4.0 moles of argon and 5.0 moles of PCl$$_5$$ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K$$_p$$ for the reaction is [Given: R = 0.082 L atm K$$^{-1}$$ mol$$^{-1}$$]

We need to find K_p for the decomposition of PCl₅ at equilibrium.

The equilibrium reaction is $$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$$.

Using the ideal gas law with a total pressure of 6.0 atm, a volume of 100 L, a temperature of 610 K, and R = 0.082 L atm K⁻¹ mol⁻¹, the total number of moles at equilibrium is calculated as $$n_{\text{total}} = \frac{PV}{RT} = \frac{6.0 \times 100}{0.082 \times 610} = \frac{600}{50.02} = 11.99 \approx 12 \text{ moles}$$.

Argon is inert and remains unchanged at 4.0 moles, so the moles of reacting gases at equilibrium are 12 − 4 = 8 moles. Let x moles of PCl₅ dissociate: PCl₅: $$5 - x$$, PCl₃: $$x$$, Cl₂: $$x$$. The total reacting moles are $$(5 - x) + x + x = 5 + x = 8$$, which gives $$x = 3$$.

The total moles remain 12, so the partial pressures are calculated as follows: $$p_{\text{PCl}_5} = \frac{5-3}{12} \times 6 = \frac{2}{12} \times 6 = 1$$ atm; $$p_{\text{PCl}_3} = \frac{3}{12} \times 6 = 1.5$$ atm; and $$p_{\text{Cl}_2} = \frac{3}{12} \times 6 = 1.5$$ atm.

Substituting these values into the expression for K_p yields $$K_p = \frac{p_{\text{PCl}_3} \times p_{\text{Cl}_2}}{p_{\text{PCl}_5}} = \frac{1.5 \times 1.5}{1} = 2.25$$.

The answer is Option A: 2.25.