The hypothetical reaction $\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB} ;$ follows the…

The hypothetical reaction $\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB} ;$ follows the following mechanism $\mathrm{A}_{2} \stackrel{\text { Fast }}{\longrightarrow} \mathrm{A}+\mathrm{A}$, $\mathrm{A}+\mathrm{B}_{2} \stackrel{\text { Slow }}{\longrightarrow} \mathrm{AB}+\mathrm{B}, \mathrm{A}+\mathrm{B} \stackrel{\text { Fast }}{\longrightarrow} \mathrm{AB}$ The order of the overall reaction is
  1. 0
  2. 1
  3. 2
  4. $3 / 2$

Solution

$\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}$;
$\mathrm{A}_{2} \longrightarrow \mathrm{A}+\mathrm{A}$ (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

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