For reaction XeF6+H2O⇌XeOF4+2HF; K1=4XeO4+XeF6⇌XeOF4+XeO3F2; K2=100Find equilibrium constant of 12XeO4+HF⇌12XeO3F2+12H2O

The equilibriums constant for the reaction is given as,Keq=K2×1K1 Keq=100×14 Keq=5

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For reaction
$\begin{array}{ll} {XeF}_{6} + H_{2} O ⇌ {XeOF}_{4} + 2 HF; K_{1} = 4 \\ {XeO}_{4} + {XeF}_{6} ⇌ {XeOF}_{4} + {XeO}_{3} F_{2}; K_{2} = 100 \end{array}$
Find equilibrium constant of
$\frac{1}{2} {XeO}_{4} + HF ⇌ \frac{1}{2} {XeO}_{3} F_{2} + \frac{1}{2} H_{2} O$

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answer is 5.

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The equilibriums constant for the reaction is given as,

$K_{e q} = \sqrt{K_{2}} \times \frac{1}{\sqrt{K_{1}}} K_{e q} = \sqrt{100} \times \frac{1}{\sqrt{4}} K_{eq} = 5$

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