chemistry
posted by lasya .
at an equilibrium mixture of PCl5, PCl3, and Cl2 has partial pressures of 217.0 Torr, 13.2 Torr and 13.2 Torr respectively. a quantity of Cl2 is injected into the mixture, and the total pressure jumps to 263.0 Torr (at the moment of mixing). the system then reequilibrates. calculate the new partial pressures after equilibrium is reestablished.

chemistry 
DrBob222
..........PCl5 > PCl3 + Cl2
E.........217.......13.2 13.2
Kp = pPCl3*pCl2/pPCl5.
Substitute the E line into Kp expression and solve for Kp.
Cl2 added.
Total P = 217 + 13.2 + 13.2 = 243.4
New total P = 263
Cl2 added = 263243.4= 19.6 torr
New Cl2 = 19.6 + 13.2 = 32.8 torr
...........PCl5 ==> PCl3 + Cl2
I..........217......13.2...32.8
C...........x.......x......x
E.........217+x....13.2x..32.8x
Substitute the E line into Kp expression and evaluate x and the others.
Post your work if you get stuck.
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