At a certain temperature, an equilibrium mixture in a 1.00 L container consists of 0.00400 M PCl5(g), 0.100 M PCl3(g), and 0.200 M Cl2(g).

Question

6.) At a certain temperature, an equilibrium mixture in a 1.00 L container consists of 0.00400 M PCl5(g), 0.100 M PCl3(g), and 0.200 M Cl2(g). (a) Calculate KC for PCl5(g) ⇌ PCl3(g) + Cl2(g); (b) To this mixture is added 0.100 mol of PCl3(g). Predict whether the new equilibrium concentrations will be higher than, lower than, or equal to the original equilibrium concentrations after the reaction returns to equilibrium; (c) Calculate the new equilibrium concentrations of the three gases. Are your answers consistent with your predictions?

8.) The volume of the container with the original equilibrium mixture in exercise 6 is doubled. (a) Calculate QC after the container changes in size but before any shift occurs; (b) Does the equilibrium shift? Which way? (c) What are the new equilibrium concentrations? Compare them with the original equilibrium concentrations.

Answer 1

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

0.004.................0.100.........0.200.....Equilibrium

Kc = [PCl3][Cl2] / [PCl5] = (0.1)(0.2) / (0.004) = 5

b) According to Le Chatelier, adding PCl3 should shift the equilibrium to the left toward reactant. After returning to equilibrium, the new equilibrium concentrations should be higher than original.

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

0.004.................0.200.........0.200....Initial

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

0.004+x..........0.2-x...........0.2-x.....Equilibrium

5 = (0.2-x)(0.2-x) / (0.004+x)

0.02 + 5x = x² - 4x + 0.04

x² - 9x + 0.02 = 0

x = 0.0037

New equilibrium concentrations:

[PCl5] = 0.004 + 0.0037 = 0.0077 M

[PCl3] = 0.2 - 0.0037 = 0.1963 M

[Cl2] = 0.2 - 0.0037 = 0.1963 M

Try #8 yourself. Doubling the volume will halve the concentrations. Compare Q to K. If Q >K, it will shift left and vice versa.