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2 August, 08:02

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

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Answers (1)
  1. 2 August, 11:58
    0
    Option C is correct.

    Rate = K [P]² [Q]

    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]
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