$1.0 m o l e$ of nitrogen and $3.0 m o l e s$ of $P {C l}_{5}$ are placed in $100 L$ vessel heated to $227^{o} C$ . The equilibrium pressure is $2.05 a t m$ . Assuming ideal behaviour, calculate the degree of dissociation for $P {C l}_{5}$ and $K_{p}$ for the reaction, $P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)$ .

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Solution

$P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)$
Initial: $3$ $0$ $0$
At equilibrium: $3 (1 - x)$ $3 x$ $3 x$
( $x =$ degree of dissociation)

Total moles $= 3 (1 - x) + 3 x + 3 x = 3 (1 + x)$

$1 m o l e$ of nitrogen is present, hence actual total number of moles at equilibrium $= 3 (1 + x) + 1$

According to gas equation,

$P V = n R T$

Given, $P = 2.05 a t m, V = 100 l i t r e s, R = 0.082 a t m \times L / (m o l \times K)$
and $T = (273 + 227) = 500 K$

So, $n = \frac{2.05 \times 100}{0.082 \times 500} = 5$
or $3 (1 + x) + 1 = 5$

$3 x = 1$ or $x = 0.333$

Degree of dissociation is 0.333.

At equilibrium, $p_{P {C l}_{5}} = \frac{3 (1 - x)}{(3 x + 4)} \times 2.05 a t m$

$p_{P {C l}_{3}} = p_{{C l}_{2}} = \frac{3 x}{(3 x + 4)} \times 2.05 a t m$

$K_{p} = \frac{\frac{3 x}{(3 x + 4)} \times 2.05 \times \frac{3 x}{(3 x + 4)} \times 2.05}{\frac{3 (1 - x)}{(3 x + 4)} \times 2.05} = \frac{3 \times {(0.333)}^{2} \times 2.05}{(5) (0.667)} = 0.204 a t m$

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Similar questions

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 $P C l_{5} ⇌ P C l_{3} + C l_{2}$ .

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: