QJEEM25Chb8fb: JEE Main 2025 Chemistry - Class 12 JEEM25Chb8fb Solution: The reaction follows the mechanism: (fast) (slow) (fast) The…

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