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?
forgot to include the equation required to do this problem, The equation is PCl5 (g) <-> PCl3(g) + Cl2(g)
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%).
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?
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.
To solve parts b1, b2, and b3, we need to consider the changes in pressure and the equilibrium constant (Kp) as the volume of the gas mixture is decreased from 2.00 atm to 0.200 atm at 250°C.
First, let's recall the chemical equation for the reaction:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
(a1) from the given data, we know the equilibrium gas contains 40.7% Cl2 by volume. Since the volume ratios of PCl3, Cl2, and PCl5 in the balanced equation are 1:1:1, we can say the volume ratio of Cl2 to the total gas volume is also 40.7%. Therefore, the volume ratio of PCl5 to the total gas volume is (100 - 40.7)% = 59.3%.
Now, since the equilibrium pressure is 0.200 atm, we can calculate the partial pressures:
For Cl2:
Partial pressure of Cl2 = 0.200 atm * (40.7 / 100) = 0.0814 atm
(a2) To calculate Kp at 250°C, we need to use the ideal gas law. The expression for Kp is the ratio of the partial pressure of PCl3 and Cl2 (products) to the partial pressure of PCl5 (reactant), each raised to the power corresponding to their stoichiometric coefficients:
Kp = (P(PCl3) * P(Cl2)) / P(PCl5)
Using the values from part (a1), we have:
Kp = (0.814 atm * 0.814 atm) / 0.372 atm ≈ 1.78
Now, let's move on to solve b1, b2, and b3.
(b1) To calculate the percent of PCl5 that would be dissociated at equilibrium, we need to consider the change in pressure from 2.00 atm to 0.200 atm at 250°C. Since pressure is directly proportional to concentration for gases, we can assume that the percent dissociation is the same as the percent reduction in pressure.
Percent dissociation = (2.00 atm - 0.200 atm) / 2.00 atm * 100% ≈ 94.8%
(b2) The percent by volume of Cl2 at equilibrium can be calculated by dividing the partial pressure of Cl2 by the total pressure, then multiplying by 100%:
Percent by volume of Cl2 = (Partial pressure of Cl2 / Total pressure) * 100%
Using the values from part (a1), we get:
Percent by volume of Cl2 = (0.0814 atm / 0.200 atm) * 100% ≈ 40.7%
(b3) Lastly, to calculate the partial pressure of Cl2 at equilibrium, we need to consider the change in pressure from 2.00 atm to 0.200 atm, just like in part (b2).
Partial pressure of Cl2 at equilibrium = 0.200 atm * (40.7 / 100) = 0.0974 atm
Therefore, the answers are:
(b1) The percent of PCl5 that would be dissociated at equilibrium is approximately 94.8%.
(b2) The percent by volume of Cl2 at equilibrium is approximately 40.7%.
(b3) The partial pressure of Cl2 at equilibrium is approximately 0.0974 atm.