OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2024: Question 9

1 mark · Medium difficulty · Multiple Choice

Calculate the activation energy for a reaction given the gradient of a ln(k) versus 1/T graph.

Practise this question

Question

Multiple choice question 9 asks for the activation energy in kJ mol^-1 given that a graph of ln(k) plotted against 1/T has a gradient with a numerical value of -16000. Four options are provided: A (+1.92), B (+133), C (+1920), and D (+133000), along with a box for the answer and a mark allocation of [1].
Question text

9 A graph of ln(k) is plotted against 1 /T for a reaction.

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

The gradient has the numerical value of –16 000.

What is the activation energy, in kJ mol–1, for this reaction?

A +1.92

B +133

C +1920

D +133 000

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 9 is option B, worth 1 mark.

9 B 1 ALLOW +133 (correct numerical answer)

How to answer it

Calculating Activation Energy from an Arrhenius Graph

What this question tests:

This question assesses your ability to apply the linear form of the Arrhenius equation, manipulate graphical parameters (gradient), convert units between joules and kilojoules, and correctly handle negative signs when determining activation energy.

Question 9 — Multiple Choice [1 Mark]

Arrhenius Gradient and Activation Energy

✅ Correct Answer

B (+133)

The correct numerical value for the activation energy in kJ mol⁻¹ is 133.

💡 Key Knowledge

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

🧠 Exam Technique

  • Always write down the relevant algebraic formula from your data sheet before plugging in numbers.
  • Pay extreme attention to units: activation energy is typically calculated in J mol⁻¹ first, but exam questions frequently demand the final answer in kJ mol⁻¹ (requiring division by 1000).

❌ Common Errors

  • Forgetting to divide by 1000: Choosing option D ( +133 000 ) by leaving the answer in J mol⁻¹.
  • Sign confusion: Forgetting that the gradient is negative ( -16 000 ), meaning -Ea / R = -16 000 , which makes Ea a positive value.
  • Incorrect reciprocal usage: Multiplying incorrectly by R instead of dividing or rearranging wrongly.

📐 Step-by-Step Calculation

  1. Identify the gradient relationship:
    Gradient ( m ) = -16 000 = -Ea / R
  2. Rearrange for Activation Energy (Ea):
    Ea = - (Gradient × R)
    Ea = - (-16 000 × 8.314)
    Ea = +133 024 J mol⁻¹
  3. Convert units to kJ mol⁻¹:
    133 024 ÷ 1000 = 133.024 kJ mol⁻¹
  4. Apply appropriate significant figures:
    Given the data in the stem (-16 000 has 2 significant figures, but standard constants allow 3), rounding to 3 significant figures yields +133 kJ mol⁻¹.
Examiner Note: The mark scheme accepts the correct numerical answer (+133) for 1 mark. Distractor A (+1.92) catches students who might have divided R by the gradient instead, whilst C (+1920) and D test various unit conversion failures.

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), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.