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JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard
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Analyzing +4 | -1
The reaction A2+B2→2AB\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB} follows the mechanism

A2⇌k−1k1A+A (fast) A+B2→k2AB+B (slow) A+B→AB (fast) \begin{aligned} & {{\text{A}}_2}\mathop \rightleftharpoons \limits_{{{\text{k}}_{ - 1}}}^{{{\text{k}}_1}} {\text{A}} + {\text{A (fast) }} \\ & \mathrm{A}+\mathrm{B}_2 \xrightarrow{\mathrm{k}_2} \mathrm{AB}+\mathrm{B} \text { (slow) } \\ & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB} \text { (fast) } \end{aligned}

The overall order of the reaction is:
View answer and explanation
Correct option: A
1.5
Option A (1.5) is correct because the rate law is determined by the slow step, and the concentration of the intermediate A is found using the equilibrium of the fast step, resulting in an overall order of 1.5.

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