Question

PCl5 is introduced into an evacuated chamber and comes to equilibrium (see Problem 16.37), at 250◦C and 2.00 atm. The equilibrium gas contains 40.7% Cl2 by volume. (a1) What are the partial pressures of the gaseous components at equilibrium? (a2) From these data, calculate Kp at 250◦C on the basis of the 1 atm standard state for the reaction written in Problem 16.37. If the volume of the gas mixture is increased so that it is at 0.200 atm at 250◦C, calculate: (b1) The percent of PCl5 that would be dissociated at equilibrium. (b2) The percent by volume of Cl2 at equilibrium. (b3) The partial pressure of Cl2 at equilibrium

The answers are: (a1) P(Cl2) = P(PCl3) = 0.814 atm, P(PCl5) = 0.372 atm; (a2) 1.78 (b1) 94.8%; (b2) 48.7%; (b3) 0.0974 atm

I was able to solve parts a1 and a2, but b1 and beyond had me stumped. could someone explain how to solve b1,b2, and b3?

Answer 1

forgot to include the equation required to do this problem, The equation is PCl5 (g) <-> PCl3(g) + Cl2(g)

Answer 2

I agree with the answers for a1 and a2. Here is b2 and b3.
..........PCl5 ==> PCl3 + Cl2
I.........0.2.......0......0
C........-2p........p......p
E........0.2-2p.....p......p

Kp = 1.78 = p*p/(0.2-2p)
Solve for p. I obtained 0.0973 which is pCl2 for b(3). b(2) is volume % = XCl2 = 0.0973/0.2 = 0.487 or 48.7%.
I did not work b1 (at least I didn't get 94.8%).

Answer 3

hello drbob222, I was just wondering why you subtracted by 2p instead of just p in the C section of your ICE table. wouldn't it just be p?

Answer 4

pCl2 = pPCl3 = p
Ptotal = pCl2 + pPCl3 + pPCl5
0.2 = p + p + pPCl5
0.2-2p = pPCl5

a. Didn't you use 2-2p to arrive at the correct answer of 1.78 for Kp?
pCl2 is 0.814. pPCl3 = 0.814. pPCl5 = 0.372 so (0.814)(0.814)/0.372 = 1.78 = Kp.

b. Try it with 0.2-p and you get p = 0.181. Then Total P will be 0.181 + 0.181 + pPCl5 BUT 2*0.181 (= 0.362) is already more than you started with of 0.2.