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

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 \[2S(s)+3{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g);{{K}_{2}}={{10}^{129}}\] 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.