OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 12

1 mark · Easy difficulty · Multiple Choice

Calculate the half-life of a first-order reaction given a rate constant of 6.19 × 10⁻³ s⁻¹.

Practise this question

Question

Question 12 states: 'The rate constant, k, for a first order reaction is 6.19 × 10⁻³ s⁻¹. What is the half-life, in s, for the reaction?' with four multiple choice options: A 4.29 × 10⁻³, B 49, C 112, D 323. A box is provided for the student's answer, worth 1 mark.

Mark scheme

Show the mark scheme Mark scheme table row showing question number 12, correct answer C, and 1 mark.

How to answer it

Calculating Half-Life from the Rate Constant

What this question tests

This multiple-choice question assesses your ability to recall and apply the mathematical relationship linking the rate constant ( k ) and half-life ( t₁/₂ ) for a first-order reaction, and correctly handle exponential and logarithmic calculator functions under timed conditions.

Multiple Choice • 1 Mark

Question 12: Half-Life Calculation

✅ Correct Answer

C  (112 s)

Substituting the given value into the first-order half-life relationship yields exactly 111.97… s, which rounds to 112 s (3 significant figures).

💡 Key Knowledge

  • For any first-order reaction, half-life is constant and completely independent of the starting concentration.
  • The governing equation given in the OCR Data Sheet is:
    k = ln(2) / t₁/₂  or  t₁/₂ = ln(2) / k
  • ln(2) ≈ 0.69315 (always use the ln button on your calculator for accuracy).
  • Units must match: since k is in s⁻¹ , t₁/₂ will directly be evaluated in seconds ( s ).

📐 Step-by-Step Calculation

Step 1: State the rearranged formula
We need half-life ( t₁/₂ ) in terms of the rate constant ( k ):
t₁/₂ = ln(2) / k
Step 2: Substitute the numerical value
Given k = 6.19 × 10⁻³ s⁻¹ :
t₁/₂ = 0.69315 / (6.19 × 10⁻³)
Step 3: Calculate and round
t₁/₂ = 111.978… s
To 3 significant figures (matching the data provided): t₁/₂ = 112 s

🧠 Exam Technique & Strategy

  • Formula Sheet Lookup: The formula is provided in the OCR A data booklet under the Physical Chemistry section. Always check the exact form if you have a momentary mental block.
  • Quick Sanity Check: Because k is around 6 × 10⁻³ , 1 / k is roughly 1 / 0.006 ≈ 160 . Multiplying by ~0.7 gives a number around 110. This immediately rules out A, B, and D!

❌ Anatomy of Distractors

  • A (4.29 × 10⁻³): The candidate multiplied by ln(2) instead of dividing: ln(2) × k = 4.29 × 10⁻³ .
  • B (49): The candidate mistakenly used base-10 log ( log₁₀ ) instead of the natural log ( ln ): log₁₀(2) / k = 0.301 / 0.00619 ≈ 48.6 ≈ 49 .
  • D (323): The candidate divided by 2 rather than taking the natural log of 2: 2 / k = 2 / 0.00619 ≈ 323 .
Mark Scheme Breakdown: 1 mark awarded for selecting option C. No partial marks available for multiple-choice questions.

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 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.