Chemistry
posted by Kat on .
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!

Yes, I would find Kp first.

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? 
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 = 263243 = 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.0.01737p.0.04316p..0.2855+2p
Substitute into Kp expression with these ICE chart values and solve for p. 
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 : 0.01737946.3 = 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? 
ohh i'm sorry i understand now thank you Dr. Bob!