AQA A-Level Chemistry Paper 3, June 2022: Question 1

18 marks · Medium difficulty · State/Explain/Numerical

Define enthalpy of lattice dissociation, calculate values from enthalpy cycles, determine enthalpy of solution from calorimetry data and evaluate uncertainty, and calculate Gibbs free-energy change and feasibility temperature.

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Question text

01 A value for enthalpy of solution can be determined in two ways:

• from a cycle, using lattice enthalpy and enthalpies of hydration

• from the results of a calorimetry experiment.

01.1 Define the term enthalpy of lattice dissociation.

[2 marks]

01.2 The enthalpy of solution for ammonium nitrate is the enthalpy change for the reaction

shown.

NH NO (s) + aq → NH +(aq) + NO −(aq) ∆H = +26 kJ mol−1

43 4 3

Table 1

NH +(g) NO −(g)

Enthalpy of hydration H / kJ mol 1 −307 −314

hyd

Draw a suitably labelled cycle and use it, with data from Table 1, to calculate the

enthalpy of lattice dissociation for ammonium nitrate.

[3 marks]

4 −1

Enthalpy of lattice dissociation kJ mol

01.3 A student does an experiment to determine a value for the enthalpy of solution for

ammonium nitrate.

*03* The student uses this method.

• Measure 25.0 cm3 of distilled water in a measuring cylinder.

• Pour the water into a beaker.

• Record the temperature of the water in the beaker.

• Add 4.00 g of solid NH4NO3 to the water in the beaker.

• Stir the solution and record the lowest temperature reached.

Table 2 shows the student’s results.

Table 2

Initial temperature / °C 20.2

Lowest temperature / °C 12.2

Calculate the enthalpy of solution, in kJ mol−1, for ammonium nitrate in this

experiment.

Assume that the specific heat capacity of the solution, c = 4.18 J K−1 g−1

Assume that the density of the solution = 1.00 g cm−3

[3 marks]

5 −1

Enthalpy of solution kJ mol

01.4 The uncertainty in each of the temperature readings from the thermometer used in

this experiment is ±0.1°C

Calculate the percentage uncertainty in the temperature change in this experiment.

[1 mark]

Percentage uncertainty

01.5 Suggest a change to the student’s method, using the same apparatus, that would

reduce the percentage uncertainty in the temperature change.

Give a reason for your answer.

[2 marks]

Change

Reason

01.6 Another student obtained a value of +15 kJ mol−1 using the same method.

Suggest the main reason for the difference between this experimental value for the

enthalpy of solution and the correct value of +26 kJ mol−1

[1 mark]

01.7 Table 3 shows some entropy data at 298 K

Table 3

Entropy S / J K 1 mol 1

NH4NO3(s) 151

NH +(aq) 113

NO −(aq) 146

Calculate a value for the Gibbs free-energy change (∆G), at 298 K, for the reaction

when ammonium nitrate dissolves in water.

NH NO (s) + aq → NH +(aq) + NO −(aq) ∆H = +26 kJ mol−1

43 4 3

Use data from Table 3 and the value of ∆H from the equation.

Assume for the solvent, water, that the entropy change, ∆S = 0

Explain what the calculated value of ∆G indicates about the feasibility of this reaction

at 298 K

[4 marks]

∆G kJ mol−1

Explanation

01.8 Ammonium nitrate decomposes as shown.

*06* 1 −1

NH4NO3(s) → N2(g) + O2(g) + 2H2O(g) ∆H = +123 kJmol

The entropy change (∆S) for this reaction is +144 J K–1 mol−1

Calculate the temperature at which this reaction becomes feasible.

[2 marks]

Temperature K

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidelines Mark

Ignore standard states / conditions 1

The enthalpy change / ΔH when one mole of a (solid) ionic compound Allow heat change at constant pressure when…

Ignore heat change (alone) / energy change

01.1

dissociates (fully) into gaseous ions M2 Allow suitable equation with state symbols for 1

ions (2 x AO1)

Not one mole of gaseous ions

(26)

Allow + water or +aq

(−307) M1 = cycle (3 ‘corners’ with formulae and state 1

(LE) (−314)

symbols and suitable arrows)

Allow equivalent Born-Haber style energy cycle

Not ecf to M2 and M3 from incorrect cycle

01.2

LE = 26 + 307 + 314 M2 = working e.g. 26 = LE – 307 – 314 1

= (+)647 M3 = answer (+)647 gets 3/3 if M1 given or 2/3 if 1

not

–647 = 2/3 if M1 given or 1/3 if not

+595 / –595 = 2/3 if M1 given or 1/3 if not

–621/+621 = 1/3 if M1 given

Not ecf for M3 from incorrect expression in M2 (3 x AO2)

– A-LEVEL CHEMISTRY – – JUNE 2022

(q = mc T =) 25.0 × 4.18 × (20.2–12.2) OR 25.0 × 4.18 × 8 Not if m = 29 1

(= 836 (J) or 0.836 (kJ)) Ignore sign of q

4.00 g NH4NO3 = 4.00/80 OR 0.0500 mol 1

Ho = 836/0.05 = 16720 = (+)16.7(2) (kJ mol−1) Allow ecf from M1 and/or from M2 1

soln

–16.7(2) = 2/3 (3 x AO2)

01.3

+19.4 = 2/3 (using m = 29 in M1)

–19.4 = 1/3

+2.68 = 2/3

–2.68 = 1/3

+587 or +588 = 2/3

–587 or –588 = 1/3

Allow 2 sig figs or more

(2 × 0.1/8) × 100 = 2.5% Allow ecf from T in 01.3 1

01.4

(AO2)

Marking points are independent

use a larger mass/amount of NH4NO3 / solid Allow smaller volume of water / less water 1

Allow use more NH4NO3

Not larger volume of water

01.5 Ignore higher concentration (of NH4NO3)

Ignore any references to changing apparatus e.g.

insulation 1

so temperature change/decrease is greater Allow temperature increase is greater (2 x AO3)

OR final temperature is lower Not final temperature is higher

– A-LEVEL CHEMISTRY – –

12 heat gain (from the surroundings) / incomplete dissolving Allow incomplete reaction 1

Allow thermal energy gain (AO3)

01.6 Not heat loss

Ignore energy gain

Ignore references to mistakes in method

S = (113 + 146) − 151 = +108 (J K−1 mol−1) 1

G = H − T S OR 26 − (298 × 108 × 10−3) Allow ecf 26 – (298 x M1 x 10–3) 1

Allow ecf 26 – (298 x M1)

Allow M2 for 26000 – (298 x 108)

Allow M2 for 26 – (298 x 108)

G = −6.184 / −6.18 / −6.2 –32158 = M1 and M2

01.7

–32.2 = M1 and M2

–6184 = M1 and M2

(+)58.2 = M2 and M3 (ecf if -108 for M1)

negative value for G indicates reaction is feasible/spontaneous Allow positive value for G indicates reaction is 1

NOT feasible/spontaneous (3 x AO2,

Allow <0 or >0 as appropriate 1 x AO3)

M4 is standalone

Converting H into J OR S into kJ 1

01.8 (T = H/ S = 123/144 × 10−3 OR 123000/144) = 854(.1666666) (K) 0.854 (K) = 1/2 1

0.00117 (K) = 1/2 (calculation upside down) (2 x AO2)

2SF minimum

How to answer it

Enthalpy, Born-Haber Cycles & Feasibility

What this question tests

Lattice Enthalpy: Exact definition of lattice dissociation enthalpy and setting up an enthalpy of solution Hess's Law cycle.
Calorimetry: Calculating aqueous enthalpy change using q = mcΔT , accounting for masses, stoichiometric moles, and signs.
Apparatus & Errors: Thermometer percentage uncertainties, method modifications to reduce uncertainty, and heat exchange sources.
Thermodynamics: Determining entropy change ( ΔS ), Gibbs free-energy change ( ΔG = ΔH - TΔS ), and the temperature of feasibility threshold.

Question 01.1: Defining Lattice Dissociation Enthalpy 2 Marks

Definition recall (AO1)

✅ Model Answer

The enthalpy change (or heat energy change at constant pressure) when one mole of a solid ionic compound [1 mark]
dissociates completely into gaseous ions [1 mark] .

❌ Common Errors

Saying "one mole of gaseous ions is formed" (this will lose the mark).
Confusing dissociation (endothermic, bond-breaking) with formation (exothermic).
Omitting the solid state of the compound or gaseous state of the ions.

Question 01.2: Enthalpy Cycle Calculation 3 Marks

Lattice dissociation enthalpy from hydration and solution data (AO2)

🧠 How to Construct the Hess's Law Cycle

NH₄NO₃(s) ──────── +26 kJ mol⁻¹ (ΔH_soln) ────────▶ NH₄⁺(aq) + NO₃⁻(aq) │ ▲ │ │ ΔH_L (dissoc) ΔH_hyd(NH₄⁺) + ΔH_hyd(NO₃⁻) │ = (-307) + (-314) ▼ │ NH₄⁺(g) + NO₃⁻(g) ──────────────────────────────────────────┘

📐 Step-by-Step Calculation

Step 1: Set up the cycle equation
ΔH_soln = ΔH_L(dissociation) + ΣΔH_hyd
+26 = ΔH_L + (-307) + (-314) [1 mark]

Step 2: Rearrange to solve for ΔH_L
ΔH_L = +26 - (-307) - (-314) = 26 + 307 + 314 [1 mark]

Step 3: Final value with sign
ΔH_L = +647 kJ mol⁻¹ [1 mark]

❌ Sign Traps

Remember: Lattice dissociation is always endothermic (+)! If your calculation outputs a negative number ( -647 ), you calculated lattice formation. Double-check your arrow directions.

Question 01.3: Calorimetry Calculation 3 Marks

Calculating enthalpy of solution from experimental data (AO2)

📐 Step-by-Step Calculation

Step 1: Calculate heat energy absorbed (q)
ΔT = 20.2 - 12.2 = 8.0 °C
m = 25.0 g (volume of water is 25.0 cm³, density = 1.00 g cm⁻³)
q = m × c × ΔT = 25.0 × 4.18 × 8.0 = 836 J = 0.836 kJ Mark 1

Step 2: Calculate moles of NH₄NO₃ dissolved
M_r(NH₄NO₃) = (14.0 × 2) + (1.0 × 4) + (16.0 × 3) = 80.0 g mol⁻¹
n = 4.00 / 80.0 = 0.0500 mol Mark 2

Step 3: Calculate ΔH_soln
Temperature decreases, so the process is endothermic (+ΔH):
ΔH = +q / n = +(0.836 kJ / 0.0500 mol) = +16.7 kJ mol⁻¹ (or +16.72 ) Mark 3

❌ The Dreaded Mass Error

Do NOT use m = 29.0 g ( 25.0 + 4.00 ). The mark scheme penalises using 29 g because standard procedure assumes heat is absorbed only by the liquid solvent volume (25.0 cm³ of water).

🧠 Sign Check

Always re-read the thermometer values. Temp went down from 20.2 °C to 12.2 °C, meaning heat was taken in from the water. ΔH must be positive ( + )!

Questions 01.4 & 01.5: Uncertainties and Experimental Improvement 3 Marks Total

Apparatus uncertainty analysis (AO2 & AO3)

📐 01.4: Percentage Uncertainty 1 Mark

Temperature change is calculated from two readings (initial and lowest), so the total uncertainty is doubled:

% uncertainty = (2 × 0.1 / 8.0) × 100 = 2.5%

✅ 01.5: Method Modification 2 Marks

Change: Use a larger mass of solid NH₄NO₃ (or use a smaller volume of water). [1 mark]

Reason: The temperature change ( ΔT ) will be greater (the final temperature will be lower). [1 mark]

❌ Ignored & Incorrect Answers in 01.5

"Use a polystyrene cup / add a lid": The question explicitly says "using the same apparatus", so changing the equipment scores 0.
"Use a larger volume of water": This would decrease ΔT , increasing the percentage uncertainty.

Question 01.6: Evaluating Experimental Discrepancy 1 Mark

Explaining experimental value differences (AO3)

✅ Model Answer

Heat gain from the surroundings OR incomplete dissolving of the ammonium nitrate.

❌ Common Error: "Heat Loss"

Because the reaction is endothermic, the solution is colder than room temperature. Heat moves from warmer surroundings into the beaker, reducing the temperature drop and giving a lower experimental value (+15 vs +26). Writing "heat loss to surroundings" is awarded zero.

Question 01.7: Gibbs Free-Energy & Feasibility 4 Marks

Thermodynamic feasibility calculation (AO2/AO3)

📐 Step-by-Step Calculation

Step 1: Calculate entropy change (ΔS)
ΔS = ΣS(products) - ΣS(reactants)
ΔS = (113 + 146) - 151 = 259 - 151 = +108 J K⁻¹ mol⁻¹ Mark 1

Step 2: Convert units and apply ΔG = ΔH - TΔS
Remember to convert ΔS to kJ K⁻¹ mol⁻¹ (divide by 1000):
ΔS = +0.108 kJ K⁻¹ mol⁻¹
ΔG = 26 - (298 × 0.108) Mark 2

Step 3: Solve for ΔG
ΔG = 26 - 32.184 = -6.18 kJ mol⁻¹ (accepts -6.18 to -6.2 ) Mark 3

Step 4: Explanation of feasibility
Since ΔG is negative (< 0), the reaction is feasible / spontaneous at 298 K. Mark 4

Question 01.8: Temperature of Feasibility 2 Marks

Determining the decomposition temperature threshold (AO2)

📐 Step-by-Step Calculation

Step 1: Feasibility condition & unit conversion
Reaction becomes feasible when ΔG ≤ 0 , meaning T = ΔH / ΔS .
Convert ΔH to J or ΔS to kJ:
ΔH = +123 kJ mol⁻¹ = +123 000 J mol⁻¹ Mark 1

Step 2: Calculate temperature (T)
T = 123 000 / 144 = 854 K (or 854.2 K) Mark 2

💡 Examiner Tips

Temperature must be in Kelvin ( K ). Never write °C unless explicitly requested.
Inverting the division ( 144 / 123 000 = 0.00117 ) gets only 1/2 marks. Always remember: T = ΔH / ΔS .
Give your answer to at least 2 significant figures.

Topics

Physical Chemistry · Required Practicals · 3.1.4 Energetics · 3.1.8 Thermodynamics · Required Practical 2: Measurement of an enthalpy change

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.