AP EQUILIBRIUM 5
A 0.0220 mole sample of phosphorus pentachloride was placed in a 1-Liter flask and allowed to
decompose to phosphorus trichloride and chlorine at 250.0C.
PCl5(g)  PCl3(g) + Cl2(g)
a) What would have been the pressure, in kPa, if the phosphorus pentachloride had NOT
decomposed? {95.7 kPa}
PV = nRT
P=
nRT
V
(0.0220 n) (8.314 kPa L /K mol)(523.15 K)
(1 L)
= 95.68832  95.7 kPa
b) But the actual, observed pressure was 103.8 kPa? What are the equilibrium partial pressures, in
kPa, of the three gases at 250C? {PCl5 = 87.6 kPa; PCl3 = Cl2 = 8.1 kPa}
PCl5
95.7
-X
95.7-X

PCl3
0
+X
X
Cl2
0
+X
X
PCl5
95.7
-8.1
87.6

PCl3
0
+8.1
8.1
Cl2
0
+8.1
8.1
(95.7-X) + X + X = 103.8 ; X = 8.1
c) What is the value of Kp at 250C? {0.75}
Kp =
(PCl3)(Cl2)
(PCl5)
=
(8.1)2
(87.6)
= 0.74897  0.75
d) What would be the pressures of the three gases in kPa at 250C if another 0.0220 moles of PCl5
were to be added? This may require that you solve a quadratic.
{PCl5 = 179.8 kPa; PCl3 = Cl2 = 11.6 kPa}
PCl5
87.6 + 95.7
-X
183.3-X
Kp =

PCl3
8.1
+X
8.1 + X
(PCl3)(Cl2)
(PCl5)
=
Cl2
8.1
+X
8.1 + X
(8.1+ X)2
(183.3 - X)
65.61 + 16.2X + X2 = 137.475 – 0.75X
X2 + 16.95 X - 71.865 = 0
X = 3.5
PCl5
87.6 + 95.7
-3.5
179.8
0.75

PCl3
8.1
+3.5
11.6
Cl2
8.1
+3.5
11.6
OR
P=
nRT
V
PCl5
191
-X
191-X
Kp =
(0.0440 n) (8.314 kPa L /K mol)(523.15 K)
(1 L)

PCl3
0
+X
X
(PCl3)(Cl2)
(PCl5)
Cl2
0
+X
X
=
(X)2
(191-X)
X2 = (191-X)(0.74897)
X2 +0.75X – 143.55 = 0
X = 11.6
PCl5
191
-11.6
179

PCl3
0
+11.6
11.6
Cl2
0
+11.6
11.6
= 0.74897  0.75
= 191.3766  191 kPa