Calculate K p for the reaction PCl₅ (g) PCl 3 (g) + Cl₂ (g) at 25 C. The values are 0.0 (Cl₂ ), - 268 kJ/mol (PCl 3 ), and - 305 kJ/mol (PCl₅ ). R = 8.314 J/mol K

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is: Kp = (PCl3)^1(Cl2)^1 / (PCl₅) To solve for Kp, we need to calculate the partial pressures of the gases at equilibrium. Let x be the amount (in moles) of PCl3 and Cl2 that react to form PCl5. Then, the equilibrium partial pressures can be expressed in terms of x as: PCl3 = (1-x)Ptotal Cl2 = (1-x)Ptotal PCl5 = xPtotal where Ptotal is the total pressure at equilibrium. The value of Kp can then be calculated as: Kp = [(1-x)Ptotal]^2 / (xPtotal) Kp = [(1-x)/x]^2 * Ptotal We can find the value of x by using the given ΔH° values and the equation: ΔH° = Σ(nΔHf°products) - Σ(nΔHf°reactants) For the reaction PCl5 (g) ⇌ PCl3 (g) + Cl2 (g), we have: ΔH° = [1(-268) + 1(0)] - [1(-305)] = 37 kJ/mol Since ΔH° is positive, the reaction is endothermic and heat is a reactant. This means that increasing the temperature will shift the equilibrium to the right (toward PCl5). At 25°C, we have: ΔH° = 37 kJ/mol = 37,000 J/mol T = 25°C = 298 K Using the equation: ΔG° = ΔH° - TΔS° where ΔG° is the standard free energy change and ΔS° is the standard entropy change, we can solve for ΔS°: ΔS° = (ΔH° - ΔG°)/T At equilibrium, ΔG° = 0, so: ΔS° = ΔH°/T ΔS° = 37,000 J/mol / 298 K ΔS° = 124 J/mol K Now we can use the equation: ΔG° = -RT ln Kp to solve for Kp: Kp = e^(-ΔG°/RT) Kp = e^(-37,000 J/mol / (8.314 J/mol K * 298 K)) Kp = 3.27 × 10^-7 Therefore, the best choice is C) 3.27 × 10^-7.

Calculate K P for the Reaction PCl5 (G) PCl

Calculate K P for the Reaction PCl₅ (G) PCl