# Given the following reaction and its intermediate steps, calculate the relation between the rate of reaction of the resultant reaction and the rates of reaction of the intermediate steps.

Step-1: $2\;NO \begin{array} {c} \overset{k_1}\; \\ \large \rightleftharpoons \\\overset{k_2} \;\end{array} N_2O_2$

Step-2: $N_2O_2 + O_2 \overset{k_3} \rightarrow 2NO_2$

Overall: $2NO + O_2 \overset{k} \rightarrow 2NO_2$

$\begin{array}{1 1} k = \large\frac{k_1 k_3}{k_2} \\ k = \large\frac{k_1 k_2}{k_3} \\ k = \large\frac{k_1}{k_2 k_3} \\ k = \large\frac{k_3}{k_1k_2}\end{array}$

Answer: $k = \large\frac{k_1 k_3}{k_2}$
Step-1: $2\;NO \begin{array} {c} \overset{k_1}\; \\ \large \rightleftharpoons \\\overset{k_2} \;\end{array} N_2O_2$
A rapid equilibrium is established in the first step; the reaction intermediate, $N_2O_2$, is then slowly consumed in the second, rate-determining step: $N_2O_2 + O_2 \overset{k_3} \rightarrow 2NO_2$
Rate of the reaction $= k_3 [N_2O_2][O_2]$
The rate law cannot include a reaction intermediate, so $[N_2O_2]$ must be eliminated.
If the first, reversible step is established quickly, then rate of forward reaction = rate of reverse reaction
$\Rightarrow k_1[NO]^2= k_2[N_2O_2]$ or $[N_2O_2] = \large\frac{k_1}{k_2}$$[NO]^2 Substitute into the rate law, Rate of reaction =k_3 \large\frac{k_1}{k_2}$$ [NO]^2 [O_2]$ (1)
Now if we consider the Overall reaction: $2NO + O_2 \overset{k} \rightarrow 2NO_2$
The rate of reaction $=k [NO]^2[O_2]$ (2)
Comparing (1) and (2) $\rightarrow k = \large\frac{k_1 k_3}{k_2}$
edited Jul 25, 2014