Ammonium carbonate decomposes as ${\rm{N}}{{\rm{H}}_2}{\rm{COON}}{{\rm{H}}_4}\left( s \right)\leftrightharpoons2{\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right)$ For the reaction, ${K_p} = 2.9 \times {10^{ - 5}}\;{\rm{at}}{{\rm{m}}^{ - 3}}$ If we start with 1 mole of the compound, the total pressure at equilibrium would be

NEET UG CHEMISTRY MCQ: Equilibrium

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Ammonium carbonate decomposes as

\mathrm{N}{\mathrm{H}_2}\mathrm{COON}{\mathrm{H}_4}\left( s \right)\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)

For the reaction,

{K_p} = 2.9 \times {10^{ - 5}}\;\mathrm{at}{\mathrm{m}^{ - 3}}

If we start with 1 mole of the compound, the total pressure at equilibrium would be

View answer and explanation

Correct option: B

0.0582\;\mathrm{ }atm

Option B (0.0582 atm) is correct because it represents the total pressure at equilibrium calculated using the Kp expression.

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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}