Reaction Kinetics — A/L Chemistry

How do I practise Reaction Kinetics?

1 question with a worked solution

Check before you answer: Order vs Stoichiometry: Proof that reaction order (m, n) is experimentally determined and cannot be deduced from overall stoichiometric coefficients.

1. For the reaction 2A + B → C, initial rate experiments show that doubling [A] at constant [B] doubles the rate, while doubling [B] at constant [A] quadruples the rate. What is the overall reaction order and the derived SI units of the rate constant k?
   1. A) Overall order = 3, with rate constant units mol⁻² dm⁶ s⁻¹
   2. B) Overall order = 2, with rate constant units mol⁻¹ dm³ s⁻¹
   3. C) Overall order = 3, with rate constant units mol² dm⁻⁶ s⁻¹
   4. D) Overall order = 1, with rate constant units s⁻¹
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   Correct answer: A) Overall order = 3, with rate constant units mol⁻² dm⁶ s⁻¹

   Rate ∝ [A]^m [B]ⁿ. Doubling [A] doubles rate ⇒ 2^m = 2 ⇒ m = 1 (first order in A). Doubling [B] quadruples rate ⇒ 2ⁿ = 4 ⇒ n = 2 (second order in B). Overall order = m + n = 1 + 2 = 3. Rate law: Rate = k [A] [B]². Units of k = (Rate) / ([A][B]²) = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)(mol dm⁻³)²) = mol⁻² dm⁶ s⁻¹.

   Show worked solution

   Step 1: Determine order with respect to A: Rate₂ / Rate₁ = ([A]₂ / [A]₁)^m ⇒ 2 = 2^m ⇒ m = 1. Step 2: Determine order with respect to B: Rate₃ / Rate₁ = ([B]₃ / [B]₁)ⁿ ⇒ 4 = 2ⁿ ⇒ n = 2. Step 3: Calculate overall order: Order = m + n = 1 + 2 = 3. Step 4: Determine units of rate constant k: Rate = k [A]¹ [B]² ⇒ k = Rate / ([A][B]²) Units = (mol dm⁻³ s⁻¹) / [(mol dm⁻³) × (mol dm⁻³)²] = (mol dm⁻³ s⁻¹) / (mol³ dm⁻⁹) = mol⁻² dm⁶ s⁻¹. Answer: Overall order = 3, with rate constant units mol⁻² dm⁶ s⁻¹.
   Modelled on: the A/L Chemistry examination pattern

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Where does Reaction Kinetics fit in the syllabus?

Grade 13 · Term 1