Ask Questions, Get Answers


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$

$\begin{array}{1 1} k = \large\frac{k_1 k_3}{k_2} \\ k = \large\frac{k_1 k_2}{k_3} \\ k = \large\frac{k_1}{k_2 k_3} \\ k = \large\frac{k_3}{k_1k_2}\end{array}$
Download clay6 mobile app

1 Answer

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) $\rightarrow k = \large\frac{k_1 k_3}{k_2}$
answered Dec 13, 2013 by sreemathi.v
edited Jul 25, 2014 by balaji.thirumalai

Related questions

Ask Question