The hypothetical reaction $\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB} ;$ follows the…
- 0
- 1
- 2
- $3 / 2$
Solution
$\mathrm{A}_{2} \longrightarrow \mathrm{A}+\mathrm{A}$ (Fast);
$\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}+\mathrm{B}$ (Slow)
Rate law $=\mathrm{k}[\mathrm{A}]\left[\mathrm{B}_{2}ight]$ put value of $[\mathrm{A}]$ from Ist reaction since $\mathrm{A}$ is intermediate
$\sqrt{\mathrm{k}\left[\mathrm{A}_{2}ight]}=\mathrm{A}$
$\therefore$ Rate law equation $=\mathrm{K} \sqrt{\mathrm{k}\left[\mathrm{A}_{2}ight]}\left[\mathrm{B}_{2}ight]$
$\therefore$ Order $=\frac{1}{2}+1=\frac{3}{2}$ ,
Asked in: JEE-TOPICTESTS-CHEMISTRY