Question

For the reaction given below,

$2O_3\left(g\right)\to 3O_2\left(g\right)$

Step l: $O_3\left(g\right)\to O_2\left(g\right)+O\left(g\right)$

Step 2: $O_3\left(g\right)+O{\to }^{slow}2O_2\left(g\right)$

Statement I: The molecularity of the first step is 1 and of the second step is 2.
Statement-II: $O\left(g\right)$ is an intermediate and rate of reaction is $K\left[O_3{]}^2\right[O_2{]}^{-1}$ and order of reaction is 1.

• A. Both statements are true Statement - II is correct explanation of Statement - I
• B. Both statements are true but Statement -II is not correct explanation of Statement - I
• C. Statement - I is true but Statement - II is false
• D. Statement - I is false but Statement - II is true

Answer 1

The correct option is A Both statements are true Statement - II is correct explanation of Statement - I

In the first step, only one molecule participates whereas, in the second step, two molecules collide to form the product.
Thus the molecularity of the first step is 1 and of the second step is 2. Hence, the statement I is correct.

$O\left(g\right)$ is an intermediate.

The expression for the equilibrium constant is $K_c=\frac{\left[O_2\right]\left[O\right]}{\left[O_3\right]}.$

The rate law expression is $rate=K\left[O_3\right]\left[O\right]=\frac{K_c\left[O_3\right]\left[O_3\right]}{\left[O_2\right]}=\frac{K^{\prime }[O_3{]}^2}{\left[O_2\right]}.$

The order of the reaction $=2-1=1.$

Thus statement II is correct.