In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl5 were introduced and equilibriated at a temperature at 250 °C. At equilibrium, the total pressure of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, a of PCl5 and KP for the reaction  PCl5  → PCl3 + Cl2  at this temperature.

Q:  In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl5 were introduced and equilibriated at a temperature at 250oC. At equilibrium, the total pressure of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, a of PCl5 and KP for the reaction  PCl5 $ \rightleftharpoons $ PCl3 + Cl2  at this temperature.

Solution:           PCl5(g) $ \rightleftharpoons $ PCl3(g)   +   Cl2(g)

Initial moles        5                   0                0

At equilibrium     5-x              x                x

Total moles    = 5 + x + 4  (including moles of Argon)

= 9 + x

Since total moles $= \frac{PV}{RT} = \frac{4.678 \times 110}{0.082 \times 523} $

total moles = 11.99 ≈12

9 + x = 12

∴ x = 3

α = 3/5  = 0.6

$\large K_p = \frac{((3/12)\times 4.678)^2}{(2/12)\times 4.678)}$

= 1.75 atm.

The degree of dissociation of PCl5 would have been 0.6 even in the absence of Argon. As one can see, the total pressure of gases  constituting equilibrium is equal to 3.11 atm. The observed equilibrium pressure is 4.678 atm which means by the addition of 4 moles of Argon, the total pressure increases. This implies that addition of Argon has been done at constant volume which doesnot result in any change in degree of dissociation.