question_answer

If \[{{K}_{1}}\] and \[{{K}_{2}}\] are the respective equilibrium constants for the two reactions:  [AIPMT 1998]

\[Xe{{F}_{6}}(g)+{{H}_{2}}O(g)~XeO{{F}_{4}}(g)+2HF(g)\]

\[Xe{{O}_{4}}(g)+Xe{{F}_{6}}(g)~XeO{{F}_{4}}(g)+Xe{{O}_{3}}{{F}_{2}}(g)\]

the equilibrium constant of the reaction

\[Xe{{O}_{4}}(g)+2HF(g)~Xe{{O}_{3}}{{F}_{2}}(g)+{{H}_{2}}O(g)\]will be :

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 :

....(i)

...(ii)

For reaction,

From Eqs. (i) and (ii)