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Dalton's Law of Partial Pressures

In the reacti...

Question

In the reaction at equilibrium, $P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)$ , they are of equal amount and the total pressure is $3 a t m$ . The equilibrium constant $K_{p}$ is:

1 a t m

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2 a t m

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3 a t m

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1.5 a t m

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Solution

The correct option is B $1 a t m$
Explanation:
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
Moles of $P C l_{5}$ at equilibrium $= 2 m o l$
Moles of $P C l_{3}$ at equilibrium $= 2 m o l$
Moles of $C l_{2}$ at equilibrium $= 2 m o l$
Mole fraction of $P C l_{5} :$
$χ_{1} = \frac{2 m o l}{2 m o l + 2 m o l + 2 m o l} = \frac{1}{3}$
Mole fraction of $P C l_{3} :$
$χ_{2} = \frac{2 m o l}{2 m o l + 2 m o l + 2 m o l} = \frac{1}{3}$
Mole fraction of $C l_{2} :$
$χ_{3} = \frac{2 m o l}{2 m o l + 2 m o l + 2 m o l} = \frac{1}{3}$
Partial pressure of $P C l_{5}$ at equilibrium $= p_{1}$
Partial pressure of $P C l_{3}$ at equilibrium $= p_{2}$
Partial pressure of $C l_{2}$ at equilibrium $= p_{3}$
Total pressure at equilibrium $= 3 a t m$
$p_{1} = p \times χ_{1} = 3 a t m \times \frac{1}{3} = 1 a t m$
$p_{2} = p \times χ_{2} = 3 a t m \times \frac{1}{3} = 1 a t m$
$p_{3} = p \times χ_{3} = 3 a t m \times \frac{1}{3} = 1 a t m$
The equilibrium constant will be given as:
$K_{p} = \frac{p_{2} \times p_{3}}{p_{1}} = \frac{1 a t m \times 1 a t m}{1 a t m} = 1 a t m$
$∴$ $K_{p} = 1 a t m$
Hence correct answer is option $A$

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