Lattice Enthalpy OCR Past Paper Questions

Question 1

QuestionsAnswers

1. The lattice enthalpy of calcium chloride can be calculated using three of the enthalpy changes below.

Which enthalpy change is not required?

enthalpy change of solution of calcium chloride
enthalpy change of hydration of Cl⁻ ions
enthalpy change of formation of calcium chloride
enthalpy change of hydration of Ca²⁺ ions

Your answer

[1]

2(a). Lattice enthalpies give an indication of the strength of ionic bonding.

How would the lattice enthalpies of magnesium chloride and calcium chloride differ?

Explain your answer.

[3]

(b). The table below shows the enthalpy changes that are needed to determine the lattice enthalpy of magnesium chloride, MgCl₂.

Letter	Enthalpy change	Energy / kJ mol⁻¹
A	1st electron affinity of chlorine	−349
B	1st ionisation energy of magnesium	+736
C	atomisation of chlorine	+150
D	formation of magnesium chloride	−642
E	atomisation of magnesium	+76
F	2nd ionisation energy of magnesium	+1450
G	lattice enthalpy of magnesium chloride

On the cycle below, write the correct letter in each box.

[3]

Use the Born–Haber cycle to calculate the lattice enthalpy of magnesium chloride.

[2]

1			C	1
			Total	1
2	a		Lattice enthalpy of MgCl₂ is more exothermic than CaCl₂ … (1) because magnesium ion / Mg²⁺ is smaller (than calcium ions / Ca²⁺) OR Mg²⁺ has a greater charge density … (1) … therefore the attraction between Mg²⁺ and Cl⁻ is greater (than between Ca²⁺ and Cl⁻) (1)	3	ora throughout allow ‘charge density’ here only allow magnesium / Mg is smaller do not allow Mg²⁺has a smaller atomic radius do not allow chlorine ions do not allow Mg has greater attraction allow ‘attracts with more force’ for greater attraction but do not allow ‘greater force’ (could be repulsion)
	b	i	F B G E D FIVE correct (3) FOUR correct (2) THREE correct (1)	3	allow 1450 736 G 76 −642 if only one or two correct, award 0 marks.
		ii	−642 − (+76 + (2 × 150) + 736 + 1450 + (2 × −349)) (1) −642 − 1864 = −2506 (1) (kJ mol⁻¹)	2	allow for 1 mark: −2705 (2 × 150 and 2 × 349 not used for Cl) −2356 (2 × 150 not used for Cl) −2855 (2 × 349 not used for Cl) +2506 (wrong sign) do not allow any other answers
			Total	8

Question 3

QuestionsAnswers

3(a). Enthalpy changes of solution can be determined both indirectly from other enthalpy changes, and directly from the results of experiments.

The table below shows the enthalpy changes that can be used to determine the enthalpy change of solution of calcium chloride, CaCl₂, indirectly.

Enthalpy change	Energy / kJ mol⁻¹
Hydration of calcium ions	−1616
Hydration of chloride ions	−359
Lattice enthalpy of calcium chloride	−2192

Explain what is meant by the term enthalpy change of solution.

[1]

(b). The diagram below shows an incomplete energy cycle that can be used to determine the enthalpy change of solution, Δ_solH, of CaCl₂.

On the three dotted lines, add the species present, including state symbols.

[3]

Calculate the enthalpy change of solution of CaCl₂.

[2]

The table shows enthalpy changes of hydration.

Ion	Enthalpy change of hydration / kJ mol⁻¹
aluminium ion	−4741
calcium ion	−1616
magnesium ion	−1963
sodium ion	−424

Explain the differences between these enthalpy changes of hydration.

[3]

(c). Student 1 carries out an experiment to determine the enthalpy change of solution, Δ_solH, of CaCl₂ directly.

The student follows the method outlined below.

Weigh an empty polystyrene cup and weigh the bottle containing CaCl₂.
Add about 50 cm³ of water to the cup and measure the temperature of the water.
Add the CaCl₂ to the cup, stir the mixture, and record the maximum temperature.
Weigh the polystyrene cup + final solution, and weigh the empty bottle.

Results

Mass of bottle + CaCl₂	28.38 g
Mass of empty bottle	22.82 g
Mass of polystyrene cup + final solution	85.67 g
Mass of polystyrene cup	35.46 g
Initial temperature of water	22.0 °C
Final temperature of solution	53.5 °C

Calculate Δ_solH, in kJ mol⁻¹, for calcium chloride.

Give your answer to an appropriate number of significant figures.

Assume that the density and specific heat capacity, c, of the solution have the same values as water.

[4]

Student 2 carries out the same experiment but uses twice the mass of CaCl₂. All other quantities are very similar to Student 1‘s experiment.

Predict any differences between the temperature change and the calculated value of Δ_solH from the experiments of the two students. Explain your reasoning.

[2]

Question 4

QuestionsAnswers

4(a). Born—Haber cycles can be used to calculate enthalpy changes indirectly.

The table below shows enthalpy changes for a Born—Haber cycle involving potassium sulfide, K₂S.

	Enthalpy change / kJ mol⁻¹
Formation of potassium sulfide, K₂S	−381
1st electron affinity of sulfur	−200
2nd electron affinity of sulfur	+640
Atomisation of sulfur	+279
1st ionisation energy of potassium	+419
Atomisation of potassium	+89

The incomplete Born—Haber cycle below can be used to determine the lattice enthalpy of potassium sulfide.

In the boxes, write the species present at each stage in the cycle.
Include state symbols for the species.

[3]

Define, in words, the term lattice enthalpy.

[2]

Using the Born—Haber cycle, calculate the lattice enthalpy of potassium sulfide.

[2]

(b). Several ionic radii are shown below.

Ion	Na⁺	K⁺	Rb⁺	Cl⁺	Br⁻	I⁻
Radius / pm	95	133	148	181	195	216

Predict the order of melting points for NaBr, KI and RbCl from lowest to highest.

Explain your answer.

[3]

Question 5

QuestionsAnswers

5. Iron(II) iodide, FeI₂, is formed when iron metal reacts with iodine.

The table below shows enthalpy changes involving iron, iodine and iron(II) iodide.

	Enthalpy change / kJ mol⁻¹
Formation of iron(II) iodide	−113
1st electron affinity of iodine	−295
1st ionisation energy of iron	+759
2nd ionisation energy of iron	+1561
Atomisation of iodine	+107
Atomisation of iron	+416

The incomplete Born−Haber cycle below can be used to determine the lattice enthalpy of iron(II) iodide.

In the boxes, write the species present at each stage in the cycle.

Include state symbols for the species.

[4]

Define the term lattice enthalpy.

[2]

Calculate the lattice enthalpy of iron(II) iodide.

[2]

Question 6

QuestionsAnswers

6(a). This question is about four enthalpy changes, A–D, that can be linked to the dissolving of potassium sulfate, K₂SO₄, in water.

	Name of enthalpy change	Enthalpy change / kJ mol⁻¹
A	lattice enthalpy of potassium sulfate	−1763
B	enthalpy change of solution of potassium sulfate	+24
C	enthalpy change of hydration of potassium ions	−320
D	enthalpy change of hydration of sulfate ions

Table 3.1

Define the term enthalpy change of hydration.

[2]

(b). The diagram below is an incomplete energy cycle linking the four enthalpy changes in Table 3.1. One of the four energy levels is missing.

Include state symbols for all species.

Complete the energy cycle as follows.
- Add the missing energy level to the diagram. Add the species on all four energy levels.
- Add arrows to show the direction of the three missing enthalpy changes. Label these enthalpy changes using the letters B–D from Table 3.1.

[5]

Calculate the enthalpy change of hydration of sulfate ions.

[1]

(c). The entropy change of solution of K₂SO₄ is +225 J K⁻¹ mol⁻¹.

Suggest, in terms of the states of the particles involved, why this entropy change is positive.

[1]

Explain, using a calculation, why K₂SO₄ dissolves in water at 25 °C, despite the enthalpy change of solution being endothermic.

[3]

7. Which equation matches the enthalpy change of atomisation of iodine?

A	I₂(g) → 2I(g)
B	½I(g) → I(g)
C	I₂(s) → 2I(g)
D	½I₂(s) → I(g)

Your answer

[1]

Question 8

QuestionsAnswers

8. A student carries out an experiment to find the enthalpy change of solution, Δ_solH, of sulfuric acid using the following method.

1. A plastic cup is weighed.

2. Approximately 100 cm³ of distilled water is added to the cup.

3. The temperature of the water in the plastic cup is measured.

4. A bottle containing concentrated sulfuric acid is weighed.

5. The sulfuric acid is poured into the plastic cup. The solution formed is stirred with the thermometer.

6. The maximum temperature reached by the solution is recorded.

7. The plastic cup containing the solution is weighed.

8. The empty bottle is weighed.

The student’s results are shown in the table below:

Mass readings

Mass of bottle + H₂SO₄/g: 25.66
Mass of empty bottle/g: 14.38

Mass of plastic cup/g: 8.74
Mass of plastic cup + solution formed/g: 122.16

Temperature readings

Maximum temperature reached by solution/°C: 32.0
Initial temperature of distilled water/°C: 21.5

Use the student’s results to calculate the enthalpy change of solution of sulfuric acid, in kJ mol⁻¹.

Assume that the specific heat capacity, c, of the solution is the same as for water.

Give your answer to an appropriate number of significant figures.

[4]

The student’s thermometer has a maximum error of ±0.5 °C.

Calculate the percentage uncertainty in the student’s temperature change.

Give your answer to one decimal place.

[1]

The student carries out a second experiment using 150 cm³ of distilled water instead of 100 cm³ of distilled water. The mass of concentrated sulfuric acid is the same as in the first experiment.

Predict and explain the effect, if any, of the larger volume of water on the following:
The temperature change, ΔT
The calculated value of Δ_solH for H₂SO₄.

[4]

8	i	FIRST, CHECK THE ANSWER ON ANSWER LINE IF Δ_solH = −43.3 (kJ mol⁻¹ ) award 4 marks ________________________________ Energy released in J OR kJ =113.42 × 4.18 × 10.5 = 4978 (J) OR 4.978 (kJ) ✓ Correctly calculates n(H₂SO₄) = 0.115 (mol) ✓ ΔH value in J OR kJ Answer *MUST* divide energy by n(H₂SO₄) OR (−)43286 (J) OR OR (−)43.3 (kJ) ✓ (Sign ignored and/or more than 3 SF) Correct Δ_solH in kJ AND − sign AND 3 SF = −43.3 (kJ mol⁻¹) ✓	4	FULL ANNOTATIONS MUST BE USED _______________________________ Calculator: 4978.0038 DO NOT ALLOW less than 3 SF IGNORE units ALLOW correctly calculated number in J OR kJ Calculator 0.1149847095 ALLOW ECF from n(H₂SO₄) AND/OR Energy Calculator from 4978 and 0.115 = 43286.95652 From unrounded values, = 43292.74581 IGNORE absence of − sign and 3 SF requirement Final mark requires − sign, kJ AND 3 SF NOTE: Use of 100 for m → 4389 J ECF available for → −38.2 kJ mol⁻¹ (3 marks)
	ii	× 100 × 2 = 9.5% ✓ One decimal place required	1
	iii	Predictions ΔT is less AND Δ_solH is the same ✓ Reason for ΔT less (same) energy/heat spread over larger volume (of water) ✓ ΔT = 7°C ✓ Reason for Δ_solH same Same energy released per mole of H₂SO₄ ✓	4	ALLOW heat spread over more water ALLOW 6−8 °C Note: m is ∼ 1/3 larger. q = mcΔT and so ΔT will be ∼ 1/3 smaller ALLOW Δ_solH is for dissolving 1 mol
		Total	9

Question 9

QuestionsAnswers

9. Two changes are described below.

For each change,

•	write an equation, including state symbols,
•	state and explain how the entropy changes.

The reaction of aqueous barium nitrate with aqueous sodium sulfate.

[2]

The change that accompanies the standard enthalpy change of atomisation of iodine.

[2]

10. Which enthalpy change(s) is/are endothermic?

1	The bond enthalpy of the C–H bond
2	The second electron affinity of oxygen
3	The standard enthalpy change of formation of magnesium

A	1, 2 and 3
B	Only 1 and 2
C	Only 2 and 3
D	Only 1

Your answer

[1]

9	i	Equation Ba(NO₃)₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaNO₃(aq) ✓ Entropy change *and* explanation entropy decreases OR entropy change negative AND (BaSO₄) solid / ppt has less disorder / more order / fewer ways of arranging energy / less freedom / less random particles / dispersal of energy ✓	2	ALLOW multiples M2 is dependent on BaSO₄(s) (even if formula is incorrect – eg Ba(SO₄)₂(s)) seen as a product in the attempted equation as long as reactants are not solid. BaSO₄ solid / ppt may be assumed from BaSO₄(s) seen in the attempted equation. Examiner’s Comments Candidates who correctly identified barium sulfate as a solid product tended to realise that entropy had decreased, although a significant number failed to state that this decrease in entropy was as a result of less disorder being created.
	ii	Equation ½ I₂(s) → I(g) ✓ state symbols required Entropy change *and* explanation entropy increases OR entropy change positive AND gas has more disorder / less order / more ways of arranging energy / more freedom / more random particles / more dispersal of energy ✓	2	DO NOT ALLOW I₂(s) → 2I(g) DEPENDENT on ½I₂(s) → I(g) OR I₂(s) → 2I(g) Examiner’s Comments Most candidates failed to produce a correct equation for the standard enthalpy change of atomisation of iodine. Of those who were able to produce the correct equation, a significant number failed to state that the increase in entropy was as a result of increased disorder being created.
		Total	4
10		B	1	Examiner’s Comments D was the common distractor given as the answer by many candidates, suggesting confusion with the first electron affinity and second electron affinity of oxygen.
		Total	1

Question 11

QuestionsAnswers

11. This question is about enthalpy changes.

Table 16.1 shows enthalpy changes that can be used to determine the enthalpy change of hydration of fluoride ions, F⁻.

Enthalpy change	Energy / kJ mol⁻¹
Hydration of Ca²⁺	−1609
Solution of CaF₂	+13
Lattice enthalpy of CaF₂	−2630

Explain what is meant by the term enthalpy change of hydration.

[2]

The enthalpy change of hydration of F⁻ can be determined using the enthalpy changes in Table 16.1 and the incomplete energy cycle below.
On the dotted lines, add the species present, including state symbols.

[4]

Calculate the enthalpy change of hydration of fluoride ions, F⁻.

[2]

Predict how the enthalpy changes of hydration of F⁻ and Cl⁻ would differ.

Explain your answer.

[2]

Question 12

QuestionsAnswers

12. This question is about copper(II) sulfate, CuSO₄, and sodium thiosulfate, Na₂S₂O₃.

The enthalpy change of reaction, Δ_rH, for converting anhydrous copper(II) sulfate to hydrated copper(II) sulfate is difficult to measure directly by experiment.

The enthalpy changes of solution of anhydrous and hydrated copper(II) sulfate can be measured by experiment. The reactions are shown below.

In the equations, ‘aq’ represents an excess of water.

Experiment 1
A student carries out an experiment to find Δ_solH(CuSO₄(s)) for reaction 5.2.

Student’s method

•	Weigh a bottle containing CuSO₄(s) and weigh a polystyrene cup.
•	Add about 50 cm³ of water to the polystyrene cup and measure its temperature.
•	Add the CuSO₄(s), stir the mixture, and measure the final temperature.
•	Weigh the empty bottle and weigh the polystyrene cup with final solution.

Mass readings

Mass of bottle + CuSO₄(s) / g	28.04
Mass of empty bottle / g	20.06
Mass of polystyrene cup / g	23.43
Mass of polystyrene cup + final solution / g	74.13

Temperature readings

Initial temperature of water / °C	20.5
Temperature of final solution / °C	34.0

Experiment 2
The student carries out a second experiment with CuSO₄•5H₂O (reaction 5.3). The student uses the same method as in Experiment 1.

The student calculates Δ_solH(CuSO₄•5H₂O(s)) as +8.43 kJ mol⁻¹.

*Calculate Δ_solH(CuSO₄(s)) for reaction 5.2 and determine the enthalpy change of reaction 5.1, Δ_rH.

Assume that the specific heat capacity, c, of the solution is the same as for water.

Show your working, including an energy cycle linking the enthalpy changes.

[6]

The thermometer had an uncertainty in each temperature reading of ± 0.1 °C.

The student calculates a 20% uncertainty in the temperature change in Experiment 2.

Calculate the temperature change in Experiment 2.

[1]

12	i	Please refer to the marking instructions on page 4 of this mark scheme for guidance on how to mark this question. Level 3 (5–6 marks) Calculates CORRECT enthalpy change with correct – signs for Δ_solH (CuSO⁴(s)) for reaction 5.2 AND Δ_rH, for reaction 5.1. There is a well-developed line of reasoning which is clear and logically structured. The information presented is relevant and substantiated. Level 2 (3–4 marks) Calculates a value of Δ_solH (CuSO₄(s)) for reaction 5.2 from the: Energy change AND Amount in mol of CuSO₄. There is a line of reasoning presented with some structure. The information presented is relevant and supported by some evidence. Level 1 (1–2 marks) Processes experimental data to obtain the: Energy change from mcΔT OR Amount in mol of CuSO₄. There is an attempt at a logical structure with a line of reasoning. The information is in the most part relevant. 0 marks – No response or no response worthy of credit	6	Indicative scientific points may include: 1. Processing experimental data Energy change from mcΔT Energy in J OR kJ Using 50.70 g, 50.0 g = 50.70 × 4.18 × 13.5 = 2861 (J) OR 2.861 (kJ) 3SF or more (2.861001 unrounded) OR 50.0 × 4.18 × 13.5 = 2821.5 (J) OR 2.8215 (kJ) ___________________________________ Amount in mol of CuSO₄ • ———————————————- 2. ± value of Δ_solH(CuSO₄(s)) for reaction 5.2 From m = 50.70 g = = ±57.22 (kJ mol^–1) (–57.22002 unrounded) From m = 50.0 g = = ±56.43 (kJ mol^–1) ———————————————- 3. CORRECT enthalpy changes for reactions 5.2 and 5.1 with signs (using 50.70 g ONLY) Reaction 5.2 = –57.22 (kJ mol^–1) 3SF or more with correct – sign Reaction 5.1 Δ_rH = Δ_solH(CuSO₄(s)) – Δ_solH(CuSO₄•5H₂O(s)) = –57.22 – 8.43 = –65.65 (kJ mol^–1) 3SF or more with correct – sign NOTE: A clear and logically structured response would include an energy cycle ALLOW omission of trailing zeroes ALLOW minor slips Examiner’s Comments This question was assessed by level of response (LoR). Candidates were required to process raw experimental results to determine one enthalpy change, and then to determine a second enthalpy change by constructing and using an energy cycle. Levels were determined by the accuracy of the candidates’ processing, calculations and use of the energy cycle. This question discriminated extremely well. Level 3 candidates used the mass of solution as 50.7 g with mcΔT to obtain an energy change of 2861 J. They then divided this value by the moles of CuSO₄ that reacted (0.05 mol) to obtain the enthalpy change of –57.22 kJ mol^–1. Finally, they constructed an energy cycle which they then used to obtain the second enthalpy change of –65.65 kJ mol^–1. Level 2 candidates determined the first enthalpy change but may have used the approximate mass of 50 g for the mass of solution from the experimental method. Their energy cycle was often incorrect or absent, with the second enthalpy change calculated incorrectly. Level 1 candidates often calculated the initial energy change using mcΔT but made little further correct progress. A significant number of lower ability candidates used the solid mass of copper sulfate in their calculation. Overall, mathematical skills were displayed well but some basic errors were made, particularly with subtractions. This may have been the result of mis-keying values into a calculator and believing the answer displayed. An example was the mass of the solution (74.13 – 23.43) being seen as 49.7 rather than 50.7. A significant number of candidates added or subtracted the mass of copper sulfate from 50.7 or 50 for their value of m in mcΔT, using for example 57.98 and 42.72. This limited the level that they could reach. Exemplar 4 is an excellent response from a Level 3 candidate. All stages can be clearly followed, the initial energy change using mcΔT, the first enthalpy change using the energy change with the moles of copper sulfate, and the second enthalpy change including a clear energy cycle. Numbers are clearly shown and unrounded until the final values. This is an excellent response. Exemplar 4
	ii	Temperature change = =1(.0)ºC ✓	1	IGNORE direction of temperature change Working NOT required Examiner’s Comments Just over half the candidates obtained the correct temperature change of 1°C. A common error was a temperature change of 0.5°C, the result of not considering that two temperature readings are made when calculating a temperature change.
		Total	7

Question 13

QuestionsAnswers

13(a). The table below shows enthalpy changes involving potassium, oxygen and potassium oxide, K₂O.

	Enthalpy change / kJ mol⁻¹
formation of potassium oxide	–363
1st electron affinity of oxygen	–141
2nd electron affinity of oxygen	+790
1st ionisation energy of potassium	+419
atomisation of oxygen	+249
atomisation of potassium	+89

The incomplete Born–Haber cycle below can be used to determine the lattice enthalpy of potassium oxide.

In the boxes, complete the species present in the cycle.
Include state symbols for the species.

[4]

Calculate the lattice enthalpy of potassium oxide.

[2]

(b). Sir Humphry Davy discovered several elements including sodium, potassium, magnesium, calcium and strontium.

A similar Born–Haber cycle to potassium oxide in the part above can be constructed for sodium oxide.

The first ionisation energy of sodium is more endothermic than that of potassium.

Explain why.

[2]

The lattice enthalpy of sodium oxide is more exothermic than that of potassium oxide.

Explain why.

[2]

Question 14

QuestionsAnswers

14. This question is about magnesium, bromine and magnesium bromide.

The enthalpy change of hydration of bromide ions can be determined using the enthalpy changes in Table 16.2.

Enthalpy change	Energy / kJ mol⁻¹
1st ionisation energy of magnesium	+736
2nd ionisation energy of magnesium	+1450
atomisation of bromine	+112
atomisation of magnesium	+148
electron affinity of bromine	−325
formation of magnesium bromide	−525
hydration of bromide ion	to be calculated
hydration of magnesium ion	−1926
solution of magnesium bromide	−186

An incomplete energy cycle based on Table 16.2 is shown below.

On the dotted lines, add the species present, including state symbols.

[2]

Using your completed energy cycle in (i), calculate the enthalpy change of hydration of bromide ions.

[2]

Write the equation for the lattice enthalpy of magnesium bromide and calculate the lattice enthalpy of magnesium bromide.

[3]

14	i	Mg²⁺(g) + 2Br(g) + 2e^– ✓ Mg(s) + Br₂(l) ✓	2	State symbols required. CARE: Liquid state symbol for Br₂ Examiner’s Comments Many candidates got the second ionisation energy equation. Very few candidates got the correct state symbol on the lower line for Br₂, with solid being the common response.
	ii	FIRST CHECK THE ANSWER ON ANSWER LINE If answer = –346.5 award 2 marks ————————————————– 2ΔH hyd = – 525 – 186 – (2 x 112) – 148 – 736 – 1450 + (2 x -325) + 1926 OR – 525 – 186 – 224 – 148 – 736 – 1450 + 650 + 1926 OR = – 693 ✓ ΔH hyd = –346.5 (kJ mol^–1) ✓	2	ALLOW -347 (kJ mol^–1) for 2 marks. ALLOW for 1 mark ONE error with sign OR use of 2: –693 (not divided by 2 at the end) 346.5 (wrong sign on answer) Common errors for 1 mark –2272.5 (–1926 instead of 1926) -1386 (2 x -693 instead of -693) –996.5 (–650 instead of 650) –509 (2 × 325 not used) –290.5 (2 × 112 not used) –198.5 (148 instead of –148) –160.5 (186 instead of –186) –122.5 (224 instead of –224) 178.5 (525 instead of –525) 389.5 (736 instead of –736) 1103.5 (1450 instead of –1450) For other answers, check for a single transcription error or calculation error which could merit 1 mark DO NOT ALLOW any answer which involves two errors e.g. –453 (2 × 325 not used AND 2 x 112 not used)
	iii	Equation: Mg²⁺(g) + 2Br^–(g) → MgBr₂(s) ✓ CHECK THE ANSWER ON ANSWER LINE If answer = –2433 award 2 marks ———————————————– Lattice enthalpy = Δ_hyH(Mg²⁺) + 2 × Δ_hyH(Br^–) – Δ_solH(MgBr₂) OR -1926 + (2 x -346.5) – (-186) OR Δ_fH(MgBr₂) – 2Δ_atH(Br) – Δ_atH(Mg) – 1st IE(Mg) – 2nd IE(Mg) – 2Δ_eaH(Br) OR -525 – (2 x 112) – 148 – 736 – 1450 – (2 x -325) ✓ Lattice enthalpy = –2433 kJ mol^–1 ✓	3	State symbols required For other answers, check for a single transcription error or calculation error which could merit 1 mark DO NOT ALLOW any answer which involves two errors ALLOW ECF from incorrect answer to (ii) Examiner’s Comments This Born Haber cycle combined the traditional cycle with the enthalpy solution/hydration cycle. Many candidates were successful in calculating the values from the information given and showed full working to complete the calculation. The commonest errors were not doubling the atomisation and electron affinity values (112 and 325) and not dividing the enthalpy of hydration by two. Some candidates did not show working or just listed numbers. Candidates should remember to provide written indications of what it is they are working out – presenting the calculations without any annotations can make it harder for error carried forward marks to be given if there is an error in their calculation. Some candidates did not produce an equation for (iii) and wrote the lattice enthalpy sum in its place.
		Total	7

Question 15

QuestionsAnswers

15. This question is about magnesium and magnesium halides.

The enthalpy change of solution for magnesium fluoride, MgF₂, can be determined indirectly using an energy cycle based on the enthalpy changes below.

Enthalpy change	Energy / kJ mol^–1
Lattice enthalpy of magnesium fluoride	–2926
Hydration of magnesium ions	–1920
Hydration of fluoride ions	–506

Explain what is meant by enthalpy change of solution.

[1]

On the dotted lines, add the species present, including state symbols.

[2]

Calculate the enthalpy change of solution of MgF₂.

[1]

The enthalpy changes of solution of the magnesium halides show a trend from MgF₂ to MgI₂.

Explain why it is difficult to predict whether the enthalpy change of solution becomes more exothermic or less exothermic down the group from MgF₂ to MgI₂.

[4]

15	i	(enthalpy change for) 1 mole of a compound/substance/solid/solute dissolving ✓	1	IGNORE ‘energy released’ OR ‘energy required’ For dissolving, ALLOW forms aqueous/hydrated ions IGNORE ionic OR covalent DO NOT ALLOW dissolving elements DO NOT ALLOW response that implies formation of 1 mole of aqueous ions
	ii	Mg²⁺(aq) + 2F =(g) ✓ Mg²⁺(aq) + 2F=(aq) ✓	2	ALLOW Mg²⁺(g) + 2F^–(aq) ALLOW MgF₂(aq)
	iii	–6 (kJ mol^–1) ✓ Δ_solH (MgF₂) = – (–2926) + (2 × –506) + (–1920)	1	1 mark ONLY Examiner’s Comments This was an enthalpy solution/hydration cycle. Many candidates were successful in calculating the value from the information given.
	iv	Ionic radius Halide ion gets larger down the group ✓ Lattice enthalpy Lattice enthalpy is less exothermic down group OR halide ion has less attraction for Mg²⁺ ✓ Hydration enthalpy Hydration enthalpy is less exothermic down group OR halide ion has less attraction for H₂O ✓ Enthalpy of solution Difficult to predict whether lattice enthalpy or hydration enthalpy has bigger effect ✓	4	ALLOW ORA throughout ALLOW ions closer together in MgF₂ OR further apart in MgI₂ DO NOT ALLOW atomic radius ALLOW MgI₂ is less exothermic than MgF₂ for LE and hydration enthalpy -as trend ‘down the group’. ALLOW less negative/more positive BUT IGNORE is smaller/less Examiner’s Comments This question involved the qualitative explanation of the effect of ionic charge and ionic radius on the exothermic value of a lattice enthalpy and enthalpy change of hydration. These then needed to be linked to enthalpy change of solution. Only a few candidates were able to gain credit describing lattice enthalpy and hydration enthalpy becoming more exothermic down the group. A significant number of candidates described trends in ionisation energies, reactivity, intermolecular bonding and/or electronegativity. They only scored the mark for the trend in ionic radius going down the group. Candidates should focus on the use of correct terminology as atomic radius is incorrect in this context. Some candidates are still using ‘smaller/less’ to describe ‘less exothermic’. Misconception Some candidates linked ionisation energy rather than lattice enthalpy and enthalpy change of hydration.
		Total	8

Question 16

QuestionsAnswers

16. This question is about energy changes.

* A student plans to determine the enthalpy change of hydration of calcium ions.

The student finds the information below from data tables.

Enthalpy change	ΔΗ / kJ mol^-1
Lattice enthalpy of calcium chloride	-2223
Enthalpy change of hydration of chloride ions	-378

The student carries out an experiment to find the enthalpy change of solution of calcium chloride.

Student’s method:

•	Weigh a bottle containing calcium chloride and weigh a polystyrene cup.
•	Add water from a measuring cylinder to the polystyrene cup and measure its temperature.
•	Add the calcium chloride, stir the mixture, and measure the maximum temperature of the final solution.
•	Weigh the empty bottle and weigh the polystyrene cup with the final solution.

Mass readings

Mass of bottle + calcium chloride / g	27.45
Mass of empty bottle / g	18.17
Mass of polystyrene cup / g	21.24
Mass of polystyrene cup + final solution / g	127.84

Temperature readings

Initial temperature of water / °C	21.0
Maximum temperature of final solution / °C	39.5

Calculate the enthalpy change of solution of calcium chloride and determine the enthalpy change of hydration of calcium ions.

Show your working, including an energy cycle linking the energy changes.

Assume that the density and specific heat capacity, c, of the solution are the same as for water.

[6]

17. Which equation represents the change that accompanies the standard enthalpy change of atomisation of bromine?

A	½ Br₂(l) → Br(g)
B	Br₂(l) → 2Br(g)
C	½ Br₂(g) → Br(g)
D	Br₂(g) → 2Br(g)

Your answer

[1]

16	Level 3 (5-6 marks) Calculates correct enthalpy change with correct – sign for Δ_hyH (Ca²⁺), allowing for acceptable errors. There is a well-developed line of reasoning which is clear and logically structured. The information presented is relevant and substantiated. Level 2 (3-4 marks) Calculates a value of Δ_SolH (CaCl₂(s)) from the: Energy change AND Amount in mol of CaCl₂. There is a line of reasoning presented with some structure. The information presented is relevant and supported by some evidence. Level 1 (1-2 marks) Processes experimental data to obtain the: Energy change from mcΔT OR Amount in mol of CaC/₂. There is an attempt at a logical structure with a line of reasoning. The information is in the most part relevant. 0 marks – No response or no response worthy of credit.	6	Indicative scientific points may include: 1. Processing experimental data Energy change from mcΔT Energy in J OR kJ = 106.6 × 4.18 × 18.5 = 8243.378 (J) OR 8.243378 (kJ) 3SF or more Amount in mol of CaC/₂ 0.08352835284 unrounded ———————————————— 2. ± value of ΔsolH(CaCl2(s)) ———————————————— 3. CORRECT Δ_hyH(Ca²⁺) calculated with signs Δ_hyH(Ca²⁺) = L.E. + Δ_SolH(CaCl₂) – 2 Δ_hyH(Cl^–) = -2223 + (-98.7) – (2 × -378) = -1566 (kJ mol^-1) 3SF or more with correct – sign From unrounded values, -1565.689579 ———————————————— See next page for examples of acceptable errors Acceptable errors ALLOW omission of trailing zeroes ALLOW minor slips in rounding, transcription errors, etc throughout ALLOW one small error, e.g. subtracting mass of CaCl₂ for m m = 106.60 – 9.28 = 97.32 q = 7.5257556 (kJ) ΔsolH = 90.09821629 (kJ mol^-1) ΔhyH(Ca²⁺) = -1557 (kJ mol^-1) OR adding mass of CaCl₂ for m m = 106.60 + 9.28 = 115.88 q = 8.9610004 kJ Δ_solH = 107.2809423 (kJ mol^-1) Δ_hyH(Ca²⁺) = -1574 (kJ mol^-1) Examiner’s Comments This question was assessed by level of response (LoR). Candidates were required to process raw experimental results to determine one enthalpy change, and then to determine a second enthalpy change by using an energy cycle. Levels were determined by the accuracy of the candidates’ processing of the results, calculations and use of the energy cycle. Marks within a level were determined by communication. This question discriminated extremely well. Level 3 candidates used the mass of solution as 106.6 g with mcΔT to obtain an energy change of 8.24 kJ. They then divided this value by the moles of CaCl₂ that reacted (0.0835 mol) to obtain the enthalpy change of -98.7 kJ mol^-1. Finally, they constructed an energy cycle which they then used to obtain the second enthalpy change of -1566 kJ mol^-1. Level 2 candidates determined the first enthalpy change but may have used the approximate mass of 100 g for the mass of solution from the experimental method. Their energy cycle was often incorrect or absent, with the second enthalpy change calculated incorrectly. Level 1 candidates often calculated the initial energy change using mcΔT but made little further correct progress. Less successful responses used the solid mass of calcium chloride (9.28 g) instead of the mass of the solution in their mcΔT calculation. Overall, mathematical skills were displayed well but some basic errors were made, particularly with subtractions. This may have been the result of mis-keying values into a calculator and believing the answer displayed. Exemplar 3 Exemplar 3 is a Level 2 response. The candidate has calculated the initial energy change, the moles of calcium chloride and the first enthalpy change using a correct method. This response has not been given marks for the communication strand of Level 3 because there is nothing to indicate what the numbers refer to. The response could be summarised as a mass of numbers scrawled across the page. Unfortunately, this is the pattern of many responses. So, this response was given 3/6 marks.
	Total	6
17	A	1	Examiner’s Comments This question proved to be difficult, with fewer candidates selecting the correct answer of A. Option C was the most common distractor as many candidates did not know the standard state of bromine. Option B was selected by those candidates who confused the definitions of average bond enthalpy with the enthalpy change of atomisation.
	Total	1

Question 18

QuestionsAnswers

18. Which compound requires the most energy to convert one mole into its gaseous ions?

A	NaF
B	Na₂O
C	MgF₂
D	MgO

Your answer

[1]

19. This question is about energy changes.

Lattice enthalpies can be determined indirectly using Born-Haber cycles.

The table below shows the energy changes that are needed to determine the lattice enthalpy of barium iodide, BaI₂.

Energy term	Energy change / kJ mol^–1
formation of barium iodide	–602
1st electron affinity of iodine	–296
1st ionisation energy of barium	+503
2nd ionisation energy of barium	+965
atomisation of iodine	+107
atomisation of barium	+180

The diagram below shows an incomplete Born-Haber cycle that can be used to calculate the lattice enthalpy of barium iodide.

On the dotted lines, add the species present, including state symbols.

[4]

Question 20

QuestionsAnswers

20. These questions are from different areas of chemistry.

This question is about two salts of rubidium (atomic number 37): RbClO₃ and RbClO₄.

The oxidation number of chlorine is different in the two rubidium salts, RbClO₃ and RbClO₄.

What is the name of RbClO₄?

[1]

A student carries out an experiment to determine the enthalpy change of solution of RbClO₃ using the method below.

•	A 2.00 g sample of solid RbClO₃ is added to water in a well-insulated container. The initial temperature is 23.0 °C.
•	The mixture is stirred until all the RbClO₃ has dissolved. The final temperature is 21.5 °C. The final solution has a mass of 102 g.

Determine the enthalpy change of solution, Δ_sol H, of RbClO₃ in kJ mol^–1.

Assume that the specific heat capacity of the solution is the same as that of pure water.

[3]

20	i	Rubidium chlorate(VII) ✓	1	ALLOW Rubidium(I) chlorate(VII) Rubidium chloroate(VII) IGNORE Rubidium (VII)chlorate Rubidium chlorate(IIV) Rubidium chlorate (7) Rubidium perchlorate Examiner’s Comments Candidates had difficulty in naming a compound using Roman numerals for an element which can have different oxidation numbers. For the name of RbClO₄, many omitted the number entirely, showing just rubidium chlorate. Many inventive names such as rubidium chlorotetraoxide were seen. Some candidates wrote the correct VII before chlorate and many different Roman oxidation numbers were seen. Roman numerals’ use in naming compounds is part of chemical nomenclature, included in the specification.
	ii	FIRST CHECK THE ANSWER ON ANSWER LINE If answer = 54.0 OR 54.1 OR 54.2 (kJ mol^–1) award 3 marks ————————————————– Energy change from mcΔT Energy in J OR kJ = 102 × 4.18 × 1.5 OR 639.54 (J) OR 0.63954 (kJ) ✓ ————————————————– Amount in mol of RbClO₃ OR 0.0118…….. (mol) ✓ ————————————————– Δ_solH(RbClO₃) (+) 54.0 ✓ From unrounded values, ΔH = 54.04113 Examples of mixed acceptable intermediate rounding, e.g. ΔH = 54.237 → 54.2 ΔH = 54.06 → 54.1	3 (AO 2.8 ×3)	ALLOW ECF throughout IGNORE sign IGNORE RE and SF in 1st 2 marks 0.01183431953 unrounded ALLOW 54 (from 54.0) CARE 54.00 is a rounding error ————————————————– COMMON ERRORS 52.98 OR 53.14 2 marks 100 instead of 102: Energy = 100 × 4.18 × 1.5 = 627 J From unrounded n, ΔH = = 52.98 kJ mol^–1 OR 53.0 (3SF) OR 53 From rounded 0.0118, ΔH = = 53.14 OR 53.1 ————————————————– 0.02078 OR 0.0208 1 mark 102 and 2 swapped: Energy = 2 × 4.18 × 1.5 = 12.54 J n = = 0.60355…… ECF ΔH = = 0.0208 kJ mol^–1 ———————————– 1.06 2 marks 102 for n instead of 2.00: n = = 0.60355…… ΔH = = 1.06 kJ mol^–1 OR 2 for energy instead of 102 Energy = 2 × 4.18 × 1.5 = 12.54 J ΔH = = 1.06 kJ mol^–1 ———————————– 107.4 – 107.7 2 marks 8.314 for c instead of 4.18: Energy = 102 × 8.314 × 1.5 = 1272 J Energy = 102 × 8.31 × 1.5 = 1271.4 J ΔH = 107.4 – 107.7 kJ mol^–1 depends on intermediate rounding *CHECK* ———————————– Apply ECF for any other comparable responses. If in doubt contact TL Examiner’s Comments This question was a good discriminator, producing marks across the whole 3 mark range. More successful candidates correctly calculated the energy change, moles of RbClO₃ and enthalpy change of solution. However, there were pitfalls for many including the following: calculating the energy change using the mass of water rather than the mass of the solution. This was despite the suppled information that the specific heat capacity of the solution is the same as for water. Candidates should understand that m in mcΔT is the mass of the substance that produces ΔTcalculating an incorrect value for the molar mass of RbClO₃. Instead of 169, this was often seen as 120.5 (using the atomic number of 37 for Rb, rather than the mass number of 85.5) and 185 (for RbClO₄)using values of m at the wrong stages in the calculation. e.g. 2 g with the energy change and 102 g or 100 g with the moles calculationcalculating the correct numerical value for the enthalpy change of solution, but then placing a ‘–‘ sign in front of the value, despite ΔT being for a decrease in temperature. Finally, as with all multi-step calculations, candidates are advised to use calculator values throughout. Any intermediate rounding introduces rounding errors in the final value. The final value can be rounded either to the significant figures demanded by the question or to the lowest number of significant figures used in the provided data.
		Total	4

3	a		solution: (enthalpy change for) 1 mole of a compound / substance / solid / solute dissolving in water	1	IGNORE ‘energy released’ OR ‘energy required’ For dissolving, ALLOW forms aqueous / hydrated ions DO NOT ALLOW dissolving elements IGNORE ionic OR covalent DO NOT ALLOW response that implies formation of 1 mole of aqueous ions
	b	i		3	Correct species AND state symbols required for each mark. (mark independently) On middle line, ALLOW Ca²⁺(g) + 2Cl⁻(aq) (i.e. Cl⁻ hydrated before Ca²⁺) On bottom line, ALLOW CaCl₂(aq)
		ii	FIRST CHECK THE ANSWER ON ANSWER LINE IF answer = −142 (kJ mol⁻¹) award 2 marks ………………………………………………….. Δ_solH (CaCl₂) = [−1616 + (2 × −359)] − (−2192) OR −2334 + 2192 = −142 (kJ mol⁻¹)	2	IF there is an alternative answer, check to see if there is any ECF credit possible using the working shown. IF ALL 3 relevant values from the information at the start of Q3 have NOT been used, award zero marks unless one number has a transcription error, where 1 mark can be awarded ECF
		iii	Comparison of size Ca²⁺ > Mg²⁺ Comparison of charge Na⁺ < Mg²⁺ < Al³⁺ Comparison of attraction between ions size AND charge linked to greater attraction to H₂O ✔	3	IGNORE comparison of size: Na⁺ > Mg²⁺ > Al³⁺
	c	i	FIRST CHECK THE ANSWER ON ANSWER LINE IF answer = −132 (kJ mol⁻¹) award 4 marks ………………………………………………….. Correctly calculates energy released in J OR kJ = 50.21 × 4.18 × 31.5 = 6611 (J) OR 6.611 (kJ) Correctly calculates n(CaCl₂) Correctly calculates ΔH value in J OR kJ In J: OR (−)132,220 (J) OR In kJ: OR (−)132.22 (kJ) (Sign ignored and / or more than 3 SF) Correct Δ_solH in kJ AND sign AND 3SF = −132 (kJ mol⁻¹)	4	FULL ANNOTATIONS MUST BE USED ………………………………………………….. ALLOW calculator value of 6611.1507 down to 3SF value of 6610 DO NOT ALLOW fewer than 3 SF IGNORE units for this mark, i.e. just ALLOW correctly calculated number in either J or kJ ALLOW ECF from n(CaCl₂) AND / OR Energy released IGNORE absence of − sign and 3 SF requirement Final mark requires − sign, kJ AND 3 SF
		ii	Temperature change is double / × 2 / 63 °C AND Δ_solH is the same Twice the energy produced in the same volume AND ratio of energy produced to mass or number of moles is the same is the same	2	ALLOW temperature reached would be 85 °C
			Total	15

4	a	i		3	Mark each marking point independently Correct species AND state symbols required for each mark For S²⁻, DO NOT ALLOW S⁻² For e⁻, ALLOW e For e⁻ only, IGNORE any state symbols added ALLOW k and s It can be very difficult distinguishing K from k; S from s Examiner’s Comments Many candidates successfully completed the Born-Haber cycle to obtain all three marks. The species including any ionic charges and state symbols were almost always correct but sometimes one or more state symbols had been omitted. The commonest error was in the number of electrons in the middle stage; some showed two electrons and the electron was more often omitted entirely. Candidates are advised to check carefully between stages in the cycle to ensure that all species charges and state symbols are accounted for and included.
		ii	(The enthalpy change that accompanies) the formation of one mole of a(n ionic) compound from its gaseous ions (under standard conditions) ✔ ✔ Award marks as follows. 1st mark: formation of compound from gaseous ions 2nd mark: one mole for compound only DO NOT ALLOW 2nd mark without 1st mark Note: A definition for enthalpy change of formation will receive no marks	2	IGNORE ‘Energy needed’ OR ‘energy required’ ALLOW one mole of compound is formed / made from its gaseous ions ALLOW as alternative for compound: lattice, crystal, substance, solid IGNORE: 2K⁺(g) + S²⁻(g) → K₂S(s) (question asks for words) ALLOW 1 mark (special case) for absence of ‘gaseous’ only, i.e. the formation of one mole of a(n ionic) compound from its ions (under standard conditions) ✔ Examiner’s Comments The majority of candidates had learnt the definition for lattice enthalpy and scored two marks. When fewer marks were awarded, the more common reasons were for responses in terms of a mole of gaseous ions, or omission of the mole altogether. Occasionally some weaker candidates confused the definition with that for the enthalpy change of formation and so referred to forming a mole of the ionic compound from its constituent elements. Such responses gained no credit.
		iii	FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = −2116 (kJ mol⁻¹) award 2 marks −381 − (2 × +89 + 279 + 2 × +419 −200 + 640) ✔ −381 − 1735 = − 2116 ✔ (kJ mol⁻¹)	2	IF there is an alternative answer, check to see if there is any ECF credit possible using working below. See list below for marking of answers from common errors ALLOW for 1 mark ONE mistake with sign OR use of 2: −2027 (2 × 89 not used for K) −1697 (2 × 419 not used for K) −2516 (+200 rather than −200 for S 1st electron affinity) (+)2116 (wrong sign) −1354 (+381 instead of −381) (+)1354 (+1735 instead of −1735) −836 (−640 instead of +640) −1558 (−279 instead of +279) −1760 (−2 × 89 instead of +2 x 89) −439 (−2 × 419 instead of +2 x 419) −2120 (rounded to 3SF) For other answers, check for a single transcription error or calculator error which could merit 1 mark DO NOT ALLOW any other answers, e.g. −1608 (2 errors: 2 × 89 and 2 × 419 not used for K) −846 (3 errors:) Examiner’s Comments Most candidates correctly calculated the lattice enthalpy using a correct sequence of enthalpy values. The commonest mistakes were omission of ‘2’ for either the atomisation or ionisation of potassium or use of incorrect signs. Candidates are advised to check carefully that any balancing numbers are linked to the correct enthalpy changes in the cycle. Answer = −s2116 kJ mol⁻¹
	b		Lowest melting point KI RbCl Highest melting point NaBr Correct order ✔ Mark 2nd and 3rd marking points independently Attraction and ionic size linked: Greater attraction from smaller ions / closer ions / larger charge density ✔ Comparison needed Energy AND attraction / breaking bonds linked: More energy / heat to overcome attraction (between ions) OR More energy / heat to break (ionic) bonds ✔	3	FULL ANNOTATIONS MUST BE USED ORA throughout Response must clearly refer to ions for explanation marks 2nd and 3rd marking point must be comparative DO NOT ALLOW incorrect named particles, e.g. ‘atoms’, ‘molecules’, Na, Cl, Cl₂, ‘atomic’, etc DO NOT ALLOW responses using nuclear size or attraction DO NOT ALLOW responses linked with loss of electrons IGNORE larger electron density ALLOW smaller sum of radii gives a greater ionic attraction IGNORE NaBr has greater ionic attraction IGNORE NaBr has smallest ionic radius (not focussing on size of *each* ion) ASSUME bonds broken are ionic unless otherwise stated DO NOT ALLOW incorrect named particles, e.g. ‘atoms’, ‘molecules’, Na, Cl, Cl₂, ‘atomic’, etc Note: Comparison for energy only (i.e. link between more energy and breaking bonds / overcoming attraction) Examiner’s Comments This descriptive part caused more problems. Candidates were expected to apply their knowledge and understanding of lattice enthalpies to supplied data. The predicted order of melting points based on the sum of the ionic radii gave the easiest mark, although some showed the order the opposite way round. The explanation proved to be much harder. Candidates were expected to relate ionic size with attraction and then to the energy required to overcome the attractive force. Precision in language is always essential here and many candidate spoilt their response by use of incorrect particles. It was very common to see terms such as ‘atomic radius, molecules, van der Waals’ forces and ionic radius of NaBr. Some candidates simply compared the radii, the skill required for the first marking point.
			Total	10

5	i	Mark each marking point independently	4	Correct species AND state symbols required for each marks ALLOW e for e⁻ TAKE CARE: In top left box, e⁻ may be in centre of response and more difficult to see than at end. There is only ONE correct response for each line From the gaps in the cycle, there is NO possibility of any *ECF* Examiner’s Comments Many candidates completed the Born-Haber cycle to obtain three out of the four available marks. Strangely, very few candidates showed the correct species in the bottom box for the elements under standard conditions. Almost invariable, iodine was shown incorrectly, usually as I₂(g) or 2I(g). The other three boxes were usually correct although sometimes state symbols had been omitted or electrons had been included together with the gaseous ions in the top right box. Candidates are advised to check carefully between stages in the cycle to ensure that all species charges and state symbols are included and accounted for.
	ii	(The enthalpy change that accompanies) the formation of one mole of a(n ionic) compound from its gaseous ions (under standard conditions) ✔✔ Award marks as follows. 1st mark: formation of compound from gaseous ions 2nd mark: one mole for compound only DO NOT ALLOW 2nd mark without 1st mark DO NOT ALLOW any marks for a definition for enthalpy change of formation BUT note the two concessions in guidance	2	IGNORE ‘Energy needed’ OR ‘energy required’ ALLOW one mole of compound is formed / made from its gaseous ions ALLOW as alternative for compound: lattice, crystal, substance, solid IGNORE: Fe²⁺(g) + 2I⁻(g) → FeI₂(s) (Part of cycle) ALLOW 1 mark for absence of ‘gaseous’ only, i.e. the formation of one mole of a(n ionic) compound from its ions (under standard conditions) ✔ ALLOW 1 mark for ΔH_f definition with ‘gaseous’: the formation of one mole of a(n ionic) compound from its gaseous elements (under standard conditions) ✔ Examiner’s Comments The majority of candidates had learnt the definition for lattice enthalpy and scored two marks. When fewer marks were awarded, the more common reasons were for responses in terms of a mole of gaseous ions, or omission of the mole altogether. Occasionally, some weaker candidates confused the definition with that for the enthalpy change of formation and so referred to forming a mole of the ionic compound from its constituent elements. Such responses gained no credit.
	iii	FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = −2473 (kJ mol⁻¹) award 2 marks ………………………………………………….. (−113) = 416 + (2 × +107) + 759 + 1561 + (2 × −295) + ΔH_LE(FeI₂) OR ΔH_LE(FeI₂) = −113 − ( 416 + (2 × +107) + 759 + 1561 + (2 × −295)) OR −113 − 2360 ✔ = − 2473 ✔ (kJ mol⁻¹)	2	IF there is an alternative answer, check to see if there is any ECF credit possible using working below. See list below for marking of answers from common errors ………………………………………………….. ALLOW for 1 mark: +2473 wrong sign −2661 107 and −295 used instead of 2 × 107 and 2 × −295 −2366 +107 used instead of 2 × 107 −2768 −295 used instead of 2 × −295 −3653 wrong sign for 295 −2247 wrong sign for 113 −1641 wrong sign for 416 −2045 wrong sign for 2 × 107 −955 wrong sign for 750 +649 wrong sign for 1561 −3653 wrong sign for 2 × −295 Any other number: CHECK for ECF from 1st marking point for expressions with ONE error only e.g. one transcription error: e.g. +461 instead of +416 Examiner’s Comments Most candidates correctly calculated the lattice enthalpy using a correct sequence of enthalpy values. The commonest mistake was the omission of ‘2’ for either the atomisation or ionisation of iodine, leading to answers of −2366 or −2661 kJ mol⁻¹, or use of incorrect signs. There were more transcription errors seen than in previous session, most notably, −113 shown as −133. Candidates are advised to check carefully that any balancing numbers are linked to the correct enthalpy changes in the cycle and to double check values for possible transcription errors. Answer = −2473 kJ mol⁻¹
		Total	8

6	a		(enthalpy change for) 1 mole of gaseous ions OR 1 mole of hydrated ions / aqueous ions ✔ gaseous ions forming aqueous / hydrated ions ✔	2	one mole can be stated just once EITHER with gaseous ions OR with aqueous ions, e.g. 1 mole of gaseous ions forms hydrated ions / aqueous ionsGaseous ions form 1 mole of hydrated ions / aqueous ions ALLOW 1 mol for 1 mole IGNORE ‘energy released’ OR ‘energy required’ For 2nd mark IGNORE gaseous ions are hydrated IGNORE gaseous ions dissolve in water *Particles formed not stated* ALLOW 1 mark for: 1 mole of gaseous IONS forms aqueous / hydrated atoms / particles / molecules Examiner’s Comments This question assessed enthalpy changes in aqueous solutions. Most candidates were awarded both marks for a clear definition stating that 1 mole of gaseous ions formed 1 mole of aqueous ions. Some candidates instead gave a definition for enthalpy change of solution.
	b	i	4 marks for species AND state symbols on all 4 energy levels (including added energy level) 1 mark for B, C AND D labels OR enthalpy values AND arrow directions correct ✔ ALLOW K₂SO₄(aq) for 2K⁺(aq) + SO₄²⁻(aq) ALLOW arrows not touching lines. Direction is important: FROM 2K⁺(g) + SO₄²⁻(g) lineFROM K₂SO₄(s) line Extra energy line placed ABOVE top line 3 out of 4 marks awarded for energy lines and species. Top arrow is shown FROM 2K⁺(g) + SO₄²⁻(g) and arrow directions correct. Letter labels correct so last mark is awarded. 4/5 marks Extra energy line placed BELOW bottom line 3 out of 4 marks awarded for energy lines and species. Top arrow is shown FROM K₂SO₄(s) and arrow directions correct. Letter labels correct so last mark is awarded. 4/5 marks ‘2 ×’ is NOT required – part of calculation mark	5	IF extra energy level is above top line OR below bottom line, DO NOT ALLOW mark for species on this line. Same as left-hand response BUT top arrow shown TO 2K⁺(g) + SO₄²⁻(g) so last mark not awarded 3/5 marks Same as left-hand response BUT bottom arrow shown TO K₂SO₄(s) so last mark not awarded 3/5 marks ALLOW C and D with associated labels, the other way round: State symbols are essential IF no extra energy level is shown with C and D combined forming 2K⁺(aq) + SO₄²⁻(aq), No mark for the extra energy level with speciesNo mark for labels as C and D are combined Therefore 3 max for species on energy levels provided Examiner’s Comments This question assessed enthalpy changes in aqueous solutions. Many candidates successfully completed the energy cycle to obtain all marks. Correct species and state symbols are essential in such cycles and marks could not be awarded for species such as K⁻(g) or SO₄⁻(g). Poorly-prepared candidates often scored no marks at all, having shown random species on the energy levels. Candidates are advised to ensure that the state symbols (s) and (g) are clearly distinguished.
		ii	ΔH(hydration) SO₄²⁻ = −1099 (kJ mol⁻¹) ✔	1	ONLY correct answer Examiner’s Comments This question assessed enthalpy changes in aqueous solutions. Although many candidates correctly calculated the lattice enthalpy, errors were common. A common error was use of −320 (instead of 2 × −320) giving −1419. Answer = −1099 kJ mol⁻¹
	c	i	Aqueous particles are more disordered than solid (particles) OR Solid particles are more ordered than aqueous (particles) ✔	1	For particles, ALLOW ions DO NOT ALLOW molecules / atoms ALLOW ‘When the state changes from solid to aqueous, disorder increases’ For more disordered, ALLOW less ordered / more freedom / more ways of arranging energy / more random For aqueous particles, ALLOW particles in solution IGNORE dissolved Examiner’s Comments This question assessed enthalpy changes in aqueous solutions. In this part, candidates needed to recognise that solid particles are forming aqueous particles with an increase in disorder. Many candidates incorrectly used ‘liquid’ instead of aqueous and others started from gaseous particles, perhaps confusing enthalpy change of solution with hydration.
		ii	*Calculation (2 marks)* ΔG = 24 − (298 × 0.225) OR 24 − 67.05 (in kJ) OR 24000 − (298 × 225) OR 24000 − 67050 (in J) ✔ Calculation of ΔG (IGNORE UNITS) ΔG = −43 (kJ mol⁻¹) OR −43000 (J mol ⁻¹) ✔ Subsumes 1st calculation mark Reason for solubility Calculated value of ΔG that is negative AND Statement that: ΔG is negative OR ΔG < 0 OR −43 < 0 OR ΔH − TΔS < 0 OR TΔS > ΔH ✔	3	Contact TL if solely entropy approach rather than ΔG ALLOW −43.1 OR −43.05 (calculator value) ALLOW 1 calculation mark (IGNORE units) for −67.(026) OR −67026 *ECF* from 225 instead of 0.225 18.(375) OR +18.375 *ECF* from 25 instead of 298 ALLOW other ECF from ONE error in 1st step of calc, e.g. incorrect value for ΔH such as −1099 from 3bii → −1166.05 TAKE CARE that same units used for ΔH and ΔS NO reason mark from a +ve value of ΔG Examiner’s Comments This question assessed enthalpy changes in aqueous solutions. The majority of candidates recognised that the Gibbs’ equation was required. Usually the correct enthalpy change of +24 kJ mol⁻¹ was used to obtain a negative value for ΔG. The majority then went on to link the negative value to feasibility for the dissolving process. A significant number of candidates used the wrong enthalpy change (or no enthalpy change at all) or mixed units of J and kJ. Answer: ΔG = −43 kJ mol⁻¹
			Total	12
7			D	1
			Total	1

11	i	(enthalpy change when)✓ 1 mole of gaseous ions react✓ OR 1 mole of hydrated/aqueous ions are formed ✓ gaseous ions dissolve in water OR gaseous ions form aqueous/hydrated ions ✓	2	IGNORE ‘energy released’ OR ‘energy required’ Examiner’s Comments Most candidates were able to state that one mole of gaseous ions was dissolved into water. Common errors seen included candidates referring to the dissolving of one mole of substance (i.e. the enthalpy change of solution); the use of the generic term ‘solvent’ rather than water; and dissolving of gaseous ions into one mole of water.
	ii		4	Correct species AND state symbols required for each mark. (mark independently) On 2nd line, ALLOW Ca²⁺(g) + 2F⁻(aq) (i.e. F⁻ hydrated before Ca²⁺) On 3rd line, ALLOW CaF₂(aq) DO NOT ALLOW when first seen but ALLOW ECF for ‘ 2’ missing and for use of the following ions Fl⁻ F2^✓ Ca^+/3+ Examiner’s Comments Most candidates were able to score some marks on this question with more able candidates scoring all 4 marks. Common errors included the omission of state symbols and the use of only one F−(g) ion (often despite a correct formula of CaF2 being seen on the bottom line). For some reason, many candidates write their lower case ‘s’ in an identical way to their lower case ‘g’. This means the examiner cannot distinguish between these state symbols on such scripts. Centres may wish to stress this point to prevent candidates losing marks unnecessarily.
	iii	FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = −504 (kJ mol⁻¹) award 2 marks IF answer = −1008 (kJ mol⁻¹) award 1 mark ——————————————————— 2 × Δ_hydH(F⁻) = [−2630 + 13] − (−1609) OR −2617 + 1609 OR −1008(kJ mol⁻¹) ✓	2	IF alternative answer, check to see if there is any ECF credit possible using working below. ‘−‘ sign is needed. COMMON ERRORS for 1 mark: (+)2694: signs all reversed −2113: sign wrong for −1609 −2126: sign wrong for 2630 −517: sign wrong for 13 +504: sign wrong IF ALL 3 relevant values from the information at the start of Q16a(iii) have NOT been used, award zero marks unless one number has a transcription error, where 1 mark can be awarded ECF Examiner’s Comments Most candidates were able to do this relatively straightforward calculation by rearranging the values of the enthalpy changes associated with the energy cycle diagram. However, a significant number forgot to divide –1008 by 2 to score full marks.
	iv	Correct comparison of Δ_hyd linked to sizes Δ_hydH(F⁻) more negative/exothermic (than Δ_hydH(CΓ)) AND F⁻ has smaller size (than Cl⁻) ✓ *Comparison of attraction between ions and water* F⁻ OR smaller sized ion linked to greater attraction to H₂O ✓	2	ORA IGNORE ‘atomic’ before radius when comparing size of ions IGNORE charge density IGNORE electronegativity IGNORE nuclear attraction DO NOT ALLOW ‘forms stronger hydrogen bonds with water’ OR ‘forms stronger van der Waals’ forces with water’ ALLOW ‘forms bonds’ for attraction’ DO NOT ALLOW F⁻ greater attraction to H₂O if given as larger ion Assume ‘F’ / ‘Fluorine’ means ‘ions’ but DO NOT ALLOW ‘F molecules’ Examiner’s Comments When comparing enthalpy changes candidates need to be aware that descriptions such as ‘bigger’ or ‘smaller’ are meaningless as there are often negative signs involved. The correct description required here was that the enthalpy change of hydration of F⁻ ions would be more negative than that of Cl⁻ ions. Although some candidates wrote in terms of charge density, it was those candidates who related the smaller size of the F⁻ ion to the difference in enthalpy change of hydration who received credit and went on to say that this was as a consequence of greater attraction to water molecules.
		Total	10

13	a	i		4	Mark each marking point independently Correct species AND state symbols required for each mark For e^–, ALLOW e For e^– only, IGNORE any state symbols added Examiner’s Comments Some candidates wrote illegible state symbols where (g) and (s) were impossible to tell apart. Also, many candidates choose to write state symbols as a very small sub-script e.g. 2K_(s) + ½O_2(g). The convention is to use lower case letters of normal size e.g. 2K(s) + ½O₂(g) Exemplar 1 In this exemplar it was impossible to tell if K had (s) or (g) as a state symbol. Consequently, no marks could be given.
		ii	FIRST CHECK THE ANSWER ON ANSWER LINE If answer = –2277 (kJ mol^-1) award 2 marks –363 – (2 × +89 +249 + 2 × 419 – 141 + 790) ✓ –363 – 1914 = –2277 ✓ (kJ mol^–1)	2	IF there is an alternative answer, check to see if there is any ECF credit possible using working below See list below for marking of answers from common errors ALLOW for 1 mark ONE mistake with sign OR use of 2 ×: +2277 (wrong sign) –601 (2 × –419 instead of 2 × +419) –697 (–790 instead of +790) –1551 (+363 instead of –363) –1858 (2 × +419 not used for K) –1921 (2 × –89 instead of 2 × +89) –2152.5 or –2153 (+249 ÷ 2) –2188 (2 × +89 not used for K) –2280 (rounded to 3SF) –2559 (+141 instead of –141) For other answers, check for a single transcription error or calculator error which could merit 1 mark Examiner’s Comments Most candidates scored both marks. Candidates tended to forget the mole ratio of K meant its values should be multiplied by 2 in two places during this calculation.
	b	i	For sodium atomic radius smaller OR fewer shells ✓ nuclear attraction increases OR (outer) electron(s) experience more attraction ✓	2	ALLOW ‘Na/sodium is smaller’ IGNORE smaller radius / fewer shells / less shielding if applied to ions but DO NOT ALLOW responses which refer to ions losing electrons DO NOT ALLOW molecules ALLOW energy levels for shells IGNORE fewer orbitals OR fewer sub–shells ALLOW less (electron) shielding OR electron repulsion between shells IGNORE just ‘shielding’ ALLOW more/stronger/bigger nuclear attraction etc IGNORE ‘pull’ for attraction IGNORE electrons more tightly held IGNORE ‘nuclear charge’ for ‘nuclear attraction’ IGNORE more energy (in question) ALLOW reverse argument for potassium throughout Examiner’s Comments Candidates coped well with this question which was based on the AS part of the specification. Some candidates gave vague and unnecessarily long responses to this question.
		ii	*Comparison of size of cations* For sodium ions ionic radius of sodium / Na⁺ is smaller ✓ *Comparison of attraction of cation and anion* Na⁺ has stronger attraction to O^2- ✓	2	comparison of IONS is essential ALLOW Na⁺ has a larger charge density IGNORE ‘Na has smaller atomic radius’ but DO NOT ALLOW contradictory sentences eg ‘Na+ ions have smaller atomic radius’ IGNORE pull for attraction ALLOW ‘sodium ion’ and ‘oxygen ion’ IGNORE just ‘oxygen’ or just ‘O’ for oxygen ion ALLOW stronger attraction between oppositely charged ions Examiner’s Comments Many candidates did not achieve full marks, often through ambiguous statements. A typical example is shown in the exemplar below. Exemplar 4 The exemplar here uses the vague term ‘due its smaller ionic size’. It is unclear which of the ions the candidate is referring to (or if they’re assuming that the metal oxide entity is an ion) and so they could not be given marks.. Also, it is oxygen ions that ‘bond’ – the reference to oxygen bonding in the response is not specific enough.
			Total	10

18		D	1	Examiner’s Comments This question was another challenging idea. The correct answer was D. Many candidates chose A or C, possibly due to fluorine’s high electronegativity.
		Total	1
19	i		4	Examiner’s Comments Some candidates wrote illegible state symbols where (g) and (s) were impossible to tell apart. Also, many candidates choose to write state symbols as a very small sub-script, e.g. Ba(s) or I₂(s). The convention is to use lower case letters of normal size, e.g. Ba(s) or I₂(g). The most common errors were the iodine state symbol, with both (g) and (l) being used, and the use of 2I for I₂. Some candidates missed state symbols in one species, missed electrons with the two ions, or gave a charge on the top left iodine.
	ii	FIRST CHECK THE ANSWER ON ANSWER LINE If answer = –1872 award 2 marks ————————————————– ΔH lattice = 2(+ 296) – 965 – 503 – 180 + 2(–107)– 602 ✓ ΔH lattice = –1872 (kJ mol^–1) ✓	2	ALLOW for 1 mark +1872 (wrong sign on answer) Common errors for 1 mark –3056 (–296 x 2 instead of 296 x2) –2168 (296 x 1 instead of 296 x2) –1765 (–107 x 1 instead of –107 x 2) –1512 (180 instead of –180) –1444 (107 x 2 instead of –107 x 2) – 866 (503 instead of –503) – 668 (602 instead of –602) +58 (965 instead of -965) For other answers, check for a single transcription error or calculation error which could merit 1 mark if all values have been used. DO NOT ALLOW any answer which involves two errors Examiner’s Comments The correct was answer seen frequently, along with lots of the common errors. Candidates tended to forget the mole ratio and did not multiply either -107 or +296 by two. Some candidates applied the cycle incorrectly and therefore used the wrong sign for an enthalpy change, leading to them attaining 1 mark.
		Total	6