For the reaction${{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{HI}}\left( {\rm{g}} \right)$, the equilibrium constants expressed in terms of concentrations ${K_c}$ and in terms of partial pressure ${K_p}$, are related as

${K_p} = {K_c}{\left( {RT} \right)^2}$

${K_p} = {K_c}{\left( {RT} \right)^{ - 2}}$

${K_c} = {K_p}\left( {RT} \right)$

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

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Understanding +4 | -1

For the reaction

{\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{HI}\left( \mathrm{g} \right)

, the equilibrium constants expressed in terms of concentrations

{K_c}

and in terms of partial pressure

{K_p}

, are related as

View answer and explanation

Correct option: C

{K_p} = {K_c}

Option C (

{K_p} = {K_c}

) is correct because for the given reaction, the change in the number of moles of gas (Δn) is zero, making Kp equal to Kc.

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}