Question

The chemical reaction , $2O_3\to 3O_2$ proceeds as follows:

Step 1: $O_3\rightleftharpoons O_2+O$ .......... (fast)
Step2: $O+O_3\to 2O_2$ ............... (slow)

The rate law expression should be:

• A. $r=k^{\prime }\left[O_3\right]\left[O_2\right]$
• B. $r=k^{\prime }\left[O_3{]}^2\right[O_2{]}^{-1}$
• C. $r=k^{\prime }[O_3{]}^2$
• D. unpredictable

Answer 1

The correct option is B $r=k^{\prime }\left[O_3{]}^2\right[O_2{]}^{-1}$

The slow step (step-2) is the rate-determining step

$r=k\left[O_3\right]\left[O\right]$...............(i)

From step-1,

$K_{eq}=\frac{\left[O_2\right]\left[O\right]}{\left[O_3\right]}$

or $\left[O\right]=\frac{K_{eq}\left[O_3\right]}{\left[O_2\right]}$..........(ii)

From eq.(i) and (ii) ,

$r=k{K}_{eq}\frac{[O_3{]}^2}{\left[O_2\right]}=k^{\prime }\left[O_3{]}^2\right[O_2{]}^{-1}$

$[\because k^{\prime }=kKeq]$