NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

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Analyzing +4 | -1

For the reaction in equilibrium,

I.

\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{CO}_2(g)

II.

2 \mathrm{CO}(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}_2(g)

\begin{array}{cccccccccccccccccccc} {{\text{Equilibrium}}\,\,{\text{constant}}}&{{\text{Free}}\,\,{\text{energy}}\,\,{\text{charge}}} \\ {{K_1}}&{\Delta {G_1}} \\ {{K_2}}&{\Delta {G_1}} \end{array}

Then

View answer and explanation

Correct option: C

K_2=K_1^2, \Delta G_2=2 \Delta G_1

Option C (

K_2=K_1^2, \Delta G_2=2 \Delta G_1

) is correct because doubling the reaction squares the equilibrium constant and doubles the free energy change.

More Questions from Equilibrium

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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}