For the reaction, ${{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right) \to 2{\rm{HI}}\left( {\rm{g}} \right)$, the equilibrium constant ${K_p}$ changes with

The amount ${{\rm{H}}_2}\;{\rm{and}}\;{{\rm{I}}_2}$

NEET UG CHEMISTRY MCQ: Equilibrium

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Understanding +4 | -1

For the reaction,

{\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right) \to 2\mathrm{HI}\left( \mathrm{g} \right)

, the equilibrium constant

{K_p}

changes with

View answer and explanation

Correct option: D

Temperature

Option D (Temperature) is correct because the equilibrium constant

{K_p}

is temperature-dependent. Changing the temperature affects the equilibrium position and thus the value of

{K_p}

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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}