Mechanism of a hypothetical reaction X_2 + Y_2 → 2XY is given below:(i) X_2 leftharpoons X + X (fast)(ii) X + Y_2 → XY + Y (slow)(iii) X + Y → XY (fas - Chemistry MCQ NEET 2017 2026 | ExamTest.live

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Mechanism of a hypothetical reaction X_{2} + Y_{2} \to 2 XY is given below:(i) X_{2} ⇌ X + X (fast)(ii) X + Y_{2} \to XY + Y (slow)(iii) X + Y \to XY (fast)The overall order of the reaction will be:

Solution & Step-by-step Explanation

1. The slow step is the rate-determining step: R a t e = k_{2} [X] [Y_{2}] .2. From the fast equilibrium step (i): K_{e q} = \frac{[ X ] ^{2}}{[ X _{2} ]} ⟹ [X] = K_{e q} [X_{2}] = K_{e q}^{1/2} [X_{2}]^{1/2} .3. Substitute [X] in the rate law: R a t e = k_{2} (K_{e q}^{1/2} [X_{2}]^{1/2}) [Y_{2}] = K^{'} [X_{2}]^{0.5} [Y_{2}]^{1} .4. Overall order = 0.5 + 1 = 1.5 .

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Mechanism of a hypothetical reaction X_{2} + Y_{2} \to 2 XY is given below:(i) X_{2} ⇌ X + X (fast)(ii) X + Y_{2} \to XY + Y (slow)(iii) X + Y \to XY (fast)The overall order of the reaction will be:

1

2

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1.5

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