A sample of \(\mathrm{S}_{8}(g)\) is placed in an otherwise empty rigid container at 1325 \(\mathrm{K}\) at an initial pressure of \(1.00 \mathrm{atm},\) where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$ At equilibrium, the partial pressure of \(\mathrm{S}_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at 1325 \(\mathrm{K}\) .

We're given the reaction: \[\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)\] The expression for the reaction's Kp is: \[K_{p} = \frac{(P_{\mathrm{S_2}})^4}{(P_{\mathrm{S_8}})}\] where \(P_{\mathrm{S_2}}\) and \(P_{\mathrm{S_8}}\) are the equilibrium partial pressures of S₂ and S₈, respectively.

Initially, the partial pressure of S₂ is 0 atm, since the container only has S₈. At equilibrium, the partial pressure of S₈ is 0.25 atm, which means the pressure change of S₈ during the reaction is: \[ \mathrm{∆P_{S_8}} = P_{\mathrm{Initial}} - P_{\mathrm{Equilibrium}} = 1.00 \mathrm{atm} - 0.25 \mathrm{atm} = 0.75 \mathrm{atm}\] Since 1 mole of S₈ produces 4 moles of S₂, we can find the pressure change of S₂ as a result of the reaction using the stoichiometry of the balanced equation: \[ \mathrm{∆P_{S_2}} = 4 \times \mathrm{∆P_{S_8}} = 4 \times 0.75 \mathrm{ atm} = 3.00 \mathrm{ atm}\] So, the equilibrium partial pressure of S₂ is: \[P_{\mathrm{S_2}} = \mathrm{Initial} + \mathrm{∆P_{S_2}} = 0 \mathrm{atm} + 3.00 \mathrm{atm} = 3.00 \mathrm{atm}\]

Now, we have the equilibrium partial pressures of both species. They are: \(P_{\mathrm{S_8}} = 0.25 \mathrm{atm}\) and \(P_{\mathrm{S_2}} = 3.00 \mathrm{atm}\) Substitute these values into the Kp expression: \[K_{p} = \frac{(P_{\mathrm{S_2}})^4}{(P_{\mathrm{S_8}})} = \frac{(3.00 \mathrm{atm})^4}{(0.25 \mathrm{atm})}\]

Calculate Kp: \[K_{p} = \frac{(3.00 \mathrm{atm})^4}{(0.25 \mathrm{atm})} = \frac{81 \mathrm{atm^3}}{0.25 \mathrm{atm}} = 324 \mathrm{atm^3}\] So, the equilibrium constant Kp for this reaction at 1325 K is 324 atm³.