SATHEE: Equilibrium - Result Question 18

Equilibrium - Result Question 18

####19. If $K_{1}$ and $K_{2}$ are the respective equilibrium constants for the two reactions

$X e F_{6} (g) + H_{2} O (g) \to X e O F_{4} (g) + 2 H F (g)$

$X e O_{4} (g) + X e F_{6} (g) \to X e O F_{4} (g) + X e O_{3} F_{2} (g)$

the equilibrium constant of the reaction

$X e O_{4} (g) + 2 H F (g) \to X e O_{3} F_{2} (g) + H_{2} O (g)$ will be

[1998]

(a) $K_{1} / (K_{2})^{2}$

(b) $K_{1} \cdot K_{2}$

(c) $K_{1} / K_{2}$

(d) $K_{2} / K_{1}$

Show Answer

Solution:

(d) For the reaction

$X e F_{6} (g) + H_{2} O (g) \to X e O F_{4} (g) + 2 H F (g)$

$K_{1} = \frac{[X e O F_{4}] [H F]^{2}}{[X e F_{6}] [H_{2} O]}$

and for the reaction

$X e O_{4} (g) + X e F_{6} (g) \to X e O F_{4} (g) + H e O_{3} F_{2} (g)$

$K_{2} = \frac{[X e O F_{4}] [X e O_{3} F_{2}]}{[X e O_{4}] [X e F_{6}]}$

For reaction :

$X e O_{4} (g) + 2 H F (g) \to X e O_{3} F_{2 (g)} + H_{2} O (g)$ $K = \frac{[X e O_{3} F_{2}] [H_{2} O]}{[X e O_{4}] [H F]^{2}}$

$∴$ From eq. no. (a) and (b)

$K = K_{2} / K_{1}$

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