For the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ at \(600 . \mathrm{K}\), the equilibrium constant, \(K_{\mathrm{p}}\), is \(11.5 .\) Suppose that \(2.450 \mathrm{~g} \mathrm{PCl}_{5}\) is placed in an evacuated \(500 .-\mathrm{mL}\) bulb, which is then heated to \(600 . \mathrm{K}\). a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the degree of dissociation of \(\mathrm{PCl}_{5}\) at equilibrium?

Given the mass of PCl5 and the volume of the bulb, we can calculate the initial pressure using the ideal gas equation: \[ PV = nRT \] We need to first convert the mass of PCl5 to moles. Molar mass of PCl5 = 1 × (30.97 g/mol) + 5 × (35.45 g/mol) = 208.25 g/mol Therefore, moles of PCl5 = 2.450 g / 208.25 g/mol = 0.01176 mol Now we can solve for the pressure: \[ P = \frac{nRT}{V} \] Given: R = 0.0821 L atm / (K mol), T = 600 K, and V = 500 mL = 0.5 L a. \( P_{\mathrm{PCl}_5} = \frac{(0.01176 \, \text{mol})(0.0821 \, \frac{\text{L atm}}{\text{K mol}})(600 \, \text{K})}{0.5 \, \text{L}} = 11.5 \, \text{atm} \)

Let x be the change in the moles of PCl5 as it dissociates. Then the change in moles of PCl3 and Cl2 will be x (as per the reaction stoichiometry). At equilibrium: PCl5 = 0.01176 - x moles PCl3 = x moles Cl2 = x moles Now, we can express their partial pressures, \(P_{\mathrm{PCl}_5}\), \(P_{\mathrm{PCl}_3}\), and \(P_{\mathrm{Cl}_2}\) in terms of x. \[P_{\mathrm{PCl}_5} = \frac{(0.01176 - x) \cdot RT}{V} = \frac{(0.01176 - x) \cdot 0.0821 \cdot 600}{0.5}\] \[P_{\mathrm{PCl}_3} = \frac{x \cdot 0.0821 \cdot 600}{0.5}\] \[P_{\mathrm{Cl}_2} = \frac{x \cdot 0.0821 \cdot 600}{0.5}\] Next, we can write the equation for the equilibrium constant: \[ K_{\mathrm{p}} = \frac{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}} \]

Given Kp = 11.5, substitute the expressions for the partial pressures in the equilibrium constant equation and solve for x: \[ 11.5 = \frac{\frac{x \cdot 0.0821 \cdot 600}{0.5} \cdot \frac{x \cdot 0.0821 \cdot 600}{0.5}}{\frac{(0.01176 - x) \cdot 0.0821 \cdot 600}{0.5}} \] \[ x = 0.00462 \, \text{mol} \] Now, we can find the partial pressures at equilibrium: b. \( P_{\mathrm{PCl}_5} = \frac{(0.01176 - 0.00462) \cdot 0.0821 \cdot 600}{0.5} = 4.34 \, \text{atm} \) \[ P_{\mathrm{PCl}_3} = \frac{0.00462 \cdot 0.0821 \cdot 600}{0.5} = 3.58 \, \text{atm} \] \[ P_{\mathrm{Cl}_2} = \frac{0.00462 \cdot 0.0821 \cdot 600}{0.5} = 3.58 \, \text{atm} \] c. The total pressure in the bulb at equilibrium = \( P_{\mathrm{PCl}_5} + P_{\mathrm{PCl}_3} + P_{\mathrm{Cl}_2} = 4.34 + 3.58 + 3.58 = 11.5 \, \text{atm} \)

The degree of dissociation (α) is the ratio of the change in moles of PCl5 at equilibrium to the initial moles of PCl5: \[ \alpha = \frac{x}{0.01176} \] d. Degree of dissociation, α = \( \frac{0.00462}{0.01176} = 0.3927 \) In conclusion, the answers for the given problem are: a. The pressure of PCl5 if it did not dissociate is 11.5 atm. b. The partial pressure of PCl5 at equilibrium is 4.34 atm. c. The total pressure in the bulb at equilibrium is 11.5 atm. d. The degree of dissociation of PCl5 at equilibrium is 0.3927.