Question

The depletion of ozone involves the following steps:

Step 1 :$O_3\stackrel{K_1}{\underset{k_2}{\rightleftharpoons }}{O}_2+O$ (fast)

Step 2:$O_3+O\underset{k}{\overset{}{\to }}2O_2$(slow)

The predicted order of the reaction will be:

Answer 1

$\begin{array}{cccccc}& K_1& & & & \\ O_3& \rightleftharpoons & O_2& +& O& \left(fast\right)\\ & K_2& & & & \end{array}----\left(I\right)$

$O_3+O\underset{}{\overset{K}{\to }}2O_2\left(slow\right)----\left(II\right)$

Rate $19w$ would be equal to the rate of the second step because

Rate $=K\left[O_3\right]\left[O\right]----\left(III\right)$

Since $O$ is intermediate so it should be eliminated.

From equation $\left(1\right)$,

$k_1\left[O_3\right]=k_2\left[O_2\right]\left[O\right]$

$\left[O\right]=\frac{k_1}{k_2}\frac{\left[O_3\right]}{\left[O_2\right]}=\frac{K^{\prime }\left[O_3\right]}{\left[O_2\right]}$; where $k^{\prime }=\frac{k_1}{k_2}$

Put the value of $\left[O\right]$ in equation $\left(III\right)$

Rate $=K\left[O_3\right]k^{\prime }\frac{\left[O_3\right]}{\left[O_2\right]}$

$=\frac{k^{\prime \prime }[O_3{]}^2}{\left[O_2\right]}$; where $k\times k^{\prime }=k^{\prime \prime }$

$=k^{\prime \prime }\left[O_3{]}^2\right[O_2{]}^{-1}$

overall order $=2-1=1$