In the reaction, ${{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{HCl}}\left( {\rm{g}} \right)$

NEET UG CHEMISTRY MCQ: Equilibrium

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Understanding +4 | -1

In the reaction,

{\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{HCl}\left( \mathrm{g} \right)

View answer and explanation

Correct option: B

{K_p} = {K_c}

Option B (

{K_p} = {K_c}

) is correct because the change in the number of moles of gas (

\Delta n

) is zero in this reaction, making

K_p

equal to

K_c

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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}