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Law of Mass Action

If K 1 and K ...

Question

If $K_{1}$ and $K_{2}$ are the respective equilibrium constants for the two reactions:
(i) $X e F_{6} (g) + H_{2} O (g) ⇌ X e O F_{4} (g) + 2 H F (g)$
(ii) $X e O_{4} (g) + X e F_{6} (g) ⇌ X e O F_{4} (g) + X e O_{3} F_{2} (g)$
then the equilibrium constant for the reaction:
$X e O_{4} (g) + 2 H F (g) ⇌ X e O_{3} F_{2} (g) + H_{2} O (g)$ will be:

\frac{K_{1}}{K_{2}}

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K_{1} . K_{2}

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\frac{K_{1}}{K_{2}}

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\frac{K_{2}}{K_{1}}

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Solution

The correct option is D $\frac{K_{2}}{K_{1}}$
The equilibrium constant for the reaction $X e O_{4} (g) + 2 H F (g) ⇌ X e O_{3} F_{2} (g) + H_{2} O (g)$ can be obtained by reversing equation (i) and adding it to equation (ii).

Hence, the equilibrium constant (K) would be,
$K = K_{2} \times \frac{1}{K_{1}} = \frac{K_{2}}{K_{1}}$

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