OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 8

1 mark · Easy difficulty · Multiple Choice

Calculate the rate constant for a first-order reaction given its half-life is 80 seconds.

Practise this question

Question

Multiple-choice question 8 asking to find the rate constant k in s^-1 for a first-order reaction with a half-life of 80 s. Four options are given: A 8.66 x 10^-3, B 0.0125, C 55.5, D 115. A box for 'Your answer' is shown at the bottom with a 1-mark allocation.
Question text

8 The half-life for a first order reaction is 80 s.

What is the rate constant k, in s–1, for this reaction?

A 8.66 × 10–3

B 0.0125

C 55.5

D 115

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating question number 8 has the correct answer A, worth 1 mark.

8 A 1 1.2

How to answer it

Calculating the Rate Constant from Half-Life

What this question tests: This question assesses your ability to recall and apply the quantitative relationship between the half-life ( t₁ₔ₂ ) and the rate constant ( k ) for a first-order reaction, as well as rearranging equations and evaluating numerical values to appropriate significant figures.
Question 8 (Multiple Choice)

Exam Breakdown

✅ Correct Answer: A

8.66 × 10⁻³ s⁻¹ is the correct rate constant calculated using the first-order half-life equation.

Marks: 1 / 1

💡 Key Knowledge

  • For first-order reactions, half-life is constant and independent of concentration.
  • The defining equation is: t₁ₔ₂ = ln(2) / k
  • Rearranging for the rate constant gives: k = ln(2) / t₁ₔ₂ (or 0.693 / t₁ₔ₂ ).

🧠 Exam Technique

Always double-check which order the reaction is specified as before choosing a half-life formula. Zero-order, first-order, and second-order reactions all have completely different half-life relationships!

📐 Step-by-Step Calculation

  1. Identify given data: t₁ₔ₂ = 80 s (First-order)
  2. Select formula: k = ln(2) / t₁ₔ₂
  3. Substitute values: k = 0.693147... / 80
  4. Calculate final answer: k = 0.008664... = 8.66 × 10⁻³ s⁻¹

❌ Common Errors & Distractors

  • Distractor B (0.0125): Calculated incorrectly as 1 / 80 (treating it as a rate formula 1/t rather than using ln(2) ).
  • Inverted calculations: Dividing the half-life by ln(2) instead of dividing ln(2) by the half-life.

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 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.