If ${K_1}$ and ${K_2}$ are equilibrium constants for reactions (I) and (II) respectively for, ${{\rm{N}}_2} + {{\rm{O}}_2}\leftrightharpoons2{\rm{NO}}\,\,\,\,\, \ldots \left( {\rm{i}} \right)$ $\frac{1}{2}{{\rm{N}}_2} + \frac{1}{2}{{\rm{O}}_2}\leftrightharpoons{\rm{NO}}\,\,\,\, \ldots \left( {{\rm{ii}}} \right)$ Then:

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

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Analyzing +4 | -1

{K_1}

and

{K_2}

are equilibrium constants for reactions (I) and (II) respectively for,

{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}\,\,\,\,\, \ldots \left( \mathrm{i} \right)

\frac{1}{2}{\mathrm{N}_2} + \frac{1}{2}{\mathrm{O}_2}\leftrightharpoons\mathrm{NO}\,\,\,\, \ldots \left( {\mathrm{ii}} \right)

Then:

View answer and explanation

Correct option: B

{K_2} = \sqrt {{K_1}}

Option B (

{K_2} = \sqrt {{K_1}}

) is correct because reaction (II) is half of reaction (I), so its equilibrium constant is the square root of the first.

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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}