The equilibrium composition for the reaction is -

$P C l_{3} + C l_{2} ⇌ P C l_{5}$

0.20 0.10 0.40 mole/litre

If 0.20 mole of $C l_{2}$ is added at same temp. Find equilibrium concentration of $P C l_{5}$ $K_{C} = 20$

0.48 moles

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0.38 moles

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0.56 moles

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1.20 moles

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Solution

The correct option is A
0.48 moles

$P C l_{3} + C l_{2} ⇌ P C l_{5}$

0.20 0.10 0.40 moles/litre

If 0.20 mole of $C l_{2}$ is added then at equilibrium

0.20 - x 0.30 - x 0.40 + x

$K c \frac{[P C l_{5}]}{[P C l_{3}] [C l_{2}]} 20 = \frac{(0.40 + x)}{(0.20 - x) (0.30 - x)} x = 0.08 [P C l_{5}] = 0.4 + 0.08 = 0.48 m o l e s$

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Q. The equilibrium composition for the reaction is -

P C l_{3} 0.20 + C l_{2} (g) 0.10 ⇌ P C l_{5} (g) 0.40 m o l e s / l i t r e

0.20 m o l e

C l_{2}

is added at same temp. Find equilibrium concentration of

P C l_{5} (K_{C} = 20)

The equilibrium composition in 1 litre container for the reaction is
$P C l_{3} + C l_{2} ⇌ P C l_{5} 0.20 0.08 0.40$

0.22 mole of $C l_{2}$ is added at the same temperature and a new equilibrium is established. If the sum of the new equilibrium moles of $P C l_{3}, C l_{2} a n d P C l_{5}$ is p, then value of p is