Question

For the following reactions, equilibrium constants are given:
$$S(s)+O_2(g)\rightleftharpoons S{O}_2(g);K_1={10}^{52}$$
$$2S(s)+3O_2(g)\rightleftharpoons 2S{O}_3(g);K_2={10}^{129}$$
The equilibrium constant for the reaction,
$2S{O}_2(s)+O_2(g)\rightleftharpoons 2S{O}_3(g)$ is:
(A) ${10}^{181}$
(B) ${10}^{154}$
(C) ${10}^{25}$
(D) ${10}^{77}$

Answer 1

Hint: When reactions are added or subtracted, their respective equilibrium constants or formation constants are multiplied or divided respectively. A cumulative constant can always be expressed as the product of stepwise constants.

Complete answer:
For a given set of reaction conditions, the equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture.
Let say this equation $$S(s)+O_2(g)\rightleftharpoons S{O}_2(g);K_1={10}^{52}$$ as eqn (i), and $$2S(s)+3O_2(g)\rightleftharpoons 2S{O}_3(g);K_2={10}^{129}$$ as eqn (ii).
Now on multiplying eqn (i) with 2, we get,
     $$2S(s)+2O_2(g)\rightleftharpoons 2S{O}_2(g);K_1\prime ={10}^{104}$$ …… eqn (iii)
Now we will subtract equation (iii) from equation (ii), we get,
     $$2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$$
The equation obtained is same as that of equilibrium reaction, therefore, the equilibrium constant of equilibrium reaction will be derived , by dividing the equilibrium constant of eqn (ii) with eqn (iii), we get,
     $$K_{eq}={\displaystyle \frac{K_2}{K_{1^{\prime }}}}={10}^{(129-104)}={10}^{25}$$
Therefore, the equilibrium constant of the reaction will be ${10}^{25}$ .

Hence the correct answer is the C option.

Note:
The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.