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The following results were obtained during kinetic studies of the reaction; 2A + B → Products
Experiment I: [A] in mol L−1 = 0.1, [B] in mol L−1 = 0.2, Initial Rate of reaction (in mol L−1 min−1) = 6.93 × 10−3
Experiment II: [A] in mol L−1 = 0.1, [B] in mol L−1 = 0.25, Initial Rate of reaction (in mol L−1 min−1) = 6.93 × 10−3
Experiment III: [A] in mol L−1 = 0.2, [B] in mol L−1 = 0.3, Initial Rate of reaction (in mol L−1 min−1) = 1.386 × 10−2
The time (in minutes) required to consume half of A is
- 5
- 10
- 1
- 100
Correct answer: 10
Solution
Exp I vs II ([A] fixed, [B] changes, rate unchanged) -> order in B = 0. Exp I vs III ([A] doubled, rate doubled) -> order in A = 1. So rate = k[A]; k = 6.93e-3/0.1 = 0.0693 min^-1. t1/2 = 0.693/0.0693 = 10 min.
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