If ${K_1}$ and ${K_2}$ are the respective equilibrium constants for the two reactions, ${\rm{Xe}}{{\rm{F}}_6}\left( {\rm{g}} \right) + {{\rm{H}}_2}{\rm{O}}\left( {\rm{g}} \right)\;\leftrightharpoons\;{\rm{XeO}}{{\rm{F}}_4}\left( {\rm{g}} \right) + 2{\rm{HF}}\left( {\rm{g}} \right)$ ${\rm{Xe}}{{\rm{O}}_4}\left( {\rm{g}} \right) + {\rm{Xe}}{{\rm{F}}_6}\left( {\rm{g}} \right)\;\leftrightharpoons\;{\rm{XeO}}{{\rm{F}}_4}\left( {\rm{g}} \right) + {\rm{Xe}}{{\rm{O}}_3}{{\rm{F}}_2}\left( {\rm{g}} \right)$ The equilibrium constant for the reaction, ${\rm{Xe}}{{\rm{O}}_4}\left( {\rm{g}} \right) + 2{\rm{HF}}\left( {\rm{g}} \right)\;\leftrightharpoons\;{\rm{Xe}}{{\rm{O}}_3}{{\rm{F}}_2}\left( {\rm{g}} \right) + {{\rm{H}}_2}{\rm{O}}\left( {\rm{g}} \right)$ is:

NEET UG CHEMISTRY MCQ: Equilibrium

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{K_1}

and

{K_2}

are the respective equilibrium constants for the two reactions,

\mathrm{Xe}{\mathrm{F}_6}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{XeO}{\mathrm{F}_4}\left( \mathrm{g} \right) + 2\mathrm{HF}\left( \mathrm{g} \right)

\mathrm{Xe}{\mathrm{O}_4}\left( \mathrm{g} \right) + \mathrm{Xe}{\mathrm{F}_6}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{XeO}{\mathrm{F}_4}\left( \mathrm{g} \right) + \mathrm{Xe}{\mathrm{O}_3}{\mathrm{F}_2}\left( \mathrm{g} \right)

The equilibrium constant for the reaction,

\mathrm{Xe}{\mathrm{O}_4}\left( \mathrm{g} \right) + 2\mathrm{HF}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{Xe}{\mathrm{O}_3}{\mathrm{F}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)

is:

View answer and explanation

Correct option: C

{K_2}/{K_1}

The correct answer is C (

{K_2}/{K_1}

) because it represents the equilibrium constant obtained by manipulating the given reactions to match the target reaction.