At 250ºC, the equilibrium constant K_p for the reaction PCl₅(g) $\rightleftharpoons$ PCl₃(g) + Cl₂(g) is 1.80. Sufficient PCl₅ is put into a reaction vessel to give an initial pressure of 2.74 atm at 250ºC. Calculate the pressure of PCl₅ after the system has reached equilibrium.

Question

Quizplus · Accepted Answer

Let's denote the initial pressure of PCl5 as P0. At equilibrium, the pressure of PCl5 will decrease by x, while the pressures of PCl3 and Cl2 will both increase by x. Using the equilibrium constant expression: K_p = [PCl3][Cl2] / [PCl5] 1.80 = x^2 / (P0 - x) Since the value of x is expected to be small compared to P0, we can simplify the above equation to: 1.80 ≈ x^2 / P0 Therefore, x = sqrt(1.80 P0) ≈ 1.24 atm So the pressure of PCl5 at equilibrium is approximately P0 - x = 2.74 - 1.24 = 1.50 atm. Therefore, the correct answer is B.

At 250ºC, the Equilibrium Constant Kp for the Reaction PCl5(g) ⇌\rightleftharpoons⇌

At 250ºC, the Equilibrium Constant K_p for the Reaction PCl₅(g) $\rightleftharpoons$