Chemistry - Chemical Kinetics Question with Solution | TestHub

The rate law for the reaction $2\mathrm{NO}+2{\mathrm{H}}_2\to {\mathrm{N}}_2+2{\mathrm{H}}_2\mathrm{O}$ is

$\mathrm{r}=\mathrm{k}{\left[{\mathrm{H}}_2\right]}^{\mathrm{m}}{\left[\mathrm{NO}\right]}^{\mathrm{n}}$

The powers of the concentration terms m, n are the order with respect to hydrogen and nitric oxide respectively.

$$\begin{gathered} {\mathrm{r}}_1={\left[65.6\right]}^{\mathrm{m}}{\left[40.0\right]}^{\mathrm{n}}=0.135 \\ {\mathrm{r}}_2={\left[65.6\right]}^{\mathrm{m}}{\left[20.1\right]}^{\mathrm{n}}=0.033 \\ {\mathrm{r}}_3={\left[38.6\right]}^{\mathrm{m}}{\left[65.6\right]}^{\mathrm{n}}=0.214 \\ {\mathrm{r}}_4={\left[19.2\right]}^{\mathrm{m}}{\left[65.6\right]}^{\mathrm{n}}=0.106 \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{{\displaystyle {\left[65.6\right]}^{\mathrm{m}}}{\displaystyle {{\displaystyle \left[40.0\right]}}^{\mathrm{n}}}}{{\left[65.6\right]}^{\mathrm{m}}{\left[20.1\right]}^{\mathrm{n}}}=\frac{0.135}{0.033}=4 \\ \Rightarrow 2^{\mathrm{n}}=4 \\ \Rightarrow \mathrm{n}=2 \end{gathered}$$

Hence, the order with respect to $\mathrm{NO}$ is $2.$