For the reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 85°C, Kp = 1.19
If one starts with 2.00 atm pressure of PCl₃, 1.00 atm pressure of Cl₂ and no PCl₅, what is the partial pressure of PCl₅(g) at equilibrium?

Question

Quizplus · Accepted Answer

To find the partial pressure of PCl5 at equilibrium, set up the expression for Kp: Kp = (P_PCl5) / (P_PCl3 * P_Cl2). Let x be the change in pressure at equilibrium for PCl5. Then, P_PCl5 = x, P_PCl3 = 2.00 - x, and P_Cl2 = 1.00 - x. Substitute these into the Kp expression and solve for x: 1.19 = x / ((2.00 - x)(1.00 - x)). Solving this equation gives x ≈ 0.621 atm.

For the Reaction: PCl3(g) + Cl2(g) ⇌ PCl5(g) at 85°C

For the Reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 85°C