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

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

The trick with a percentage-dissociation problem is to start from 1 mole. Nothing depends on the actual amount, so the algebra stays clean and the mole fractions fall straight out.

(a) Partial pressures in terms of P

Start with 1 mol PCl5 and let 0.900 mol dissociate:

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

Notice the total is greater than 1 - one molecule became two, which is the whole reason the pressure and the position of equilibrium are linked here.

Partial pressure = mole fraction x total pressure:

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

Check they add to P: 1/19 + 9/19 + 9/19 = 19/19. Good.

(b) The Kp expression and the total pressure

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

Substitute the expressions from part (a):

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

The units work out too: pressure squared over pressure leaves pressure, which is why Kp here is quoted in kPa rather than being dimensionless.

Now set it equal to 160 kPa:

81P/19 = 160
P = 160 x 19 / 81 = 3040/81

P = 37.5 kPa

(c) The individual partial pressures

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

Two checks, both worth doing in an exam:

They sum to 1.98 + 17.8 + 17.8 = 37.6 kPa, which is P.

And back-substituting: (17.8)(17.8)/1.98 = 160 kPa, which is Kp.

The general result behind this

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

Kp = alpha^2 P / (1 - alpha^2)

Putting alpha = 0.900 in gives Kp = 0.81P/0.19 = 4.26P, the same relation as above. Rearranged, P = Kp(1 - alpha^2)/alpha^2.

That form makes the physics visible. As alpha approaches 1 the required pressure approaches zero - to push a dissociation to completion you drop the pressure, exactly as Le Chatelier predicts for a reaction that increases the number of gas molecules. Ninety percent dissociation of PCl5 needs a low pressure, and 37.5 kPa is indeed well below atmospheric.