• question_answer If ${{K}_{1}}\,and\,\,{{K}_{\text{2}}}$ are the respective equilibrium constants for the two reactions $Xe{{F}_{6}}(g)+{{H}_{2}}O(g)\,=\,\,XeO{{F}_{4}}\,(g)\,\,+2HF\,(g)$ $Xe{{O}_{4}}\,(g)+Xe{{F}_{6}}\,(g)\,\,\rightleftharpoons \,\,XeO{{\text{F}}_{\text{4}}}(g)\,\,+\,\,Xe{{O}_{3}}{{F}_{2}}(g)$ the equilibrium constant of the reaction $Xe{{O}_{4}}(g)+2HF\,(g)\rightleftharpoons \,Xe{{O}_{3}}{{F}_{2}}\,(g)+{{H}_{2}}O(g)$ will be A) ${{K}_{1}}/{{({{K}_{2}})}^{2}}$      B)        ${{K}_{1}}\,.\,{{K}_{\text{2}}}$ C) ${{K}_{1}}/\,{{K}_{\text{2}}}$                       D) ${{K}_{2}}/\,{{K}_{\text{1}}}$

Solution :

For the reaction ${{\operatorname{XeF}}_{6}}(g)+{{H}_{2}}O(g)\,\,\,XeO{{F}_{4}}\,(g)+2\,HF\,(g)$ ${{K}_{1}}=\frac{\left[ XeO{{F}_{4}} \right]{{\left[ HF \right]}^{2}}}{\left[ Xe{{F}_{6}} \right]\left[ {{H}_{2}}O \right]}$                        ? (i) and for the reaction $\operatorname{Xe}{{O}_{4}}\,(g)\,\,+\,\,Xe{{F}_{6}}(g)\,=\,\,\,\,XeO{{F}_{4}}(g)\,\,+\,\,Xe{{O}_{3}}{{F}_{2}}(g)$${{K}_{2}}=\,\,\frac{[XeO{{F}_{4}}][Xe{{O}_{3}}{{F}_{2}}]}{\left[ Xe{{O}_{4}} \right]\left[ Xe{{F}_{6}} \right]}$ For reaction: $\operatorname{Xe}{{O}_{4}}(g)+2HF(g)\,\to \,\,Xe{{O}_{3}}{{F}_{2}}\,(g)+{{H}_{2}}O\,(g)$ $\operatorname{K}=\,\,\frac{[Xe{{O}_{3}}{{F}_{2}}][{{H}_{2}}O]}{\left[ Xe{{O}_{4}} \right]{{\left[ HF \right]}^{2}}}$ $\therefore$ From eq. no. (i) and (ii) $\operatorname{K}=K{{ & }_{2}}/{{K}_{1}}$

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