How much PCl5 must be added to a one litre vessel kept at 2500C in order to obtain 0.1 mole of Cl2 gas. KC for PCl5g⇌PCl3g+Cl2gis 0.0414 mol/L

Given Moles of Cl2(g) at equilibrium = 0.1 Volume of the vessel = 1 litre Due to the decomposition of PCl5 equal moles of PCl3 and Cl2 are formed moles of PCl3(g) at equilibrium = 0.1 Now PCl5(g) PCl3(g) + Cl2(g) Moles at equilibrium a - 0.1 0.1 0.1 Equilibrium concentration Number of moles of PCl5 = 0.341

How much PCl5 must be added to a one litre vessel kept at 2500C in order to obtain 0.1 mole of Cl2 gas. KC for PCl5g⇌PCl3g+Cl2gis 0.0414 mol/L

Given Moles of Cl2(g) at equilibrium = 0.1 Volume of the vessel = 1 litre Due to the decomposition of PCl5 equal moles of PCl3 and Cl2 are formed moles of PCl3(g) at equilibrium = 0.1 Now PCl5(g) PCl3(g) + Cl2(g) Moles at equilibrium a - 0.1 0.1 0.1 Equilibrium concentration Number of moles of PCl5 = 0.341

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How much PCl₅ must be added to a one litre vessel kept at 250⁰C in order to obtain 0.1 mole of Cl₂ gas. K_Cfor ${PCl}_{5 (g)} ⇌ {PCl}_{3 (g)} + {Cl}_{2 (g)}$
is 0.0414 mol/L

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0.0341 mole

0.341 mole

0.241 mole

0.024 mole

answer is B.

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Detailed Solution

Given

$large large PC{l_{5left( g right)}}, rightleftharpoons ,PC{l_{3left( g right)}} + ,C{l_{2left( g right)}};K_C=0.0414M$

Moles of Cl₂(g) at equilibrium = 0.1

Volume of the vessel = 1 litre

Due to the decomposition of PCl₅ equal moles of PCl₃ and Cl₂ are formed

$large therefore$ moles of PCl₃(g) at equilibrium = 0.1

Now

	PCl₅(g)	$large rightleftharpoons$	PCl₃(g) +	Cl₂(g)
Moles at equilibrium	a - 0.1		0.1	0.1
Equilibrium concentration	$frac{a-0.1}{1}$		$large frac{0.1}{1}$	$large frac{0.1}{1}$