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 re-equilibrates. calculate the new partial pressures after equilibrium is reestablished.

Answer 1

..........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 = 263-243.4= 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....13.2-x..32.8-x

Substitute the E line into Kp expression and evaluate x and the others.
Post your work if you get stuck.