For the reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 70.5°C, Kp = 1.05.
If one starts with 1.80 atm pressure of PCl₃(g) , 1.72 atm pressure of Cl₂(g) , and no PCl₅(g) , what is the partial pressure of PCl₅(g) at equilibrium?

Question

Quizplus · Accepted Answer

To find the partial pressure of PCl5 at equilibrium, use the expression for Kp: Kp = (P_PCl5) / (P_PCl3 * P_Cl2). Let x be the change in pressure at equilibrium. Then, P_PCl5 = x, P_PCl3 = 1.80 - x, and P_Cl2 = 1.72 - x. Substitute these into the Kp expression and solve for x, which gives approximately 0.856 atm.

For the Reaction: PCl3(g) + Cl2(g) ⇌ PCl5(g) at 70

For the Reaction: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 70