Equilibrium - Result Question 18
####19. If $K_1$ and $K_2$ are the respective equilibrium constants for the two reactions
$XeF_6(g)+H_2 O(g) \rightarrow XeOF_4(g)+2 HF(g)$
$XeO_4(g)+XeF_6(g) \rightarrow XeOF_4(g)+XeO_3 F_2(g)$
the equilibrium constant of the reaction
$XeO_4(g)+2 HF(g) \rightarrow XeO_3 F_2(g)+H_2 O(g)$ will be
[1998]
(a) $K_1 /(K_2)^{2}$
(b) $K_1 \cdot K_2$
(c) $K_1 / K_2$
(d) $K_2 / K_1$
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Solution:
- (d) For the reaction
$XeF_6(g)+H_2 O(g) \rightarrow XeOF_4(g)+2 HF(g)$
$K_1=\frac{[XeOF_4][HF]^{2}}{[XeF_6][H_2 O]}$
and for the reaction
$XeO_4(g)+XeF_6(g) \rightarrow XeOF_4(g)+HeO_3 F_2(g)$
$K_2=\frac{[XeOF_4][XeO_3 F_2]}{[XeO_4][XeF_6]}$
For reaction :
$XeO_4(g)+2 HF(g) \to XeO_3 F _{2(g)}+H_2 O(g)$ $K=\frac{[XeO_3 F_2][H_2 O]}{[XeO_4][HF]^{2}}$
$\therefore$ From eq. no. (a) and (b)
$K=K_2 / K_1$
