What is the rate law for the following mechanism: A-B+C=D (fast equilibrium; rate of forward reaction: k1, rate of reverse reaction: k.1) A+D -E (slow, rate: k2) E +G - H (fast, rate: k3) Rate= k2 (k1 k.j) [Aj°{B][C] OA Rate= k2 (k1 k.1)[A][D] OB. oc Rate= k1 [A][B][C] Rate= k2 [AJ°{B][C] OD. O E. Rate= k2 (k1/k.1) [A][B][C]

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Chapter1: Chemical Foundations
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What is the rate law for the following mechanism:
A+B +C2D (fast equilibrium; rate of forward reaction: k1, rate of reverse reaction: k_1)
A+D -E (slow, rate: k2)
E +G - H
(fast, rate: k3)
OA Rate= k2 (k1 k.j) [A]²{B][C]
Rate= k2 (k1/k-1)[A]D]
oc Rate= k1 [A][B][C]
OD, Rate= k2 [A]²B][C]
O E. Rate= k2 (k1 /k-1) [A][B][C]
Transcribed Image Text:What is the rate law for the following mechanism: A+B +C2D (fast equilibrium; rate of forward reaction: k1, rate of reverse reaction: k_1) A+D -E (slow, rate: k2) E +G - H (fast, rate: k3) OA Rate= k2 (k1 k.j) [A]²{B][C] Rate= k2 (k1/k-1)[A]D] oc Rate= k1 [A][B][C] OD, Rate= k2 [A]²B][C] O E. Rate= k2 (k1 /k-1) [A][B][C]
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