Asked by Kat
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has a partial pressure of 217.0 torr, 13.2 torr, and 13.2 torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 torr. The appropriate chemical equation is PCl3(g) + Cl2(g) <=> PCl5(g)
Calculate the new partial pressures of PCl3, Cl2, and PCl5
Am i supposed to find Kp first..? please help me!
Calculate the new partial pressures of PCl3, Cl2, and PCl5
Am i supposed to find Kp first..? please help me!
Answers
Answered by
DrBob222
Yes, I would find Kp first.
Answered by
Kat
Hi lol
Is this correct?:
Kp = PPCl5/(PPCl3)(PCl2) = 217.0 torr/(13.2 torr)(13.2 torr) = 1.245
x/(263 torr - x) = 1.245
x = 327.435 - 1.245x
2.245x = 327.435
x = 145.85 torr = PPCl5
263.0 total torr - 145.85 PCl5 torr = 117.15 torr - 13.2 torr = 103.95 torr for Cl2
13.2 torr for Cl3?
Is this correct?:
Kp = PPCl5/(PPCl3)(PCl2) = 217.0 torr/(13.2 torr)(13.2 torr) = 1.245
x/(263 torr - x) = 1.245
x = 327.435 - 1.245x
2.245x = 327.435
x = 145.85 torr = PPCl5
263.0 total torr - 145.85 PCl5 torr = 117.15 torr - 13.2 torr = 103.95 torr for Cl2
13.2 torr for Cl3?
Answered by
DrBob222
I would do this. First I would convert torr to atm.
PPCl5 = 217/760 = 0.2855 atm
PPCl3 = 13.2/760 = 0.01737 atm.
PCl2 = 13.2/760 = 0.01737 atm
Total P = 243.4/760 = 0.3203 atm
Cl2 added = 263-243 = 19.6 and 19.6/760 = 0.02579
............PCl3 + Cl2 ==> PCl5
initial..0.01737.0.01737..0.2855
add.............0.02579..........
change.......-p....-p.......+2p
equil.0.01737-p.0.04316-p..0.2855+2p
Substitute into Kp expression with these ICE chart values and solve for p.
PPCl5 = 217/760 = 0.2855 atm
PPCl3 = 13.2/760 = 0.01737 atm.
PCl2 = 13.2/760 = 0.01737 atm
Total P = 243.4/760 = 0.3203 atm
Cl2 added = 263-243 = 19.6 and 19.6/760 = 0.02579
............PCl3 + Cl2 ==> PCl5
initial..0.01737.0.01737..0.2855
add.............0.02579..........
change.......-p....-p.......+2p
equil.0.01737-p.0.04316-p..0.2855+2p
Substitute into Kp expression with these ICE chart values and solve for p.
Answered by
Kat
Hi Dr. Bob. I'm coming up with outrageous numbers for when I try to use atm's with my Kp value.. please look!
Kp = PPCl5/PPCl3PCl2 = (0.2855)/(0.01737)(0.01737) = 946.3
PCl3 : 0.01737-946.3 = -946.28 atm?!
Cl2 : 0.04316 - 946.3 = -946.26 atm o.o
PCl5 : 0.2855 + 2(946.3) = 1,892.8855 >.<
when i tried it with torr it gave me 11.955, 31.555, and 219.49 >.< am i reading your directions correctly?
Kp = PPCl5/PPCl3PCl2 = (0.2855)/(0.01737)(0.01737) = 946.3
PCl3 : 0.01737-946.3 = -946.28 atm?!
Cl2 : 0.04316 - 946.3 = -946.26 atm o.o
PCl5 : 0.2855 + 2(946.3) = 1,892.8855 >.<
when i tried it with torr it gave me 11.955, 31.555, and 219.49 >.< am i reading your directions correctly?
Answered by
Kat
ohh i'm sorry i understand now thank you Dr. Bob!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.