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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

The equilibri...

Question

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

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Solution

$P C l_{3} + C l_{2} ⇌ P C l_{5} 0.20 0.08 0.40$ moles/litre

Kc = 25
If 0.22 mole of $C l_{2}$ is added then at new equilibrium
0.20 – x 0.30 – x 0.40 + x
$25 = \frac{0.40 + x}{(0.20 - x) (0.30 - x)}$
or x=0.1
$[P C l_{3}] = 0.2 - 0.1 = 0.1$ moles
$[C l_{2}] = 0.3 - 0.1 = 0.2$ moles
$[P C l_{5}] = 0.4 + 0.1 = 0.5$ moles

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The equilibrium composition in 1 litre container for the reaction is PCl3 + Cl2⇌PCl50.20 0.08 0.40 0.22 mole of Cl2 is added at the same temperature and a new equilibrium is established. If the sum of the new equilibrium moles of PCl3,Cl2 and PCl5 is p, then value of p is

PCl3 + Cl2⇌PCl50.20 0.08 0.40 moles/litre Kc = 25 If 0.22 mole of Cl2 is added then at new equilibrium 0.20 – x 0.30 – x 0.40 + x 25=0.40+x(0.20−x)(0.30−x) or x=0.1 [PCl3]=0.2−0.1=0.1 moles [Cl2]=0.3−0.1=0.2 moles [PCl5]=0.4+0.1=0.5 moles