For the gaseous reaction, $A + B ⇌ C + D,$ the initial concentration of ‘A’ and ‘B’ are equal. The equilibrium concentration of ‘C’ is half the initial concentration of ‘A’. The equilibrium constant for the reaction is

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\frac{1}{4}

\frac{1}{9}

answer is A.

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Detailed Solution

Given

$\large {\left[ A \right]_{initial}} = {\left[ B \right]_{initial}}$

$\large {\left[ C \right]_{eq}} = \frac{1}{2}{\left[ A \right]_i}$

Let 'x' moles each of A and B react to give 'x' moles each of C & D

Stoichiometry	$\large \mathop {{A}\left( g \right)}\limits^{1\,mole} +$	$\large \mathop {{B}\left( g \right)}\limits^{1\,mole}$	$\rightleftharpoons$	$\large \large \mathop {C}\limits^{1\,mole} \left( g \right)+$	$\large \mathop {D}\limits^{1\,mole} \left( g \right)$
Initial moles	1	1		0	0
Moles at equilibrium	(1 - x)	(1 - x)		x	x
Equilibrium concentration	$\large \left( {\frac{{1 - x}}{V}} \right)$	$\large \left( {\frac{{1 - x}}{V}} \right)$		$\large \left( {\frac{{x}}{V}} \right)$	$\large \left( {\frac{{x}}{V}} \right)$