The equilibrium constant for ${N_2}(g) + {O_2}(g)\leftrightharpoons2NO\left( g \right)$ is ${K_1}$ and that for $NO(g)\leftrightharpoons\frac{1}{2}{N_2}(g) + \frac{1}{2}{O_2}(g)$ is ${K_2}$, then ${K_1}$ and ${K_2}$ related as

${K_2} = {\left( {\frac{1}{{{K_1}}}} \right)^2}$

NEET UG CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

The equilibrium constant for

{N_2}(g) + {O_2}(g)\leftrightharpoons2NO\left( g \right)

{K_1}

and that for

NO(g)\leftrightharpoons\frac{1}{2}{N_2}(g) + \frac{1}{2}{O_2}(g)

{K_2}

, then

{K_1}

and

{K_2}

related as

View answer and explanation

Correct option: A

{K_1} = {\left( {\frac{1}{{{K_2}}}} \right)^2}

The correct answer is A (

{K_1} = {(\frac{1}{{{K_2}}})^2}

) because it correctly relates the equilibrium constants of the forward and reverse reactions.

Practice All Chapter MCQs →

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}