$20.85 g$ of $PCl X_{5}$ (molar mass $208.5 g/mol$ ) is heated in a $5.00 L$ vessel at $500 K$ and the equilibrium total pressure is $1.50 atm$ . Use $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ . Find $K_{c}$ for $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ in mol/L, to two decimal places.

Numerical answer type, answer in mol/L, tolerance 0.01.

Answer: 0.08 mol/L

n_{0} = 0.100

mol.

n_{t o t} = P V / RT = 1.50 \times 5.00/ (0.0821 \times 500) = 0.1827

mol

= n_{0} (1 + α)

, so

α = 0.8270

K_{c} = α^{2} n_{0} / [V (1 - α)] = 0.6840 \times 0.100/ (5.00 \times 0.1730) = 0.0791

mol/L, i.e. 0.08.