Question

For the following reactions, equilibrium constants are given:

$S\left(s\right)+O_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right);K_1={10}^{52}$

$2S\left(s\right)+3O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);K_2={10}^{129}$
Find the equilibrium constant for
$2S{O}_2+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);K_3$

<!--td {border: 1px solid #cccccc;}br {mso-data-placement:same-cell;}--> JEE MAIN 2019

• A. ${10}^{154}$
• B. ${10}^{181}$
• C. ${10}^{25}$
• D. ${10}^{77}$

Answer 1

The correct option is C ${10}^{25}$

$S\left(s\right)+O_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right);K_1={10}^{52}............\left(1\right)$

$2S\left(s\right)+3O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);K_2={10}^{129}............\left(2\right)$

$2S{O}_2+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);K_3=x$

Multiplying equation (1) by 2;

$2S\left(s\right)+2O_2\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right);{K^{\prime }}_1=K_1^2={10}^{104}.......\left(3\right)$

$\Rightarrow$ Substracting (3) from (2); we get

$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)K_{eq}=\frac{K_2}{{K^{\prime }}_1}={10}^{(129-104)}={10}^{25}$