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Standard XII

Chemistry

First Law of Thermodynamics

PCl5 dissocia...

Question

$P C l_{5}$ dissociates 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: $R = 0.082 L a t m K^{- 1} m o l^{- 1}$ ; Assume ideal gas behaviour)

1107.0

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1107

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1107.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
$\begin{matrix} Initial & 5 & 0 & 0 moles Equilibrium & 5 - x & x & x moles \end{matrix}$
Number of moles of $N_{2} = 2$
Equilibrium pressure $= 2.46$ atm
Total no. of moles = moles of $N_{2}$ (inert gas) + equilibrium moles,
$n = 2 + (5 - x) + x + x$

$P_{e q} = \frac{(7 + x) \times 0.082 \times 600}{200} = 2.46$
On solving, $x = 3$
$∴ K_{p} = \frac{(\frac{3 P}{10}) (\frac{3 P}{10})}{(\frac{2 P}{10})} = \frac{9 \times 2.46}{20}$
$= 1107 \times 10^{- 3}$ atm

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