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Question

A reversible reaction: $S O_{2} (g) + N O_{2} (g) ⇋ S O_{3} (g) + N O (g)$ .. $(1)$ , takes place in two reversible steps as given below with equilibrium constant values $2.0$ and $0.45$ respectively.

$S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇋ S O_{3} (g), K = 2.0...... (2)$
$N O_{2} (g) ⇋ N O (g) + \frac{1}{2} O_{2} (g), K = 0.45...... (3)$

The equilibrium constant $K_{c}$ for reaction $(1)$ is:

0.9

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\frac{400}{9}

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\frac{9}{400}

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\frac{1}{9}

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Solution

The correct option is A $0.9$
The reaction is $S O_{2} (g) + N O_{2} (g) \Leftrightarrow S O_{3} (g) + N O (g) . . . (1)$
It is obtained by adding the following two reactions:
$S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇋ S O_{3} (g), K_{1} = 2.0 . . . (2)$
$N O_{2} (g) ⇋ N O (g) + \frac{1}{2} O_{2} (g), K_{2} = 0.45 . . . (3)$
The value of the equilibrium constant of reaction $(1)$ is obtained by multiplying the values of the equilibrium constants of the reactions $(2)$ and $(3)$ .
Thus, $K = K_{1} \times K_{2} = 2.0 \times 0.45 = 0.9$ .

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