NEET UG CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

Xe{F_6} + {H_2}O\underset{{}}{\overset{{}}{\leftrightharpoons}}XeO{F_4} + 2HF

equilibrium constant

{K_1}

Xe{O_4} + Xe{F_6}\underset{{}}{\overset{{}}{\leftrightharpoons}}XeO{F_4} + Xe{O_3}{F_2}

equilibrium constant

{K_2}

then equilibrium constant for the reaction

Xe{O_4} + 2HF\underset{{}}{\overset{{}}{\leftrightharpoons}}Xe{O_3}{F_2} + {H_2}O

will be

View answer and explanation

Correct option: C

\frac{{{K_2}}}{{{K_1}}}

Option C (

\frac{{{K_2}}}{{{K_1}}}

) is correct because the equilibrium constant for the target reaction can be obtained by dividing the second reaction's constant by the first.

More Questions from Equilibrium

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Mg{(HC{O_3})_2}

MgC{O_3}

Mg{(OH)_2}

MgC{O_3}

K_{\mathrm{P}}

\mathrm{CaCO}_3

4.5 \times 10^{-2}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

K_{\mathrm{P}}

2.0 \mathrm{~atm}.

\mathrm{CaO}

\mathrm{CaCO}_3

+

\leftrightharpoons

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}

{K_{\mathrm{eq}}} = 0.41

\mathrm{p}{K_a}