At $600\,,\,{K_p}$ for the following reaction is 1 atm. $X\left( {\rm{g}} \right)\,\,\,\leftrightharpoons\,\,\,Y\left( {\rm{g}} \right) + Z\left( {\rm{g}} \right)$ At equilibrium, $50\,\% $ of $X\left( {\rm{g}} \right)$ is dissociated. The total pressure of the equilibrium system is $p$ atm. What is the partial pressure (in atm) of $X\left( {\rm{g}} \right)$ at equilibrium?

NEET UG CHEMISTRY MCQ: Equilibrium

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600\,,\,{K_p}

for the following reaction is 1 atm.

X\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,Y\left( \mathrm{g} \right) + Z\left( \mathrm{g} \right)

At equilibrium,

50\,\%

X\left( \mathrm{g} \right)

is dissociated. The total pressure of the equilibrium system is

p

atm. What is the partial pressure (in atm) of

X\left( \mathrm{g} \right)

at equilibrium?

View answer and explanation

Correct option: A

Option A is correct because, at equilibrium, the partial pressure of X is equal to 1 atm, given the total pressure and the degree of dissociation.

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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}