At same temperature and under a pressure of $4 a t m$ , $P {C l}_{5}$ is $10$ % dissociation. Calculate the pressure at which $P {C l}_{5}$ will be $20$ % dissociated, temperature remaining same.

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Solution

$P C l_{5} \to P C l_{3} + C l_{2}$
Let P atm be the initial pressure of $P C l_{5}$
10% of it or $P \times \frac{10}{100} = 0.1 P$ atm of it will dissociate to form 0.1P atm of $P C l_{3}$ and 0.1P atm of $C l_{2}$ . The equilibrium pressure of $P C l_{5} = P - 0.1 P = 0.9 P$ atm.
Total pressure $= 0.9 P + 0.1 P + 0.1 P = 1.1 P = 4$ atm
$P = \frac{4}{1.1} = 3.64$ atm
The equilibrium constant $K_{P} = \frac{P_{P C l 3} P_{C l 2}}{P_{P C l 5}}$
$K_{p} = \frac{0.1 P \times 0.1 P}{0.9 P}$
$K_{P} = 0.01111 P$
$K_{P} = 0.01111 \times 3.64$
$K_{p} = 0.0404$
20% of $P C l_{5}$ or $P \times \frac{20}{100} = 0.2 P$ atm of it will dissociate to form 0.2P atm of $P C l_{3}$ and 0.2P atm of $C l_{2}$ .
The equilibrium pressure of $P C l_{5} = P - 0.2 P = 0.8 P$ atm.
Total pressure $= 0.8 P + 0.2 P + 0.2 P = 1.2 P$ atm
The equilibrium constant $K_{P} = \frac{P_{P C l 3} P_{C l 2}}{P_{P C l 5}}$
$0.0404 = \frac{0.2 P \times 0.2 P}{0.8 P}$
$0.0404 = 0.05 P$
$P = 0.808$
Total pressure $= 1.2 P = 1.2 \times 0.808 = 0.96$ atm.

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