Year 12 Chemistry Work Pack Phase IV 0.Pdf

Year 12 Chemistry Ark Globe Academy Remote Learning Pack Phase IV Monday 8 June- Friday 19 June

Year 12 Chemistry

Session Title Work to be completed Resource Outcome On-Line provided Support 1 Recap of AS Read through content to Work pack IV Complete: Session 1 Recap of Energetics recap AS Energetics and Page(s) 3 – 8 Task 1 AS Energetics (VLE) apply to related recap questions in task 1

2 Definitions Review the table of key Work pack IV Complete: definitions and use these Page(s) 9 – 10 Task 2 to complete task 2. Textbook Page(s): 258 – 260 3 MS Teams Live lesson: Work pack IV Complete: Lesson Enthalpy of Solution Pages(s) 11 – 13 1 - DO NOW Wednesday Calculations Textbook page(s): 2 - Checking for 267 – 268 Understanding Question(s) 3 - Application Question(s) 4 Enthalpy of Complete task 3 Work pack IV Complete: Session 4 Enthalpy Solution questions using Pages(s) 14 Task 3 of Solution Calculations (VLE) Calculations information from Textbook page(s): previous session 267 – 268

5 MS Teams Live lesson: Work pack IV Complete: Lesson Born Haber Cycle Pages(s) 15 – 17 1 - DO NOW Friday Textbook page(s): 2 - Checking for 261 – 266 Understanding Question(s) 3 - Application Question(s) 6 Born Haber Complete task 4 Work pack IV Complete: Session 6 Born- Cycle questions using Pages(s) 18 Task 4 Haber Cycle Calculations (VLE) Calculations information from Textbook page(s): previous session 261 – 266

7 Application Complete a series of PPQs Work pack IV Complete: of New on the theory you have Pages(s) 19 – 22 Past paper exam questions, Theory covered since session 2 Q. 1 to Q. 5

8 MS TEAMS Live lesson: Work pack IV Complete: Lesson Comparing Experimental Pages(s) 23 - 24 1 - DO NOW Wednesday and Theoretical Enthalpy Textbook page(s): 2 - Checking for Values 267 – 268 Understanding Question(s) 3 - Application Question(s)

9 Lattice Complete task 5 Work pack IV Complete: Session 9 Lattice Enthalpy questions using Pages(s) 25 Task 5 Enthalpy Calculations (VLE) Calculations information from Textbook page(s): previous session 267 – 268

10 MS TEAMS Live lesson: Work pack IV Complete: Lesson Work Pack IV Revision Pages(s) 26 1 - DO NOW Friday lesson 2 - Checking for Understanding Question(s) 3 - Application Question(s) Tasks 6 and 7 NOTE: After completing each session, check all task/PPQ answers using answers at the back.

Session 1 - Recap of AS Energetics

1) What is enthalpy?

• What is enthalpy? It is a measure of the heat content of a substance

• Enthalpy change (ΔH) = Change in heat content at constant pressure

• Standard conditions (ΔHº) = 100 kPa and a stated temperature

[Note that the symbol for standard conditions should be a circle with a horizontal line through it, but it is safer here to put a circle with no line as otherwise the character that may appear on the computer the document is from may substitute some random symbol!]

2) Reaction profiles

3) Standard enthalpy change of reaction (ΔrH°) (“enthalpy of reaction”)

This is the enthalpy change for a reaction with the quantities shown in the chemical equation. This means that the value should always be quoted along with the equation.

In this example, the second equation contains half the molar quantities of the first and so the rH value is half as much.

-1 e.g. H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) ΔrH° = –114.2 kJ mol

-1 ½ H2SO4(aq) + NaOH(aq) → ½ Na2SO4(aq) + H2O(l) ΔrH° = –57.1 kJ mol

the value of –114.2 kJ mol-1 in the first equation means that 114.2 kJ of heat energy is released when 1 mole of H2SO4 reacts with 2 moles of NaOH.

the value of –57.1 kJ mol-1 in the second equation means that 57.1 kJ of heat energy is released when ½ mole of H2SO4 reacts with 1 mole of NaOH.

4) Standard enthalpy change of formation (ΔfH°) (“enthalpy of formation”)

Enthalpy change when 1 mole of a substance is formed from its constituent elements with all reactants and products in standard states under standard conditions. e.g.

CH4(g) C(s) + 2 H2(g) → CH4(g)

H2O(l) H2(g) + ½ O2(g) → H2O(l)

NH3(g) ½ N2(g) + 3/2 H2(g) → NH3(g)

C2H5OH(l) 2 C(s) + 3 H2(g) + ½ O2(g) → C2H5OH(l)

CH3Br(l) C(s) + 3/2 H2(g) + ½ Br2(l) → CH3Br(l)

Na2O(s) 2 Na(s) + ½ O2(g) → Na2O(s)

Note: re ΔfHº of an element in its standard state = 0 by definition

5) Standard enthalpy change of combustion (∆cH°) (“enthalpy of combustion”)

Enthalpy change when 1 mole of a substance is completely burned in oxygen with all reactants and products in standard states under standard conditions. e.g.

CH4(g) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

H2(g) H2(g) + ½ O2(g) → H2O(l)

C2H6(g) C2H6(g) + 3½ O2(g) → 2 CO2(g) + 3 H2O(l)

C2H5OH(l) C2H5OH(g) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

Na(s) Na(s) + ¼ O2(g) → ½ Na2O(s)

C6H14(l) C6H14(g) + 9½ O2(g) → 6 CO2(g) + 7 H2O(l)

6) Standard enthalpy change of neutralisation (ΔneutH°) (“enthalpy of neutralisation”)

Enthalpy change when 1 mole of water is formed in a reaction between an acid and alkali under standard conditions. e.g.

HCl(aq) + NaOH(aq) NaOH(aq) → NaCl(aq) + H2O(l)

H2SO4(aq) + NaOH(aq) ½H2SO4(aq) + NaOH(aq) → ½ Na2SO4(aq) + H2O(l)

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

HNO3(aq) + Ba(OH)2(aq) HNO3(aq) + ½Ba(OH)2(aq) → ½ Ba(NO3)2(aq) + H2O(l)

H2SO4(aq) + Ba(OH)2(aq) ½H2SO4(aq) + ½Ba(OH)2(aq) → ½BaSO4(aq) + H2O(l)

Hess’s Law Calculations

The enthalpy change for a reaction is independent of the route taken

A B e.g. the enthalpy change to go from A → B direct is the same as going from A → C → B C

1) Calculations involving enthalpies of formation (“Type 1 questions”)

• If the enthalpy of formation for the reactants and products in a reaction are known, the overall enthalpy change is easy to calculate.

ΔH = [SUM of ΔfH products] – [SUM ΔfH reactants]

• Remember that ΔfH of all elements in their standard states is zero. • Watch for the very frequent mistake of doing reactants – products, rather than products – reactants. • If the overall enthalpy change for a reaction is known along with the enthalpy of formation of all but one of the reactants/products, then this equation can be used to find the missing enthalpy of formation.

2) Calculations involving enthalpies of combustion (“Type 2 questions”)

• Questions that involve enthalpies of combustion can usually be done using the cycle shown. • The reaction involved across the top is often an enthalpy of formation (from elements to a compound). • The sum of the clockwise arrows equals the sum of the anticlockwise arrows.

• Be careful when drawing your cycle to ensure that arrows are going in the right direction and the number of moles is correct. • If you use a cycle like this, there is no need to worry about getting the number of oxygen molecules in the downward arrows.

3) Calculations involving bond enthalpies (“Type 3 questions”)

• Bond enthalpy is the enthalpy change to break one mole of covalent bonds in the gas phase.

• For most bonds (e.g. C-H, C-C, C=O, O-H, etc.) the value for the bond enthalpy is an average taken from a range of molecules as the exact value varies from compound to compound. For some bond enthalpies (e.g. H-H, H-Cl, O=O, etc) the value is exact as only one molecule contains that bond.

• Mean bond enthalpy is the enthalpy change to break one mole of covalent bonds in the gas phase averaged over several compounds.

• Enthalpies of reaction that have been calculated using mean bond enthalpies are not as accurate as they might be because the values used are averages and not the specific ones for that compound.

• The following cycle works for any question that involves bond enthalpies, whether to find a bond enthalpy or ΔH for a reaction.

• Remember that substances must be in the gas state before bonds are broken, and so ΔH to go to the gas state is needed for solids and liquids. (Note - ΔH vaporisation is the enthalpy change to convert a liquid to a gas)

• As with other cycles, the sum of the clockwise arrows equals the sum of the anticlockwise arrows. Be careful to ensure that arrow directions and number of moles are correct.

4) Calorimetry calculations

• The enthalpy change for a reaction can be found by measuring the temperature change in a reaction. • The heat energy given out (or taken in) is used to heat (or cool) a known mass of water. We know that it takes 4.18 J of energy to raise the temperature of 1 g of water by 1°C (i.e. 1 K). • The amount of energy needed to make 1 g of a substance 1°C (1 K) hotter is called the specific heat capacity (measured in J g-1 K-1).

• The following equation is then used to find the amount of heat energy give out (or absorbed).

• To find the enthalpy change in terms of J (or kJ) per mole, the following expression is needed: (THINK kJ per mole!)

• Heat loss is a major problem with calorimetry and can lead to errors in the results. The techniques used in calorimetry are designed to reduce heat loss (one way to reduce errors from heat loss is to measure the heat capacity of the calorimeter as a whole).

TASK 1 – AS Energetics Revision

1 Find ΔfH of butane given that the following data.

-1 ΔcH: C4H10(g) = –2877, C(s) = –394, H2(g) = –286 kJ mol

2 Find ΔH for the following reaction using the data below.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) -1 ΔfH: C3H8(g) = –104, CO2(g) = –394, H2O(l) = –286 kJ mol

3 Find ΔH for the following reaction using the bond enthalpy data below.

C2H6(g) + 3½ O2(g) → 2 CO2(g) + 3 H2O(g) C-C = 348, C-H = 412, O=O = 496, C=O = 743, O-H = 463 kJ mol-1

4 Find ΔcH of propan-2-ol given that the following data.

-1 ΔfH: CH3CH(OH)CH3(l) = –318, ΔcH: C(s) = –394, H2(g) = –286 kJ mol

5 Calculate ΔfH of CCl4(l) given the following data.

-1 -1 CCl4(l) → CCl4(g) = +31 kJ mol C(s) → C(g) = +715 kJ mol Bond enthalpy (Cl-Cl) = +242 kJ mol-1 Bond enthalpy (C-Cl) = +338 kJ mol-1

6 Find ΔH for the hydrogenation of propene using the data below.

CH3CH=CH2(g) + H2(g) → CH3CH2CH3(g) -1 ΔcH: CH3CH=CH2(g) = –2059, H2(g) = –286, CH3CH2CH3(g) = –2220 kJ mol

0.55 g of propanone was burned in a calorimeter containing 80 g of water. The temperature rose by 47.3ºC. 7 -1 -1 Calculate ΔHc for propanone given the specific heat capacity of water is 4.18 J mol K .

8 25.0 cm3 of 2.00 mol dm-3 nitric acid was reacted with 25.0 cm3 of 2.00 mol dm-3 potassium hydroxide is an insulated cup. The temperature rose from 20.2ºC to 33.9ºC. Calculate ΔH for the reaction given the specific heat capacity of water is 4.18 J mol-1 K-1.

9 The engines of the lunar module of Apollo 11 used methylhydrazine (CH3NHNH2) and dinitrogen tetraoxide. They react as follows:

4 CH3NHNH2(l) + 5 N2O4(l) → 4 CO2(g) + 12 H2O(l) + 9 N2(g) Calculate the enthalpy change for the reaction using the following data: ΔfHº: -1 CH3NHNH2(l) = +53, N2O4 (l) = –20, CO2(g) = –393, H2O(l) = –286 kJmol

10 Calculate the average C-C bond enthalpy in benzene (C6H6) given the following data. -1 ΔfH (C6H6(l)) = +49 kJmol C(s) → C(g) ΔH = +715 kJmol-1 -1 H2(g) → 2 H(g) ΔH = +436 kJmol -1 C6H6(l) → C6H6(g) ΔH = +31 kJmol E (C-H) = +413 kJmol-1

9 Session 2 - Definitions

TASK 2 – Enthalpy Change Definitions: Write equations for reactions described below using definitions given in the table on the previous page.

1. ΔfH of C6H6(l) ………………………………………………………………...…………………………..

2. ΔfH of CH3COOH(l) ……………………………………………………………………………………...

3. ΔcH of H2(g) ………………………………………………………………………………………………

4. ΔcH of CH3COOH(l) ……………………………………………………………………………………..

5. 1st ionisation energy of aluminium ……………………………………………………………….……

6. 2nd ionisation energy of aluminium ………………………………………………………………...…

7. 3rd ionisation energy of aluminium ……………………………………………………………………

8. 1st electron affinity of chlorine …………………………………………………………………………

9. lattice enthalpy of formation of sodium oxide ……………………………………………………......

10. lattice enthalpy of dissociation of aluminium oxide …………………………………………………

11. ΔhydH of sodium ions …………………………………………………………………………………..

12. Enthalpy of vaporisation of bromine …………………………………………………………......

13. ΔsolH of sodium hydroxide …………………………………………..………...... ……

14. Enthalpy of fusion of sodium chloride ………………………………………………...... ……

15. Bond dissociation enthalpy of water ………………………………………………...... …

16. Bond dissociation enthalpy of hydrogen …………………………………………...... ……………

17. ΔatmH of bromine ………………………………………………………...... ……

18. Bond dissociation enthalpy of bromine ……………………………………………………...... …

19. 1st electron affinity of bromine ……………………………………………………...... ……

20. 2nd electron affinity of sulphur …………………………………………………………...... …

Session 3 – MS Teams Lesson – Wednesday: Enthalpy of Solution Calculations

The notes in this section will help supplement what you will cover in the MS Teams lesson and allow you to successfully complete the tasks in the sessions to follow.

The enthalpy change of solution is the enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.

Enthalpies of solution may be either positive or negative - in other words, some ionic substances dissolved endothermically (for example, NaCl); others dissolve exothermically (for example NaOH).

An infinitely dilute solution is one where there is a sufficiently large excess of water that adding any more doesn't cause any further heat to be absorbed or evolved.

So, when 1 mole of sodium chloride crystals are dissolved in an excess of water, the enthalpy change of solution is found to be +3.9 kJ mol-1. The change is slightly endothermic, and so the temperature of the solution will be slightly lower than that of the original water.

Thinking about dissolving as an energy cycle

Why is heat sometimes evolved and sometimes absorbed when a substance dissolves in water? To answer that it is useful to think about the various enthalpy changes that are involved in the process.

You can think of an imaginary process where the crystal lattice is first broken up into its separate gaseous ions, and then those ions have water molecules wrapped around them. That is how they exist in the final solution.

The heat energy needed to break up 1 mole of the crystal lattice is the lattice dissociation enthalpy.

The heat energy released when new bonds are made between the ions and water molecules is known as the hydration enthalpy of the ion.

The hydration enthalpy is the enthalpy change when 1 mole of gaseous ions dissolve in sufficient water to give an infinitely dilute solution. Hydration enthalpies are always negative.

Factors affecting the size of hydration enthalpy

Hydration enthalpy is a measure of the energy released when attractions are set up between positive or negative ions and water molecules.

With positive ions, there may only be loose ion-dipole attractions between the δ- oxygen atoms in the water molecules and the positive ions, or there may be formal dative covalent (co-ordinate covalent) bonds.

With negative ions, ion-dipole attractions are formed between the negative ions and the δ+ hydrogens in water molecules.

The size of the hydration enthalpy is governed by the amount of attraction between the ions and the water molecules.

• The attractions are stronger the smaller the ion. For example, hydration enthalpies fall as you go down a group in the Periodic Table. The small lithium ion has by far the highest hydration enthalpy in Group1, and the small fluoride ion has by far the highest hydration enthalpy in Group 7. In both groups, hydration enthalpy falls as the ions get bigger.

• The attractions are stronger the more highly charged the ion. For example, the hydration enthalpies of Group 2 ions (like Mg2+) are much higher than those of Group 1 ions (like Na+).

Estimating enthalpies of solution from lattice enthalpies and hydration enthalpies

The hydration enthalpies for calcium and chloride ions are given by the equations:

The following cycle is for calcium chloride and includes a lattice dissociation enthalpy of +2258 kJ mol-1.

We have to use double the hydration enthalpy of the chloride ion because we are hydrating 2 moles of chloride ions. Make sure you understand exactly how the cycle works.

So . . .

ΔHsol = +2258 - 1650 + 2(-364)

ΔHsol = -120 kJ mol-1

Whether an enthalpy of solution turns out to be negative or positive depends on the relative sizes of the lattice enthalpy and the hydration enthalpies. In this particular case, the negative hydration enthalpies more than made up for the positive lattice dissociation enthalpy.

Session 4 – Enthalpy of Solution Calculations

• This cycle works for questions involving enthalpies of Gas ions solution. It is a simple cycle from a solid to gas ions to dissolved ions.

• Beware of whether the lattice enthalpy is formation or dissociation.

TASK 3 – Enthalpy of solution calculations

1 Calculate the enthalpy of solution of NaCl given that the lattice enthalpy of formation of NaCl is –771 kJmol-1 and the enthalpies of hydration of sodium and chloride ions are –406 and –364 kJmol-1 respectively.

2 Calculate the enthalpy of hydration of bromide ions given that the hydration enthalpy of barium -1 -1 ions is –1360 kJmol , the lattice enthalpy of formation for BaBr2 is –1937 kJmol and the -1 enthalpy of solution of BaBr2 = –38 kJmol .

3 Calculate the lattice enthalpy of formation of calcium iodide given that its enthalpy of solution is –120 kJmol-1 and the enthalpies of hydration of calcium and iodide ions are –1650 and –293 kJmol-1 respectively.

-1 4 Calculate the enthalpy of solution of the ammonium chloride using this data: ΔHhyd (kJ mol ): + - NH4 –301; Cl –364; Lattice enthalpy of dissociation (kJ mol ): ammonium chloride +640 kJ mol .

Session 5 – MS Teams Lesson – Friday: Born Haber Cycle

The notes in this section will help supplement what you will cover in the MS Teams lesson and allow you to successfully complete the tasks in the sessions to follow.

Lattice enthalpy

• Lattice enthalpy represents the enthalpy change when the ions in one mole of a solid ionic compound are broken apart (lattice enthalpy of dissociation) or brought together (lattice enthalpy of formation). • The lattice enthalpy of a compound is an indication of the strength of the ionic bonding – the greater the magnitude of the lattice enthalpy, the stronger the bonding.

• Generally speaking, compounds with smaller ions and/or ions with higher charges have stronger attractions and so greater lattice enthalpy.

e.g. NaCl has a higher lattice enthalpy (and therefore stronger ionic bonding) than KCl as they Na+ ion is smaller than the K+ ion. e.g. MgCl2 has a higher lattice enthalpy (and therefore stronger ionic bonding) than NaCl as the Mg2+ ion has a higher charge and is smaller than the Na+ ion. Measuring lattice enthalpy

• The lattice enthalpy of a compound can be found using a Born-Haber cycle – this value is often called the ‘experimental value’ as the data used in the Born-Haber cycle is determined by experiments.

• A Born-Haber cycle is a cycle that includes all the enthalpy changes in the formation of an ionic compound.

• In these cycles, lattice enthalpy is usually shown as lattice enthalpy of formation (with –ve value) – this is so that an equation can be written where the enthalpy of formation of the ionic compound (not to be confused with the lattice enthalpy of formation) equals the sum of all the other enthalpy changes.

Note that when drawing Born-Haber cycles:

• draw a separate step for every enthalpy change (e.g. for atomisation of the metal atoms separately from the non-metal atoms, for each individual ionisation enthalpy, for each individual electron affinity)

• second and third electron affinities are endothermic and shown be drawn going up not down

• It is best to write the numerical values of the enthalpy changes on each step (you may also be asked to write the names of each step)

Example 1 – Find the lattice enthalpy of formation of calcium oxide using a Born-Haber cycle and these enthalpy changes: 1st ionisation enthalpy of calcium = +590 kJ mol-1 atomisation enthalpy of calcium = +193 kJ mol-1 2nd ionisation enthalpy of calcium = +1150 kJ mol-1 atomisation enthalpy of oxygen = +248 kJ mol-1 1st electron affinity of oxygen = –142 kJ mol-1 enthalpy of formation of calcium oxide = –635 kJ mol-1 2nd electron affinity of oxygen = +844 kJ mol-1

Example 2 – Find the enthalpy of formation of magnesium bromide using a Born-Haber cycle and these enthalpy changes: 1st ionisation enthalpy of magnesium = +736 kJ mol-1 2nd ionisation enthalpy of magnesium = +1450 kJ mol-1 1st electron affinity of bromine = –342 kJ mol-1 atomisation enthalpy of magnesium = +150 kJ mol-1 atomisation enthalpy of bromine = +112 kJ mol-1 lattice enthalpy of formation of magnesium bromide = –2394 kJ mol-1

Session 6 – Born Haber Cycle Calculations

Task 4 – Born Haber Cycle Calculations

Use this data in the questions that follow.

kJmol-1 Na K Ca Al Co Cu Br I O S Cl

enthalpy of atomisation +107 +90 +193 +314 +427 +112 +107 +248 +279 +121

1st ionisation enthalpy +496 +418 +590 +577 +757 +745

2nd ionisation enthalpy +4562 +3070 +1150 +1820 +1640 +1960

3rd ionisation enthalpy +6910 +4600 +4940 +2740 +3230 +3550

1st electron affinity –342 –142 –200 –364

2nd electron affinity +844 +649

1 Calculate the enthalpy of formation of potassium chloride given that the lattice enthalpy of formation of potassium chloride is –710 kJmol-1.

2 Calculate the lattice enthalpy of formation of sodium sulfide given that the enthalpy of formation of sodium sulfide is –370 kJmol-1.

3 Calculate the enthalpy of formation of calcium bromide given that the lattice enthalpy of formation of calcium bromide is –2125 kJmol-1.

4 Calculate the lattice enthalpy of formation of aluminium oxide given that the enthalpy of formation of aluminium oxide is –1669 kJmol-1.

5 Calculate the first electron affinity of iodine given that the lattice enthalpy of dissociation of calcium iodide is +2054 kJmol-1 and its enthalpy of formation is –535 kJmol-1.

6 Calculate the enthalpy of atomisation of copper given that the enthalpy of formation of CuO is – 155 kJmol-1 and its lattice enthalpy of formation is –4149 kJmol-1.

7 The lattice enthalpy of formation of the three possible chlorides of cobalt are given:

-1 CoCl –700; CoCl2 –2624; CoCl3 –5350 kJmol .

a) Using Born-Haber cycles, calculate the enthalpy of formation of each chloride.

b) Which of these chlorides is energetically stable with respect to their elements under standard conditions?

c) Which compound is likely to be formed when cobalt and chlorine react under normal conditions?

______Session 7 – Application of New Theory ______

Q1. (2) Which equation represents the process that occurs when the standard enthalpy of atomisation of iodine is measured? (c) Solubility is the measure of how much of a substance can be dissolved in water to make a saturated solution. A salt solution is saturated when an A undissolved solid is in equilibrium with its aqueous ions.

I2(s) → I(g) Use your answer to part (b) to deduce how the solubility of MgCl2 changes as the temperature is increased. B I2(s) → 2I(g) Explain your answer.

______C ______

I2(g) → I(g) ______D I2(g) → 2I(g) ______(Total 1 mark) ______

(3) Q2. (Total 6 marks) This question is about magnesium chloride.

(a) Write the equation, including state symbols, for the process corresponding Q3. to the enthalpy of solution of magnesium chloride. Thermodynamics can be used to investigate the changes that occur when substances such as calcium fluoride dissolve in water. ______(1) (a) Give the meaning of each of the following terms.

(b) Use these data to calculate the standard enthalpy of solution of magnesium (i) enthalpy of lattice formation for calcium fluoride chloride. ______

–1 ______Enthalpy of lattice dissociation of MgCl2 = +2493 kJ mol Enthalpy of hydration of magnesium ions = –1920 kJ mol–1 ______Enthalpy of hydration of chloride ions = –364 kJ mol–1 ______(2) ______

(ii) enthalpy of hydration for fluoride ions ______Q4. ______This question is about lattice enthalpies. ______(a) The diagram shows a Born–Haber cycle for the formation of magnesium (1) oxide. (b) Explain the interactions between water molecules and fluoride ions when the fluoride ions become hydrated. Complete the diagram by writing the missing symbols on the appropriate energy levels. ______

(2)

Use these data to calculate a value for the enthalpy of solution for CaF2 ______

(2) (Total 7 marks)

(3)

Q5. (b) The table contains some thermodynamic data. (a) Define the term electron affinity for chlorine.

______Enthalpy change / ______kJ mol–1 ______Enthalpy of formation for magnesium oxide –602 ______Enthalpy of atomisation for magnesium +150 ______

First ionisation energy for magnesium +736 (2)

Second ionisation energy for magnesium +1450 (b) Complete this Born−Haber cycle for magnesium chloride by giving the missing species on the dotted lines. Include state symbols where Bond dissociation enthalpy for oxygen +496 appropriate.

First electron affinity for oxygen –142 The energy levels are not drawn to scale.

Second electron affinity for oxygen +844

Calculate a value for the enthalpy of lattice formation for magnesium oxide.

Enthalpy of lattice formation ______kJ mol–1 (3) (Total 6 marks) (6)

Table 1 Table 2

Enthalpy change / kJ Enthalpy change / kJ mol−1

mol−1 Enthalpy of hydration of Mg2+ ions −1920 Enthalpy of atomisation of magnesium +150 Enthalpy of hydration of Na+ ions −406 Enthalpy of atomisation of chlorine +121 Enthalpy of hydration of Cl− ions −364 First ionisation energy of magnesium +736

Second ionisation energy of magnesium +1450 (i) Explain why there is a difference between the hydration enthalpies of the magnesium and sodium ions. ______Enthalpy of formation of magnesium chloride −642 ______Lattice enthalpy of formation of magnesium −2493 ______chloride ______Use your Born−Haber cycle from part (b) and data from Table 1 to calculate a value for the electron affinity of chlorine. (2) ______(ii) Use data from Table 1 and Table 2 to calculate a value for the ______enthalpy change when one mole of magnesium chloride ______dissolves in water. ______(3) ______(2)

(Total 15 marks)

Session 8 – MS Teams Lesson – Wednesday: Comparing Experimental and Theoretical Enthalpy Values

Experimental lattice Theoretical lattice enthalpy (the perfect ionic

enthalpy model) How is it calculated? Using a Born-Haber cycle By a theoretical calculation that considers (all the other H values are the size, charge and arrangement of ions in the found by accurate lattice. It is assumed that the structure is measurement in perfectly ionic. experiments) This is a complex calculation and makes use of the Born-Landé equation (you do not need to know this equation)

Which is the real The real value value?

• There is often some distortion of the ions in an ionic compounds (i.e. they are polarised) – this means that the ions are not perfectly spherical. If there is a lot of distortion then the ions are said to have some covalent character. This does not mean that the compound is covalent – it is still ionic, but the ions are not perfectly spherical.

• Positive ions (cations) that are small and/or highly charged are very good at distorting (i.e. they are very good at polarising) negative ions.

• Negative ions (anions) that are large and/or highly charged are easier to distort (i.e they are polarisable).

• Ionic compounds that have some covalent character often have low solubility in water (or are insoluble) and their melting points and electrical conductivity may not be as high as expected.

• Generally, the bigger the difference between the experimental and theoretical values of lattice enthalpy, the greater the covalent character of an ionic compound. If the difference is small the compound will have almost spherical ions, but if it is significant then the compound has some covalent character due to distorted ions.

Substance NaCl LiCl LiI MgO AgCl Al2O3 Experimental value (from Born- –771 –846 –744 –3513 –905 –15421 Haber cycle) / kJ mol-1 Theoretical value / kJ mol-1 –766 –833 –728 –3477 –770 –14910 Difference / kJ mol-1 5 13 16 36 135 511 % Difference 0.6% 1.5% 2.2% 1.0% 14.9% 3.3%

• For NaCl, the difference is only 0.6% indicating that the bonding is almost pure ionic with very little covalent character at all.

• For LiCl, the difference is 1.5% indicating that the bonding is still very ionic but with a little more covalent character than NaCl, as the Li+ is very smaller than Na+ and so Li+ is better at polarising negative ions.

• For LiI, the difference is 2.2% indicating that the bonding is still very ionic but with a little more covalent character than LiCl, as the I– is bigger than Cl– and so more easily polarised.

• For AgCl, the difference is 14.9% indicating significant covalent character.

Note that: • The magnitude of the lattice enthalpy indicates the overall strength of the ionic bonding.

• The difference between the experimental and theoretical value indicates the amount of covalent character.

Compounds Strongest ionic bonding (the biggest Most covalent character (biggest being compared magnitude of lattice enthalpy) difference between experimental and theoretical value

Al2O3 and MgO Al2O3 Al2O3

Al2O3 and AgCl Al2O3 AgCl

Session 9 – Lattice Enthalpy Calculations

Task 5 – Lattice Enthalpy Calculations

1 Use the data in the table below to answer the questions that follow. The data refers to values of lattice enthalpies.

Substance LiBr NaBr AgBr

-1 Experimental value / kJ mol -800 -733 -890 -1 Theoretical value / kJ mol -787 -732 -758

a) Are these values lattice enthalpies of dissociation or formation? Explain your answer.

b) Which of these compounds has the strongest ionic bonding? How can you tell?

c) Which of these compounds has the most covalent character? How can you tell?

d) Silver bromide is insoluble in water, but sodium bromide is soluble. Suggest why this is the case using this data.

2 The lattice enthalpy of dissociation of calcium oxide calculated using a Born-Haber cycle is 3513 kJ mol-1. The value calculated using the perfect ionic model is 3477 kJ mol-1.

a) Calculate the percentage difference between these two values as a percentage of the actual value.

b) Calcium oxide is insoluble in water. Suggest why.

3 The theoretical value for the lattice enthalpy of formation of silver iodide is –736 kJ mol-1.

a) Construct a Born-Haber cycle to calculate the lattice enthalpy of formation of silver iodide.

-1 1st electron affinity of iodine = –314 kJ mol 1st ionisation enthalpy of silver = +732 kJ mol-1 enthalpy of formation of silver iodide = –62 kJ mol-1 atomisation enthalpy of silver = +289 kJ mol-1 atomisation enthalpy of iodine = +107 kJ mol-1

b) How is the theoretical value for the lattice enthalpy of formation of silver iodide calculated? c) Which is the true value of the lattice enthalpy of formation of silver iodide?

d) Silver iodide is insoluble in water. Explain why this may be the case.

Session 10 – MS Teams Lesson – Friday: Work Pack IV Revision Lesson

TASK 6 – Work pack IV mind-map

In the space below, or on a separate piece of paper, produce a mind-map summarise the new content covered in this work pack.

TASK 7 – Revision Flash Cards

In preparation for your year 13 exams and for our eventual return to school, produce 20 revision flash cards on area of development you have identified, while working through this work pa

WORK PACK IV ANSWERS Task 1

Task 2 Task 3

Task 4

Session 7 – PPQs

Q1. A [1]

Q2. 2+ - (a) MgCl2(s) → Mg (aq) + 2Cl (aq) State symbols essential

Do not allow this equation with H2O on the LHS Ignore + aq on the LHS

Allow H2O written over the arrow / allow equation written as an equilibrium 2+ Allow correct equations to form [Mg(H2O)6] ions 1 2+ – (b) ∆Hsoln MgCl2 = LE + ( ∆HhydMg ) + 2( ∆HhydCl )

∆Hsoln MgCl2 = 2493 – 1920 + (2 × -364)

= –155 (kJ mol-1) M1 for expression in words or with correct numbers Ignore units, but penalise incorrect units 1 1

M2: the enthalpy of solution is exothermic / reaction is exothermic / backwards reaction is endothermic

M3: (According to Le Chatelier) the equilibrium moves to absorb heat/reduce temperature/oppose the increase in temperature (in the endothermic direction)

If M1 is incorrect then CE=0/3 (b) water is polar / H on water is / is electron deficient / is unshielded If answer to (b) is a +ve value, allow: 1 M1: Solubility increases (as temp increases) penalise H+ on water 1 mark M2: Enthalpy of solution is endothermic etc

M3: (According to Le Chatelier) the – equilibrium moves to absorb heat/reduce the (F ions) attract water / on H / hydrogen temperature/oppose the increase in allow H on water forms H-bonds with F– temperature (in the endothermic direction) allow fluorine ions 1 1 penalise co-ordinate bonds for M2 1 penalise attraction to O for M2 [6] 1

(c) ΔH = –(–2611) – 1650 +2x – 506 Q3. ignore cycles (a) (i) (Enthalpy change for formation of) 1 mol (of CaF2) from its ions M1 is for numbers and signs correct in allow heat energy change expression do not allow energy or wrong formula for 1 CaF2 = –51 (kJ mol–1) penalise 1 mol of ions correct answer scores 2 CE=0 if atoms or elements or molecules mentioned ignore units even if incorrect 1 ignore conditions 1 [7]

ions in the gaseous state ions can be mentioned in M1 to score in M2 Q4. allow fluorine ions (a) 2+ – Ca (g) + 2F (g) → CaF2 scores M1 and M2 1

(ii) (enthalpy change when) 1 mol of gaseous (fluoride) ions (is converted) into aqueous ions / an aqueous solution allow F–(g) → F–(aq) (ignore + aq) do not penalise energy instead of enthalpy allow fluorine ions do not allow F– ions surrounded by water 1

ignore standard conditions One mark for each level with correct state 1 symbols 1 (b) Mg2+(g) + 2e− + 2Cl(g) (1) 1 (M5) 1

(b) ∆fH = ∆aH (Mg) + ½ ∆BDH (O2) + ∆1st IEH (Mg) + ∆2nd IEH (Mg) + 1

∆1st EAH (O) + ∆2nd EAH(O) + ∆LEH (MgO)

- 602 = 150 + (½ × 496) + 736 +1450 – 142 + 844 + ∆LEH (MgO) 1

–1 ∆LEH (MgO) = -3888 / -3890 (kJ mol ) Allow answers to 2sf or more 1 mark for +3888 or +3890 1 mark for -4136 or -4140 (not 496 × ½) 1 [6] Allow e for electrons (i.e. no charge) State symbols essential If no electrons allow M5 but not M3,M4 Q5. If incorrect 1 / 2 Cl2 used allow M3 and M4 (a) The enthalpy change / heat energy change / ΔH for the formation of one mole for correct electrons (scores 2 / 6) of (chloride) ions from (chlorine) atoms 6 Allow enthalpy change for Cl + e− → Cl− st nd Do not allow energy change (c) −ΔHf(MgCl2) + ΔHa(Mg) + 1 IE(Mg) + 2 IE(Mg) +2ΔHa(Cl)= −2EA(Cl) ionisation energy description is CE=0 − LE(MgCl2) Allow enthalpy change for the addition of 1 Allow Enthalpy of Formation = sum of other mol of electrons to Chlorine atoms enthalpy changes (incl lattice formation) 1 penalise Cl2 and chlorine molecules CE = 0 allow chlorine ions −2EA(Cl) = 642 + 150 + 736 + 1450 + 242 − 2493 = 727 1 1

Atoms and ions in the gaseous state EA(Cl) = −364 (kJ mol−1 ) Or state symbols in equation Allow −363 to −364 Cannot score M2 unless M1 scored Allow M1 and M2 for −727 except allow M2 if energy change rather Allow 1 (1 out of 3) for +364 or +363 but than enthalpy change award 2 if due to arithmetic error after

correct M2 Also allow 1 for −303 TASK 5 Units not essential but penalise incorrect units 1 Look for a transcription error and mark as a) formation – as energy is released when electrostatic attractions between ions form AE−1 b) silver bromide – greatest magnitude of lattice enthalpy (using real values, i.e. the 1 experimental values) c) silver bromide – greatest difference between experimental and theoretical lattice enthalpy (d) (i) Magnesium (ion) is smaller and more charged (than the sodium values ion) d) silver bromide has significant covalent character OR magnesium (ion) has higher charge to size ratio / charge 2 density a) 100 x 36/3513 = 1.02% Do not allow wrong charge on ion if given b) high lattice enthalpy (NB – not due to covalent character as it is only small) Do not allow similar size for M1 Do not allow mass / charge ratio 3 a) 1

(magnesium ion) attracts water more strongly . Mark independently Mention of intermolecular forces, (magnesium) atoms or atomic radius CE = 0 1

(ii) Enthalpy change = −LE(MgCl2) + Σ(ΔHhydions)

= 2493 + (−1920 + 2 × −364) 1

= −155 (kJ mol−1) Units not essential but penalise incorrect units 1

[15]

b) using the charge and size of ions assuming that the structure is perfectly ionic

c) –876 kJ mol-1 d) significant covalent character