IB Chemistry Online EQ Ans

1 Answers to examination questions in Chapters 1 to 21 1 Stoichiometric relationships 4 a Amount of hydrochloric acid 13.95dm 3 = × 2.00 mol dm−3 = 0.0279 mol [1] 1000 Paper 1 b Amount of sodium carbonate in 20.00 cm3 of 1 D 14 A solution = 0.01395 mol [1] 2 B 15 D c Concentration of sodium carbonate 1000 dm 3 3 D 16 C = × 0.01395 mol = 0.6975 mol [1] 20 4 C 17 B d Molar mass of hydrated sodium carbonate 200.00 g 5 D 18 D = = 286.74 g mol−1 [1] 0.6975mol 6 D 19 A e Molar mass of anhydrous sodium carbonate,

7 D 20 B Na2CO3 = (22.99 × 2) + 12.01 + (3 × 16.00) = 105.99 g mol−1 8 C 21 B Molar mass of water = 286.74 g mol−1 − 105.99 g mol−1 9 B 22 D = 180.75 g mol−1 10 D 23 A 180.75gmol−1 Hence, n = = 10 [2] −1 11 B 24 D 18.02gmol + 3− 12 D 25 A 5 a 3Ag (aq) + XO4 (aq) → Ag3XO4(s) [1] b i Amount of silver ions = 13 B 26 D 0.2040 mol dm−3 × 0.04118 dm3 = 8.401 × 10−3 mol [1] 1 ii Amount of Ag XO = × 8.401 × 10−3 mol Paper 2 3 4 3 = 2.800 × 10−3 mol [1] 1 a C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) [2] −3 3 b Carbon dioxide produced = 20 cm iii 2.800 × 10 mol of Ag3XO4 weighs 1.172 g. Hence one mole weighs Oxygen remaining = 10 cm3 [2] 1.172 g −1 −3 = 418.6 g mol [2] 2 a Zn(s) + I2(s) → ZnI2(s) [1] 2.800 × 10 mol 100g b Amount of zinc = –1 = 1.530 mol iv (3 × 107.87) + X + 4 (16.0) = 418.6; 65.37gmol hence X = 30.99 and X is phosphorus (P). [2] 100g (1 g) Amount of iodine = –1 = 0.3940 mol 6 11 mg × = 0.0011 g 253.8gmol (1000 mg) The reacting molar ratio is 1 : 1 therefore the zinc is mass of solute(g) ppm = × 106 ppm present in excess. [3] mass of solventand solute(g) c Amount of zinc iodide = amount of iodine used 0.011g = × 106 ppm/1 = 5.5 ppm [2] 100 (2000 g+0.011g) Amount of zinc iodide = = 0.3940 mol 253.8 − − + 7 IO3 (aq) + 5I (aq) + 6H (aq) → 3I2(aq) + 3H2O(l) 100 KIO 0.6125g Mass of zinc iodide = × (65.37 + 253.8) Amount of KIO /IO 3– = = 0.00286 mol 235.8 3 3−−1 IO3 214.00 gm ol in 250 cm3 = 125.8 g [1] – 3 Amount of KIO3/IO3 in 25.00 cm = 0.000286 mol V V 955cm 3 V 3 a 1 = 2 ; = 2 ; V = 1041 cm3 2 Amount of I = 8.58 × 10−4 mol T1 T2 278K 303K 2 2− − 2– The balloon will burst. [3] 2 SO23(aq) + I2(aq) → 2I (aq) + SO46 (aq) 2− −3 3 b The increase in temperature increases the average Amount of SO23(aq) = 1.716 × 10 mol in 24.50 cm kinetic energy of the nitrogen molecules and hence 1000 × 1.716 × 10−3 Concentration of Na S O = increases their average speed. They collide more 2 2 3 24.50 frequently and energetically with each other and = 0.07 mol dm−3 the walls of the balloon. [2]

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2 Atomic structure electrons move from high energy levels to lower energy levels. Light of a particular frequency is released during this process. [2] Paper 1 5 a Weighted average mass of naturally occurring 1 D 12 B 1 isotopes of an element compared to g of 2 C 13 D carbon-12. 12 [1] 3 A 14 D b Diagram of mass spectrometer containing in the 4 D 15 D correct sequence: 5 C 16 A vaporization/vaporized sample; 6 D 17 A ionization/electron gun; 7 C 18 B acceleration/oppositely charged plates; 8 A 19 A deflection/magnetic field; 9 C 20 D detection [5] 10 C 21 C c 63x + 65(1 − x) = 63.55 63 65 11 A 22 A Cu = 72.5% and Cu = 27.5% [2] d Cobalt-60 (60Co), iodine-131 (131I) or Paper 2 iodine-125 (125I) [1] 1 a An atom of 79Br An ion of 81Br− [3] Protons 35 35 3 Periodicity Neutrons 44 46 Electrons 35 36 Paper 1 1 A 13 C b 79Br because its relative isotopic mass is closer to 79.90. [1] 2 D 14 A 2 a Atoms of the same element but with different mass 3 A 15 B numbers and hence different number of neutrons. [1] 4 C 16 B b 47 protons, 46 electrons and 60 neutrons. [2] 5 C 17 D c Carbon-12 (12C). [1] 6 B 18 D n = ∞ 3 n = 6 7 B 19 B n = 5 8 A 20 C n = 4 9 A 21 B A n = 3 10 A 22 B 11 C 23 C B n = 2 12 C 24 D

Paper 2 n = 1 1 a i The minimum energy required for the removal a Line starting at n = 1; of a mole of electrons from a mole of gaseous atoms to form a mole of unipositive gaseous Line finishing at n = ∞ (not above ∞); ions. [2] Upward arrow [2] ii Al(g) → Al+(g) + e− [1] b Line from n = 3 to n = 2; b The magnesium atom has a greater nuclear charge Arrow pointing downward (in any transition) [2] due a higher number of protons. Consequently, the atomic radius is smaller and there is a stronger 4 The emission spectrum consist of a series of sharp or attraction for the electrons by the nucleus. [3] discrete coloured lines on a black background. [1] c 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) The lines converge together at high energy (high frequency). The lines are generated when excited Chemistry for the IB Diploma Second Edition © Christopher Talbot, Richard Harwood and Christopher Coates 2015 3

There will be effervescence: release of gas at the 4 The ions formed across period 3 would be Na+, Mg2+, surface. Bubbles of a colourless gas will be observed. Al3+ and P3−, S2− and Cl−. The lithium moves around the surface and is There is a decrease in ionic radii from Na+ to Al3+: all converted into a soluble product. Heat and sound the ions have the electron arrangement of 2, 8 (i.e. energy are released. [3] they are isoelectronic); however, there is a progressive 2 a i The bromine atom has a greater number of increase in the nuclear charge due to the additional protons and hence a greater nuclear charge: 35+ protons: the sodium ion has 11 protons, the magnesium versus 34+. Hence the outer or valence electrons ion has 12 protons and the aluminium ion has 13 are attracted more strongly. The bromide ion is protons. All of the electrons experience a higher formed by the addition of one electron to the effective nuclear charge and the electrons in the two bromine atom; the selenide ion is formed by shells are pulled progressively towards the nucleus. the addition of two electrons to the selenium There is a decrease in ionic radii from P3−, S2− to Cl−: all atom. The additional electrons cause electron– the ions have the electron arrangement of 2, 8, 8 (i.e. electron repulsion which increase radii. The they are isoelectronic); however, there is a progressive effect is highest in the selenide ion. [2] increase in the nuclear charge due to the additional ii The electronegativity is the ability of atom to protons: the phosphide ion has 15 protons, the sulfide attract a bonding pair of electrons in a covalent ion has 16 protons and the chloride ion has 17 protons. bond. Fluorine has a smaller radius and hence (All of the electrons experience a higher effective the electrons are closer to a nucleus in a lower nuclear charge and the electrons in the three shells are energy level. The electrons are repelled by pulled progressively towards the nucleus.) [6] fewer inner electrons and there is a decreased 5 a Ionization energy increases. shielding effect. [3] Nuclear charge increases/electron is closer (on b i Brown solution formed; Cl2 + 2KI → I2 + 2KCl; − − average) to the nucleus/atomic radius decreases. [2] Cl2 + 2I → I2 + 2Cl [2] b Aluminium is lower than expected; sulfur is lower ii A cream or off-white precipitate (solid) is than expected. formed: Ag+ + Br− → AgBr [2] In aluminium the electron removed is in a p sub- c The reaction in (i) is redox. The chlorine is shell/at a higher energy than in magnesium. reduced/gains electrons/decreases its oxidation number. The iodide ion is oxidized/loses electrons/ The s electron removed is in an orbital that increases its oxidation number. [4] contains a second electron that repels it/change to a half-filled p sub-shell. [4] The reaction in (ii) is not redox. There is no electron transfer/change in oxidation number. 6 Across period 3 of the periodic table, the oxides of the elements change from basic to amphoteric to acidic 3 a The size of an atom is determined by the attraction in nature due to the increasingly electronegativity of of the nucleus for electrons and the shielding of the elements, which gradually changes the bonding outer electrons from the nucleus by inner shells of from ionic to covalent. electrons. [3] Sodium and magnesium oxide react to form hydroxide b i This is because the sodium ion, Na+, has one ions: less shell of electrons compared to the sodium

atom. Consequently, there is a decrease in Na2O + H2O → 2NaOH; MgO + H2O → Mg(OH)2 shielding and an increase in the nuclear charge Aluminium oxide is amphoteric and reacts with both experienced by each electron. [2] acids and bases: ii This is because with increasing proton number, Al O + 6H+ → 2Al3+ + 3H O; each successive positive ion contains an 2 3 2 Al O + 2OH− + 3H O → 2Al(OH) − additional shell of electrons. The additional 2 3 2 4 screening or shielding effect outweighs the Silicon dioxide is an acidic oxide and reacts with hot, effect of increased nuclear charge. [2] concentrated alkali:

− 2− c The magnesium ion has 12 protons and 10 electrons; SiO2 + 2OH → SOi 3 + H2O the sodium ion has 11 protons and 10 electrons. The The highest oxides of the remaining non-metallic nuclear charge of the magnesium ion (+12) is greater elements react with water to form strongly acidic than the sodium ion (+11). Consequently, all the solutions: electrons in the magnesium ion will experience a

greater force and be located nearer to the nucleus. [2] P4O10 + 6H2O → 4H3PO4; SO3 + H2O → H2SO4; 3− Cl O + H O 2HClO [3] d i N < N [1] 2 7 2 → 4 ii Fe3+ < Fe2+ < Fe [1]

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4 Chemical bonding and b The C–Cl bond is more polar than the N–Cl bond. Nitrogen is more electronegative than structure carbon. NCl3 is polar; CCl4 is polar. The NCl3 molecule is not symmetrical and dipole moments do not cancel; CCl is symmetrical and the four Paper 1 4 dipole moments cancel. [5] 1 A 15 A 3 Ethanol and water are miscible in each other due to the 2 A 16 B formation of hydrogen bonding. The ethane molecule 3 B 17 D is non-polar and is not able to form hydrogen bonds with water molecules. [2] 4 B 18 C Ethane molecules can only form weaker London 5 A 19 D (dispersion) forces with water molecules. Cholesterol 6 C 20 A has a polar alcohol group; however, it behaves as a non- polar molecule due to the relatively large size of the 7 D 21 A hydrocarbon group. [2] 8 C 22 A 4 a Potassium metal has a lattice of unipositive ions 9 B 23 B (cations) in fixed positions with a ‘sea’ of delocalized or mobile valence electrons that are free to move 10 C 24 A through the lattice. The metal is held together by 11 D 25 A attraction between the nuclei of the positive ions and the delocalized electrons. [4] 12 D 26 A b Lewis structure for fluorine: [2] 13 B 27 A F F 14 A 28 A The bonding within a fluorine molecule is a single Paper 2 covalent bond. This consists of a shared pair of 1 Sodium, magnesium and aluminium are metallic valence electrons. [1] and hence are good electrical conductors due to the The bonding between molecules arises from presence of delocalized valence electrons. From sodium, London (dispersion) forces. These involve the magnesium to aluminium, the atomic and ionic size attraction between dipoles temporarily set up by decrease and the number of valence electrons available electron polarization. [1] for delocalization increases from one to three. Thus, c The potassium atom has the configuration the metallic bonding strength increases and hence 2, 8, 8, 1. [1] the melting point increases since more heat energy is required to break the metallic bonding. In addition, The fluorine atom has the configuration 2, 7. [1] with more valence electrons for delocalization, the Potassium’s outer electron is transferred to the outer electrical conductivity also increases. [3] shell of the fluorine atom. [1] Silicon has a giant covalent structure with strong Oppositely charged ions are formed (with complete silicon–silicon single bonds. A large amount of heat octets of electrons) which attract each other due to energy is required to break all these bonds and hence it electrostatic forces of attraction. [1]

has a high melting point. Phosphorus (P4), sulfur (S8), d The ions are not free to move in the solid state. chlorine (Cl2) and argon (Ar) are simple molecular However, they can move in the molten state under covalent substances and hence are held together in the influence of a voltage. [1] the solid state by London (dispersion) forces. A small amount of thermal energy is required to break these 5 a Boron trifluoride is a simple molecular substance intermolecular forces and hence their melting points and aluminium fluoride is ionic. [6] + + are low. The strength or extent of London (dispersion) + + + F + intermolecular forces increases with molecular size + + + + + + + + F + B + F + (number of electrons). Thus, sulfur has the highest + + + + melting point followed by phosphorus, chlorine and then argon. [3] 3+ + + 2 a Electron dot structures for N2 and F2: [2] F AI F

N N F F

+ The triple bond in the nitrogen molecule is F stronger and hence harder to break. [1]

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b i A coordinate covalent bond. Both electrons d Significant heat loss to the surroundings. are supplied by the nitrogen atom of ammonia Insulate the reaction vessel and use a lid. Draw a which has a lone pair of electrons. [2] temperature versus time graph and extrapolate to

+ + calculate the rise in temperature. [2] ii + + F H [2] + F + H + + + + e The temperature change would be the same, F + B N + H FB N H + + namely, 3.5°C. This occurs because the amount of F H lithium hydroxide reacted would be the same. The F H excess hydrochloric acid would not react; lithium hydroxide is the limiting reagent. [2] 5 Energetics/thermochemistry 3 a The enthalpy change that occurs when one mole of a pure compound is formed under standard Paper 1 thermodynamic conditions from its elements in 1 C 10 A their standard states. [2] 2 D 11 A b The enthalpy change that occurs when one mole of a pure compound undergoes complete combustion 3 D 12 D in the presence of excess oxygen under standard 4 D 13 A conditions. [2] 5 A 14 C c Hess’s law states that the total enthalpy change for 6 B 15 A a reaction is independent of the route taken. It depends only on the initial and final states. [1] 7 B 16 D ΔH1 8 A 17 C d 3C(s) + 4H2(g) C3H8(g)

9 D + 5O2(g) + 5O2(g)

ΔH H Paper 2 2 Δ 3 1 a i The energy needed to break one mole of a particular covalent bond in a molecule in the 3CO2(g) + 4H2O(l) gaseous state and form gaseous atoms. According to Hess’s law, ΔH1 = ΔH2 − ΔH3 The value is averaged using those from similar ΔH = ΔHØ [C H (g)] compounds. [3] 1 f 3 8 ΔH = 3 × ΔHØ [C ] + 4 × ΔHØ [H (g)] ii It is an element and is the only species with a 2 c graphite c 2 −1 −1 F–F bond. [1] ΔH2 = 3 × (−393 kJ mol ) + 4 × (−286 kJ mol ) = −2323 kJ mol−1 b i Sum of bonds broken = 412 + 158 = 570 kJ ΔH = ΔHØ [C H (g)] = −2220 kJ mol−1 Sum of bonds formed = 484 + 562 = 1046 kJ 3 c 3 8 Hence, ΔH = −476 kJ mol−1 [3] ΔHØ [C H (g)] = (−2323 kJ mol−1) − (−2220 kJ mol−1) ii CH4 + F2 [2] f 3 8 = −103 kJ mol−1 [4] enthalpy CH3F + HF

iii About the same since the same number and 6 Chemical kinetics type of bonds are being broken and formed. [2] Paper 1 2 a It is exothermic because heat is released to the surroundings and the temperature rises. [1] 1 A 5 D b To make any heat loss as small as possible, so that 2 B 6 A all the heat will be given out very rapidly. [1] 3 C 7 C c Heat released = mass × specific heat capacity 4 C × temperature increase Amount of LiOH or HCl used = Paper 2 0.500 dm3 × 0.050 mol dm−3 = 0.025 mol 1 a i Carbon dioxide, CO [1] Heat released = 100 g × 4.18 J g−1 °C−1 × 3.5°C 2 = 1463 J = 1.463 kJ ii CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H O(l) [2] ΔH = (1.463 kJ ÷ 0.025 mol) = −58.5 kJ mol−1 [4] 2

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b 800 cm3; 8 minutes [2] b The rate of a reaction increases with temperature as the particles are moving faster (have greater kinetic c i The collision frequency between carbonate energy) and therefore collide more frequently [1] and hydrogen ions is greatest at the start, but The colliding particles are more likely to have decreases with time. The concentrations of combined kinetic energy greater than the activation the two reactants is highest at the start of energy of the reaction and so more collisions the reaction. However, as the chemicals react produce a reaction. [1] the number of reacting ions decreases the The second of these factors is the more significant/ concentrations and hence the collision rate important of the two. [1] decreases. [2] ii The mass of calcium carbonate may be less 7 Equilibrium in Y and hence greater in X. [1] The concentration of hydrochloric acid may Paper 1 be less in Y and hence higher in X. [1] 1 A 8 A 2 a i 30 cm3 [1] 2 D 9 B ii 19 cm3 [1] 3 A 10 A iii 9 cm3 [1] 4 D 11 D b As the reaction proceeds the surface area of the magnesium (the limiting reagent) decreases. The 5 C 12 C magnesium is being used up, and hence the rate 6 D 13 B decreases. The acid remains in excess and its 7 B concentration remains relatively constant. [2] c The reaction has stopped; all of the magnesium has Paper 2 been consumed. [1] 1 a Any one of the following statements: [1] d Plot an accurate graph of the results: labelling the axes, plotting points carefully, and drawing a curve Rate of forward process/reaction = rate of of best fit. [3] backwards/reverse process/reaction Parts i and ii; draw tangents to your curve and Rate of vaporization/evaporation = rate of estimate the slopes at the two points. [4] condensation; e i Initial rate increased [1] Concentrations of reactants and products remain constant; ii Initial rate decreased [1] There is no change in macroscopic properties in a iii Initial rate increased [1] closed system; f i Final volume of gas increased [1] Do not accept concentration of reactants and products ii Final volume of gas unchanged [1] are equal.

g i Use to measure one of the following: gas Accept constant colour of Br2 vapour/liquid. pressure, motion or change in mass using a 2 [SO]3 balance connected to computer. [2] b i Kc = 2 [1] [SO]2 [O2 ] Could not use: light sensor. [1] ii Two or three correct [1]; four or five correct [2]; ii Give a brief description of one use: pressure all correct [3] sensor fitted to sealed flask detecting increase Position of Value of K in pressure with time; motion sensor detecting c equilibrium movement of gas syringe barrel with time; balance detecting loss in mass of flask and Decrease in temperature Shifts to right/ Increases products contents with time. [3] Increase in pressure Shifts to right/ No effect products 3 a i Increasing the concentration of H2O2 will increase the rate of reaction as the frequency Addition of catalyst No effect No effect of collisions will increase as there are more particles present. [2] c i Minimum/least/smallest energy needed (by reactants/colliding particles) to react/start/ ii This will have no effect; the concentration initiate a reaction. [1] of NaI does not depend on the size of the crystals used to make the solution, just the Allow energy difference between reactants and mass used. [2] transition state.

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Equilibrium established when rate of forward ii T 1 reaction = rate of reverse reaction. [7] b Reaction is endothermic [1]

Kc increases with (increasing) temperature [1] T2 the forward reaction favoured/heat is used up. [1] 4 a 200°C [1] and 600 atm [1]. [2] (Allow the ‘highest pressure and the lowest Probability of molecules Probability temperature’.)

(Kinetic) energy b i Yield increases/equilibrium moves to the right/ more ammonia. [1] The x-axis label: (kinetic) energy; the y-axis label: probability/fraction of molecules/particles/ 4 (gas) molecules → 2/decrease in volume/ probability density. fewer molecules on right-hand side. [1] Allow number of molecules/particles for y-axis. ii Yield decreases/equilibrium moves to the left/ less ammonia; exothermic reaction. [2] Correct shape of a typical Maxwell–Boltzmann energy distribution curve; c High pressure expensive/greater cost of operating at high pressure/reinforced pipes, etc. needed. Do not award mark if curve is symmetric, does not start at zero or if it crosses x-axis. Lower temperature: greater yield, but lowers rate. [2] Two curves represented with second curve for (Do not award a mark for simply saying ‘compromise’.) 2 [NH]3 T2 >T1 to right of first curve, peak maximum lower d Kc = 3 [1] than first curve and after the curves cross going to [N22][H] the right, T2 curve needs to be above T1 curve as e i Artificial fertilizers/increasing crop yields; illustrated. [2] production of high explosives for mining [HI]2 2 a K = [1] [1 max]. [1] c [H ][I] 22 ii Fe/iron (allow magnetite/iron oxide). [1] b Concentration of product / HI greater The claim is not valid since catalysts do not (than [H2] and [I2]) [1] alter the yield/position of equilibrium they only c It will have no effect. [1] increase the rate of reaction. [1] d As the reaction is endothermic, increasing the temperature will shift the equilibrium position to the right (more product formed). [1] 8 Acids and bases 3 a Two curves – one labelled ‘forward’ starting up high Paper 1 up y-axis and one labelled ‘reverse’ starting from zero. The curves merge and become horizontal. 1 D 14 B 2 C 15 D 3 A 16 B forward 4 C 17 D 5 D 18 D Rate 6 A 19 A

reverse 7 A 20 A 8 D 21 C Time 9 B 22 A i Forward reaction – highest concentration, 10 D 23 C thus rate high to begin with. As the reaction proceeds, the concentrations decrease, as does 11 B 24 B the rate. 12 A 25 B ii Reverse reaction – zero rate initially at t = 0 13 D (since no products present). The rate increases as concentration of products increases.

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Paper 2 9 Redox processes 1 a i Acid X [1] ii A factor of 10 [1] Paper 1 1 A 10 D iii Y, Z and X [1] 2 D 11 D Conductivity increases with ion concentration. As the pH value decreases the concentration of 3 D 12 C hydrogen ions increases. [1] 4 A 13 B 2 a A strong acid is fully ionized or dissociated in 5 A 14 C solution. [1] 6 B 15 B HCl(aq) → H+(aq) + Cl−(aq) [1] 7 D 16 D A weak acid is partly ionized or dissociated in solution. [1] 8 D 17 A

+ 2− 9 B H2CO3(aq) ⇋ 2H (aq) + CO3 (aq) [1] or + − H2CO3(aq) ⇋ H (aq) + HCO3 (aq) Paper 2 b Universal indicator: hydrochloric acid – red; 1 a i Magnesium (Mg), iron (Fe), copper (Cu), and carbonic acid – yellow/orange. [1] gold (Au). [1] Electrical conductivity: hydrochloric acid – high; ii The loss or removal of electrons from a carbonic acid – low chemical species. [1] or Iron atoms lose electrons during their reaction Reaction with metal/metal carbonate: hydrochloric with copper(II) ions: acid – vigorous; carbonic acid – slow. [1] Fe(s) → Fe2+(aq) + 2e− [1] c 10000 : 1 or 104 : 1 or 0.1 : 0.00001 or 10−1 : 10−5 [2] iii Reduction involves a decrease in oxidation d i Base: HCO − ; conjugate acid: H CO [1] 3 2 3 number. [1] − Acid: HCl; conjugate base: Cl [1] Copper(ii) ions gain electrons during their ii Brønsted-Lowry [1] reaction with iron atoms: 3 a Both solutions have an excess of hydroxide ions, Cu2+(aq) + 2e− → Cu(s) [1] which can accept protons to form water molecules. [1] iv Magnesium (Mg) is the strongest reducing b The pH of potassium hydroxide solution would agent. [1] be higher. This is because it is a stronger base It reduces Fe2+; Fe reduces Cu2+; Cu than ethylamine. It hence undergoes greater reduces Au+. [1] dissociation. [3] + + v Gold ions, Ag [1] c C2H5NH2(l) + H2O(l) → CH25NH3 (aq) + OH−(aq) [1] Every metal present can reduce gold ions to gold atoms. [1] − + OH is a base/conjugate base and CH25NH3 is a conjugate acid/acid [1] vi Does not react. [1] + It is too low in the reactivity series and hence d C2H5NH2(aq) + HCl(aq) → CH25NH3 (aq) + Cl−(aq) [1] is less reactive than magnesium. [1] source of current [1] e It is a strong acid and hence corrosive. [1] b [4] + 4 a Proton (H ) donor [1] + – electrodes [1] − correct polarity with reference to battery [1] b HSO3 (aq) and H2O(l) [1]

c Nitric(v) acid, HNO3 [1] molten salt [1] d Leaching of minerals into freshwater lakes; killing of young fish [2] c At the cathode (negative electrode):

e S(s) + O2(g) → SO2(g); electrons are given to the cations (positive ions) [1] SO (g) + H O(l) → H SO (aq) [3] 2 2 2 3 electrons are removed from the anions + 2– H2SO3(aq) Y 2H (aq) + SO3 (aq) (negative ions). [1]

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At the anode (positive electrode): b Molar mass of empirical formula = 71. [1] copper atoms formed (at the cathode) [1] Molecular mass = 142; therefore molecular formula = 2 × empirical formula. chlorine molecules formed (at the anode). [1] Molecular formula of hydrocarbon = C H [1] 2 a At the negative electrode (cathode): 10 22 c i 2C H + 31O → 20CO + 22H O [2] K+ + e− → K 10 22 2 2 2 ii 2C H + 21O → 20CO + 22H O [2] At the positive electrode (anode): 10 22 2 2

− − Could also have an equation where C is produced. 2F → F2 + 2e d i C H [1] Electrons flow through the external circuit or wires 4 10

and the ions move through the electrolyte to the ii C5H10 [1] electrodes where they gain or lose electrons. [4] 2 a i % Oxygen = 36.4

b 0.2 mol since the K to F2 molar ratio is 2 : 1. [2] Empirical formula is C2H4O [3] c i Sodium atoms lose electrons and undergo ii Molar mass of empirical formula = 44. reduction. Aluminium ions (in aluminium chloride) gain electrons and undergo Molecular formula = 2 × empirical formula

oxidation. [2] = C4H8O2 [2]

ii Reactants: sodium metal: 0; aluminium in b i Empirical formula is CH2O aluminium chloride: 3 ii Molar mass of empirical formula = 30. [2] Products: aluminium metal: 0; sodium in sodium chloride: +1 [2] Molecular formula = 2 × empirical formula = C2H4O2 [2] 3 a i Fe2+(aq) → Fe3+(aq) + e− 3 a Full or condensed structures: − + − 2+ ii MnO4 + 8H + 5e → Mn + 4H2O CH3CH2CH2CH3, butane / CH3CH(CH3)CH3, − 2+ + 2+ 3+ 2-methylpropane. [4] iii MnO4 + 5Fe + 8H → Mn + 5Fe + 4H2O

− 11.6 b i Amount of MnO = × 0.0200 b A: CH3CH2COOH / CH3CH2CO2H propanoic acid. 4 1000 −4 B or C: CH COOCH methyl ethanoate. = 2.32 × 10 mol 3 3

2+ −4 C or B: HCOOCH CH ethyl methanoate. [6] ii Amount of Fe = 5 × 2.32 × 10 2 3 = 1.16 × 10−3 mol c i A forms hydrogen bonds with water. [1] iii Mass of Fe2+ = 55.85 × 1.16 × 10−3 A ionizes/dissociates to give H+ ions. [1] = 6.48 × 10−2 g ii C2H5COOH + NaOH → C2H5COONa −2 + H O [1] % of Fe2+ in tablet = 6.48 × 10 × 100 2 1.43 iii No C=C bond. [1] = 4.53% d A has the highest boiling point as the acid molecules have the capability of hydrogen bonding together; B 10 Organic chemistry and C have similar boiling points as they are both esters, their boiling points are lower than that of A as they cannot take part in intermolecular hydrogen Paper 1 bonding. [2] 1 D 7 D e i Ester [1] 2 B 8 C Flavourings/plasticizers/solvents/perfumes [1] 3 B 9 D ii Carboxylic acid/alkanoic acid [1]; alcohol/ 4 A 10 B alkanol [1]; water [1]. [3] 5 C 11 A f Any feasible formula containing C=C (e.g. CH CHC(OH) or HOCH=C(OH)CH ). [1] 6 B 12 C 3 2 3 Addition [1] Paper 2 Mixture is decolorized/colour change is from yellow/ orange to colourless. [1] 1 a Molar ratio = 1 : 2.2. [1] 4 a i Same general formula/differ by −CH2−/similar Therefore empirical formula is C5H11 [1] chemical properties/gradual change in physical properties. [3]

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ii [4] 73.99 CH3 CH2 CH2 CH2 OH 6 a i Number of moles of carbon = = 6.161 I 12.01 6.55 CH3 Number of moles hydrogen = = 6.49 1.01 CH3 C OH 9.09 Number of moles nitrogen = = 0.649 14.01 II CH 3 10.37 Number of moles oxygen = = 0.648 H 16.00 C : H : N : O molar ratio is 9.5 : 10 : 1 : 1 CH3 C C2H5 Empirical formula is C19H20N2O2 [3] III OH ii Mr of empirical formula = actual Mr

H Molecular formula is C19H20N2O2 [1] iii (0.5) (40 − 20 − 2) = 9 [1] CH C CH OH 3 2 b A is carboxamide (amide); B is phenyl; C is

IV CH3 carboxyl (carboxy); D is hydroxyl. [2]

b i Esterification / condensation CH COOH + 3 11 Measurement and data C H OH → CH COOC H + H O 2 5 3 2 5 2 processing O CH3 C Paper 1 O C2H5 1 B 8 B Ethyl ethanoate [4] 2 D 9 D ii Catalyst; lowers activation energy (by providing 3 C 10 C an alternate pathway) [2] 4 B 11 C iii Flavouring agents / in plasticizers / in solvents / in perfumes [1] 5 C 12 B c i II reacts with bromine / contains C=C / II is an 6 A 13 C alkene / has unsaturated R group. 7 B I contains only saturated R groups. [2] ii Addition polymerization Paper 2 1 a i The actual volume of sodium hydroxide will be CH CH 2 lowered, and the calculated concentration of O hydrochloric acid will be higher. [2] ii If there was water in the burette, the sodium CHO [2] hydroxide would be diluted, lowering the amount of sodium hydroxide and the 5 a i Propan-2-ol [1] concentration of hydrochloric acid. [2] ii Propan-1-ol [1] iii This will make the concentration of the iii 2-Methylbutan-2-ol [1] hydrochloric acid appear greater than it is. [1] iv Ethane-1,2-diol [1] iv The end point will not be accurate, causing random errors. [1] b i Secondary [1] b i Systematic error. [1] ii Primary [1] ii Measure the pH of a buffer solution of known iii Tertiary [1] pH value. [1] iv Primary [1] 2 a Mass of copper (dependent), time (independent) and current (controlled variable). [3] c i Propanone, CH3COCH3 [2] b Directly proportional. [1] ii Propanoic acid, CH3COOH [2] mass of copper(g) 1.24 g c Rate = = = 0.08 g min−1 [2] iii No oxidation [1] time(min) 15min d Interpolation, area under the graph and iv Ethane-1,2-dioic acid, HOOCCH CH COOH [2] 2 2 extrapolation. [2]

Chemistry for the IB Diploma Second Edition © Christopher Talbot, Richard Harwood and Christopher Coates 2015 11

 1  (ratio of areas or integration traces in) 3 a Percentage uncertainty in length =   × 100 1-iodobutane: 3 : 2 : 2 : 2 298 = 0.336% [1] (ratio of areas or integration traces in) 2-iodobutane: 3 : 1 : 2 : 3 [4]  1  b Percentage uncertainty in width =   × 100 b 1-fluoro-2-methylpropane has three peaks: 210  = 0.476% (ratio of areas or integration traces in) c Area = 298 mm × 210 mm = 62 580 mm2 [1] 1-iodobutane: 2 : 1 : 6 [3]  1 1  Percentage uncertainty in area = +  × 298 210  12 Atomic structure 62 580 = 508 mm2 Area = 62 580 mm2 ± 508 mm2 [2] Paper 1 4 a The bond angle (H–S–H) changes as bending 1 A 9 A occurs. [1] 2 A 10 A The lengths of the S–H bonds change as stretching occurs. [1] 3 C 11 B The dipole moment of the molecule changes. [1] 4 B 12 C 5 A 13 C b i Absorptions in NMR are due to transitions between different spin states in the nucleus 6 D 14 D (when an external magnetic field is applied). [1] 7 B 15 D IR absorptions are due to bond vibrations. [1] 8 C 16 B The nuclear transitions are at a much lower energy than bond vibrations. [1] Paper 2 There is an inversely proportional relationship 1 a The first ionization energy is the minimum between frequency and energy. [1] energy required to remove one mole of electrons from one mole of gaseous atoms to form one ii The bond in both molecules vibrates and mole of unipositive gaseous ions (under standard stretches; only the stretching in the hydrogen thermodynamic conditions). [2] bromide, H–Br, causes a change in dipole moment. [2] b In a group 2 element (C) the electron (for the first 1 ionization energy) is removed from a spin pair in an c wavenumber = ; s sub-shell. For D, a Group 13 element, the electron wavelength(cm) (for the first ionization energy) is removed from a Planck’s constant × velocity energy = ; 3p sub-shell further away from the nucleus. The 3p wavelenght sub-shell also experiences more shielding. Hence energy there is a decrease in first ionization energy when wavenumber = Planck’s constant × velocity [2] moving from C to D. The increase in nuclear charge accounts for the increase from E (np2 to F np3). In d Atomic masses increase from chlorine to iodine. [1] addition C cannot be in Group 15 (the other group Heavier atoms vibrate more slowly and hence the after which there is a slight decrease) because there stretching frequency decreases. [1] is a steady rise for the next 3 elements (D to E to F) indicating the filling of a p sub-shell. [4] 5 a P: +CH , Q: +C H and R: +C H [3] 3 2 5 3 7 c As you move down group 2 the first ionization b Information Analytical technique energies decrease. As you descend the group the nuclear charge increases due to the presence Relative atomic mass of an Mass spectrometry element of additional protons, but the shielding effect progressively increases due to the presence of an Functional groups and Infrared spectroscopy strength of bonds present in extra electron shell as you move from one period an organic molecule to the next. The effect of the extra protons is Potassium ions in a sample of Atomic absorption compensated for by the effect of the extra electrons. blood serum spectroscopy and The only factor left is the additional distance inductively-coupled between the outer electron and the nucleus. That plasma spectroscopy lowers the first ionization energy. [3] [3] Detailed structure of an Nuclear magnetic d C is in period 3 since in period 2 the Group 2 organic molecule (without resonance degrading the sample) element would have the higher ionization energy while in period 4 the Group 2 element would have 6 a Both 1-fluorobutane and 2-fluorobutane have four a lower ionization energy. This can be accounted for peaks: by the progressively increasing average distance the

Chemistry for the IB Diploma Second Edition © Christopher Talbot, Richard Harwood and Christopher Coates 2015 12

valence electron is from the nucleus. C cannot be in Paper 2 the first period (He) as p-orbitals are being filled, it cannot be in the fourth period (Ca) as (x + 8) would 1 a Ester functional group. [1] not bring you back to Group 2 again. [3] b Coordinate covalent bond. [1] e There is a slight decrease from F to G due to the c i z z presence of a spin pair in the np sub-shell for G. [2] The resulting electron–electron repulsion is greater than the effect of the increase in nuclear charge y y and reduces the ionization energy of G+. [2] 2 a The energy needed to remove one mole of valence x x electrons from a mole of (neutral) atoms; in the s orbital p orbital gaseous state: x Mg(g) → Mg+(g) + e− [3] ii 16 [1] 2 2 6 2 6 10 2 6 6 b The electron removed (from aluminium) is in a p d 1s 2s 2p 3s 3p 3d 4s 4p 4d [1] sublevel/sub-shell while that from magnesium is in e [1] an s sub-level/sub-shell. 3s 3p 4s 3d

The p sub-level is of higher energy/further from 2 a A pale blue precipitate of hydrated copper(ii) nucleus (than the s sub-level)/experiences greater hydroxide, Cu(OH)2. The precipitate reacts with shielding from nucleus. [2] excess ammonia to form a solution containing a c The third electron is more difficult to remove deep blue complex ion: because it is taken from a positively charged ion and Cu2+ + 2OH− → Cu(OH) from a (main) energy level closer to nucleus / from 2 [Cu(NH )]2+ − second energy level/n = 2 while the first electron Cu(OH)2 + 4NH3 → 3 4 + 2OH [4] comes from the third energy level/n = 3. [1] + 2− − b i 4H + 3 MnO4 → 2 MnO4 + MnO2 + 2H2O [2] d 1s2 2s2 2p6 3s2 3p6 3d7 4s2; 4s [2] ii +6, +7 and +4 [3] 3 a 4p [1] iii Catalysis [1] c The zinc ion, Zn2+ has a 3d10 electron configuration b Ge+(g) → Ge2+(g) + e− [1] (fully occupied 3d-orbitals). Hence it is not possible c The germanium atom contains four valence to have d–d transitions and it is thus colourless. [2] electrons: 4s2 and 4p2. There is a steady increase 3 a +3 and +2 [2] in ionization energy. Upon reaching the fifth ionization energy an electron must be removed from b Six cyanide ligands are coordinated octahedrally to a completely filled inner 3d sub-shell. Once filled a central iron(ii) ion. [2] these act as core electrons and this is reflected in c A transition metal is a metal that forms compounds in the large increase in ionization energy. [2] which the metal has a partially filled 3d sub-shell. [1] d Hemoglobin is an iron-containing protein found 13 The periodic table – the inside red blood cells. The iron forms a complex with oxygen, known as oxyhemoglobin, when transition metals the oxygen concentration is high. At low oxygen concentrations the oxygen is released and Paper 1 hemoglobin is re-formed. [2] hc 4 E = ; so the larger the wavelength, the smaller the 1 A 9 A λ energy, so Z > Y > X. [3] 2 A 10 A hc 3 B 11 D 5 E = ; E = (6.63 × 10−34 J s × 3.00 × 108 m s−1) λ 4 B 12 D × 1 × 10−9 m × 1/1000 × 6.02 × 1023 = 163 kJ mol−1 [4] 5 A 13 D 6 a +3 [1] 2 2 6 2 6 2 10 6 6 6 A 14 A b 1s 2s 2p 3s 3p 4s 3d 4p 4d [1] c 4 [1] 7 A 15 C d Octahedral 8 B 16 D

[5]

Chemistry for the IB Diploma Second Edition © Christopher Talbot, Richard Harwood and Christopher Coates 2015 13

14 Chemical bonding and 3 a Each correct structure (with non-bonding electron pairs where relevant) structure C2H2 linear/straight Paper 1 Angle 180° + + + + + + 1 A 11 C H C+ + C H

2 B 12 D C2Cl4 (trigonal) planar 3 C 13 C Angle 120°

+ + + + 4 D 14 D + + + + + Cl + + Cl + + + 5 A 15 B C C + + 6 C 16 C + + + + + Cl+ + Cl + 7 C 17 C + + + +

8 D 18 A SF4 K-shaped/based on trigonal bipyramid/see-saw 9 B 19 A Angle 90° 10 B 20 B Angle 120°

+ + + + + + + Paper 2 + F + + F + + + + + + S + 1 a The left-hand one. [1] + F + + + + + + + F + b +2 and 0 [2] + + [10] c The right-hand one. [1] b sp (C2H2) 4σ∗ 2 a 2 sp (C2Cl4) [2] ∗ 2π [2] c

2p 2p [2]

1π CH3 CH3

Energy 3σ OC OC

∗ 2σ O O [2] 2s 2s 4 a Sigma bonds involve an overlap of orbitals end-on/ 1σ along the molecular axis (it has axial symmetry around O O O A 2 B axis joining the two nuclei) overlap of s orbitals.

b 2s(OA) and 2s(OB) combine to form 1σ and 2σ*. Pi bonds result from overlapping of parallel p orbitals/sideways overlap. 2p (O ) and 2p (O ) combine to form 3σ and 4σ*. [2] z A z B Double bond: a sigma bond and a pi bond. c 2p (O ) and 2p (O ) combine to form one pair x A x B Triple bond: a sigma bond plus two pi bonds. [4] of 1π and 2π*. p orbitals: head-on overlap: sigma bond formation sigma molecular orbital

2py(OA) and 2py(OB) combine to form one pair of 1π and 2π*. [2] d (1σ)2 (2σ*)2 (3σ)2 (1π)4 (2π*)2. [1] s orbitals overlap: sigma bond formation sigma molecular orbital e Paramagnetic. Paramagnetic refers to the magnetic state of an atom, ion or molecule with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons’ p orbitals: sideways overlap: pi bond formation pi molecular orbital magnetic dipole moments. [2] internuclear axis f Two [1]

g If four electrons are added to O2, the bond order will reduced to zero. This will generate two oxide atomic orbitals molecular orbitals − 2− anions: O2 + 4e → 2O . [2]

Chemistry for the IB Diploma Second Edition © Christopher Talbot, Richard Harwood and Christopher Coates 2015 14

b Delocalization: when pi electron pairs are not Paper 2 confined to two adjacent bonding atoms but extend Ø over three or more atoms. [2] 1 a i ΔH f is the standard enthalpy change of formation of a substance. [1] 5 a Lewis electron dot structures for PBr and PBr : 3 5 It is the heat change (absorbed or released + + + + + + + under constant pressure) when a mole of a + Br+ Br+ + + + + compound is formed from its elements in + + + + + + P + + + + + their standard states. [1] + + + + + + Br P Br + + + + B+ Br + + + + + + + + + + + + Ø + Br+ + B + S represents the standard or absolute + + + + entropy. [1] It is related to the disorder or randomness of Molecular shapes for PBr3 and PBr5: particles. [1] Br Related to standard conditions or 298 K Br (or 25°C) and one atmosphere pressure. [1] PBr P Br ii Δ was not included because SØ has absolute Br values; SØ values can be experimentally Br Br Br [4] measured. [1] iii ΔHØ (Cu) = 0 (by definition an element is in b The bond angles present in PBr5 are 90° and 180°, f respectively. its standard state) [1] b i ΔHØ reaction = ΣΔHØ products − ΣΔHØ reactants The bond angle in PBr3 is approximately 107° since f f f a lone pair of electrons (due to its more diffuse [1] nature) produces more repulsion than a bonding Ø ΔH f reaction =([4 × (−242) + (−1084)] − [−2278]) pair, hence the bond angle is reduced below the [1] tetrahedral bond angle of 109° 28’. [4] = +226 kJ mol−1 [1] c Ion [PBr ]+ [1] 4 An endothermic process or reaction needs heat Number of electrons in (5 + 4 − 1) = 8 valency shell of central energy. [1] (adjusting for ionic phosphorus atom charge) ii ΔSØ reaction = ΣΔSØ reactants − ΣΔSØ products Number of electron pairs 4 = (4 × (189) + (150) ) − (305) [1] contributing to basic ΔSØ reaction = 601 J K−1 mol−1 or molecular shape 0.601 kJ K−1 mol−1 [2] Molecular shape Tetrahedral (+) sign/value, thus products more disordered d The hybridizations of the phosphorus in than reactants. [1] phosphorus(iii) bromide and the oxygen in water iii Gibbs free energy (ΔG). [1] are sp3 and sp3, respectively. [2] Units: kJ mol−1 [1] 15 Energetics/thermochemistry c i At 25°C, ΔGØ = ΔHØ − TΔSØ ΔGØ = (+226) kJ mol−1 − (25 + 273) K ×  601  Paper 1  kJ mol−1 K−1 [2] 1000  1 C 8 B ΔGØ = +46.9 kJ mol−1 [1] 2 D 9 D As the value of the Gibbs free energy change, 3 B 10 C ΔGØ, is positive at 25°C, the forward reaction 4 B 11 A is not spontaneous. However, this implies that the backward reaction is spontaneous. 5 A 12 C Hence, at 25°C, CuSO4.5H2O(s) is the 6 B 13 A thermodynamically more stable compound. [2] 7 D 14 C ii When both compounds have equal thermodynamic stabilities, the Gibbs free energy change will be zero, ie, ΔG= 0. ΔGØ= ΔHØ − TΔSØ

 601  0 = (+226 kJ mol−1) − T kJ K−1mol−1 [1] 3 a ΔHØ = [−394 + (−558) − (−1219)] = +267 kJ mol−1 1000  226kJmol−1 ΔSØ = [214 + 70] − [112] = +172 kJ mol−1 T = = 376 K (103°C) [1] 0.601kJKm−−11ol ΔGØ = 267 − (298 × (0.172) = +216 kJ mol−1 [6]

Hence, CuSO4.H2O(s) is more b Non-spontaneous as ΔGØ is positive. [1] thermodynamically stable above 103°C. [1] c ΔGØ = ΔHØ − T SØ; 267/0.172; 1552 K 2 a The lattice enthalpy is the energy needed to break down one mole of ionic solid into gaseous ions The factor TΔSØ predominates and ΔGØ becomes (separated to a infinite distance) under standard negative; TΔSØ must be greater than ΔH for ΔGØ to thermodynamic conditions (1 atm and 25°C). [2] be negative. [4] b The greater the charge on the ions, the higher the value of the lattice enthalpy. Larger charges 16 Chemical kinetics produce greater electrostatic forces of attraction between oppositely charged ions. [1] Paper 1 The smaller the ionic radius, the higher the value 1 C 5 D of the lattice enthalpy. Smaller ions can approach each other more closely and hence experience 2 C 6 D greater electrostatic forces of attraction. [1] 3 C 7 D

c Born-Haber cycle for rubidium oxide, Rb2O: [2] 4 D 2Rb+(g) + O2–(g)

2nd EA(O) Paper 2 2Rb+(g) + 2e– + O(g) 1 a The factor A is indicative of frequency of collisions and the probability that collisions have the proper 1st EA(O) orientation to react. [1] 2 × IE(Rb) 2Rb+(g) + e– + O–(g) b k = exp[(−87.0 × 1000 / 8.31 × 298) + ln(2.10 × 1011)] = 1.2 × 10−4 mol−1 dm3 s−1 2Rb(g) + O(g) These units are chosen as this is an SN2 reaction

∆Hat(O) which implies it is second order. [2] LE(Rb2O) 1 2 a i Change in concentration of reactant/product 2Rb(g) + O2(g) 2 with time or rate of change of concentration. [1] 2 × ∆H (Rb) at ii Concentration; particle size / surface area; light; 1 2Rb(s) + 2 O2(g) pressure. [2] iii (Measuring electrical) conductivity / H (Rb O) ∆ f 2 (measuring) pH. [1] Rb O(s) 2 b i Minimum/least/smallest energy needed (by (2 × 80.9) + 249.2 + (2 × 403.0) − 146.1 + 795.5 + reactants/colliding particles) to react/start/

−ΔHlattice (Rb2O) = −339.0 initiate a reaction [1] −ΔH (Rb O) = −339.0 − (1866.4) = ii lattice 2 T1 −2205.4 kJ mol−1 [3]

−1 ΔHlattice (Rb2O) = −2205.4 kJ mol [2]

d The ionic model is a good model for rubidium oxide T2 – the bonds have very little covalent character. The ionic model is not a good model for silver bromide – the bonds are polar covalent. [2] Probability of molecules Probability e The sodium ion is smaller than that of rubidium. It can therefore approach the oxide ion more (Kinetic) energy closely and exert higher electrostatic forces of attraction. [2]

4 3 a i k = ln 2 / t½ = 0.693 / (1.62 × 10 ) 3 a i There will be no change in pressure. [1] = 4.28 × 10−5 s−1 [1] ii The pressure will decrease. [1] ii ln k = ln A − E /RT = ln 25 − E /(8.31 × 1107) a a b i The temperature will increase. [1]

Therefore: Ea/(8.31 × 1107) = ln 25 − ln k = ln 25 − ln (4.28 × 10−5) ii The methanol concentration will increase. [1] [CHOH] −5 c i K = 3 [1] Ea = (8.31 × 1107)(ln 25 – ln (4.27 × 10 )) c 2 = 122 167 J mol−1 [CO][H2 ] −1 ii Ea = +122. 17 kJ mol [2] CO(g) + 2H (g) ⇋ CH OH(g) 4 a Step 1 since it is the slowest. [1] 2 3 Initial amount: 1.00 2.00 0 b Step 1 is the slowest step, therefore it has a higher Amount at activation energy relative to step 2. [2] equilibrium/

c Rate = k[NO2][F2] [1] mol: 1.00 − 0.85 2.00 − 1.70 0.85 [1] Equilibrium 17 Equilibrium concentration: 0.15/0.45 0.30/0.45 0.85/0.45 [1] = 0.333 mol = 0.667 mol = 1.889 mol [1] Paper 1 1.888 Kc = 2 = 12.7 [1] 1 A 5 D 0.333 × ()0.667 iii Side reactions or leaks in the system or not 2 A 6 C operating under equilibrium conditions or 3 D 7 A operating at a higher temperature or the 4 B 8 B product might be collected before equilibrium is reached. [1] Paper 2 iv No effect on K (it just speeds up the [HI]2 [HI] c 1 a K = or K = [1] reaction). [1] c c 1 1 [H22][I ] 2 2 4 a A; E [2] [H2 ][I]2 b i After 15 s (product) = 0.37 (mol dm−3); 1.50 × 10−2 b Initial [H ] = initial [I ] = 2 2 2 rate = 0.37 mol dm−3 / 15 s = 0.025 mol dm−3 s−1 [3] = 7.5 × 10−3 mol dm−3 ii At equilibrium/rates of forward and reverse H + I ⇋ 2HI 2 2 reactions are equal/ΔG = 0. [1] Initial: 7.5 × 10−3 7.5 × 10−3 0 Equilibrium: (7.5 × 10−3 − x) (7.5 × 10−3 − x) 2x [HI]2 18 Acids and bases Therefore K = c [H ][I ] 22 Paper 1 (2x)2 53 = (7.5 × 10−3 − x)2 1 C 10 A 2 B 11 D Equilibrium concentrations: [H2] = [I2] = 1.62 × 10−3 mol dm−3 3 B 12 A [HI] = 1.18 × 10−2 mol dm−3 4 A 13 C See pages 595 and 596 for examples of this 5 D 14 B type of calculation [4] 6 B 15 C c Hydrogen bonding (dipole–dipole interactions) and London dispersion forces. [1] 7 A 16 D

2 Change Shift Reason [3] 8 C 17 C Increase in Equilibrium Since forward reaction 9 C temperature position shift is exothermic (∆H is to left negative) Increase in Equilibrium Since there are fewer Paper 2 pressure position shift gas molecules on right 1 a i pH = −log [H+(aq)] [1] to right hand side Addition of No change in A catalyst affects the catalyst to equilibrium rate of forward and mixture position reverse reactions equally

ii The curve should include the following: Hence, there is only a very small change in the starting pH = 1; equivalence point: 25.0 cm3 concentrations of HA and A−. The strong acid (H+) of NaOH; pH at equivalence point = 7; pH to is replaced by a weak acid, HA. [5] finish = 12–13. [4] 7.2g b Amount of sodium benzoate = 144 gmol−1 13 = 0.05 mol The calculation assumes that all of the anion

pH 7 concentration is derived from the salt only (as that from the acid is too small to consider). It also assumes 1 that the ionization of the acid is so small it can be ignored. Standard conditions (1 atm and 298 K) are 25 3 also assumed. [3] Volume/cm [H+ (aq)] × 0.05 10−5 mol dm−3 × 6.3 × 10−5 = iii K = 10−4.76 = 1.74 × 10−5 ; 0.03 a [H+(aq)] = 3.78 × 10−5 mol dm−3 K = [H+(aq)]2 ÷ [CH COOH] / 1.74 × 10−5 a 3 pH = −log [H+(aq)] = 4.42 [2] [H+2(aq)] 10 = c A weak acid is partially or incompletely ionized 0.100 or dissociated. A monoprotic acid donates one [H+(aq)] = 1.32 × 10−3 (mol dm−3); hydrogen ion per molecule. [2] Starting pH = 2.88; 3 a Acidic because it is a hydrogen ion donor and pH at equivalence point: 8–9 [5] basic because it is a hydrogen ion acceptor. Such substances are described as amphoteric. [2] b i HIn is a weak acid: b Oxide ion, O2−. [1] HIn ⇋ H+ + In− and two colours indicated: c pH is the negative logarithm to the base ten of the In acid the equilibrium moves to the left or vice + hydrogen ion concentration. pH = −log10[H (aq)]. versa. [3] pH of pure water = 7 [2] ii Phenolphthalein / phenol red / bromothymol + − d i Kw = [H (aq)] × [OH (aq)] [1] blue; the colour change of indicator occurs within the range of pH at the equivalence ii Water is more dissociated or ionized at a higher point / on vertical part of graph. [2] temperature. Le Châtelier’s principle implies that endothermic reactions are favoured by c i Specific examples of weak base and its salt / increases in temperature, hence the dissociation

specific strong acid and weak base, e.g. NH3 or ionization of water is endothermic. [3]

and NH4Cl. [1] ii pH changes very little / most of the acid is 19 Redox processes neutralized by base, for example:

+ + Paper 1 NH3 + H → NH4 or + + NH4OH + H → NH4 + H2O [2] 1 D 8 A d Brønsted–Lowry acid is a proton donor; a Lewis 2 C 9 B acid is an electron pair acceptor; + − 3 B 10 B HI(g) + H2O(l) → H3O (aq) + I (aq); + + N2H4(g) + H (g) → N2H5 (g) 4 C 11 B e It will be acidic due to a hydrolysis reaction 5 D 12 A between the aluminium ion and water molecules; 3+ 2+ + 6 B 13 B [Al(H2O)6] → [Al(H2O)5OH] + H (aq) 7 A 2 a A buffer solution is an aqueous solution that resists a change in pH when a small amount of a strong acid or base is added. Paper 2 A buffer contains a weak acid (HA) and its 1 a Oxidation number of copper on left hand side is 0; conjugate base (A−) or a weak base and its conjugate oxidation number of nitrogen on right hand side acid. is +5. [2] Oxidation number of copper on right hand side is If hydrogen ions, H+(aq), are added, these will +2; oxidation numbers of nitrogen on right hand react with the base of the buffer. The hydrogen side are +5 and +2. [2] ions are removed as they combine with A− and the equilibrium shifts to the left. Chemistry for the IB Diploma Second Edition © Christopher Talbot, Richard Harwood and Christopher Coates 2015 18

3Cu(s) + 8HNO (aq) → 3Cu(NO ) (aq) + 2NO(g) 3 3 2 20 Organic chemistry + 4H2O(l) [1] b i The two electrodes must be dipping into aqueous Paper 1 solutions of their ions. The two electrodes must 1 C 6 A be connected via a salt bridge and an external circuit consisting of wire and a voltmeter. [3] 2 D 7 B ii Hydrogen gas at 1 atm pressure passed over 3 A 8 D a platinum electrode. The temperature of all 4 A 9 C chemicals is 298 K and all solutions have a concentration of 1 mol dm−3. [3] 5 C 10 C c i Ni2+(aq) + 2e− → Ni(s); Mn(s) → Mn2+(aq) + 2e− [2] Paper 2 Manganese is oxidized. [1] 1 a Mechanism showing: arrow from C=C double bond Manganese(ii) ions are the oxidizing agent. [1] towards I; arrow from I–Cl bond to Cl; carbocation ii Manganese is the anode. [1] showing I on first carbon and positive charge on second carbon; arrow from Cl− to carbon with The electron flow is from manganese positive charge and structure of product to nickel. [1] (see page 686 for this type of mechanism). [4] iii EØ = (−0.23 V) + (1.18 V) = +0.95 V [1] cell b 1-Chloro-2-iodobutane / 2-iodo-1-chlorobutane; ΔG is negative. [1] formed via primary carbocation, which is less stable than the secondary carbocation; free electron- d Gold is formed at the cathode releasing alkyl groups / positive charge spread (negative electrode). [1] out less. [3] Oxygen is formed at the anode 2 a Electrophilic substitution [1] (positive electrode). [1] b Mechanism showing: curly arrow from delocalized 2 a Cathode (negative electrode): electrons to Cl+; structure of intermediate showing Cl Cu(s) → Cu2+(aq) + 2e− [1] attached to ring, positive charge and the remaining Anode (positive electrode): four electrons in the benzene ring; curly arrow Cu2+(aq) + 2e− → Cu(s) [1] showing C–H bond fission and electrons moving into benzene ring (see Figure 20.38b, page 689). [3] b There is no change in the intensity of colour since

the rate of removal of copper(ii) ions at the cathode c J: C6H5CH2Cl; K: C6H5CHCl2; L: C6H5CCl3 [3] is balanced by the rate of formation of copper(ii) d Free radical substitution [1] ions at the anode. [1] e Cl → 2Cl•; sunlight / UV light [2] There is no change in the pH since hydrogen and 2

hydroxide ions are not involved in the discharge f C6H5CH2Cl + Cl• → C6H5CHCl• + HCl reactions. [1] C6H5CHCl• + Cl2 → C6H5CHCl2 + Cl• [2] + − c 2H2O(l) → O2(g) + 4H (aq) + 4e [2] g C6H5CCl2• + Cl• → C6H5CCl3 [1] d Charge passed = 0.360 A × 600 s = 216 C [1] 3 a i Show structure with the Hs on the different Amount of electrons = 216 C ÷ 96 500 C mol−1 carbon atoms diagonally opposite each other = 2.24 × 10−3 mol [1] (the trans form). [1] 2 moles of electrons are required for the discharge of ii There is restricted rotation around the axis of one mole of copper atoms. [1] the carbon–carbon double bond, so one form cannot be transformed into the other / the Mass of copper deposited presence of the pi electron cloud makes the = 0.50 × 63.55 g mol−1 × 02.24 × 10−3 mol = 0.071 g [1] molecule rigid. [2] iii Hex-2-ene [1]

b i Electrophilic addition [1] Chapter 21 Measurement and ii CH CH CHBrCH CH CH ; the third carbon 3 2 2 2 3 analysis atom in the chain is a chiral centre as it has four different substituents attached to it (note that the alternative structure with the Br Paper 1 attached to the second carbon atom is also 1 A 4 A possible and also shows optical isomerism). [2] 2 C 5 D iii Mechanism involving carbocation formation (see page 686). [4] 3 B 6 D iv In this case there are two possible secondary carbocations that could be formed and Paper 2 Markovnikov’s rule cannot distinguish between 1 a There is a stationary phase and a mobile phase. [1] them. [2] The components of the mixture dissolve, absorb, 4 a The isomers are: A 1-bromobutane; or partition to differing extents between the two B 2-bromobutane; C 2-bromo-2-methylpropane; phases. [1] D 1-bromo-2-methylpropane. [4] b i A [1] b i C; S 1: Substitution/Nucleophilic/ N ii B [1] unimolecular. [2] iii No, the two ‘oranges’ are different substances ii RBr → R+ + Br−; heterolytic fission to form the since they have different R values. [2] carbocation intermediate. [2] f 9 iv R value = = 0.41 [1] iii A; see Figure 20.3, page 673. [4] f 22 1.6 c Mechanism b(i): no change in rate; mechanism c i Rf value of compound 1 = = 0.13 [1] b(iii): rate doubles. [2] 12.6 9.2 d The rate of reaction with 1-bromobutane is R value of compound 2 = = 0.73 [1] f 12.6 faster than for 1-chlorobutane as the C–Br bond is weaker than the C–Cl bond; Br− is a better Compound 1 has a greater affinity for the thin leaving group than Cl−. [2] layer than compound 2. Since the thin layer is composed of silica gel, compound 1 is more e B shows optical isomerism as the molecule contains polar than compound 2. [1] a carbon atom with four different groups attached. Solutions of the two isomers will rotate the plane ii TLC is faster than paper chromatography. of plane-polarized light in opposite directions and The thin layer may be made from a variety of this is detected in a polarimeter (see page 709). different solids. Hence, a range of mixtures can The polarimeter is set up with water to align the be separated by using different stationary and filters to obtain maximum extinction of light. mobile phases. [1] The solutions of the two isomers of the same 2 a By measuring the difference in time between the concentration are then tested – one will rotate the injection of the sample and the centre of the peak plane to the right, the other to the left by the same for a component. [1] amount. The physical and chemical properties of the two isomers are the same. [5] b The areas under the peaks represent the amounts of the components in the mixture. [1] 5 In each case decide the priority of the different groups attached to the carbon atoms making the carbon– c i Hexan-1-ol [1] carbon double bond ii Ethanol [1] A is the E-isomer as the high priority groups are on d Flow rate of the carrier gas; temperature of the different sides of the molecule (different carbon atoms column; length and diameter of the column; the of the two forming the double bond). nature of column surface and the volatility of the B is the E-isomer as the high priority groups are on solute. [1] different sides of the molecule. e i The two compounds have relatively high C is the Z-isomer as the high priority groups are on the melting points and are involatile. [2] same side of the molecule (attached to the same carbon ii HPLC. [1] atom of the two forming the double bond). [9]

−1 −1 1 3 a 3350 cm due to O–H bond; 2900 cm due methylene (–CH2–) protons. In the high resolution H

to C–H bond. [2] NMR spectrum the methyl group (CH3CH2CH2OH) is a triplet at approximately 1 ppm and the alcohol or Award [1] if both bonds given but not linked to correct hydroxyl group (CH CH CH OH) is a broad singlet at absorption. 3 2 2 approximately 2.3 ppm. Of the two methylene groups, b Comparison of fingerprint regions/comparison with the one next to the highly electronegative oxygen atom library, database or known spectra. [1] of the –OH group is most downfield at approximately 3.6 ppm. This methylene group (CH CH CH OH) is c Same molecular ion/parent ion (M) due to 3 2 2 split into a triplet. The methylene group adjacent to the C H OH+/C H O+. [1] 4 9 4 10 methyl group appears as a sextet. + + d i (m/z = 57) C4H9 / (CH3)3C . + + + OH ii (m/z = 45) C2H5O / CH3CHOH / C2H4OH . [2] OH 2 2 OH CH Penalize missing charge once only in (c) and (d). CH 2 2 2 CH CH CH 2 3 e i Four peaks because four different hydrogen 3 OH CH CH environments/four different types of proton/ CH 3 OWTTE; 6 : 2 : 1 : 1. [2] CH

ii (CH3)3COH/more detailed formula. [1] 6 5 4 3 2 1 0 4 It has four hydrogen environments and there will Chemical shift/ppm be splitting of the methyl group and both sets of [5]