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

• A. $25M^{-1}$
• B. $5litr{e}^{1/2}mo{l}^{-1/2}$
• C. $25M$
• D. none of the above

Answer 1

The correct option is A $25M^{-1}$

The given reaction is :-

${SO}_3\left(g\right)\rightleftharpoons {SO}_2\left(g\right)+1/2O_2\left(g\right)$

$\Rightarrow K_C=\frac{\left[{SO}_2\right]{\left[O_2\right]}^{1/2}}{\left[{SO}_3\right]}=0.2\to \left(I\right)$

Now, $2{SO}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2{SO}_3\left(g\right)$

$\Rightarrow {K_C}^{{}^{\prime }}=\frac{{\left[{SO}_2\right]}^2}{{\left[{SO}_2\right]}^2\left[O_2\right]}$

$={\left(\frac{1}{K_C}\right)}^2={\left(\frac{1}{0.2}\right)}^2=\frac{1}{0.04}=25M^{-1}$