Question:
A reaction A$_2$ + B$_2$ → 2AB occurs by the following mechanism:
A$_2$ → A + A (slow)
A + B$_2$ → AB + B (fast)
A + B → AB (fast)
Its order would be:
A reaction A$_2$ + B$_2$ → 2AB occurs by the following mechanism:
A$_2$ → A + A (slow)
A + B$_2$ → AB + B (fast)
A + B → AB (fast)
Its order would be:
A$_2$ → A + A (slow)
A + B$_2$ → AB + B (fast)
A + B → AB (fast)
Its order would be:
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In multi-step reactions, the slowest step determines the reaction order, as it is the rate-limiting step.
Updated On: Feb 25, 2025
- 1
- 2
- Zero
- \( \frac{1}{2} \)
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The Correct Option is A
Solution and Explanation
Since the rate-determining step is the first step (A$_2$ → A + A), the rate law depends only on A$_2$. Thus, the reaction follows first-order kinetics with respect to A$_2$, making the overall reaction first order.
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