Question

The rate constant of the elementary reaction

$\mathit{B}\mathit{r}\mathit{O}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathit{N}\mathit{O}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{}\mathit{B}\mathit{r}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}\mathit{N}{\mathit{O}}_{\mathbf{2}}\left(g\right)$

is 1.3×10¹⁰ Lmol^-1 and its equilibrium constant is 5.0 × 10¹⁰ at this temperature. Calculate the rate constant at 25°C of the elementary reaction

$\mathit{B}\mathit{r}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathit{N}{\mathit{O}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{}\mathit{B}\mathit{r}\mathit{O}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}\mathit{N}\mathit{O}\left(g\right)$

Short answer

$Ratecons\tan tvalueis2.6\times {10}^{-1}Lmo{l}^{-1}{s}^{-1}$

Step-by-step solution

Step-1: Rate constant for equilbrium reactions:

For a reaction in chemical equilibrium rate of forward reaction is equal to the rate of backward reaction,so the rate constant of forward reaction is eqaul to rate constant of backward reaction according to the principle of detailed balance and hence equilbrium constant is the defined as the ratio between forward and reverse equilibrium constants.

$BrO\left(g\right)+NO\left(g\right)\underset{k_{-1}}{\overset{k}{⟷}}Br\left(g\right)+N{O}_2\left(g\right)......\left(1\right)$

Where the forward rate constant k is equal to the rate constant k_-1 of the backward reaction, if the equilibrium constant of the reaction is K then,

It can be defined as ratio between the forward rate constant and the rate constant of backward reaction.

$K=\frac{k}{k_{-1}}........\left(2\right)$

Step-2:   Rate constant value:

From the above given question rate conatant value of backward reaction can be sorted out in the following way

$$\begin{gathered} K=\frac{k}{k_{-1}} \\ SubstitutingK=5.0\times {10}^{10}k=1.3\times {10}^{10}inEquation-2 \\ 5.0\times {10}^{10}=\frac{1.3\times {10}^{10}Lmo{l}^{-1}{s}^{-1}}{k_{-1}} \\ {k}_{-1}=\frac{1.3\times {10}^{10}}{5.0\times {10}^{10}}Lmo{l}^{-1}{s}^{-1} \\ {k}_{-1}=2.6\times {10}^{-1}Lmo{l}^{-1}{s}^{-1} \\ Ratecons\tan tvalueis2.6\times {10}^{-1}Lmo{l}^{-1}{s}^{-1} \end{gathered}$$