OCR A-Level Chemistry Unified chemistry (03), November 2020: Question 5

9 marks · Hard difficulty · Calculations

Calculate enthalpy change and entropy change for an equilibrium using van 't Hoff equation data, a graph of ln Kp against 1/T, and the effect of temperature on equilibrium position.

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Question

Question text

5 The equilibrium constant Kp and temperature T (in K) are linked by the mathematical relationship

shown in equation 5.1 (R = Gas constant in J mol−1 K−1 and ∆H is enthalpy change in J mol−1).

∆H 1 ∆S

lnKp = – × + Equation 5.1

R T R

(a) The table shows the values of Kp at different temperatures for an equilibrium.

Complete the table by adding the missing values of and lnKp.

Temperature, T/ K 400 500 600 700 800

K 3.00 × 1058 5.86 × 1045 1.83 × 1037 1.46 × 1031 1.14 × 1026

1 −1 −3

/ K 2.50 × 10

T …

ln Kp 135

[2]

(b) State and explain how increasing the temperature affects the position of this equilibrium and

whether the forward reaction is exothermic or endothermic.

… [1]

Use your graph and equation 5.1 to determine ∆H, in kJ mol−1, for this equilibrium.

Give your answer to 3 significant figures.

∆H = … kJ mol−1 [4]

(d) Explain how ∆S could be calculated from a graph of lnKp against .

… [2]

01.0 × 10–3 2.0 × 10–3 3.0 × 10–3 4.0 × 10–3

1 –1

T /K

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

5 (a) T /K 500 600 700 800 2 1.2×2

Kp 5.86 × 1045 1.83 × 1037 1.46 × 1031 1.14 × 1026 Mark by row

T 2.00 × 10–3 1.67 × 10–3 1.43 × 10–3 1.25 × 10–3 ALLOW 2 SF or more for 1/T but ignore

/K–1 trailing zeroes

ln Kp 105 86 72 60 ALLOW whole numbers (±1) for ln K

Calculator values ALLOW 1 small slip in each row.

1/T /10–3 2.00 1.66 recurring 1.428571429 1.25

e.g. 1.66 for 1.67; 71.7 for 71.8

ln Kp 105.3844788 85.79996441 71.75857432 59.99824068 Check with calculator values below table

BUT DO NOT ALLOW whole number errors,

e.g. 85 for 86

(b) Equilibrium (position) shifts to the left 1 2.2 ALLOW ‘favours reverse reaction’

AND Implies shift to left

(forward) reaction is exothermic

ALLOW ‘shifts in endothermic direction’ BUT

only if (forward) reaction stated as exothermic

element

All points correctly plotted

AND best-fit straight line 3.1

Gradient

Correct gradient of best-fit straight line within the range

±57000 → ±63000 3.1

∆H calculation (subsumes mark for gradient)

∆H = (–) gradient × 8.31(4) OR calculated value 3.2

e.g. from ±60000, ∆H = (+)498840 (J) OR ±498.840 (kJ)

ALLOW 4 points on graph

∆H in kJ mol–1 Tolerance 1 small square

∆H correct in kJ mol–1

AND 3SF

AND – sign 3.2 ALLOW ∆H in range: –480 → –530 (kJ mol–1)

e.g. from ±498840, ∆H = –499 (kJ mol–1) This mark subsumes gradient mark

(d) Extrapolate line to (y) intercept OR Measure/Use (y) intercept 2 3.1×2 ALLOW substitute values of ln Kp, 1/T and

∆S gradient into Equation 5.1

Intercept = OR ∆S = R × (y) intercept

This statement automatically subsumes 1st mark From provided values and gradient = 60000:

∆S

= ln Kp – gradient × 1/T

NOTE: If ‘x’ intercept, DO NOT ALLOW 1st mark but 2nd mark R

OR 135 – 60000 × 2.50 × 10–3 = –15

available for × R as BOD

Total 9

How to answer it

Thermodynamics & Equilibrium Constants (Kp)

OCR A-Level Chemistry • Exam Question Breakdown

What this question tests

This question assesses your mastery of advanced chemical thermodynamics and equilibria. Specifically, it tests your ability to manipulate exponential data using natural logarithms (ln Kp), interpret linear relationships in the form y = mx + c linked to the Van 't Hoff equation, calculate enthalpy changes (ΔH) from graph gradients, apply Le Chatelier's principle to temperature changes, and understand entropy changes (ΔS) via graph intercepts.

Part (a): Completing the Data Table

Processing Temperature and Equilibrium Constant Data

✅ Correct Answers

1/T / K⁻¹:
500 K: 2.00 × 10⁻³
600 K: 1.67 × 10⁻³
700 K: 1.43 × 10⁻³
800 K: 1.25 × 10⁻³
ln Kp:
500 K: 105
600 K: 86
700 K: 72
800 K: 60

❌ Common Errors

Failing to apply reciprocal calculations correctly ( 1 / T ).
Inappropriate rounding of natural logarithm values or missing significant figures consistency.
Making arithmetic slips when working with standard form numbers on calculators.

Marks available: 2 (1 mark per completed row. 1 small slip allowed per row).

Part (b): Effect of Temperature on Equilibrium

Le Chatelier's Principle & Enthalpy Link

✅ Correct Answer

Equilibrium position shifts to the left AND the forward reaction is exothermic.

💡 Key Knowledge

As temperature increases across the table, the value of Kp decreases dramatically (from 3.00 × 10⁵⁸ down to 1.14 × 10²⁶). A smaller Kp at higher temperatures means the equilibrium shifts to oppose the temperature increase by moving in the endothermic direction. Therefore, the forward reaction must be exothermic.

Marks available: 1 (Must state both direction of shift and sign/type of enthalpy change).

Part (c): Graphical Determination of Enthalpy Change (ΔH)

Plotting, Gradients, and Calculation Steps

🧠 Exam Technique & Plotting

Ensure all 4 points from the table are plotted accurately onto the grid. Draw a best-fit straight line passing through or balancing the points. Choose a large gradient triangle that spans at least half of your plotted line to minimize reading errors.

📐 Calculation Steps for ΔH

Determine the gradient (m):
From Equation 5.1 ( ln Kp = (-ΔH/R) × (1/T) + (ΔS/R) ), the gradient of the line is equal to -ΔH / R .
Calculate ΔH in Joules:
-ΔH = gradient × R
Using standard R = 8.31 J mol⁻¹ K⁻¹ and an accepted gradient around +60000:
-ΔH = 60000 × 8.31 = 498600 J mol⁻¹
Convert to kJ mol⁻¹ and apply sign:
Divide by 1000 to get 498.6 kJ mol⁻¹ . Include the negative sign because the gradient is positive (making ΔH negative): -499 kJ mol⁻¹ (to 3 SF).

❌ Common Calculation Traps

Unit mismatch: Forgetting to divide Joules by 1000 to convert final enthalpy into kJ mol⁻¹.
Sign errors: Omitting the negative sign in front of the gradient or final ΔH value.
Significant figures: Giving answers to 2 SF or 4+ SF when the question explicitly requests 3 significant figures.

Marks available: 4 (1 for correct plotting/line, 1 for correct gradient range, 1 for calculation using R, 1 for units, sign, and 3 SF).

Part (d): Calculating Entropy Change (ΔS)

Using the Y-Intercept

✅ Correct Answer

Extrapolate the line of best fit back to the y-axis to find the y-intercept. Since the equation of the line is y = mx + c , the y-intercept represents ΔS / R . Multiply this intercept value by the gas constant R (8.31) to find ΔS.

💡 Examiner Insight

Top-level responses immediately recognized the linear equation format matching Equation 5.1. Examiners noted that students who tried to use the x-intercept instead failed to score the first mark because the x-intercept calculates ΔS / ΔH , which is mathematically invalid for a direct ΔS isolation without further complex steps.

Marks available: 2 (1 for identifying/measuring the y-intercept or substituting into Equation 5.1, 1 for multiplying by R to calculate ΔS).

Topics

Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH · 5.2 Energy

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.