The reaction \(\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}\) follows the mechanism \(\begin{aligned}…

The reaction \(\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}\) follows the mechanism
\(\begin{aligned}
& A_2 \xrightarrow[k_{-1}]{k_1} A+A \text { (fast) } \\
& A+B_2 \xrightarrow{k_2} A B+B \text { (slow) } \\
& A+B \rightarrow A B \text { (fast) }
\end{aligned}\)
The overall order of the reaction is :
  1. 2
  2. 2.5
  3. 3
  4. 1.5

Solution

$\begin{aligned}
& \text { rate }=\mathrm{k}_2[\mathrm{~A}]\left[\mathrm{B}_2\right] \ldots \ldots \\
& \left(\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}\right)=\left(\frac{[\mathrm{A}]^2}{\left[\mathrm{~A}_2\right]}\right) \\
& \Rightarrow[\mathrm{A}]=\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}} \cdot \sqrt{\left[\mathrm{~A}_2\right]}
\end{aligned}$
Substituting in (1) ; we get
$\begin{aligned} & \text { Rate }=\mathrm{k}_2 \sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}} \cdot\left[\mathrm{~A}_2\right]^{\frac{1}{2}} \cdot\left[\mathrm{~B}_2\right] \\ & \therefore \text { order }=\left(\frac{3}{2}\right)=1.5\end{aligned}$

Asked in: JEE Main 2025 (29 Jan Shift 1)

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