OCR A-Level Chemistry AS Depth in chemistry (02), June 2024: Question 3

9 marks · Medium difficulty · Extended Response

Define enthalpy change of reaction, calculate the enthalpy change for the displacement reaction between zinc and copper(II) nitrate from experimental data including assumptions and improvements, and explain why halving the volumes keeps the enthalpy change constant.

Practise this question

Question

An exam question with three parts. Part (a) asks for the definition of the term enthalpy change of reaction (1 mark). Part (b) is an extended response question detailing an experiment where zinc reacts with copper(II) nitrate solution, giving experimental data, and asks to calculate the enthalpy change in kJ mol^-1, state assumptions, and suggest improvements (6 marks). Part (c) asks why modifying the experiment by using half the volume of copper(II) nitrate yields the same enthalpy change (2 marks).
Question text

3 Enthalpy changes of reaction can be determined by experiment.

(a) What is meant by the term enthalpy change of reaction?

… [1]

(b)* A student carries out an experiment to determine the enthalpy change for the reaction between

zinc and copper(II) nitrate solution.

Zn(s) + Cu(NO3)2(aq) Zn(NO3)2(aq) + Cu(s) ΔrH Equation 3.1

The student follows the method outlined below.

• Add 100 cm3 of 0.500 mol dm–3 Cu(NO ) (aq) to a beaker.

• Measure the temperature of the solution.

• Add excess zinc to the beaker.

• Stir the mixture and record the maximum temperature.

The temperature of the solution changes from 19.5 °C to 38.1 °C.

Calculate Δ H, in kJ mol–1, for equation 3.1.

r

State any assumptions you have made in your calculation.

Suggest improvements for obtaining a more accurate value for ΔrH. [6]

Extra answer space if required

(c) The student modifies the experiment using 50 cm3 instead of 100 cm3 of 0.500 mol dm–3

copper(II) nitrate solution.

The value of ΔrH for this modified experiment is the same as in equation 3.1.

Explain why.

… [2]

Mark scheme

Show the mark scheme The mark scheme provides answers for question 3. Part (a) awards 1 mark for defining the enthalpy change for the stated equation. Part (b) uses levels of response (Levels 1 to 3) for calculating the correct enthalpy change, stating assumptions, and improvements, with guidance detailing mc delta T and n calculations resulting in -155 kJ mol^-1. Part (c) awards 2 marks for explaining that both energy and moles are halved, keeping the temperature change and enthalpy change the same.

Question Answer Marks Guidance

3 (a) (The enthalpy change) for the stated equation ✓ 1 ALLOW reaction in molar quantities/stoichiometric ratio as

shown/stated/given/in equation

IGNORE standard states or conditions

DO NOT ALLOW Energy released (can’t assume reaction is

exothermic)

(b) Please refer to the marking instructions on page 5 Indicative Scientific Points

of the mark scheme for guidance on how to mark Energy change from mcΔT

this question. Energy in J OR kJ

q = 100.0 × 4.18 × 18.6 = 7774.8(J) OR 7.7748 (kJ)

Level 3 (5–6 marks)

Calculates CORRECT enthalpy change H in kJ mol–1

AND n(Cu(NO3)2) = 0.05 (mol)

states multiple assumptions AND improvements H = – q/n = 7.7748/0.05 = –155 kJ mol–1 (3 SF)

There is a well-developed line of reasoning which is ALLOW -156 kJ mol-1 (use of 7.775 kJ)

clear and logically structured. The information ALLOW answer in J mol-1 if units are given

presented is relevant and substantiated. ALLOW a single slip/rounding errors

Level 2 (3–4 marks) Assumptions and Improvements (NOT INCLUSIVE)

Calculates CORRECT enthalpy change Assumptions

• density of solution is 1 g cm–3/same as water

OR • c of solution is same as water

Correctly calculates the moles AND attempts the

calculation of q • ignore the mass and c of zinc

AND states multiple assumptions OR • no heat escapes the system/lost to surroundings

improvements. • mass of solution remains constant

• no water lost/evaporated

There is a line of reasoning presented with some

• reaction goes to completion

structure. The information presented is relevant and

supported by some evidence. • reaction completed under standard conditions

• measurements recorded are accurate

Improvements

Level 1 (1–2 marks) • polystyrene cup /thermos flask

Attempts any part of the calculation • use a lid

AND

• more precise thermometer

states an assumption OR an improvement.

• more precise balance

OR • measure mass of solution

Correctly calculates the moles AND attempts • use burette to measure volume

calculation of q • use a cooling curve

• use standard conditions

OR

States multiple assumptions OR improvements

There is an attempt at a logical structure with a line Aspects of the communication statement might typically have been

of reasoning. The information is in the most part met when calculations have been completed in a logical order, and

relevant for L3 or L2 (where level awarded for calculation only) the use of the

correct sign with the final answer given to 3 or 4 significant figures.

0 marks

No response or no response worthy of credit.

(c) Half the energy/q OR volume/mass of solution 2 ALLOW response that links the same proportionality/ratio of

AND energy/volume/mass of solution to number of moles

half the moles ✓ ALLOW same amount of energy (released) per mole

Temperature change would be same ✓ ALLOW both marks if seen by a calculation i.e.

q = 50.0 × 4.18 × 18.6 = 3887.4(J) OR 3.8874(kJ)

n(Cu(NO3)2)= 0.025 (mol)

H = (–) q/n = 3.8874/0.025 = (–)155 kJ mol–1 ✓

Use of same temperature ✓

May need to check answer in 3b to compare

IGNORE Sign

How to answer it

Enthalpy Changes of Reaction Experiment & Calculations

What this question tests

This question assesses your understanding of fundamental thermodynamic definitions, practical calorimetry techniques, calculation of enthalpy changes from experimental temperature data, identifying assumptions and experimental improvements, and applying molar stoichiometric ratios to proportional changes.

Part (a): Defining Enthalpy Change of Reaction

Question Part (a)

✅ Correct Answer

The enthalpy change that accompanies a reaction in the stoichiometric ratio shown by the stated equation.

💡 Key Knowledge

Definitions must be precise. Examiners accept phrases like "reaction in molar quantities as shown in the equation". Standard states or conditions are not required for this specific definition.

❌ Common Errors

Students frequently lose the mark by stating "energy released", which falsely assumes every reaction is exothermic.

Part (b): Calorimetry Calculation, Assumptions, and Improvements

Question Part (b) - Extended Response (6 Marks)

📐 Step-by-Step Calculation

  1. Calculate energy change (q):
    Using q = m × c × ΔT
    m = 100.0 g (assuming density of solution = 1 g cm⁻³)
    c = 4.18 J g⁻¹ K⁻¹
    ΔT = 38.1 - 19.5 = 18.6 K
    q = 100.0 × 4.18 × 18.6 = 7774.8 J = 7.7748 kJ
  2. Calculate moles of limiting reagent (n):
    n(Cu(NO₃)₂) = concentration × volume / 1000
    n = 0.500 × (100 / 1000) = 0.0500 mol
  3. Calculate enthalpy change per mole (ΔH):
    ΔH = -q / n
    ΔH = -7.7748 / 0.05 = -155 kJ mol⁻¹ (to 3 SF)

🧠 Exam Technique & Level 3 Criteria

This is a banded "levels of response" question. To hit Level 3 (5–6 marks), you must:

  • Calculate the correct numerical enthalpy change with the minus sign and correct units ( kJ mol⁻¹ ).
  • State multiple valid assumptions (e.g., density of solution is 1 g cm⁻³, no heat loss).
  • State multiple valid improvements (e.g., use a polystyrene cup, use a lid).

💡 Valid Assumptions

  • Density of the solution is 1 g cm⁻³ (same as water).
  • Specific heat capacity of the solution is the same as water (4.18 J g⁻¹ K⁻¹).
  • No heat is lost to the surroundings or apparatus.
  • The mass and specific heat capacity of zinc are ignored.

❌ Common Calculation Traps

  • Sign Error: Forgetting to add the negative sign ( - ) to indicate an exothermic reaction.
  • Unit Mix-up: Failing to convert joules into kilojoules before dividing by moles, resulting in values off by a factor of 1000.
  • Significant Figures: Giving answers to 1 or 2 SF instead of standard 3 or 4 SF.

Part (c): Proportional Changes in Calorimetry

Question Part (c)

✅ Correct Answer

  • Half the energy ( q ) is released AND there are half the moles of reactants.
  • Therefore, the temperature change ( ΔT ) remains the same.

🧠 Examiner Insight

Top-level students recognized that scaling down both the volume of solution and the moles by the exact same factor (halving them) causes q and n to decrease proportionally. Because the ratio q / n stays constant, the calculated ΔH value does not change.

Topics

Module 3: Periodic table and energy · Practical Activity Groups · 3.2 Physical chemistry · PAG 3: Enthalpy determination

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.