At \(250^oC\) and 1 atmospheric pressure, the vapour density pf \(PCl_5\)is \(57.9\). Calculate:

(i) \(K_p\) for the reaction, \(PCl_5(g) ⇌ PCl_3 (g) + Cl_2(g)\), at \(250^oC\)

(ii) the percentage dissociation when pressure is doubled.

1.78 and 68.6%

The correct answer is option 1. 1.78 and 68.6%.

Molar mass of \(PCl_5\) \(= 208.5\)

Vapour density, \(D = \frac{208.5}{2} = 104.25\)

Observed vapour density, \(d = 57.9\)

Degree of dissociation, \(\alpha = \frac{D - d}{d} = \frac{104.25 - 57.9}{57.9} = 0.80\)

Total number of mole \(= (1 - \alpha ) + \alpha  + \alpha = 1 + \alpha = 1 + 0.80 = 1.80\)

Partial Pressure of \(PCl_5\) \(= \frac{0.2}{1.8} \times  1 = \frac{1}{9}\)

Partial Pressure of \(PCl_3\) \(= \frac{0.8}{1.8} \times  1 = \frac{4}{9}\)

Partial Pressure of \(Cl_2\) \(= \frac{0.8}{1.8} \times  1 = \frac{4}{9}\)

So,

\(K_p = \frac{p_{PCl_3} \times p_{Cl_2}}{p_{PCl_5}}\)

\(⇒ K_p = \frac{\frac{4}{9} \times \frac{4}{9}}{\frac{1}{9}}\)

\(⇒ K_p = \frac{16}{9} = 1.78\)

Let the degree of dissociation be \(\alpha \) when pressure is \(2\) atmosphere.

At equilibrium,

\(p_{PCl_5} = \frac{1 - \alpha }{1 + \alpha }P = \frac{1 - \alpha }{1 + \alpha } \times 2\)

\(p_{PCl_3} = \frac{\alpha }{1 + \alpha }P = \frac{\alpha }{1 + \alpha } \times 2\)

\(p_{Cl_2} = \frac{\alpha }{1 + \alpha }P = \frac{\alpha }{1 + \alpha } \times 2\)

\(K_p = \frac{\frac{\alpha }{1 + \alpha } \times 2 \times \frac{\alpha }{1 + \alpha } \times 2}{\frac{1 - \alpha }{1 + \alpha } \times 2}\)

or, \(\frac{\alpha ^2}{1 - \alpha ^2} \times 2 = 1.78 \)

or, \(\frac{\alpha ^2}{1 - \alpha ^2} = 0.89\)

or, \(\alpha ^2 = 0.89 - 0.89\alpha ^2\)

or, \(1.89\alpha ^2 = 0.89\)

or, \(\alpha ^2 = \frac{0.89}{1.89}\)

or, \(\alpha = 0.686\)