If the equilibrium constant for ${{\rm{N}}_{\rm{2}}}\left( {\rm{g}} \right)\,{\rm{ + }}\,{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\,\,\,\leftrightharpoons\,\,\,{\rm{2NO}}\left( {\rm{g}} \right)\,$ is k, the equilibrium constant for $\frac{{\rm{1}}}{{\rm{2}}}{{\rm{N}}_{\rm{2}}}\left( {\rm{g}} \right)\,{\rm{ + }}\,\frac{{\rm{1}}}{{\rm{2}}}\,{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)\,\,\leftrightharpoons\,\,{\rm{NO}}\left( {\rm{g}} \right)$ will be:

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

1 of 1

Application +4 | -1

If the equilibrium constant for

{\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,\mathrm{2NO}\left( \mathrm{g} \right)\,

is k, the equilibrium constant for

\frac{\mathrm{1}}{\mathrm{2}}{\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,\frac{\mathrm{1}}{\mathrm{2}}\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\leftrightharpoons\,\,\mathrm{NO}\left( \mathrm{g} \right)

will be:

View answer and explanation

Correct option: C

\,{\mathrm{K}^{\frac{1}{2}}}

Option C (

\sqrt{{K}}

) is correct because when the reaction is halved, the equilibrium constant is raised to the power of 1/2.

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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}