For the reaction: PCl₃(g)+ Cl₂(g)⇌ PCl₅(g)at 85 °C,K_p = 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 for PCl5, then at equilibrium: P_PCl5 = x, P_PCl3 = 2.00 - x, 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,Kp = 1.19

For the Reaction: PCl₃(g)+ Cl₂(g)⇌ PCl₅(g)at 85 °C,K_p = 1.19