Question

The following data were obtained for the gas-phase decomposition of di nitrogen pentoxide,

$\mathbf{2}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}\left(g\right)\mathbf{\to }\mathbf{4}{\mathbf{NO}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)$

${\left[N_2{O}_5\right]}_{\mathbf{0}}\left(\mathrm{mol}/L\right)$	Initial Rate$\left({\mathrm{molL}}^{-1}{s}^{-1}\right)$
0.0750	$\mathbf{8}\mathbf{.}\mathbf{90}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$
0.190	$\mathbf{2}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$
0.275	$\mathbf{3}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$
0.410	$\mathbf{4}\mathbf{.}\mathbf{85}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$

Where,$\mathbf{Rate}\mathbf{=}\frac{\mathbf{-}\mathbf{d}\left[N_2{O}_5\right]}{\mathbf{dt}}$

Write the rate law and calculate the value of the rate constant.

Short answer

$\mathrm{Rate}=\mathrm{k}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$

The value of the rate constant will be${\mathrm{k}}_{\mathrm{mean}}=1.19\times {10}^{-2}{\mathrm{s}}^{-1}$.

Step-by-step solution

Write the rate laws for the first 2 experiments

It is given that

$\begin{array}{rcl}\mathrm{Rate}& =& \frac{-\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{dt}}\\ \mathrm{Rate}& =& \mathrm{k}{\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}^{\mathrm{x}}\end{array}$

For the 1st two experiments:

$\begin{array}{c}8.90\times {10}^{-4}=k{\left[0.0750\right]}^x......\left(1\right)\\ 2.26\times {10}^{-3}=k{\left[0.190\right]}^x......\left(2\right)\end{array}$

Dividing (2) by (1), we get

$\begin{array}{c}2.54={\left(\frac{0.190}{0.0750}\right)}^x\\ 2.54={\left(2.53\right)}^x\\ x\approx 1\end{array}$

So, the rate of the reaction will be given by:

$\mathrm{Rate}=\mathrm{k}{\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}^{\mathrm{x}}=\mathrm{k}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$

Now calculate the rate constant.

$\mathrm{Rate}=\mathrm{k}{\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}^{\mathrm{x}}$

$\begin{array}{rcl}\therefore \mathrm{k}& =& \frac{\mathrm{Rate}}{\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}\\ & =& \frac{8.90\times {10}^{-4}{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}}{0.0750\mathrm{mol}/\mathrm{L}}\end{array}$

(From experiment 1)

So,$k=1.19\times {10}^{-2}{\text{s}}^{-1}$