OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2019: Question 11

1 mark · Medium difficulty · Multiple Choice

Calculate the activation energy in kJ mol⁻¹ given the gradient of an Arrhenius plot of ln k against 1/T.

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Question

Multiple choice question 11. The text states: A graph of ln k against 1/T (T in K) for a reaction has a gradient with the numerical value of -4420. What is the activation energy, in kJ mol⁻¹, for this reaction? Four options are given: A, -532; B, -36.7; C, +36.7; D, +5.32 x 10⁵. An answer box is provided at the bottom with 1 mark allocated.
Question text

11 A graph of lnk against T (T in K) for a reaction has a gradient with the numerical value of −4420.

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

A −532

B −36.7

C +36.7

D +5.32 × 105

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme for question 11 showing the correct answer is C.

11 C 1 AO2.6

How to answer it

Arrhenius Equation & Activation Energy

OCR A-Level Chemistry • Multiple Choice Question

What this question tests

This question assesses your understanding of the graphical form of the Arrhenius equation (ln k against 1/T), your ability to relate the gradient of this line to activation energy (Ea), and your competence in converting units between Joules and kilojoules per mole.

Question 11

Determining Activation Energy from a Gradient

✅ Correct Answer: C (+36.7)

The correct option is C because the calculation correctly accounts for the negative sign in the gradient formula and converts J mol⁻¹ into kJ mol⁻¹ by dividing by 1000.

💡 Key Knowledge

  • The linear Arrhenius equation is: ln k = (-Ea / R) × (1 / T) + ln A
  • Comparing this to the equation of a straight line ( y = mx + c ):
  • y-axis = ln k
  • x-axis = 1 / T
  • Gradient ( m ) = -Ea / R
  • Universal gas constant ( R ) = 8.314 J mol⁻¹ K⁻¹ (found in your Data Sheet).

🧠 Exam Technique

Always write out the underlying formula next to your working space. When dealing with Arrhenius gradients, watch out for double negatives: a negative gradient multiplied by a negative R yields a positive activation energy.

❌ Common Errors

  • Forgetting to divide by 1000: Leaving the answer in J mol⁻¹ instead of kJ mol⁻¹ (leads to option D).
  • Sign errors: Forgetting that gradient = -Ea / R , resulting in a negative activation energy (options A or B). Activation energy can never be negative!

📐 Step-by-Step Calculation Breakdown

  1. Identify the gradient relationship:
    Gradient = -Ea / R = -4420
  2. Rearrange to solve for Ea in J mol⁻¹:
    Ea = - (gradient × R)
    Ea = - (-4420 × 8.314)
    Ea = +36,747.88 J mol⁻¹
  3. Convert units to kJ mol⁻¹:
    Divide by 1000: 36,747.88 / 1000 = +36.74788 kJ mol⁻¹
  4. Apply appropriate significant figures:
    The gradient (-4420) is given to 3 significant figures, so round your final answer to 3 significant figures: +36.7 kJ mol⁻¹ .
Examiner Commentary: This question discriminated well between candidates who memorised the formula mechanically and those who truly understood graphical analysis. Top-level responses quickly recognised that activation energy must always be a positive value and successfully navigated the unit conversion from Joules to kilojoules.

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 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.