Edexcel A-Level Chemistry Paper 1, June 2023: Question 5

7 marks · Medium difficulty · Calculations

Plot a graph of delta G against temperature, calculate the entropy change of the system from the gradient, identify the feature representing the enthalpy change, and determine the effect of increased pressure on the equilibrium position and colour for the decomposition of dinitrogen tetroxide.

Practise this question

Question

Exam question about the decomposition of dinitrogen tetroxide containing a table of temperature and delta G values, a grid to plot a graph of delta G against temperature, multiple-choice questions on graph intercepts, and a multiple-choice question on the effect of pressure on equilibrium and colour.
Question text

5 This question is about the decomposition of dinitrogen tetroxide.

The reaction eventually reaches equilibrium.

N2O4(g) 2NO2(g)

colourless brown

(a) The table shows values of ΔG at different temperatures for this reaction.

Temperature / K ΔG / kJ mol–1

350 –4.0

400 –13

450 –22

500 –31

550 –40

(i) Plot a graph of ΔG against temperature.

(2)

ΔG / kJ mol–1

Temperature / K

(ii) Calculate the entropy change of the system, ΔS , in J K–1 mol–1, using your

system

straight line from the graph in (a)(i) and the equation shown.

8 ΔG = –TΔS + ΔH

system

(3)

(iii) What feature of the graph in (a)(i) gives the enthalpy change of the reaction?

(1)

A intercept of the x‑axis

B (intercept of the x‑axis)×–1

C intercept of the y‑axis

D (intercept of the y‑axis)×–1

(b) What happens to the position of the equilibrium and the colour of the mixture

when the pressure is increased?

The volume of the system remains constant.

N2O4(g) 2NO2(g)

colourless brown

(1)

Position of equilibrium Change in colour

A moves to the right mixture gets lighter

B moves to the right mixture gets darker

C moves to the left mixture gets lighter

D moves to the left mixture gets darker

Mark scheme

Show the mark scheme Mark scheme showing the expected graph with correct scale, plotting, and line of best fit, followed by calculation steps for the entropy change using the gradient, and the correct multiple-choice answers for parts (a)(iii) and (b).

Question

Answer Additional Guidance Mark

Number

5(a)(i) A graph that shows (2)

300 350 400 450 500 550 600

-10

-20

∆G

/ kJ mol−1

-30

-40

-50

Temperature / K

• suitable scale, points must cover at Allow break in x axis

least half the available space in both Allow x axis values at top or bottom of graph

directions (1) y axis must go in correct direction

Stand alone mark based on scale used

• all points plotted correctly and

Allow ± ½ a square

straight line of best fit (1)

Question

Answer Additional Guidance Mark

Number

5(a)(ii) Example of calculation (3)

• deduction that gradient = −ΔSsystem (1) (y = mx + c so) m = −ΔSsystem

M1 could be subsumed within award of M2

(36 ÷ (−200)) = −0.18 (kJ K−1 mol−1)

• calculation of gradient (1)

TE on graph in (i)

M2 can be subsumed within M3

Allow this left as a fraction e.g. -9/50

Penalise absence of negative sign once only in M1 and

M2

• calculation of ΔS in J K−1 mol−1 (1) −1 −1

system = (+)180 (J K mol )

TE on M2

Correct answer, with or without working, scores (3)

Ignore SF except 1 SF

Do not award incorrect units but allow J K− mol−

Note

Allow use of 2 simultaneous equations or calculation of

intercept and substitution

Question

Answer Mark

Number

5(a)(iii) The only correct answer is C (intercept of the y axis) (1)

A is not correct because the value would have units of temperature

B is not correct because the value would have units of temperature and be a negative value in K

D is not correct because this value is − ΔH

Question

Answer Mark

Number

5(b) The only correct answer is C (moves to the left, mixture gets lighter) (1)

A is not correct because there are more particles on the right-hand side, so increase in P moves equilibrium to the left

B is not correct because there are more particles on the right-hand side, so increase in P moves equilibrium to the left and N2O4 is

colourless, so mixture lightens

D is not correct because N2O4 is colourless, so mixture lighten

How to answer it

Decomposition of Dinitrogen Tetroxide Study Guide

What this question tests

This question assesses your ability to apply thermodynamics and equilibrium principles to the gas-phase decomposition of N₂O₄. Specifically, it tests graph-plotting skills, manipulating the linear Gibbs free energy equation (ΔG = -TΔSsystem + ΔH) to find entropy changes, identifying graphical intercepts, and predicting Le Chatelier's principle responses to pressure changes alongside visual colour observations.

Question 5 (a)(i)

Plotting a Graph of ΔG against Temperature

💡 Key Knowledge

  • Temperature (K) must go on the x-axis, and ΔG (kJ mol⁻¹) on the y-axis.
  • The data points follow a straight line with a negative gradient.

🧠 Exam Technique

  • Scale: Ensure plotted points cover at least half of the available grid space in both directions.
  • Line of Best Fit: Draw a single straight line passing cleanly through the trend of points. Allow a tolerance of ±½ a square for plotted points.
Mark Breakdown (2 marks):
• Mark 1: Suitable scale and points covering ≥ 50% of available grid space.
• Mark 2: All points plotted correctly and accurate straight line of best fit.
Question 5 (a)(ii)

Calculating Entropy Change of the System (ΔSsystem)

✅ Correct Answer & Calculation

Using the equation of a straight line ( y = mx + c ) matched to ΔG = (-ΔSsystem)T + ΔH :

  1. Step 1 (Deduction): Recognise that the gradient equals -ΔSsystem .
  2. Step 2 (Calculate gradient): Take two points on your line of best fit. Using examiner example points: Δy / Δx = (36 - (-200)) / ... giving a gradient of -0.18 kJ K⁻¹ mol⁻¹ (or equivalent depending on your drawn line).
  3. Step 3 (Convert units & sign): Multiply by -1 to get ΔS , and convert from kJ to J by multiplying by 1000:
    +180 J K⁻¹ mol⁻¹ (Accept range based on student line).

❌ Common Errors & Traps

  • Unit mismatch: ΔG is given in kJ mol⁻¹, but ΔSsystem is requested in J K⁻¹ mol⁻¹. Forgetting to multiply by 1000 leads to losing the final conversion mark.
  • Sign errors: Forgetting that gradient = minus ΔSsystem. Dropping negative signs during gradient calculation is a very common pitfall.
Mark Breakdown (3 marks):
 • Mark 1: Deduction that gradient = -ΔSsystem.
• Mark 2: Correct calculation of the gradient from the graph.
• Mark 3: Correct calculation of ΔSsystem in J K⁻¹ mol⁻¹ with correct sign.
Question 5 (a)(iii)

Identifying Enthalpy Change from the Graph

✅ Correct Answer: C (intercept of the y-axis)

Comparing ΔG = -ΔSsystem(T) + ΔH to the linear equation y = mx + c :

  • y-axis = ΔG
  • x-axis = T
  • Gradient (m) = -ΔSsystem
  • y-intercept (c) = ΔH

❌ Distractor Analysis

  • A & B: Involve the x-axis, which yields units of temperature (K), not energy.
  • D: Incorrectly applies a negative sign; the intercept directly reads positive ΔH.
Mark Breakdown (1 mark): 1 mark for selecting C.
Question 5 (b)

Effect of Pressure Change on Equilibrium and Colour

✅ Correct Answer: C

Position of equilibrium: moves to the left
Change in colour: mixture gets lighter

💡 Key Knowledge (Le Chatelier's Principle)

  • Reaction: N₂O₄(g) ⇌ 2NO₂(g) (1 mole gas vs 2 moles gas).
  • Increasing pressure shifts the equilibrium towards the side with fewer moles of gas (the left-hand side, producing more colourless N₂O₄).
  • Because brown NO₂ gas is consumed and colourless N₂O₄ is formed, the overall mixture gets lighter.
Mark Breakdown (1 mark): 1 mark for selecting C.

Topics

Physical Chemistry · Topic 13: Energetics II · Topic 10: Equilibrium I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.