Question

The rate of the reaction

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

was studied at a different temperature. This reaction is one step of the nitric oxide-catalyzed destruction of ozone in the upper atmosphere.

a. In one experiment ${\mathbf{NO}}_{\mathbf{2}}$was in large excess at a concentration $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}\mathbf{molecule}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$ of with the following data collected:

What is the order of the reaction with respect to oxygen atoms?

b. The reaction is known to be first order with respect to ${\mathbf{NO}}_{\mathbf{2}}$. Determine the overall rate law and the value of the rate constant.

Short answer

a. The reaction is first order with respect to oxygen

b. Overall rate law=$\mathrm{Rate}=\mathrm{K}{\left[{\mathrm{NO}}_2\right]}^1{\left[\mathrm{O}\right]}^1$ . Value of rate constant: $\mathrm{K}=1\times {10}^{-11}{\mathrm{cm}}^3/\mathrm{mole}/\mathrm{s}$

Step-by-step solution

Given reaction

Chemical kinetics is the branch of chemistry that deals with the speed/rate of reactions.

$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{NO}}_2\left(\mathrm{g}\right)\to \mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

The overall rate law for the reaction

$\mathrm{Rate}=\mathrm{K}{\left[{\mathrm{NO}}_2\right]}^{\mathrm{n}}{\left[\mathrm{O}\right]}^{\mathrm{m}}\dots \dots \left(1\right)$

Where, K= Rate constant

n =Order w.r.t $N{O}_2$

m =order of reaction w.r.t [O]

The graph between ln[O] vs t for given data

Time(s)	ln[O]
0	22.33
$1\times {10}^{-2}$	21.36
$2\times {10}^{-2}$	20.33
$3\times {10}^{-2}$	19.33

ln[O]

$\mathrm{Time}\left(\mathrm{s}\right)\times {10}^{-2}$

The graph between ln[O] versus time shows a straight line. Therefore, the reaction follows first-order kinetics.

Hence integrated rate law is as follows for the first order.

$\ln \left[O\right]=-Kt+\ln {\left[O\right]}_{\circ }$

Order of the reaction and rate constant

The graph between $\ln \left[O\right]$and t gives a straight line and the slope is equal to -K .

Since the reaction is first order with both respective oxygen and nitrogen oxide the overall rate law is as follows:

$\mathrm{Rate}=\mathrm{K}{\left[{\mathrm{NO}}_2\right]}^1{\left[\mathrm{O}\right]}^1$

$\begin{array}{rcl}\mathrm{Let},\mathrm{K}& =& {\mathrm{K}}^1\left[{\mathrm{NO}}_2\right]\\ \mathrm{Rate}& =& {\mathrm{K}}^1\left[\mathrm{O}\right]\\ {\mathrm{K}}^1& =& \mathrm{Slope}=\frac{\mathrm{\Delta ln}\left[\mathrm{O}\right]}{\mathrm{\Delta t}}\\ {\mathrm{K}}^1& =& \frac{\left(21.36-19.33\right)}{\left(1\times {10}^{-2}-3\times {10}^{-2}\right)}\\ {\mathrm{K}}^1& =& -101.4{\mathrm{s}}^{-1}\end{array}$

So, the pseudo rate constant $K^1$is equal to $101.4{\mathrm{s}}^{-1}$.

Calculation of overall rate constant reaction

$\begin{array}{rcl}{\mathrm{K}}^1& =& \mathrm{K}\left[{\mathrm{NO}}_2\right]\\ \mathrm{K}& =& \frac{{\mathrm{K}}^1}{\left[{\mathrm{NO}}_2\right]}\\ \mathrm{K}& =& \frac{101.4{\mathrm{s}}^{-1}}{1\times {10}^{13}}\\ \mathrm{K}& =& 1\times {10}^{-11}{\mathrm{cm}}^3/\mathrm{mole}/\mathrm{s}\\ & & \end{array}$