Name:____ [KEY]___________________ Date:_____________ Period:____
WS Equilibrium Calculations Practice III (with x’s)
[KEY]
H2O(g) + Cl2O(g) ↔ 2 HOCl(g)
[KEY]
Kc = 0.676
1. 1.0 mol pure HOCl is placed in a 2.0 L flask. Calculate the equilibrium concentrations of
H2O(g) and Cl2O(g).
Kc = [HOCl]2
[H2O] [Cl2O]
I
H2O
0
Cl2O
0
2 HOCl
0.50
C
+x
+x
–2x
E
x
x
0.50 – 2x
0.676 = (0.50 – 2x)2 (√ both sides and solve for x)
(x)2
0.822 = (0.50 – 2x)
x
[H2O] = 0.18 M
0.822x = 0.50 – 2x
2.822x = 0.50
2.
x = 0.18
[Cl2O] = 0.18 M
Gaseous HCl is added to a rigid vessel containing excess solid iodine at 25oC until the partial
pressure of HCl reaches 1.47 atm. The following reaction brings the system to equilibrium.
Kp = 3.91 x 10–33 at 25oC
2 HCl(g) + I2(s) ↔ 2 HI(g) + Cl2(g)
(a) Find the equilibrium partial pressures of all species at 25oC.
Kp = (PHI)2(PCl2)
(PHCl)2
2 HCl(g) +
I2(s)
↔
2 HI(g) +
Cl2(g)
I
1.47 atm
0
0
C
–2x
+2x
+x
E
1.47 – 2x
2x
x
–33
3.91 x 10
2
= (2x) (x)
(1.47 – 2x)2
(PHCl)eq = 1.47 atm
(PHI)eq = 2.56 x 10–11 atm
(PCl2)eq = 1.28 x 10–11 atm
–33
3.91 x 10
Because K is very small:
( 2x <<< 1.47 )
= 4x
2
OR
(1.47)
(1.47 – 2x = 1.47)
3
.
8.45 x 10–33 = 4x3
2.11 x 10–33 = x3
x = 1.28 x 10–11
The temperature of the system changed so that the equilibrium constant, Kp , is 8.39 x 10–27.
(b) Find the new partial pressures of all the species at equilibrium.
Kp = (PHI)2(PCl2)
(PHCl)2
2 HCl(g) +
I2(s)
↔
2 HI(g) +
Cl2(g)
I
1.47 atm
0
0
C
–2x
+2x
+x
E
1.47 – 2x
2x
x
8.39 x 10–27 = (2x)2(x)
(1.47 – 2x)2
(PHCl)eq = 1.47 atm
(PHI)eq = 3.30 x 10–9 atm
8.39 x 10–27 = 4x3
Because K is very small:
(1.47)2
(1.47 – 2x ≈ 1.47)
.
1.81 x 10–26 = 4x3
4.53 x 10–27 = x3
(PCl2)eq = 1.65 x 10–9 atm
x = 1.65 x 10–9
HCl(g) + CH3Cl(g) ↔ CH4(g) + Cl2(g)
3.
The equilibrium constant, Kp , is 6.3 x 10–15 at 1500 K for the reaction represented above. A
chemist mixes 45% HCl(g) and 55% CH3Cl(g) by moles into a container so that the total
pressure inside the container is 1.5 atm at 1500 K.
(a) Find the initial pressures of each gas.
(PHCl)in = 0.68 atm
PA = PT x XA
(PCH3Cl)in = 0.83 atm
PHCl = (1.5) x (0.45)
(PCH4)in = 0 atm
PCH3Cl = (1.5) x (0.55)
(PCl3)in = 0 atm
(b) Find the equilibrium partial pressures of each gas at 1500 K.
Kp = (PCH4)(PCl2)
(PHCl)(PCH3Cl)
I
HCl(g) + CH3Cl(g) ↔
0.68 atm
0.83 atm
CH4(g) +
0
Cl2(g)
0
C
–x
–x
+x
+x
E
0.68 – x
0.83 – x
x
x
6.3 x 10–15 =
x2
(0.68 – x)(0.83 – x)
(PHCl)eq = 0.68 atm
(PCH3Cl)eq = 0.83 atm
(PCH4)eq = 6.0 x 10–8 atm
(PCl2)eq = 6.0 x 10–8 atm
6.3 x 10–15 =
x2
(0.68)(0.83)
3.6 x 10–15 = x2
x = 6.0 x 10–8
.
Because K is very small:
(0.68 – x ≈ 0.68)
The temperature of the system changed so that the equilibrium constant, Kp , is 4.7 x 10–10.
(c) Find the new partial pressures of all gaseous species at equilibrium.
Kp = (PCH4)(PCl2)
(PHCl)(PCH3Cl)
I
HCl(g) + CH3Cl(g) ↔
0.68 atm
0.83 atm
CH4(g) +
0
Cl2(g)
0
C
–x
–x
+x
+x
E
0.68 – x
0.83 – x
x
x
4.7 x 10–10 =
x2
(0.68 – x)(0.83 – x)
(PHCl)eq = 0.68 atm
(PCH3Cl)eq = 0.83 atm
(PCH4)eq = 1.6 x 10–5 atm
(PCl2)eq = 1.6 x 10–5 atm
4.7 x 10–10 =
x2
(0.68)(0.83)
2.7 x 10–10 = x2
x = 1.6 x 10–5
.
Because K is very small:
(0.68 – x ≈ 0.68)