Chemical and Ionic Equilibrium - Result Question 60

####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:

  1. $S+O _2 \rightleftharpoons SO _2, K _1$

$\therefore \quad SO _2 \rightleftharpoons S+O _2, K _1{ }^{\prime}=\frac{1}{K _1}$

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,

$$ \begin{aligned} K _3 & =K _2 \times K _1{ }^{\prime \prime}=K _2 \times \frac{1}{K _1^{2}} \ & =10^{129} \times \frac{1}{\left(10^{52}\right)^{2}}=10^{129-104}=10^{25} \end{aligned} $$



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