Question

The following equilibrium concentrations were measured at $800K:\left[S{O}_2\right]=3.0\times {10}^{-3}M;\left[O_2\right]=3.5\times {10}^{-3}M;\left[S{O}_3\right]=5.0\times {10}^{3}M$ Calculate the equilibrium constant at $800K$ for the chemical reaction in equilibrium for both directions.
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$
($K_c$ is the equilibrium constant for the forward reaction and $K_c^{{}^{\prime }}$ is the equilibrium constant for the backward reaction)

• A. $K_c=0.793\times {10}^3,K_c^{{}^{\prime }}=1.2\times {10}^{-3}$
• B. $K_c=0.893\times {10}^3,K_c^{{}^{\prime }}=1.2\times {10}^{-3}$
• C. $K_c=0.793\times {10}^3,K_c^{{}^{\prime }}=2.2\times {10}^{-3}$
• D. $K_c=0.893\times {10}^3,K_c^{{}^{\prime }}=2.2\times {10}^{-3}$

Answer 1

The correct option is A $K_c=0.793\times {10}^3,K_c^{{}^{\prime }}=1.2\times {10}^{-3}$

In case of homogeneous or hetrogeneous equilibrium, only the concentration of pure solids and pureliquids is to be taken as 1.

$\Rightarrow K_c=\frac{[S{O}_3{]}^2}{\left[O_2\right][S{O}_2{]}^2}$
$\Rightarrow K_c=\frac{(5\times {10}^{-3}{)}^2}{3.5\times {10}^{-3}\times (3.0\times {10}^{-3}{)}^2}$
$\Rightarrow K_c=\frac{25\times {10}^{-6}}{3.5\times {10}^{-3}\times 9\times {10}^{-6}}$
$\Rightarrow K_c=793.65=0.793\times {10}^3$
$K_c^{{}^{\prime }}=\frac{1}{K_c}$
$K_c^{{}^{\prime }}=\frac{1}{0.793\times {10}^3}$
$K_c^{{}^{\prime }}=1.2\times {10}^{-3}$