At 377 °C, KC = 0.0111 M for PCl5(g) ⇌ PCl3(g) + Cl2(g).

Question

At 377 °C, KC = 0.0111 M for PCl5(g) ⇌ PCl3(g) + Cl2(g). Into a 2.00 L flask are placed 0.120 mol PCl5, 0.0600 mol PCl3, and 0.0400 mol Cl2. Are the contents of the flask at equilibrium? If not, which way will the reaction proceed?

Answer 1

PCl5(g) <==> PCl3(g) + Cl2(g)

0.120.............0.0600.......0.0400 in 2 L

0.06 M...........0.03 M........0.02 M

Calculate Q:

Q = [PCl3][Cl2] / [PCl5] = (0.0300)(0.0200) / (0.060)

Q = 0.01

Since Q is equal to Kc, the system appears to be at equilibrium (or close enough for our purposes). If you want to be more exact, Q is less than K so reaction will proceed to the right toward the products.