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 $P C l_{5} ⇌ P C l_{3} + C l_{2}$ .

10 %

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25 %

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33.3 %

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66.6 %

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Solution

The correct option is C
33.3 %

$N_{2} + P C l_{5} ⇌ P C l_{3} + C l_{2}$

\begin{array}{c c c}

\text{ Moles at t = 0 } & \text{1} & \text{3} & \text{0} & \text{0} \\

\text{Moles} & \text{1} & \text{(3-x)} & \text{x} & \text{x} \\

\end{array}

at equilibrium

Total moles present at equilibrium = 4 + x

Given total pressure at equilibrium = 2.05

PV = nRT

$2.05 \times 100 = (4 + x) \times 0.082 \times 500$

x = 1.0

Degree of dissociation $^{'} α^{'} f o r P C l_{5}$

$= \frac{M o l e s d i s s o c i a t e d}{T o t a l m o l e s p r e s e n t} = \frac{1}{3} = 0.3333 = 33.33$

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Similar questions

Q. One mole of

N_{2}

and

3 m o l e s

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}

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: