Chemical and Ionic Equilibrium 1 Question 4
4. For the following reactions, equilibrium constants are given :
$$ \begin{aligned} \mathrm{S}(s)+\mathrm{O}{2}(g) & \rightleftharpoons \mathrm{SO}{2}(g) ; K_{1}=10^{52} \ 2 \mathrm{~S}(s)+3 \mathrm{O}{2}(g) & \rightleftharpoons 2 \mathrm{SO}{3}(g) ; K_{2}=10^{129} \end{aligned} $$
The equilibrium constant for the reaction,
$$ 2 \mathrm{SO}(g)+\mathrm{O}{2}(g) \rightleftharpoons 2 \mathrm{SO}{3}(g) \text { is } $$
(2019 Main, 8 April II)
(a) $10^{25}$
(b) $10^{77}$
(c) $10^{154}$
(d) $10^{181}$
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Solution:
- $\mathrm{S}+\mathrm{O}{2} \rightleftharpoons \mathrm{SO}{2}, K_{1}$
$\therefore \quad \mathrm{SO}{2} \rightleftharpoons \mathrm{S}+\mathrm{O}{2}, K_{1}{ }^{\prime}=\frac{1}{K_{1}}$
or, $2 \mathrm{SO}{2} \rightleftharpoons 2 \mathrm{~S}+2 \mathrm{O}{2}, K_{1}{ }^{\prime \prime}=\left(K_{1}{ }^{\prime}\right)^{2}=\frac{1}{K_{1}^{2}}$
$\Rightarrow \quad 2 \mathrm{~S}+3 \mathrm{O}{2} \rightleftharpoons 2 \mathrm{SO}{3}, K_{2}$
Now, [(i) + (ii) $]$ gives
$$ 2 \mathrm{SO}{2}+\mathrm{O}{2} \rightleftharpoons 2 \mathrm{SO}{3}, K{3} $$
The value of equilibrium constant,
$$ \begin{aligned} K_{3} & =K_{2} \times K_{1}{ }^{\prime \prime}=K_{2} \times \frac{1}{K_{1}^{2}} \ & =10^{129} \times \frac{1}{\left(10^{52}\right)^{2}}=10^{129-104}=10^{25} \end{aligned} $$
