Calculate the equilibrium constant, Kp, she obtained for this reaction.

Question

A student ran the following reaction in the laboratory at 495 K:

PCl₅(g) PCl₃(g) + Cl₂(g)

When she introduced PCl₅(g) at a pressure of 0.634 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of Cl₂(g) to be 0.343 atm.

Calculate the equilibrium constant, K_p, she obtained for this reaction.

K_p =

Answer 1

PCl₅(g) <===> PCl₃(g) + Cl₂(g)

0.634...............0..............0............Initial

-x...................+x.............+x...........Change

0.634-x............x...............x...........Equilibrium

x = 0.343 atm

P_PCl5 = 0.634 - 0.343 = 0.291 atm

P_PCl3 = 0.343 atm

P_Cl2 = 0.343 atm

Kp = (0.343)(0.343) / (0.291

Kp = 0.404