Chemical and Ionic Equilibrium - Result Question 4
####4. For the following reactions, equilibrium constants are given :
$$ \begin{aligned} S(s)+O _2(g) & \rightleftharpoons SO _2(g) ; K _1=10^{52} \ 2 S(s)+3 O _2(g) & \rightleftharpoons 2 SO _3(g) ; K _2=10^{129} \end{aligned} $$
The equilibrium constant for the reaction,
$$ 2 SO(g)+O _2(g) \rightleftharpoons 2 SO _3(g) \text { is } $$
(2019 Main, 8 April II)
(a) $10^{25}$
(b) $10^{77}$
(c) $10^{154}$
(d) $10^{181}$
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Answer:
Correct Answer: 4. (c)
Solution:
- $S+O _2 \rightleftharpoons SO _2, K _1$
$\therefore \quad SO _2 \rightleftharpoons S+O _2, K _1{ }^{\prime}=\frac{1}{K _1}$
or, $2 SO _2 \rightleftharpoons 2 S+2 O _2, K _1{ }^{\prime \prime}=\left(K _1{ }^{\prime}\right)^{2}=\frac{1}{K _1^{2}}$
$\Rightarrow \quad 2 S+3 O _2 \rightleftharpoons 2 SO _3, K _2$
Now, [(i) + (ii) $]$ gives
$$ 2 SO _2+O _2 \rightleftharpoons 2 SO _3, K _3 $$
The value of equilibrium constant,
$$ \begin{aligned} K _3 & =K _2 \times K _1{ }^{\prime \prime}=K _2 \times \frac{1}{K _1^{2}} \ & =10^{129} \times \frac{1}{\left(10^{52}\right)^{2}}=10^{129-104}=10^{25} \end{aligned} $$
