Question

The equilibrium constant of the reaction; $S{O}_{3\left(g\right)}\rightleftharpoons S{O}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$; is $0.20mol{e}^{\frac{1}{2}}.litr{e}^{\frac{-1}{2}}$ at $1000K$. Calculate equilibrium constant for $2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2S{O}_{3\left(g\right)}$.

Answer 1

For $S{O}_{3\left(g\right)}\rightleftharpoons S{O}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$

$K_{c_1}=\frac{\left[S{O}_2\right][O_2{]}^{1/2}}{\left[S{O}_3\right]}=0.20....\left(1\right)$

For $2S{O}_2+O_2\rightleftharpoons 2S{O}_3$

$K_c=\frac{[S{O}_3{]}^2}{\left[S{O}_2{]}^2\right[O_2]}...\left(2\right)$

By reversing Eq. (1),

$\frac{1}{K_{c_1}}=\frac{\left[S{O}_3\right]}{\left[S{O}_2\right][O_2{]}^{1/2}}$

Squaring both sides

${\left(\frac{1}{K_{c_1}}\right)}^2=\frac{[S{O}_3{]}^2}{\left[S{O}_2{]}^2\right[O_2]}=K_c$ by Eq. (2)

$\therefore K_c={\left[\frac{1}{0.20}\right]}^2=25mo{l}^{-1}litre$.