Question

Calculate a value for the equilibrium constant for the reaction

${\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

given that

$$\begin{gathered} \mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{ }\stackrel{\mathbf{h}\mathbf{v}}{\mathbf{\rightleftharpoons }}\mathit{N}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{K}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{ }\mathbf{49}} \\ {\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{N}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{ }\mathbf{\rightleftharpoons }\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{ }\mathbf{+}\mathbf{ }{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{K}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{ }\mathbf{34}} \end{gathered}$$

(See the hint in Exercise 72.)

Short answer

The equilibrium constant value for the given reaction is $2.6\times {10}^{81}$.

Step-by-step solution

Derive the equilibrium constant expression.

The equilibrium constants for the given reactions are named as follows:

Reaction 1:${\mathrm{NO}}_2\left(\mathrm{g}\right)\stackrel{\mathrm{hv}}{\rightleftharpoons }\mathrm{NO}\left(\mathrm{g}\right) + \mathrm{O}\left(\mathrm{g}\right){\mathrm{K}}_1=6.8\times {10}^{- 49}$

Reaction 2:${\mathrm{O}}_3\left(\mathrm{g}\right) + \mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{NO}}_2\left(\mathrm{g}\right) + {\mathrm{O}}_2\left(\mathrm{g}\right){\mathrm{K}}_2=5.8\times {10}^{- 34}$

The equilibrium constant for the reaction${\mathrm{O}}_2\left(\mathrm{g}\right) + \mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{O}}_3\left(\mathrm{g}\right)$ can be calculated as follows:

$$\begin{gathered} NO\left(g\right) + O\left(g\right)\stackrel{\mathrm{hv}}{\rightleftharpoons }N{O}_2\left(g\right)1/K_1 \\ \underline{N{O}_2\left(g\right) + O_2\left(g\right)\rightleftharpoons O_3\left(g\right) + NO\left(g\right)1/K_2} \\ {O}_2\left(g\right) + O\left(g\right)\rightleftharpoons O_3\left(g\right)K=\frac{1}{K_1{K}_2} \end{gathered}$$

Calculate the equilibrium constant for the given reaction.

The derived equilibrium constant expression for the reaction${\mathrm{O}}_2\left(\mathrm{g}\right) + \mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{O}}_3\left(\mathrm{g}\right)$is as follows:

$\mathrm{K}=\frac{1}{{\mathrm{K}}_1{\mathrm{K}}_2}$

Substitute the values of equilibrium constants K₁ and K₂ into the above expression as follows:

$\begin{array}{c}\mathrm{K}=\frac{1}{6.8\times {10}^{-- 49}\times 5.8\times {10}^{-- 34}}\\ =2.6\times {10}^{81}\end{array}$

Hence, the equilibrium constant for the given reaction is $2.6\times {10}^{81}$.