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$P C l_{5}$ dissociate as : $P C l_{5} (g) ⇌ P C l_{3 (g)} + C l_{2 (g)}$
5 moles of $P C l_{5}$ are placed in a 200 litre vessel which contains 2 moles of $N_{2}$ and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant $K_{p}$ for the dissociation of $P C l_{5}$ is $____\times 10^{- 3}$ (nearest integer). (Given : R = 0.082 l atm $k^{- 1} m o l^{- 1}$ assume ideal gas behaviour).

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answer is 1107.

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Detailed Solution

$P C l_{5 (g)} ⇄ P C l_{3 (g)} + C l_{2 (g)}$

Initial moles : 5 X X

At equilibrium : $5 - X$

Total moles : $5 - X + X + X + 2 (m o l e s o f N_{2})$

$= 7 + X$

$P V = n R T ⇄ n_{T o t a l} = \frac{P V}{R T} = \frac{2.46 \times 200}{0.0821 \times 600} = 10$

$7 + X = 10 \to X = 3$

$K_{P} = \frac{[\frac{3}{10}] \times 2.46 \times (\frac{3}{10}) \times 2.46}{\frac{2}{10} \times 2.46} = \frac{3 \times 3}{2 \times 10} \times 2.46$

$= 1.107 = 1107 \times 10^{- 3}$

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