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:

<p>Reaction given:</p> <p>$A_2+B_2 \rightarrow 2 A B$</p> <p>Mechanism:</p> <p>$A_2\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{k_{-1}}^{k_1}} A+A$ (fast)</p> <p>$$\begin{aligned} & A+B_2 \xrightarrow{k_2} A B+B \text { (slow) } \\ & A+B \longrightarrow A B \text { (fast) } \end{aligned}$$</p> <p>The rate determining step in a mechanism is the slow step.</p> <p>So, late $=k_2[A]\left[B_2\right]$</p> <p>$A$ is the intermediate; to determine the concentration of $A$, the equilibrium is considered.</p> <p>$A_2\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{k_{-1}}^{k_1}} A+A$</p> <p>$$\begin{aligned} \frac{k_1}{k_{-1}} & =\frac{[A][A]}{\left[A_2\right]} \\ & =\frac{[A]^2}{\left[A_2\right]} \end{aligned}$$</p> <p>$$\begin{aligned} S_{0,}[A]^2 & =\frac{k_1}{k_{-1}}\left[A_2\right] \\ {[A] } & =\left\{\frac{k_1}{k_{-1}}\left[A_2\right]\right\}^{1 / 2} \end{aligned}$$</p> <p>Substitute $[A]$ in the late,</p> <p>late $=k_2[A]\left[B_2\right]$</p> <p>$$\begin{aligned} & =k_2\left(\frac{k_1}{k_{-1}}\right)^{1 / 2}\left[A_2\right]^{1 / 2}\left[B_2\right] \\ & =k_2\left(\frac{k_1}{k_{-1}}\right)^{1 / 2}\left[A_2\right]^{1 / 2}\left[B_2\right]^1 \end{aligned}$$</p> <p>$$\begin{aligned} \text { Overall rate } & =\frac{1}{2}+1 \\ & =\frac{3}{2} \\ & =1.5 \end{aligned}$$</p>