For the reactions, ${{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right)\leftrightharpoons{\rm{CO}}\left( {\rm{g}} \right) + {{\rm{H}}_2}{\rm{O}}\left( {\rm{g}} \right)$ if the initial concentration of $[{{\rm{H}}_2}\left] = \right[{\rm{C}}{{\rm{O}}_2}]$ and $x$ mol/L of hydrogen is consumed at equilibrium, the correct expression of${K_p}$ is

$\frac{{{x^2}}}{{{{\left( {2 + x} \right)}^2}}}$

$\frac{{{{\left( {1 + x} \right)}^2}}}{{{{\left( {1 - x} \right)}^2}}}$

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

1 of 1

Application +4 | -1

For the reactions,

{\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{CO}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)

if the initial concentration of

[{\mathrm{H}_2}\left] = \right[\mathrm{C}{\mathrm{O}_2}]

and

x

mol/L of hydrogen is consumed at equilibrium, the correct expression of

{K_p}

View answer and explanation

Correct option: A

\frac{{{x^2}}}{{{{\left( {1 - x} \right)}^2}}}

Correct Answer + Core Reason:

\frac{{{{x^2}}}}{{{{{{\left( {1 - x} \right)}^2}}}}

is correct because it represents the equilibrium constant expression based on the stoichiometry of the reaction and the amount of hydrogen consumed.