Question

For the reaction: $2H_2+2NO\rightleftharpoons 2H_{2}O+N_2$, the following rate data was obtained:

S. No.	$\left[NO\right]$ $mol$ $L^{-1}$	$\left[H_2\right]$ $mol$ $L^{-1}$	Rate: $mol$ $L^{-1}$ ${sec}^{-1}$
$1$	$0.40$	$0.40$	$4.6\times {10}^{-3}$
$2$	$0.80$	$0.40$	$18.4\times {10}^{-3}$
$3$	$0.40$	$0.80$	$9.2\times {10}^{-3}$

Calculate the following:
(1) The overall order of reaction.
(2) The rate law.
(3) The value of rate constant $\left(k\right)$.

Answer 1

(1)
From experiments (1) and (2), $\left[H_2\right]$ is kept constant at 0.40 M and $\left[NO\right]$ is doubled from 0.40 M to 0.80 M. The rate of the reaction is quadrupled from $4.6\times {10}^{-3}$ M/s to $18.4\times {10}^{-3}$ M/s. This suggests that the order of the reaction with respect to NO is 2.
From experiments (1) and (3), $\left[NO\right]$ is kept constant at 0.40 M and $\left[H_2\right]$ is doubled from 0.40 M to 0.80 M. The rate of the reaction is doubled from $4.6\times {10}^{-3}$ M/s to $9.2\times {10}^{-3}$ M/s. This suggests that the order of the reaction with respect to $H_2$ is 1.
The overall order of the reaction is $2+1=3$.
(2)
The rate law expression is
rate $=k\left[NO{]}^2\right[H_2]$
(3)
To calculate the value of the rate constant (k), substitute the data for experiment (1) in the rate law expression.
rate $4.6\times {10}^{-3}=k[0.40{]}^2\times 0.40$
$k=0.072mo{l}^{-2}{L}^2{s}^{-1}$