Question

The reaction $A_2+B_2\to 2\mathrm{AB}$, is bound to obey the following mechanism
$A_2\stackrel{k_1}{\underset{k_{-1}}{\rightleftarrows }}A+A\left(\mathrm{fast}\right)$
$A+B_2\stackrel{k_2}{\to }\mathrm{AB}+B\left(\mathrm{slow}\right)$
$A+B\to \mathrm{AB}\left(\mathrm{fast}\right)$
The order of overall reaction is

• A. 3
• B. 2
• C. 2.5
• D. 1.5

Answer 1

The correct option is D 1.5

$\mathrm{From}\mathrm{the}\mathrm{equilibrium}\mathrm{step},\frac{k_1}{k_{-1}}=\frac{{\left[A\right]}^2}{\left[A_2\right]}\Rightarrow \left[A\right]={\left(\frac{k_1\left[A_2\right]}{k_{-1}}\right)}^{\frac{1}{2}}\mathrm{Rate}\mathrm{of}\mathrm{the}\mathrm{reaction}\mathrm{is},r=k_2\left[A\right]\left[B_2\right]r=k_2\left(\frac{k_1}{k_{-1}}{)}^{\frac{1}{2}}\right[A_2{]}^{\frac{1}{2}}\left[B_2\right]\mathrm{So}\mathrm{the}\mathrm{order}\mathrm{of}\mathrm{the}\mathrm{reaction}\mathrm{is}1+0.5=1.5$