NEET UG CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

The equilibrium constant for the reaction between

C{H_4}\left( g \right)

and

{H_2}S\left( g \right)

to form

C{S_2}\left( g \right)

and

{H_2}\left( g \right)

, at 1173 K is 3.6. For the following composition of the reaction mixture,

\left[ {C{H_4}} \right] = 1.0\,M,\,\left[ {{H_2}S} \right] = 1.20M\,\,\,\left[ {C{S_2}} \right] = 2.0M,\,\,\left[ {{H_2}} \right] = 1.8M

decide which of the following option is correct?

\left[ {C{H_4}\left( g \right) + 2{H_2}S\left( g \right)C{S_2}\left( g \right) + 4{H_2}\left( g \right)} \right]

View answer and explanation

Correct option: C

Reaction will shift to form more

{H_2}S

Option C is correct because the reaction quotient Q is greater than Kc, indicating that the reaction will shift to the left, forming more H2S.

More Questions from Equilibrium

Practice All Chapter MCQs →

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}