JEE MAIN - Chemistry (2025 - 29th January Morning Shift - No. 8)
The reaction $A_2 + B_2 \rightarrow 2AB$ follows the mechanism:
$A_2 \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} A + A$ (fast)
$A + B_2 \xrightarrow{k_2} AB + B$ (slow)
$A + B \rightarrow AB$ (fast)
The overall order of the reaction is:
0
1
2
3
Explanation
$$\begin{aligned}
&\begin{aligned}
& \text { rate }=\mathrm{k}_2[\mathrm{~A}]\left[\mathrm{B}_2\right] \quad\text{.... (1)}\\
& \left(\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}\right)=\left(\frac{[\mathrm{A}]^2}{\left[\mathrm{~A}_2\right]}\right) \\
& \Rightarrow[\mathrm{A}]=\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}} \sqrt{\left[\mathrm{~A}_2\right]}
\end{aligned}\\
&\text { Substituting in (1) ; we get }\\
&\begin{aligned}
& \text { Rate }=\mathrm{k}_2 \sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}} \cdot\left[\mathrm{~A}_2\right]^{\frac{1}{2}} \cdot\left[\mathrm{~B}_2\right] \\
& \therefore \text { order }=\left(\frac{3}{2}\right)=1.5
\end{aligned}
\end{aligned}$$
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