A-Level Chemistry Equilibrium Question

Question

If solid PCl5 is heated to 200 degrees celsius, it vaporises and dissociates according to the equation:

PCl5(g) <==> PCl3(g) + Cl2(g) Kp = 160kPa at 200 degrees celsius

90.0% of the PCl5 is found to have dissociated.

(a) If the total equilibrium pressure is P, write expressions for the partial pressures of the gases present at equilibrium in terms of P.

(b) Write an expression for Kp and use it to calculate the total pressure P.

(c) Calculate the partial pressure of the gases present at equilibrium.

Answer 1

This is the same problem you posted separately, so here is the full worked solution.

PCl5(g) <==> PCl3(g) + Cl2(g), Kp = 160 kPa, 90.0% dissociated.

Start from 1 mole. Nothing in the answer depends on the actual amount taken, and starting from 1 keeps the mole fractions clean.

(a) Partial pressures in terms of P

Let 0.900 mol of the 1 mol dissociate:

PCl5 remaining = 1 - 0.900 = 0.100 mol
PCl3 formed = 0.900 mol
Cl2 formed = 0.900 mol
total = 1.900 mol

The total exceeds 1 because one molecule becomes two. That is what makes this equilibrium pressure-dependent at all.

Partial pressure = mole fraction x P:

p(PCl5) = (0.100/1.900)P = P/19 = 0.0526 P
p(PCl3) = (0.900/1.900)P = 9P/19 = 0.474 P
p(Cl2) = (0.900/1.900)P = 9P/19 = 0.474 P

They add to 19P/19 = P, as they must.

(b) Kp expression and the total pressure

Kp = p(PCl3) x p(Cl2) / p(PCl5)

Kp = (9P/19)(9P/19) / (P/19) = (81P^2/361)(19/P) = 81P/19

Setting that equal to 160 kPa:

P = 160 x 19 / 81 = 37.5 kPa

Note Kp carries units of kPa here, because pressure squared over pressure leaves pressure. If your Kp had come out dimensionless you would know a term was misplaced.

(c) The partial pressures

p(PCl5) = 37.53/19 = 1.98 kPa
p(PCl3) = 9(37.53)/19 = 17.8 kPa
p(Cl2) = 17.8 kPa

Check both ways: they sum to 37.5 kPa, and (17.8)(17.8)/1.98 = 160 kPa.

The general result worth memorising

For any A <==> B + C with degree of dissociation alpha:

Kp = alpha^2 P / (1 - alpha^2), or rearranged, P = Kp(1 - alpha^2)/alpha^2

Check it here: 0.81P/0.19 = 4.26P = 160, giving P = 37.5 kPa again.

That form shows you the physics without any arithmetic. Driving alpha toward 1 requires the pressure to fall toward zero, because the forward reaction increases the number of gas molecules and low pressure favours it. Ninety percent dissociation therefore has to correspond to a pressure well below atmospheric - and 37.5 kPa is about 0.37 atm, so the answer is self-consistent.