One mole of $N_{2}$ and $3 m o l e s$ of $P C l_{5}$ are placed in a $100$ litre vessel heated to $227^{\circ} C$ . The equilibrium pressure is $2.05$ atm. Assuming ideal behaviour, calculate degree of dissociation of $P C l_{5}$ and $K_{p}$ of the reaction:
$P C l_{5} ⇌ P C l_{3} + C l_{2}$ .

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Solution

Initial
$P C l_{5} (g) \frac{3}{100} M ⇌ P C l_{3} (g) + C l_{2} (g) (+ \frac{1}{100} \frac{m o l e s}{l i t} o f N_{2})$
At $e q -^{b}$
$P C l_{5} (g) 0.03 - 0.03 α ⇌ P C l_{3} (g) 0.03 α + C l_{2} (g) 0.03 α a t T = 500 K$

$∴ T o t a l \frac{m o l e s}{l i t} a t e q -^{b} = 0.03 - 0.03 α + 0.03 α + 0.03 α + 0.01 = 0.04 + 0.03 α$

$∴$ Total moles = 4 + 3x
$∴ P_{P C l_{3}} = \frac{3 α}{4 + 3 α} \times 2.05 a t m$
$P c l_{2} (g) = \frac{3 α}{4 + 3 α} \times 2.05 a t m$
$P_{P C l_{5})} (g) = \frac{3 - 3 α}{4 + 3 α} \times 2.05 a t m$

From PV=nRT
$2.05 \times 100 = (4 + 32) \times 0.0821 \times 500$
$α = 0.33$

$K p = \frac{P_{P C l_{3}} . P_{C l_{2}}}{P_{P_{C l_{5}}}} = 0.205 a t m$

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One mole of $N_{2} a n d 3$ moles of $P C l_{5}$ are placed in a 100 litre vessel heated to $227^{\circ} C$ . The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate degree of dissociation of $P C l_{5}$ for the reaction $P C l_{5} ⇌ P C l_{3} + C l_{2}$ .

At $227^{o} C$ one mole of $C l_{2}$ and 3 moles of $P C l_{5}$ are placed in a 100 litre container and allowed to reach equilibrium. It is found that the equilibrium pressure is 2.05 atmospheres. Assuming ideal behaviour, the degree of dissociation 'x' for $P C l_{5}$ and $K_{p}$ for the reaction $P C l_{5}$ (g) $⇌$ $P C l_{3}$ (g) + $C l_{2}$ (g) is: