The reaction A_2 + B_2 rightarrow 2AB follows | vedclass

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Question

(A) The rate of the reaction is determined by the slow step: $	ext{Rate} = k_2[A][B_2] \dots (1)$From the fast equilibrium step,$A_2 ightleftharpoo

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(A) The rate of the reaction is determined by the slow step: $	ext{Rate} = k_2[A][B_2] \dots (1)$From the fast equilibrium step,$A_2 ightleftharpoons 2A$,the equilibrium constant is $K_{eq} = \frac{[A]^2}{[A_2]} = \frac{k_1}{k_{-1}}$.Thus,$[A] = \sqrt{\frac{k_1}{k_{-1}}} [A_2]^{1/2}$.Substituting this into equation $(1)$,we get $	ext{Rate} = k_2 \sqrt{\frac{k_1}{k_{-1}}} [A_2]^{1/2} [B_2]^1$.The overall order of the reaction is the sum of the exponents of the concentration terms: $0.5 + 1 = 1.5$.

The reaction $A_2 + B_2 \rightarrow 2AB$ follows the mechanism $A_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2A$ (fast),$A + B_2 \xrightarrow{k_2} AB + B$ (slow),$A + B \rightarrow AB$ (fast). The overall order of the reaction is:

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