2 mole of PCl5 is heated in a one litre vessel. If PCl5 dissociates to the extent of 80%, the equilibrium constant for the dissociation of PCl5 is

Vvessel = 1 Degree of decomposition, α = For every 1 mole of PCl5 taken 0.8 moles decomposes. For every 2 mole of PCl5 taken 1.6 moles decomposes. 1 mole of PCl5 decomposes to give 1 mole of PCl3 and 1 mole of Cl2 1.6 mole of PCl5 decomposes to give 1.6 mole of PCl3 and 1.6 mole of Cl2 Stoichiometry Initial moles 2 - - Moles at equilibrium (2 - 1.6) 1.6 1.6 Equilibrium concentration

2 mole of PCl5 is heated in a one litre vessel. If PCl5 dissociates to the extent of 80%, the equilibrium constant for the dissociation of PCl5 is

Vvessel = 1 Degree of decomposition, α = For every 1 mole of PCl5 taken 0.8 moles decomposes. For every 2 mole of PCl5 taken 1.6 moles decomposes. 1 mole of PCl5 decomposes to give 1 mole of PCl3 and 1 mole of Cl2 1.6 mole of PCl5 decomposes to give 1.6 mole of PCl3 and 1.6 mole of Cl2 Stoichiometry Initial moles 2 - - Moles at equilibrium (2 - 1.6) 1.6 1.6 Equilibrium concentration

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2 mole of PCl₅ is heated in a one litre vessel. If PCl₅ dissociates to the extent of 80%, the equilibrium constant for the dissociation of PCl₅ is

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2 $\times$ 10^-2

0.67

6.4

0.32

answer is B.

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Detailed Solution

$\large {\left( {{n_{PC{l_5}}}} \right)_i} = 2$

V_vessel = 1

Degree of decomposition, α = $\large \frac{{80}}{{100}} = 0.8$

For every 1 mole of PCl₅ taken 0.8 moles decomposes.

For every 2 mole of PCl₅ taken 1.6 moles decomposes.

1 mole of PCl₅ decomposes to give 1 mole of PCl₃ and 1 mole of Cl₂

1.6 mole of PCl₅ decomposes to give 1.6 mole of PCl₃ and 1.6 mole of Cl₂

Stoichiometry	$\large \mathop {PC{l_{5(g)}}}\limits^{1{\mkern 1mu} mole}$	$\rightleftharpoons$	$\large \mathop {PC{l_{3(g)}}}\limits^{1{\mkern 1mu} mole} +$	$\large \mathop {C{l_{2(g)}}}\limits^{1{\mkern 1mu} mole}$
Initial moles	2		-	-
Moles at equilibrium	(2 - 1.6)		1.6	1.6
Equilibrium concentration	$\large \frac{{0.4}}{1}$		$\large \frac{{1.6}}{1}$	$\large \frac{{1.6}}{1}$