Question

For the reaction, $S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons {SO}_3\left(g\right)$, the value of $K_c$ is $1.7\times {10}^{12}$ at $300K$. Calculate the equilibrium constants for the following reactions at $300K$:
(I) $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2{SO}_3\left(g\right)$
(II) ${SO}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

Answer 1

The given reaction is :-

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

$\Rightarrow K_C=\frac{\left[{SO}_3\right]}{\left[{SO}_2\right]{\left[O_2\right]}^{1/2}}=1.7\times {10}^{12}\to \left(A\right)$

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

$\Rightarrow {K_C}^{{}^{\prime }}=\frac{{\left[{SO}_3\right]}^2}{{\left[{SO}_2\right]}^2\left[O_2\right]}={\left(K_C\right)}^2={(1.7\times {10}^{12})}^2$

$\Rightarrow {K_C}^{{}^{\prime }}=2.89\times {10}^{24}$

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

$\Rightarrow {K_C}^{{}^{\prime }}=\frac{\left[{SO}_2\right]{\left[O_2\right]}^{1/2}}{\left[{SO}_3\right]}$

$=\frac{1}{K_C}=\frac{1}{1.7\times {10}^{12}}$

$\Rightarrow {K_C}^{{}^{\prime }}=0.588\times {10}^{-12}$