Question

Consider the following hypothetical reaction: 2 P + Q → 2 R + S

The following mechanism is proposed for this reaction:
P + P → T (fast)
Q + T → R + U (slow)
U → R + S (fast)
Substances T and U are unstable intermediates. What rate law is predicted by this mechanism?
a. Rate = k[P][Q]2
b. Rate = k[U]
c. Rate = k[P]2[Q]
d. Rate = k[P][Q]
e. Rate = k[P]2

Answer 1

Option C is correct.

Rate = K [P]² [Q]

Step-by-step explanation:

P + P → T (fast)

Q + T → R + U (slow)

U → R + S (fast)

Overall reaction is

2 P + Q → 2 R + S

Let the rate constant for equations 1, 2 and 3 in the mechanism be k₁, k₂ and k₃.

The rate law is usually obtained from the slow reaction because it is theoretically the rate determining step.

Rate = k₂ [Q] [T]

But for the other reactions in the mechanism, the rate constants can be written as

k₁ = [T]/[P][P] = [T]/[P]²

[T] = k₁ [P]²

k₃ = [U]/[R][S]

[U] = k₃ [R][S]

But, since only [T] is the intermediate in the in the rate determining equation, we substitute for [T] in the rate determining equation.

Rate = k₂ [Q] [T] = k₂ [Q] k₁ [P]² = k₁k₂ [Q] [P]²

k₁k₂ = K

Rate = K [P]² [Q]