NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

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Analyzing +4 | -1

The following equilibrium constants were determined at

1120 {~K}

\begin{aligned} & 2 {CO}(g) \rightleftharpoons {C}({s})+{CO}_2({~g}) ; K_{p_1} =10^{-14} {~atm}^{-1} \\ & {CO}(g)+{Cl}_2(g) \rightleftharpoons {COCl}_2({~g}) ; K_{p_2} =6 \times 10^{-3} {~atm}^{-1} \end{aligned}

What is the equilibrium constant

K_c

for the following reaction at

1120 {~K}

{C}({s})+{CO}_2({~g})+2 {Cl}_2({~g}) \rightleftharpoons 2 {COCl}_2({~g})

View answer and explanation

Correct option: A

3.31 \times 10^{11} {M}^{-1}

Option A (

3.31 \times 10^{11} {M}^{-1}

) is correct because it is calculated by manipulating the given equilibrium constants and converting

K_p

K_c

using the ideal gas law.

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}