of methanol. 2 To dilute the reactants so the equilibrium reactions are effectively stopped. 3 To rinse the flask so no organic compounds are left behind 4 With the stopper held firmly in place by the palm of the hand, the funnel is inverted and the tap carefully opened. 5 The water is added to the funnel, which is then shaken thoroughly. After shaking (then releasing any pressure) the contents of the funnel are allowed to settle then the lower aqueous layer is run off. The water removes any remaining methanol and sulfuric acid. 6 The sodium carbonate removes any remaining benzoic acid. 7 The layer goes clear when all the water has been removed. 8 M of C H COOH = 122.0; 8 g = 8 g = 0.0656 mol r 6 5 122.0 g mol−1 Density of methanol = 0.79 g cm−3; M = 32.0 r Mass of methanol = 15 × 0.79 = 11.85 g 11.85 g −1 ∴ amount of methanol = 32.0 g mol = 0.370 mol, which is excess. 9 M of methyl benzoate = 136.0 r Theoretical yield of methyl benzoate = 0.0656 × 136.0 = 8.92 g % yield = 4.8 × 100% = 53.8% = 54% (2 s.f.) 8.92

10 Concentrated sulfuric acid is corrosive – wear gloves and eye protection Methanol is flammable – keep away from naked flames Methyl benzoate and methanol are harmful by inhalation – work in fume cupboard. Benzoic acid is irritating to eyes – wear eye protection.

Activity: Preparation and purification of N-phenylamide (page 588)

1 C H NH (l) + (CH CO) O(l) → CH CONHC H (s) + CH COOH(l) 6 5 2 3 2 3 6 5 3 2 Ethanoyl chloride produces corrosive hydrogen chloride rather than ethanoic acid. 3 The reaction is exothermic so is controlled better by addition of small amounts to a cooled mixture. 4 To provide a rough surface on which small bubbles of vapour can form. 5 i) The flask should be rinsed with some of the cold water. ii) The beaker of water should be cooled in ice to reduce the solubility of the product.

iii) The beaker should be rinsed with a little cold water to move all the crystals into the Buchner funnel. 6 The crystals should be compressed in the Buchner funnel with a cork. The suction should be turned off, then a little cold water poured onto the crystals. When the water has soaked through all of the solid the suction can then be turned on again. 7 Silica gel or anhydrous calcium chloride. 8 5 cm3 phenylamine has mass 5.0 cm3 × 1.02 g cm−3 = 5.1 g 5.1 g −1 Amount of phenylamine = 93.0 g mol = 0.0548 mol Amount of N-phenylethanamide = 0.0548 mol Theoretical yield of N-phenylethanamide = 0.0548 × 135.0 g mol−1 = 7.40 g Yield is 70.0%, so mass formed = 0.70 × 7.40 = 5.18 g 9 Phenylamine is toxic by skin absorption and inhalation. Ethanoic anhydride (and ethanoic acid formed) are corrosive and harmful by inhalation. Wear gloves and work in a fume cupboard or in a well-ventilated room. Both reagents are flammable, so keep away from naked flames.

Core practical 15 (part 2) (page 575)

1 Test Observation for positive result 2,4-Dinitrophenylhydrazine Orange precipitate

Iodine with sodium hydroxide Pale yellow precipitate

Acidified potassium dichromate(vi) Orange solution turns to a green solution

2 A and B contain a C=O group in an aldehyde or ketone, but C does not. 3 B and C contain CH3C=O or CH CH(OH) groups. 3 4 A and C can be oxidised so are aldehydes or primary/secondary alcohols. 5 A could be either CH CH CH CHO or (CH ) CHCHO. 3 2 2 3 2 B could be CH COCH CH . 3 2 3 C could be CH CH(OH)CH CH . 3 2 3 6 D will not react with iodine with sodium hydroxide. Oxidation of D will form CH CH COCHO, then CH CH COCOOH. 3 2 3 2 7 E is an acid: CH CH CH COOH or (CH ) CHCOOH. 3 2 2 3 2 8 F is an ester: CH CH COOCH or CH COOCH CH or HCOOCH CH CH or HCOOCH(CH ) . 3 2 3 3 2 3 2 2 3 3 2

Exam practice questions (pages 590–2)

1 a) i) A is immiscible with water. This suggests that part or all of the molecule is non-polar. [1] ii) The cream precipitate is silver bromide [1]; so, A contains bromine. [1] A might be a bromoalkane which has reacted with OH– ions in NaOH(aq) to form an alcohol and Br– ions. [1] (Maximum 2 marks) b) i) Charring is the formation of carbon [1]; the vapour produced on heating is water [1]; this suggests that B may be hydrated/contain water of crystallisation. [1] (Maximum 2 marks) ii) B is acidic. [1]

iii) B is sufficiently acidic to produce CO2 [1] from Na2CO3(aq). So the acid group present is not a phenol. [1] iv) The sweet-smelling product is probably an ester [1], which has been formed by reaction

of a carboxylic acid [1] with ethanol plus concentrated H2SO4 as catalyst. c) i) The very smoky flame suggests C has an aryl group/high C : H ratio. [1] ii) C is not acidic/a carboxylic acid. [1] iii) C contains an −OH group [1]; which reacts with sodium to produce hydrogen. [1] 2 a) The diagram for refluxing should show: ● liquids heated in a round-bottomed or pear-shaped flask [1] ● Liebig (water) condenser arranged vertically [1] ● water in at the bottom, out at the top. [1] Subtract [1] if condenser is closed at the top. b) A Refluxing for 45 minutes as the reaction is slow [1]; the reflux condenser prevents

escape of volatile reagents such as ethanol. [1] Conc. H2SO4 is the catalyst. [1] B Collecting all the distillate below 84 °C ensures that all the ethyl ethanoate is collected. [1] C Sodium carbonate neutralises any acid impurity in the distillate. [1] D Anhydrous sodium sulfate or anhydrous calcium chloride are added to remove water [1] from the organic product. E Collecting liquid that boils between 75 and 79 °C ensures that most of the ethyl ethanoate is collected [1], but other impurities with boiling temperatures above or below this range are not collected. [1] (Total 8 marks) 3 a) Phenol [1]; carboxylic acid [1]

b) C7H6O3 [1]

c) i) ii)

(Allow one or two Br atoms at any positions on the ring.) iii)

4 a) To spread the heat from the Bunsen over the bottom of the flask. [1] b) So that a larger surface area of the rind is exposed to hot water and steam. [1] c) Steam distillation [1] Use a dropping pipette to remove the upper oily layer or use a separating funnel [1] d) Leave the limonene in a specimen tube with a suitable drying agent such as anhydrous calcium chloride, anhydrous sodium sulfate, anhydrous magnesium sulfate or silica gel. [1] e) Because the temperature required to vaporise the limonene would char the peel and decompose the limonene. [1] f) The alkene group. [1] Show that the limonene will decolourise yellow/orange bromine water

[1] and will also decolourise dilute acidified purple potassium manganate(VII). [1]

5 a) C13H20O3 [1] b) Alkene [1]; carbonyl [1]; ester [1] c) i) and ii)

[2] [2]

iii) The stereoisomers with the skeletal formula shown will all be Z isomers. [1] There will be (+) and (−) isomers at both the chiral centres [1]; therefore four stereoisomers [1]; with the skeletal formula shown.

6 a) F is CH3CH2CH2OH [1]; propan-1-ol. [1]

G is CH3CHOHCH3 [1]; propan-2-ol. [1]

H is CH3CH2CHO [1]; propanal [1]

J is CH3COCH3 [1]; propanone [1] b) i) The orange [1]; mixture turns green. [1] The products are propanoic acid [1]

and chromium(III) ions Cr3+/chromium(III) sulfate. [1] ii) Reduction half-equation:

2– + – 3+ Cr2O7 (aq) + 14H (aq) + 6e → 2Cr (aq) + 7H2O(l) [1] Oxidation half-equation:

+ – CH3CH2CHO(aq) + H2O(l) → CH3CH2COOH(aq) + 2H (aq) + 2e [1] Overall ionic equation:

2– + 3+ Cr2O7 (aq) + 8H (aq) + 3CH3CH2CHO(aq) → 2Cr (aq) + 3CH3CH2COOH(aq) + 4H2O(l) [1]

(Accept CH3CH2CHO + [O] → CH3CH2COOH for [1]) 12.0 7 a) Mass of C in 0.36 g of the compound = 0.88 × g = 0.24 g [1] 44.0 2.0 Mass of H in 0.36 g of the compound = 0.36 × g = 0.04 g [1] 18.0 Mass of O in 0.36 g of the compound = (0.36 − 0.28) g = 0.08 g 0.24 0.04 0.08 Ratio of moles of C : H : O = : : [1] 12.0 1.0 16.0 = 4 : 8 : 1

Empirical formula of X = C4H8O [1] Relative molecular mass of X ≈ 70 Relative molecular mass of empirical formula = 72

∴ Molecular formula of X = C4H8O [1] b) X contains a carbonyl group. [1]

c) CH3CH2CH2CHO butanal [1]

2 methylpropanal [1]

CH3CH2COCH3 butanone [1] d) Butanal and butanone have four different types of carbon so either could be X. [1] Take the melting temperature of X [1] and compare with values on the data sheet of melting temperatures of 2,4-dinitrophenylhydrazine derivatives of carbonyl compounds. [1]

8 a) Warm (reflux) with aqueous NaOH. [1] Acidify with HNO3(aq) and then add AgNO3(aq). [1] Cl → white precipitate [1], I → yellow precipitate. [1] b) Iodine and aqueous sodium hydroxide. [1] Pentan-2-one will give a pale yellow precipitate of

CHI3. [1] There will be no precipitate with pentan-3-one. [1] c) Add 2,4-dinitrophenylhydrazine at room temperature. [1] Ketone → phenylhydrazone, yellow solid [1] Ester → no reaction [1]

d) Acidified potassium dichromate(VI). [1] Phenol – no reaction. [1] Primary alcohol – solution turns from orange to green. [1] e) Dip a glass rod into concentrated hydrochloric acid and hold over open bottles of the compounds. [1] Amide – no effect. [1]

+ – Amine – clouds of white smoke formed as CH3CH2CH2CH2NH3 Cl (s) forms. [1] 12.0 9 a) Mass of carbon = 7.92 × = 2.16 g [1] 44.0 2.0 Mass of oxygen = 1.44 × = 0.16 g [1] 18.0 Mass of oxygen in Y = (2.64 − 2.16 − 0.16) g = 0.32 g [1] ∴ In Y, ratio of masses C : H : O = 2.16 : 0.16 : 0.32 2.16 0.16 0.32 Ratio of moles C : H : O = : : [1] 12.0 1.0 16.0 = 0.18 : 0.16 : 0.02 = 9 : 8 : 1

Empirical formula of Y = C9H8O [1] Relative molecular mass of Y = 132

Relative molecular mass of C9H8O = 132

∴ Molecular formula of Y = C9H8O [1] b) Observations Inferences • Burns with a smoky flame • Y is probably aromatic and contains a benzene ring [1] • The yellow/orange colour of • Y is probably an alkene [1] bromine water is decolourised • A yellow/orange precipitate is • Y probably contains a carbonyl group [1] produced • A silver mirror or dirty grey solid • Y is probably an aldehyde [1] appears

c) Possible structures for Y would be

or 1,2 or 1,3 isomers [1] 10 a) Indicative content: • Weigh a sample of D. • Heat (reflux) with NaOH(aq).

• Cool and add HNO3 to neutralise excess NaOH.

• Add an excess of AgNO3(aq). • Filter the cream precipitate of AgBr. • Wash, dry and weigh. b) Reaction with NaOH(aq) converts the halogenoalkane to an alcohol + sodium bromide. [1]

[1]

− HNO3 neutralises excess NaOH and the AgNO3 precipitates Br (aq) ions in the aqueous solution as cream-coloured AgBr(s). [1] A [Br] Mass of Br in D = mass of AgBr × r [1] Ar[AgBr] mass of Br in D % of Br in D = × 100% [1] mass of D

11 a) Step 1 CH3CH2CH2Br with KCN in ethanol [1] forms CH3CH2CH2CN. [1]

Step 2 CH3CH2CH2CN with Ni/H2 [1] forms CH3CH2CH2CH2NH2. [1]

Step 3 CH3CH2CH2CH2NH2 with CH3COCl [1] forms CH3CONHCH2CH2CH2CH3. [1]

b) Step 1 C2H4 with HBr [1] forms CH3CH2Br. [1]

Step 2 CH3CH2Br with NaOH(aq) [1] forms CH3CH2OH. [1]

Step 3 CH3CH2OH with acidified K2Cr2O7 [1] and distil off the product CH3CHO immediately it forms. [1]

Step 4 CH3CH2Br with Mg in ethoxyethane [1] forms CH3CH2MgBr. [1]

Step 5 CH3CH2MgBr with CH3CHO [1] then hydrolysis [1] forms CH3CH2CH(OH)CH3. [1]

Step 6 CH3CH2CH(OH)CH3 with conc. H2SO4 [1] forms CH3CH=CHCH3. [1] 12 a) Ratio of masses C : H : Al = 50 : 12.5 : 37.5 50 12.5 37.5 Ratio of moles C : H : Al = : : [1] 12.0 1.0 27.0 = 4.17 : 12.5 : 1.39 = 3 : 9 : 1

Empirical formula of A = C3H9Al [1]

b) [1]

c) 0.24 dm3 of B has a mass of 0.16 g ∴ 24 dm3 of B has a mass of 16 g ∴ Molecular mass of B = 16 g mol–1 [1]

d) B is methane, CH4. [1] C is aluminium hydroxide, Al(OH)3. [1]

e) i) 3CH3Cl(g) + Al/3Na(s) → Al(CH3)3(l) + 3NaCl(s) [1]

3 ii) 0.24 g of A produce 0.24 dm of B (CH4)

3 ∴ 24 g of A produce 24 dm of B (CH4), i.e. 1 mole of CH4 [1]

1 mole of C3H9Al (72.0 g) produces 3 moles of CH4. [1]

Al(CH3)3(l) + 3H2O(l) → Al(OH)3(s) + 3CH4(g) [1]

iii) Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l) [1]

iv) Al(OH)3(s) + NaOH(aq) → NaAl(OH)4(aq) or

− – Al(OH)3(s) + OH (aq) → Al(OH)4 (aq) [1]

Test yourself (page 596)

2 m/z = 136: the parent ion [C6H5COOCH3]+ m/z = 105: [C6H5CO]+ m/z = 77: [C6H5]+ m/z = 51: [C4H3]+ – this is one of the characteristic fragments from phenyl groups. 3 a) Chlorine consists of two isotopes: chlorine-35 and chlorine-37, with abundances in the ratio 3 : 1. In a molecular ion of C2H2Cl2 with two chlorine atoms the possibilities are: most likely two atoms of chlorine-35 (96); next one of chlorine-35 and one of chlorine-37 (98); and least likely two atoms of chlorine-37 (100). The adjacent peaks differ by two mass units and the expected ratio is 9 : 6 : 1. b) These two peaks are fragments formed by the loss of one chlorine atom leaving a single chlorine atom in the fragment, so the ratio of abundances corresponds to the ratio of the abundance of the isotopes. 4 Left spectrum: chloroethane. Molecular ion peaks at 64 and 66 are in the ratio 3 : 1 suggesting that a single chlorine atom is present with chlorine-35 and chlorine-37 present in this ratio.

+ The same ratio of peak heights is seen for CH2Cl at 49 and 51. The peak at 29 corresponds to

+ C2H5 . Loss of further hydrogen atoms gives the peaks at 28, 27 and 26. Right spectrum: bromomethane. Peaks at 94 and 96 are the molecular ion peaks with two isotopes of bromine which occur which roughly equal abundance: bromine-79 and bromine-

+ 81. The peak at 15 is CH3 . The peaks at 79 and 81 correspond to ions of the individual bromine isotopes 79Br+ and 81Br+.

5 Compound is C2H6O

Other Mr: CH2O2 = 46.0253 CH6N2 = 46.0715

Test yourself (page 597)

6 a) Peak A: N−H stretch; Peaks B and C: C=O stretch

b) Left spectrum: ethanamide; right spectrum: ethyl ethanoate

7 a) P: C6H5CHO; Q: C6H5COOH b) Benzaldehyde gives an orange precipitate with 2,4-dinitrophenylhydrazine. Benzoic acid is soluble in hot water to give an acidic solution which releases carbon dioxide when mixed with sodium carbonate.

Test yourself (page 599)

8 a) 13C atoms have one extra neutron compared to 12C.

b) CDCl3 has a deuterium atom instead of the hydrogen in CHCl3. 9 Inert so doesn’t react with samples. Low boiling temperature so easily removed from a sample after the NMR test has been run. Non-toxic. 10 All four carbons are in identical environments.

Test yourself (page 600)

11 a) 3 b) 5 c) 5 d) 1 e) 5 12 Propanal has 3 peaks but propanone has two. 13 10 peaks

Test yourself (page 601)

14 δ for C–C bonds in range 0–60 ppm; δ for C–X 30–70 ppm; and δ for C=O above 170 ppm.

15 a) CH3 δ = 18 ppm; CH2OH δ = 58 ppm b) Using letters from Figure 19.12: A 30 ppm; B 209 ppm; C 53 ppm; D 25 ppm; E 23 ppm 16 Structures below:

ethyl ethanoate cyclohexene a) 4 peaks: A δ = 21 ppm; B δ = 171 ppm; C δ = 60 ppm; D δ = 14 ppm. b) 3 peaks: A δ = 127 ppm: B δ = 25 ppm; C δ = 23 ppm.

Test yourself (page 603)

17 a) In methyl ethanoate, CH COOCH , there are two different chemical environments for hydrogen 3 3 atoms.

b) The peak just above 2 is due to H C−CO−, while the peak at 3.7 is due to −O−CH 3 3

c) Both chemical environments have three H atoms so the step size in the integration trace is the same for both peaks giving a ratio of 1 : 1. 18 a) 1 b) 2 c) 3 d) 2 e) 4 f) 3 19 a) There are two distinct chemical environments. b) The ratio is about 9 : 1 c) The peak at 8.4 is close to the range for −CHO. The peak close to 1.0 is in the range for R−CH . 3 Three methyl groups contain nine hydrogen atoms and these all have to be in the same chemical environment. A possible structure is (CH3)3CCHO. d) Any two of these tests:

• add a solution of 2,4-dinitrophenylhrazine – formation of a bright orange precipitate confirms the presence of a carbonyl group • add fresh Tollens’ reagent and warm – a silver mirror confirms the presence of an aldehyde group • add fresh Fehling’s reagent and warm – an orange-red precipitate confirms the presence of an aldehyde group. 20 a) b)

Test yourself (page 606)

21 First way: all three aligned with the field (one possibility). Second way: two aligned with the field and one against (three possibilities with three protons). Third way: one aligned with the field and two against (three possibilities with three protons). Fourth way: all three aligned against the field (one possibility). 22 a) X−CH2CH2−Y CH2 are non-equivalent so will split each other forming two triplets b) X−CH CH −X CH are equivalent so no splitting occurs and one singlet will be seen. 2 2 2 23 C H CH CH . Five hydrogen atoms at a shift of just over 7 suggest the five hydrogen atoms of 6 5 2 3 a benzene ring with one substituent. There are two hydrogen atoms with a chemical shift corresponding to hydrogen atoms bonded to a carbon atom which is attached to a benzene ring; these are split into a quartet by CH . 3 There are three hydrogen atoms which might be part of a methyl group; these are split into a triplet by CH . 2

24 a) b)

Test yourself (page 607)

25 CH CH OH + D O CH CH OD + HDO 3 2 2 3 2 ⇌ 26 a δ = 11.8 ppm is H in COOH.

δ = 2.9 ppm is CH attached to COOH split into a triplet by the adjacent CH . 2 2 δ = 25 ppm is CH attached to Cl split into a triplet by the adjacent CH . 2 2 b) Peak at δ =11.8 for COOH would disappear. c) Compound is CH CHClCOOH 3 Singlet in reg ion δ = 10–12, with relative integration 1, for COOH. Doublet in region δ = 0–2, with relative integration 3, for 3 protons in CH split by single H in 3 the adjacent CHCl. Quartet in region δ = 2–4.5, with relative integration 1, for proton in CHCl split by three H in the adjacent CH . 3 27 There are three hydrogen atoms with a chemical shift value consistent with R–CH . The peak is 3 a triplet so these hydrogens are next to a carbon atom with two hydrogen atoms. There are two hydrogen atoms with a chemical shift consistent with –CH – next to –COOH. The 2 peak is a quartet and so these hydrogen atoms are next to a carbon atom with four hydrogen atoms. There is one hydrogen atom attached to oxygen in a carboxylic acid group that is not split by hydrogens on the neighbouring carbon atom. Compound is: CH CH COOH 3 2 28 The public is generally nervous of ‘nuclear technologies’ because of the association in their minds of nuclear power with nuclear weapons or the bad press associated with the disposal of waste from nuclear power stations. Test yourself (page 611)

29 a) Powdered chalk b) Hydrocarbon solvent 30 a) Green b) Yellow 31 Two possibilities: • measure the melting temperatures of solids • record IR spectra and compare with reference spectra.

Test yourself (page 613)

32 Trace amino acids on the fingers would contaminate the chromatogram. 33 To prevent the solvent evaporating from the plate 34 a) Q b) R and S c) P 35 R = 0.75/3.0 = 0.25 f 36 The stationary phase, the mobile phase, the temperature. Test yourself (page 616)

37 Helium and argon are noble gases and inert; nitrogen is very unreactive. There is no danger that they will react with the chemicals in the mixture being analysed. 38 a) There were 7 chemicals in the mixture. b) 60 seconds. Proportions of chemicals in the mixture are indicated by the areas under the peaks. If the peaks are narrow, the peak height can be taken as a measure of peak size. 39 The liquid of the stationary phase must not vaporise at the temperature of the column. The column is heated in an oven. 40 This means that it is not possible to identify compounds by the retention times, even if GC is carried out under carefully standardised conditions. 41 There is no data on the retention time of a new chemical when analysed by gas chromatography. This means that GC cannot be used to identify a novel compound.

Activity: Analysing a perfume chemical (pages 608-609)

1 a) The parent ion shows that the relative molecular mass is 164.

b) The molecular formula is the same as the empirical formula, C11H16O. c) There must be four double bonds or rings. If fully saturated there would be 24 hydrogen atoms in the molecule, not 16. 2 a) 11 b) The peak at 209 ppm indicates one C=O carbon. Four peaks between 170 and 120 ppm indicate 4 C=C carbons. Six peaks between 40 and 10 ppm indicate 6 C–C carbons. c) Three double bonds in total indicate that there must be one ring. 3 The peak at 1645 cm−1 is consistent with the presence of C=C. The peak at 1700 cm−1 suggests the presence of C=O in a ketone. The peaks at around 2850–3100 cm−1 indicate the presence of C–H bonds as in alkanes and alkenes.

4 A structure consistent with the given information is:

This structure (which is the correct structure) has the molecular formula C11H16O. It has one C=O ketone double bond, two C=C double bonds and a five-membered ring. The structure has the Z- or cis configuration for the C=C bond in the longer side chain attached to the five- membered ring. 5 a) Jasmone is expected to decolourise an orange solution of bromine. Bromine adds to C=C bonds. b) Jasmone is not expected to react with Tollens’ reagent. Unlike aldehydes, ketones do not react. c) Jasmone is expected to give an orange-yellow precipitate with 2,4-dintrophenylhydrazine. This is characteristic of ketones and aldehydes.

Activity: Using TLC to investigate the aspirin produced in Core practical 16 (page 613)

1 The ink would dissolve in the solvent and contaminate the chromatogram. 2 A is recrystallised aspirin B is commercial sample of aspirin C is crude product

3 The recrystallised product was pure and confirmed to be aspirin as it had the same R as the f commercial sample. The crude aspirin was impure and contained aspirin and at least two other substances. 4 The R value was small as the solvent was not sufficiently polar. The proportion of ethyl f ethanoate in the mixture should be increased. 5 If the initial spots were too big, they would spread, possibly overlap, and lead to inaccurate R f values. Forensic investigations of arson (pages 616-617)

1 A fuel used to accelerate a fire is likely to have burned or distilled away from any area that is completely burned. 2 The soil in pots is moist and stays cooler so long as there is water present. Also, soil is not flammable. This means that traces of chemicals used to accelerate a fire are more likely to be detected if spilled on the soil.

3 The chromatogram for petrol has many more peaks with shorter retention times. The fraction from crude oil distillation that is used to make petrol has a lower boiling range than the fraction used to make paraffin. Hydrocarbons in petrol are generally more volatile than the hydrocarbons in paraffin and the lighter components pass through a GC column faster. 4 The chromatogram for partially evaporated petrol has lost the peaks for chemicals with shorter retention times. Chemicals in the fuel with low boiling temperatures are more likely to be lost when petrol is first evaporated. 5 Sample A shows no sign of any fuel that might have been used to accelerate the fire. If an arsonist used a fuel it was not in or near the front door. 6 Sample B produces a GC trace that resembles that of partially evaporated paraffin. This means that there is a possibility that paraffin was used to accelerate the fire and that this paraffin was partly but not completely evaporated in the region near the window. 7 Further investigations would include the need to establish: • whether or not there was a paraffin heater in the room that might have been the source of the fuel • whether or not analysis of other samples from the house showed signs of the presence of paraffin • whether or not there were other clues that an intruder had been present in or near the property at the time the fire started. Activity: Solving a pollution problem with GC-MS (page 619)

1 a) Ionisation converts the molecules into positive ions that can be accelerated by an electric field into the analyser region. b) The analyser region with electric and/or magnetic fields separates the ions according to their mass-to-charge ratio. c) The positive ions produced from molecules in a mass spectrometer commonly fragment. Some of the fragments are positive ions; some are uncharged free radicals. All the positive ions are detected and show up as peaks in the mass spectrum. + 2 This fragment is C H CH . This fragment is observed in the mass spectra of methylbenzene, 6 5 2 dimethylbenzenes and related compounds. 3 The computer compares the mass spectrum recorded with reference spectra in a database. It identifies spectra that are very similar and shows the probability that the match is correct. 4 GC–MS is a very sensitive method of analysis that can get results from very small samples. Not only does the method separate all the chemicals in a mixture, but it also identifies them. This means that it positively identifies the specific chemicals present. 5 GC–MS cannot be used with chemicals that are non-volatile. This includes compounds that are ionic (such as salts) and large biochemical molecules that decompose before they vaporise.

Exam practice questions (pages 621–5)

1 a) C4H10O Mr = 74.1212 [1]

C3H6O2 Mr = 74.0783 and C2H6N2O Mr = 74.0816 [1]

b) Isomer 1 – 4 peaks Isomer 2 – 4 peaks

Isomer 3 – 2 peaks Isomer 4 – 3 peaks

c) If the compound in part (a) is an alcohol that is unaffected by acidified potassium

dichromate(VI), it must be a tertiary alcohol. [1] Isomer 3 shown in the answer to part (b) is the tertiary alcohol, 2-methylpropan-2-ol. [1]

2 a) C6H6 + 2HNO3 → C6H4(NO2)2 + 2H2O [1] b)

[1] [1] [1] c) 13C NMR would confirm that the major product, 1,3-dinitrobenzene, has four peaks in its spectrum. [1] The minor products are 1,2-dinitrobenzene with three peaks [1] and 1,4-dinitrobenzene with two peaks. [1]

+ + 3 a) Top spectrum: peak at m/z = 105: C6H5CO [1]; peak at 77: C6H5 [1]

+ Bottom spectrum: peak at 93: HO–C6H4 [1] b) Top spectrum: benzoic acid [1] can produce peaks at m/z = 105 and 77 but disubstituted 3-hydroxybenzaldehyde cannot. [1] c) Tollens’ reagent [1]: gives silver mirror with 3-hydroxybenzaldehyde [1]; no reaction with benzoic acid. [1] Sodium hydrogen carbonate (aqueous) [1]: gives effervescence with benzoic acid [1]; no reaction with 3-hydroxybenzaldehyde. [1] 4 a) i) Three environments. [1] ii) From the integration trace data: 3 : 3 : 2 [1]

b) Peak at δ = 1.0: R−CH3 [1]; peaks at δ = 2.1 and 2.5: consistent with H3C−CO− and RCH2CO−. [1] c) Peak at δ = 1.0: triplet – two protons on an adjacent carbon atom. [1] Peak at δ = 2.5: quadruplet – three protons on an adjacent carbon atom. [1]

d) The compound is a carbonyl compound. [1] It is a ketone and not an aldehyde. [1] There is no −OH group in the molecule so it is not an alcohol or acid. [1] e) Butan-2-one [1]

5 a) The larger the molecules, the more strongly they are adsorbed by the stationary phase. [1] Larger molecules are less volatile because they are more polarisable and London forces can act over a larger area. [1] b) A more compact, branched molecule [1] is more volatile and less strongly adsorbed than an unbranched molecule. [1] c) i) The chemicals in the beer sample: ethanol, propan-2-ol (or 2-methylpropan-1-ol), propanone, ethanoic acid, propanal (or butan-2-ol), butanal and butanoic acid. 6–7 correct [3]; 4–5 correct [2]; 2–3 correct [1] ii) The peaks where there are chemicals with very similar retention times are hard to identify [1] – see the answer to part (i). [1]

d) The peak is propanal (CH3CH2CHO) and not butan-2-ol (CH3CH2CH(OH)CH3). [1] Reasons: • the molecular ion has a mass-to-charge ratio of 58 (not 74) [1]

+ • the peak at 57 corresponds to CH3CH2CO [1] and the large peak at 29 to

+ + CHO or CH3CH2 . [1] 6 a) Original spot of hydrolysed dipeptide [1]; solvent front [1]; correct length for distance B moved (0.60 × distance solvent moved) [1]; correct length for distance C moved (0.26 × distance solvent moved). [1] b) i) [1]

ii) C5H11O2N [1]

iii) or branched chain [2]

c) OR

[2]

In either case, the C3H7– chain could also be branched. d) Spray the chromatogram with a solution of ninhydrin [1] in propanone. Heat the paper in an oven at about 100 °C. [1] e) Make a solution of B in water. Use this in a polarimeter [1] and show that it rotates the plane of plane-polarised light. [1] 7 a) Pentan-2-one pentan-3-one

[1] [1]

b) The two isomers have the same molar mass. [1] Their molecular ion peaks have the same value for m/z. [1]

+ c) i) m/z = 29: CH3CH2 [1]

+ m/z = 43: CH3CO [1]

+ m/z = 57: CH3CH2CO [1]

+ m/z = 71: CH3CH2CH2CO [1] ii) Ketone X is pentan-2-one [1]; ketone Y is pentan-3-one. [1] d) i) 5 peaks [1] C=O (ketone) 200–220 ppm (actually 209 ppm). [1] The other peaks in the region 0–60 [1] (actually 46 for C3 and 30 for C1 either side of the carbonyl group, 17 for C4 and, 14 for C5). ii) Two environments for hydrogen [1], one with 6 atoms and the other 4. [1] One peak at a chemical shift in the range 1.8 to 3.0 (actually 2.4) [1], split into four by the neighbouring methyl groups. [1] A second peak in the range 0.2 to 1.8 (actually 1.1) [1], split into three by the adjacent

CH2 groups. [1]

8 Mass spectrum shows Mr = 72. [1] 13C NMR spectrum shows three carbon environments. [1] 172 ppm means C=O. [1] Two peaks [1] in range 120–140 ppm means C=C. [1] Infrared broad peak in range 3300–2500 cm–1 means O–H carboxylic acid. [1]

COOH has mass 45, so this leaves 27 for the rest of the molecule [1], which is H2C=CH. [1] Compound is propenoic acid. [1] Displayed formula:

[1] 9 a) The presence of pairs of lines of roughly equal height separated by 2 m/z units strongly suggests that there is a single bromine atom in the molecule of Z. [1] The bromine-79 and bromine-81 isotopes are equally abundant. [1] b) There are four carbon atoms in the molecule plus Br, giving a mass of 48 + 79 = 127. This

leaves 166 – 127 =39 [1], which is H7O. So a possible molecular formula consistent with the

relative molecular mass is C4H7O2Br. [1] c) The ratios of the step heights are about 30 mm : 20 mm : 20 mm, or 3 : 2 : 2. [1] This shows that there are three hydrogens associated with the singlet peak and two hydrogens with each of the triplet peaks. [1] d) The chemical shift at 3.7 ppm is consistent with H−C−O. The singlet peak is not split and so

not next to a carbon atom with hydrogen atoms attached, so CH3–O–. [1] The chemical shift at 3.5 ppm fits with the range for CHBr. This peak is a triplet so the two hydrogens must be next to another carbon atom with two hydrogens. [1] The chemical shift at 2.9 ppm is consistent with H−C−C=O. This, too, is a triplet, so must also

be next to another carbon atom with two hydrogens [1]. So –CH2–CH2– is present. [1]

e) The structure for Z that fits this information is:

[1]

+ f) Fragmentation to produce [BrCH2CH2] gives fragments of masses 107 and 109, depending on which bromine isotope is present. [1]

10 a) i) Four chemical environments. [1] ii) Ratios: 3 : 2 : 6 : 1 [1] (which gives 12 hydrogens in total, which is consistent with the molecular formula). b) The structure includes a carbonyl group and is a ketone and there are no −OH groups. [1] The formula shows that there can only be one double bond in each molecule and so only one carbonyl group. [1]

c) Chemical shift = 2.2 ppm (three H atoms): H3C−CO− [1]

Chemical shift = 2.7 ppm (two H atoms): RCH2CO− [1]

Chemical shift = 3.4 ppm (six H atoms): −O−CH3 (twice) [1] d) i) The hydrogen atoms corresponding to chemical shift 2.7 ppm have one proton on a neighbouring atom. [1] The hydrogen atom corresponding to chemical shift 4.8 ppm has two protons on a neighbouring atom. [1] ii) The hydrogen atoms corresponding to chemical shifts 2.2 ppm and 3.4 ppm have no protons on neighbouring atoms. [1] e) The skeletal formula of a compound consistent with all this information is:

[1] 4,4-dimethoxybutan-2-one 11 a) Indicative content: • IR shows absorption at 1750 so C=O (but no O–H) possibly ester [1] • 1H NMR spectrum shows three types of proton [1]

• quartet at δ = 4.12 ppm for H–C–O– (possibly ester) next to CH3 [1]

• singlet at δ = 2.04 ppm for H–C–C=O– so CH3–C=O [1]

• triplet at δ = 1.26 ppm for H–C–C–next to CH2 [1]

• compound A is CH3COOCH2CH3 (ethyl ethanoate). [1] b) This question assesses a student’s ability to show a coherent and logically structured answer with linkages and fully sustained line of reasoning. Assess the quality of the answer taking into account both the key points made (up to 6 marks) and the logic and coherence of the discussion (up to 3 marks). Points to make in the answer: • IR shows absorption at 1710 cm–1 so C=O (possibly ester) [1] • absorption at 3400 cm–1 so –O–H alcohol [1] • 1H NMR spectrum shows four types of proton [1] • with integration ratio 2 : 1 : 2 : 3 [1] • singlet (integration 1) at δ = 3.4 ppm is H–O of alcohol [1]

• triplet (integration 2) at δ = 3.84 ppm for H–C–O– [1] so –O –CH2 (next to CH2 [1])

• triplet (integration 2) at δ = 2.70 ppm for H–C–C=O so –CH2–C=O (next to CH2 [1])

• singlet (integration 3) at δ = 2.20 ppm for H–C–C=O [1] so CH3–C=O

• compound B is CH3COCH2CH2OH (4-hydroxybutan-2-one). [1]