Chemistry - Chemical Equilibrium Question with Solution | TestHub

For the reaction,

$\mathrm{\text{S}}{\mathrm{\text{O}}}_2\left(\mathrm{\text{g}}\right)+\frac{1}{2}{\mathrm{\text{O}}}_2\left(\mathrm{\text{g}}\right)\rightleftharpoons \mathrm{\text{S}}{\mathrm{\text{O}}}_3\left(\mathrm{\text{g}}\right)$

Equilibrium constant,${\text{K}}_1=\frac{\left[\mathrm{\text{S}}{\mathrm{\text{O}}}_3\right]}{\left[\mathrm{\text{S}}{\mathrm{\text{O}}}_2\right]{\left[{\mathrm{\text{O}}}_2\right]}^{1/2}}$...(i)

For the reaction,

$2{\mathrm{\text{S}}\mathrm{\text{O}}}_3\left(\mathrm{\text{g}}\right)\rightleftharpoons 2{\mathrm{\text{S}}\mathrm{\text{O}}}_2\left(\mathrm{\text{g}}\right)+{\mathrm{\text{O}}}_2\left(\mathrm{\text{g}}\right)$

Equilibrium constant,${\text{K}}_2=\frac{{\left[\mathrm{\text{S}}{\mathrm{\text{O}}}_2\right]}^2\left[{\mathrm{\text{O}}}_2\right]}{{\left[\mathrm{\text{S}}{\mathrm{\text{O}}}_3\right]}^2}$...(ii)

On squaring both sides in Eq.(i), we get

${\text{K}}_1^2=\frac{{\left[\mathrm{\text{S}}{\mathrm{\text{O}}}_3\right]}^2}{{\left[\mathrm{\text{S}}{\mathrm{\text{O}}}_2\right]}^2\left[{\mathrm{\text{O}}}_2\right]}$...(iii)

Eq. (ii)$\times$Eq (iii), we get

${\text{K}}_1^2\times {\text{K}}_2=1$

or${\text{K}}_2=\frac{1}{{\text{K}}_1^2}$

or${\text{K}}_1^2=\frac{1}{{\text{K}}_2}$