Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\). A 7.5-L gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at \(450 \mathrm{~K}\). At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{~atm}\) \(P_{\mathrm{Cl}_{2}}=0.157 \mathrm{~atm},\) and \(P_{\mathrm{PCl}_{5}}=1.30 \mathrm{~atm}\) (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at \(450 \mathrm{~K}\).

To calculate the value of Kp, use the equilibrium pressures of the reactants and products. The formula for Kp in this reaction is: Kp = \(\frac{P_{PCl_{5}}}{P_{PCl_{3}} \times P_{Cl_{2}}}\) We have given: \(P_{PCl3} = 0.124 \, atm \), \(P_{Cl2} = 0.157 \, atm \), and \(P_{PCl5} = 1.30 \, atm \). Now, just plug the values into the formula: \( Kp = \frac{1.30}{0.124 \times 0.157} \) Calculate the value of Kp. (b)

To determine whether the equilibrium favors reactants or products, compare the calculated Kp value with 1: - If Kp > 1, the equilibrium favors products. - If Kp < 1, the equilibrium favors reactants. - If Kp = 1, the equilibrium is equal between reactants and products. Compare the calculated value of Kp with 1 to see whether the equilibrium favors reactants, products, or is equal. (c)

To calculate Kc, use the following equation: Kc = \(\frac{[PCl_{5}]}{[PCl_{3}] \times [Cl_{2}]}\) where \([...\]\) denotes concentration in moles per liter. We can convert the pressures of the reactants and products to concentrations using the ideal gas law: PV = nRT dividing both sides by RT gives: \([...] = \frac{P}{RT}\) Now, calculate the concentrations of all three gases at 450 K. Keep in mind that R must be in the same unit as the pressure (L*atm/(K*mol)) which is 0.0821 L*atm/(K*mol): \([PCl3] = \frac{0.124}{0.0821 \times 450}\) \([Cl2] = \frac{0.157}{0.0821 \times 450}\) \([PCl5] = \frac{1.30}{0.0821 \times 450}\) Now, substitute the concentrations in the equation of Kc: \(Kc = \frac{\frac{1.30}{0.0821 \times 450}}{\frac{0.124}{0.0821 \times 450} \times \frac{0.157}{0.0821 \times 450}}\) Notice that we can simplify by canceling out the \(0.0821 \times 450\)'s from the equation. Then, calculate the value of Kc, which will be the same value as Kp.