A) \[{{K}_{2}}/{{({{K}_{2}})}^{2}}\]
B) \[{{K}_{1}}\,.\,{{K}_{2}}\]
C) \[{{K}_{1}}/{{K}_{2}}\]
D) \[{{K}_{2}}/{{K}_{1}}\]
Correct Answer: D
Solution :
\[Xe{{F}_{6}}(g)+{{H}_{2}}O(g)\rightleftharpoons XeO{{F}_{4}}(g)+2HF(g)\] \[{{K}_{1}}=\frac{[XeO{{F}_{4}}]\,{{[HF]}^{2}}}{[Xe{{F}_{6}}]\,[{{H}_{2}}O]}\] ....(i) \[Xe{{O}_{4}}(g)+Xe{{F}_{6}}(g)\rightleftharpoons XeO{{F}_{4}}(g)+Xe{{O}_{3}}{{F}_{2}}(g)\] \[{{K}_{2}}=\frac{[XeO{{F}_{4}}]\,[Xe{{O}_{3}}{{F}_{2}}]}{[Xe{{O}_{4}}]\,[Xe{{F}_{6}}]}\] ...(ii) For reaction, \[Xe{{O}_{4}}(g)+2HF(g)\rightleftharpoons Xe{{O}_{3}}{{F}_{2}}(g)+{{H}_{2}}O\,(g)\] \[K=\frac{[Xe{{O}_{3}}{{F}_{2}}]\,[{{H}_{2}}O]}{[Xe{{O}_{4}}]\,{{[HF]}^{2}}}\] \[\therefore \,\] From Eqs. (i) and (ii) \[K=\frac{{{K}_{2}}}{{{K}_{1}}}\]You need to login to perform this action.
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