Consider the following reversible chemical reactions $\begin{array}{l} {A_2}(g) + {B_2}(g)\underset{{}}{\overset{{{K_1}}}{\leftrightharpoons}}2AB(g).....\,\left( {\rm{i}} \right)\\ 6AB(g)\underset{{}}{\overset{{{K_2}}}{\leftrightharpoons}}3{A_2}(g) + 3{B_2}(g)\,.....\,\left( {{\rm{ii}}} \right) \end{array}$ The relation between ${K_1}$ and ${K_2}$ is

NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

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Analyzing +4 | -1

Consider the following reversible chemical reactions

\begin{array}{l} {A_2}(g) + {B_2}(g)\underset{{}}{\overset{{{K_1}}}{\leftrightharpoons}}2AB(g).....\,\left( \mathrm{i} \right)\\ 6AB(g)\underset{{}}{\overset{{{K_2}}}{\leftrightharpoons}}3{A_2}(g) + 3{B_2}(g)\,.....\,\left( {\mathrm{ii}} \right) \end{array}

The relation between

{K_1}

and

{K_2}

View answer and explanation

Correct option: C

{K_2} = K_1^{ - 3}

The correct answer is C (

{K_2} = K_1^{ - 3}

) because it correctly relates the equilibrium constants of the forward and reverse reactions with appropriate powers.

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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}