Answer

Solution: Write the balanced reaction, write the mass action expression for K_p (the data is provided in atm so this is the natural value to use for the problem), set up a Table of Concentrations (ICE table) using pressure units. Use the equilibrium partial pressure to determine all of the other equilibrium values, plug these into the mass action expression to find K_p, and then convert K_p to K_c using the relation K_c = K_p(RT)^–Δn, where Δn = +1.

1) PCl₅(g) → ← PCl₃(g) + Cl₂(g)

K_p = P_PCl3eP_Cl2e                       P_PCl5e

Initial1.00 0 0

Change– x +x +x

Equilibrium1.00 – x x x

The partial pressure of PCl₅ at equilibrium = 0.30 atm = 1.00 – x so x = 0.70 atm.

This allows the calculation of K_p = x²/(1.0 – x) = (0.70)²/(0.30) = 1.6

Then, K_c = K_p(RT)^–Δn, where K_p = 1.6, R = 0.0821, Δn = 1, and T = 250 + 273 = 523 K.

K_c = (1.6)(0.0821×523)^–1 = 0.037.

 

2) COCl₂(g) → ← CO(g) + Cl₂(g)

K_p = P_COeP_Cl2e                      P_COCl2e

Initial1.00 0 0

Change– x +x +x

Equilibrium1.00 – x x x

The partial pressure of COCl₂ at equilibrium = 0.42 atm = 1.00 – x so x = 0.58 atm.

This allows the calculation of K_p = x²/(1.0 – x) = (0.58)²/(0.42) = 0.80

Then, K_c = K_p(RT)^–Δn, where K_p = 0.80, R = 0.0821, Δn = 1, and T = 527 + 273 = 800 K.

K_c = (0.80)(0.0821×800)^–1 = 0.012.