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In the reaction $\large A\, + \,2B\, \Leftrightarrow \,2C\, + \,D$ , the initial concentration of B is two times that of 'A'. But equilibrium concentration of B & C were found to be equal. Then the K_C for the above system is

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0.27

0.18

0.11

answer is D.

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Detailed Solution

Given

A + 2B $⇌$ 2C + D

from stoichiometry

x moles A react with 2x moles of B to give 2x moles of C and X moles of D

Substituting the value of X = $\frac{1}{2}$

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Stoichiometry	$\large \mathop {{A}\left( g \right)}\limits^{1\,mole} +$	$\large \mathop {{2B}\left( g \right)}\limits^{1\,mole}$	$\rightleftharpoons$	$\large \mathop {2C}\limits^{2\,mole} \left( g \right)$	$\large \mathop {{D}\left( g \right)}\limits^{1\,mole}$
Initial moles	1	2		0	0
Moles at equilibrium	(1 - x)	(2 - 2x)		2x	x
Equilibrium concentration	$\large \left( {\frac{{1 - x}}{{V}}} \right)$	$\large \left( {\frac{{2 - 2x}}{{V}}} \right)$		$\large \left (\frac{{2x}}{{V}} \right )$	$\large \left (\frac{{x}}{{V}} \right )$