$K_{p}$ for $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ is $1.80$ at $523 K$ . The density of the equilibrium mixture in $g {l i t}^{- 1}$ at a total pressure of $100 k P a$ will be : (nearest integer value)

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Solution

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
At t = 0 P 0 0
At equilibrium $(1 - α) P 100 - 2 x α P x α P x$

$\sum P = P (1 + α) = 100$

$∴ K_{p}^{0} = \frac{P_{P C l_{3}} \times P_{C l_{2}}}{P_{P C l_{5}}}$

$= \frac{x \times x}{100 - 2 x}$ (Given, $P_{T} = 100 k P a$ )

$1.80 = \frac{x^{2}}{(100 - 2 x)}; x^{2} + 3.6 x - 180 = 0$

$∴ x = 11.74 k P a$

Initial pressure $P_{i} = 100 - x = 100 - 11.74 = 88.26 k P a$

Degree of dissociation, $α = \frac{11.74}{88.26} = 0.133$

$∴ E x p . M_{P C l_{5}} = \frac{208.5}{1.133} = 184.02 g {m o l}^{- 1}$

Now, $P V = \frac{w}{M_{P C l_{5}}} \times R T$

$∴$ Density $= \frac{w}{V} = \frac{P \times M}{R T}$

$P = 100 k P a \approx 1 a t m$

Density $= \frac{1 \times 184.02}{0.0821 \times 523} = 4.280 \approx 4 g / l i t r e$

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