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A reaction P -> R proceeds by the mechanism: P <-> 2Q (fast equilibrium); 2Q + P -> R (slow). What is the rate law for the overall reaction P -> R?
- k[P][Q]
- k[P]
- (k1k3[P]²)/k2
- k[P]³
Correct answer: (k1k3[P]²)/k2
Solution
The slow step rate = k3[Q]²[P]; substituting [Q]² = (k1/k2)[P] from the fast equilibrium gives rate = (k1k3/k2)[P]².
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