MCQ
For the gaseous reaction, equilibrium constant is given

$XeF_6 + H_2O \rightleftharpoons  XeOF_4 + 2HF, \,\,\,\,K_1$

$XeO_4 + XeF_6 \rightleftharpoons  XeOF_4 + XeO_3F_2,\,\,\,\, K_2$

The equilibrium constant for the reaction

$XeO_4 + 2HF \rightleftharpoons  XeO_3F_2 + H_2O$ will be

  • A
    $\frac{{{K_1}}}{{{K_2}}}$
  • B
    $\frac{{{K_1}}}{{K_2^2}}$
  • C
    $\frac{{K_1^2}}{{{K_2}}}$
  • $\frac{{{K_2}}}{{{K_1}}}$

Answer

Correct option: D.
$\frac{{{K_2}}}{{{K_1}}}$
d
$\mathrm{XeF}_{6}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{XeOF}_{4}+2 \mathrm{HF}, \mathrm{K}_{1}......(i)$

$\mathrm{XeO}_{4}+\mathrm{XeF}_{6} \rightleftharpoons \mathrm{XeOF}_{4}+\mathrm{XeO}_{3} \mathrm{F}_{2}, \mathrm{K}_{2}......(ii)$

$(\mathrm{ii})+(\mathrm{i})$

$\mathrm{XeO}_{4}+2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_{3} \mathrm{F}_{2}+\mathrm{H}_{2} \mathrm{O}, \mathrm{K}=\mathrm{K}_{2} / \mathrm{K}_{1}$

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