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The following results were obtained during kinetic studies of the reaction: 2A + B → Products
Experiment I: [A] = 0.10 mol L⁻¹, [B] = 0.20 mol L⁻¹, Initial Rate of reaction = 6.93 × 10⁻³ mol L⁻¹ min⁻¹
Experiment II: [A] = 0.10 mol L⁻¹, [B] = 0.25 mol L⁻¹, Initial Rate of reaction = 6.93 × 10⁻³ mol L⁻¹ min⁻¹
Experiment III: [A] = 0.20 mol L⁻¹, [B] = 0.30 mol L⁻¹, Initial Rate of reaction = 1.386 × 10⁻² mol L⁻¹ min⁻¹
The time (in minutes) required to consume half of A is -
- 5
- 1
- 100
- 10
Correct answer: 10
Solution
Exp I vs II: changing [B] leaves rate unchanged -> order in B = 0. Exp I vs III: doubling [A] doubles rate -> order in A = 1. So rate = k[A], k = 6.93e-3/0.10 = 0.0693 min^-1. t(1/2) = 0.693/k = 10 min.
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