# Chemistry

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

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1. Yes, I would find Kp first.

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2. 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?

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3. 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
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.

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4. 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?

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5. ohh i'm sorry i understand now thank you Dr. Bob!

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