OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 8

1 mark · Medium difficulty · Multiple Choice

Calculate the activation energy in kJ mol⁻¹ given the gradient of a ln(k) against 1/T graph.

Practise this question

Question

Multiple choice question 8 asking to find the activation energy in kJ mol⁻¹ given a gradient of -55 000 for a ln(k) against 1/T graph. Four options are provided: A (+1.5 x 10⁻⁷), B (+2.22 x 10⁻⁶), C (+6.62), and D (+457). An answer box is shown at the bottom left with a [1] mark allocation at the bottom right.
Question text

8 A graph is plotted of ln(k) against 1 /T.

(k = rate constant, T = temperature in K)

The gradient has the numerical value of −55 000.

What is the activation energy, in kJ mol−1?

A +1.5 × 10−7

B +2.22 × 10−6

C +6.62

D +457

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme showing question number 8 with correct answer D and 1 mark.

8 D 1 2.6

How to answer it

Arrhenius Equation & Graphical Analysis

📌 What this question tests

This question assesses your understanding of the Arrhenius equation in logarithmic form, specifically how to interpret the gradient of an ln(k) against 1/T graph to calculate the activation energy ( Ea ), including essential unit conversions from Joules to kilojoules.

Question 8: Multiple Choice Solution

Determining Activation Energy from a Graph Gradient

✅ Correct Answer: D (+457)

The correct option is D. The calculated activation energy comes out to approximately +457 kJ mol⁻¹ after converting from J mol⁻¹ to kJ mol⁻¹.

Awarded 1 mark for selecting D.

💡 Key Knowledge

  • The linear Arrhenius equation is: ln(k) = (-Ea / R) × (1/T) + ln(A)
  • This matches the straight-line equation format y = mx + c .
  • Therefore, the gradient m = -Ea / R .
  • The ideal gas constant R = 8.314 J mol⁻¹ K⁻¹ .

🧠 Exam Technique

  • Always write down the relevant formula from your data booklet first, even in multiple-choice questions.
  • Watch out for negative signs: both the gradient and the Arrhenius formula contain negative components that cancel out for Ea .
  • Pay extreme attention to unit prefixes ( J vs kJ ).

❌ Common Errors

  • Forgetting to divide by 1000: Leaving the answer in J mol⁻¹ instead of converting to kJ mol⁻¹.
  • Sign confusion: Dropping or mismanaging the negative sign of the gradient, leading to negative activation energy values (which are chemically impossible).
  • Incorrect R value usage: Using incorrect constants or forgetting that activation energy must be positive.

📐 Step-by-Step Calculation

  1. Identify the relationship:
    Gradient = -Ea / R
  2. Substitute known values:
    -55000 = -Ea / 8.314
  3. Rearrange to solve for Ea in J mol⁻¹:
    Ea = 55000 × 8.314 = 457270 J mol⁻¹
  4. Convert to kJ mol⁻¹ (divide by 1000):
    Ea = 457.27 kJ mol⁻¹
  5. Round to appropriate significant figures:
    Given the data points in the stem, rounding to 3 significant figures gives +457 kJ mol⁻¹ (Option D).

Topics

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

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.