Question

The equilibrium constant of the reaction $S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons S{O}_3\left(g\right)$ is $4\times {10}^{-3}at{m}^{-1/2}.$ The equilibrium constant of the reaction $2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$ would be:

• A. 250 atm
• B. $4\times {10}^3$ atm
• C. $0.25\times {10}^4$ atm
• D. $6.25\times {10}^4$ atm

Answer 1

The correct option is D $6.25\times {10}^4$ atm

(1) $S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons S{O}_3\left(g\right)K_p=4\times {10}^{-3}at{m}^{-1/2}$
$\Rightarrow$$K_P=\frac{\left[P_{S{O}_3}\right]}{\left[P_{O_2}{]}^{1/2}\right[P_{S{O}_2}]}$
$\Rightarrow$ $K_p^{\prime }=\frac{1}{K_P}=\frac{\left[P_{O_2}{]}^{1/2}\right[P_{S{O}_2}]}{\left[P_{S{O}_3}\right]}...\left(i\right)$

(2) $2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$
$\Rightarrow K_P{}^{\prime \prime }=(K_P^{\prime }{)}^2=(\frac{1}{K_P}{)}^2=\frac{\left[P_{O_2}\right][P_{S{O}_2}{]}^2}{[P_{S{O}_3}{]}^2}...\left(ii\right)$
from equation $\left(i\right)$ and $\left(ii\right)$, we get

$K_P{}^{\prime \prime }=(\frac{1}{K_P}{)}^2=\frac{1}{(4\times {10}^{-3}{)}^2}$
$\Rightarrow K_P{}^{\prime \prime }=\frac{1}{16\times {10}^{-6}}\text{atm}=6.25\times {10}^4\text{atm}$
Therefore the correct option is (d)