ICE

5
Q = P PCL5 = 0.10 = 0.5 P PCL3 * P Cl2 0.30 * 0.60 Q (0.58) > K (0.0870) therefore reverse PCl 3(g) + Cl 2(g) PCl 5(g) 0.30 atm 0.60 atm 0.10 atm +X +X -X 0.30 + X 0.60 + X 0.10-X

description

Q = P PCL5 = 0.10 = 0.5 P PCL3 * P Cl2 0.30 * 0.60 Q (0.58) > K (0.0870) therefore reverse PCl 3(g) + Cl 2(g)  PCl 5(g). - PowerPoint PPT Presentation

Transcript of ICE

Page 1: ICE

Q = PPCL5 = 0.10 = 0.5 PPCL3 * PCl2 0.30 * 0.60

Q (0.58) > K (0.0870) therefore reverse

PCl3(g) + Cl2(g) PCl5(g)

0.30 atm 0.60 atm 0.10 atm

+X +X -X

0.30 + X 0.60 + X 0.10-X

Page 2: ICE

K = PPCL5 = (0.10-X) PPCL3 * PCl2 (0.30+X) (0.60+X)

PCl3(g) + Cl2(g) PCl5(g)0.30 atm 0.60 atm 0.10 atm

+X +X -X

0.30 + X 0.60 + X 0.10-X

Page 3: ICE

0.0870= (0.10-X) (0.30+X) (0.60+X)

(0.10-X) (0.18 + 0.90X + X2)

Page 4: ICE

0.0870X2 + 0.0783X + 0.0157 = 0.10-X

0.0870X2 + 1.0783X - 0.0843 = 0

X= -1.0783 + (1.0783)2 -4(0.0870)(-0.0843) 2(0.0870)

Page 5: ICE

X= -1.0783 + (1.0783)2 -4(0.0870)(-0.0843) 2(0.0870)

X= -1.0783 + 1.0918 = 0.078 0.174

PCL3 = 0.30 + 0.078 = 0.378 atmCl2 = 0.60 + 0.078 = 0.678 atmPCl3 = 0.10 – 0.078 = 0.022 atm