Question

At 127°C and 1 atm pressure, PCl₅(g) is partially dissociated into PCl₂(g) and Cl₂(g) as PCl₅(g)⇌PCl₂(g) + Cl₂(g). The density of the equilibrium mixture is 3.5 g/L. Percentage dissociation of PCl₅ is (R=0.08 L-atm/K-mol, P = 31, Cl = 35.5)

Answer 1

86.16%

Answer 2

1. Molar mass of PCl₅ ($M_0$): $M_0 = 31 + 5 \times 35.5 = 208.5$ g/mol

2. Molar mass of the equilibrium mixture ($M_{mix}$): Using the ideal gas law, $\rho = \frac{PM}{RT}$, so $M_{mix} = \frac{\rho RT}{P}$. $M_{mix} = \frac{(3.5 \text{ g/L}) \times (0.08 \text{ L-atm/K-mol}) \times (127+273 \text{ K})}{1 \text{ atm}} = 112$ g/mol

3. Degree of dissociation ($\alpha$): For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), $n=2$. The relation is $M_{mix} = \frac{M_0}{1+(n-1)\alpha}$. $112 = \frac{208.5}{1+\alpha}$ $1+\alpha = \frac{208.5}{112} \approx 1.8616$ $\alpha \approx 0.8616$

4. Percentage dissociation: $\alpha \times 100 \approx 86.16\%$