Question

For the following reactions, equilibrium constants are given:
$S(s)+{{O}_{2}}(g)\rightleftharpoons S{{O}_{2}}(g);{{K}_{1}}={{10}^{52}}$
$2S(s)+3{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g);{{K}_{2}}={{10}^{129}}$
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

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 $2S(s)+3{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g);{{K}_{2}}={{10}^{129}}$ as eqn (ii).
Now on multiplying eqn (i) with 2, we get,
$2S(s)+2{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{2}}(g);{{K}_{1}}\prime ={{10}^{104}}$ …… 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,
${{K}_{eq}}=\dfrac{{{K}_{2}}}{{{K}_{1'}}}={{10}^{(129-104)}}={{10}^{25}}$
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.