Question

The rate constant of the elementary reaction

$BrO\left(g\right)+NO\left(g\right)\to Br\left(g\right)+N{O}_2\left(g\right)$

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.

$\text{Br}\left(g\right)+N{O}_2\left(g\right)\to BrO\left(g\right)+NO\left(g\right)$

Short answer

Rate constasnt value of backward reaction is$2.6\times {10}^{-1}{\text{Lmol}}^{-1}{\text{s}}^{\text{-1}}$

Step-by-step solution

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.

$BrO\left(g\right)+NO\left(g\right)\rightleftharpoons Br\left(g\right)+N{O}_2\left(g\right)\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(1\right)$

Above the arrow there is k- which is rate constant for forward reaction

Below the arrow there is k -1 which is rate constant for backward reaction

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.

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

Rate constant value:

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

$$\begin{gathered} K=\frac{k}{k_{-1}}.......................\left(2\right) \\ {\text{SubstitutingK=5.0×10}}^{10}k=1.3\times {10}^{10}ineqaution2 \\ {k}_{-1}=\frac{1.3\times {10}^{10}}{5.0\times {10}^{10}} \\ {k}_{-1}=2.6\times {10}^{-1}{\text{Lmol}}^{\text{-1}}{\text{s}}^{\text{-1}} \end{gathered}$$

Rate constant for back ward reaction is found to be$2.6\times {10}^{-1}{\text{Lmol}}^{\text{-1}}{\text{s}}^{\text{-1}}$