Starting from 'a' moles of H2 and 'b' moles of I2 an equilibrium is established with 2x moles of HI. The equilibrium constant KC is

Stoichiometry Initialmoles a b 0 Moles at equlilibrium (a - x) (b - x) 2x Equilibrium concentration

Stoichiometry Initialmoles a b 0 Moles at equlilibrium (a - x) (b - x) 2x Equilibrium concentration

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Starting from 'a' moles of H₂ and 'b' moles of I₂ an equilibrium $large {H_2}, + ,{I_2}, Leftrightarrow ,2HI$ is established with 2x moles of HI. The equilibrium constant K_C is

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$frac{{4{x^2}}}{{ab}}$

$frac{{4{x^2}}}{{left( {a - x} right)left( {b - x} right)}}$

$frac{{2{x^2}}}{{left( {a - x} right)left( {b - x} right)}}$

$frac{{4{x^2}}}{{left( {a - 2x} right)left( {b - 2x} right)}}$

answer is B.

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Stoichiometry	$large mathop {{H_2}left( g right){mkern 1mu} }limits^{1mole} {kern 1pt} +$	$large {mkern 1mu} {kern 1pt} mathop {{I_2}left( g right)}limits^{1{text{ mole}}} {mkern 1mu}$	$large rightleftharpoons {mkern 1mu} {kern 1pt}$	$large mathop {2HIleft( g right)}limits^{2{text{ moles}}}$
Initialmoles	a	b		0
Moles at equlilibrium	(a - x)	(b - x)		2x
Equilibrium concentration	$large left( {frac{{a - x}}{V}} right)$	$large left( {frac{{a - x}}{V}} right)$		$large frac{{2x}}{V}$