CHAPTER 6 - CHEMICAL EQUILIBRIUM

LEARNING OUTCOMES 6.2

• At the end of this topic, students should be able
to :

(c) use the equations, Kp = Kc (RT)∆n to solve
equilibrium problems.

(d) calculate Kc, Kp or quantities of species
present at equilibrium.

51

The Relationship between Kc and Kp
• For general reaction :

aA (g) + bB (g) cC (g) + dD (g)

Kc  Cc Dd and Kp  PC c PD d
Aa Bb PA a PB b

• The relationship is derived from the PV = nRT

equation. PB  nB RT  [B] RT
V
PA  nVA RT  [A] RT

PC  nC RT  [C] RT PD  nD RT  [D]RT
V V

52

• Substituting this relationship into the expression for

Kp,

Kp  PC c PD d  [C] RTc [D] RTd
PA aPB b [A] RTa [B] RTb

 [C]c [D]d x RTc RTd
[A]a [B]b RTa RTb

 Kc (RT)(cd) - (ab)

 Kp  Kc (RT)Δn

n = (number of moles gaseous products) -
(number of moles of gaseous reactants)

= (c+d) – (a+b)

53

Exercise 11 :

The equilibrium constant Kp for the reaction

4NH3 (g) + 7O2 (g) 4NO2 (g) + 6H2O (g)

is 1.8 x 10-5 at 350°C. What misolK-1c for this
reaction ? ( R= 0.08206 L atm K-1 )

54

Answer

KP  KC RTn
1.8 x 105  KC 0.08206 x 623 1011

Kc  9.2 x 104

55

Exercise 12 :

Calculate 1K0c-4f.or the following reaction at STP if
Kp= 2.1 x

CaCO3 (s) CaO (s) + CO2 (g)

( R = 0.08206 L atm mol-1 K-1 )

56

Answer

KP  KC RTn
2.1x 104  KC 0.08206 x 2731

Kc  9.37 x 106

57

Determination of the Kp, Kc & quantities of
species at equilibrium.

For a reversible reaction :

A (g) + 2B (g) 2C (g)

If a moles of reactant A and b moles of reactant B
was allowed to achieve equilibrium in V liter
container.

… the ‘ICE΄ Table may help..

Let x = amount (mol) of reactant that dissociates (change)
n eq = n initial –n change

58

A (g) + 2B (g) 2C (g)

Initial mole (mol)

Changes (mol)

Mole at
equilibrium (mol)

Concentration at
equilibrium (M)

59

OR A (g) + 2B (g) 2C (g)

Initial
concentration (M)
Changes (M)

Concentration at
equilibrium (M)

60

Example 1

Natural gas methane, CH4 reacts with steam to produce
hydrogen gas:

CH4 (g) + H2O(g) CO (g) + 3H2(g)

In an experiment, 3.00 mol of methane and 0.60 of steam

were introduced into a 1 dm3 vessel. At equilibrium, 0.50 mol

of carbon dioxide is present in the mixture.

a) Calculate the number of moles of CH4 , H2O and H2 in the
equilibrium mixture.

b) Write an expression for the equilibrium constant, Kc for
the reaction.

c) Calculate the value of Kc at the temperature of the
experiment.

61

SOLUTION : CH4 (g) + H2O(g) CO (g) + 3H2(g)
CO 3H2
ninitial CH4 H2O
n 00
3.00 0.60
+ x + 3x
-x -x

nat equilibrium 3.00 - x 0.60 - x x = 0.50 3x

[ ]at equilibrium 3.00 - x 0.60 – x x = 0.50 3x
1 dm3 1 dm3 1 dm3
1 dm3

62

a) Calculate the number of moles of CH4 , H2O and
H2 in the equilibrium mixture.
Ans : CH4 = 2.5 mol ; H2O = 0.10 mol and H2 = 1.5
mol

b) Write an expression for the equilibrium constant,
Kc for the reaction.

c) Calculate the value of Kc at the temperature of the
experiment.

Ans : Kc = 6.75

63

Example 2

0.020 mol of SO3 is placed into a closed 1.52L
vacuum container and heated to 900K until

equilibrium is achieved. 0.0142 mol of SO3
exist at equilibrium. Calculate Kc and Kp at
900K.

2SO (g) 2SO (g) + O (g)
3 22

64

SOLUTION :

2SO3 2SO2 O2

n initial 0.020 00
n
nat equilibrium -2x +2x +x
[ ]at equilibrium
0.0142 +2x +x

 0.0142  2x x
1.52 1.52 1.52

 9.3421 x103 M  2 x 2.9 x 10-3  2.9 x 10-3
1.52 1.52

 3.8158 x 10-3M 1.9079 x 10-3M

65

Calculate Kc
Given at equilibrium

neq SO3 = 0.0142
0.020-2x = 0.0142

x = 2.9 x 10-3

KC  SO2 2O2 
SO3 2

     
3.8158 x 10-3M 2 1.9079 x 10-3M
9.3421 x103 M 2

 3.1830x10 4

66

Calculate Kp

KP  KCRTn

 3.1830x104 (0.08206 x 900 )1
 2.3508 x 102

67

Exercise

1. The equal amount of hydrogen and iodine are

injected into a 1.50 L reaction flask at a fixed

temperature.

H2(g) + I2(g) 2HI(g)

At equilibrium, analysis shows that the flask
contains 1.80 mole of H2, 1.80 mole of I2, and
0.520 mole of HI. Calculate Kc

Ans : 0.083

68

SOLUTION :

H2(g) + I2(g) 2HI(g)

nat equilibrium 1.80 1.80 0.520
[ ]at equilibrium
 1.8  1.8  0.520
1.5 1.5 1.5

 1.2 M  1.2 M  0.3467 M

KC  HI2 69
H2 I2 

 0.3467 2
1.21.2

 0.083

2. At a fixed temperature, 0.200 mole of
hydrogen halide is injected into a 2.00L of
reaction flask and form hydrogen and iodine
gas.

2HI(g) H2(g) + I2(g)

If [ HI ]=0.078 M at equilibrium calculate the
value of Kc.

Ans : 0.019

70

SOLUTION : H2(g) + I2(g)

2HI(g)

n initial 0.2 0 0
n -2x
nat equilibrium 0.2 – 2x +x +x
[ ]at equilibrium
0.078 x x

H2   x I2   x
2 2

 0.022  0.022
2 2

 0.011  0.011

71

SOLUTION : KC  H2I2 
HI2
0.2  2x  0.078
2.0
x  0.022

 0.0112
0.0782

 0.019

72

3. At 440C, the equilibrium constant Kc for
reaction,

H2(g) + I2(g) 2HI(g)

has a value of 49.5. If 0.200 mole of H2 and
0.200 mole of I2 are placed into a 10.0 L
vessel and permitted to react at this

temperature, what will be the concentration

of each substance at equilibrium ?

Ans : [H ] = [I ] = 4.42 X 10-3 M
[HI] = 0.031 M

73

SOLUTION :

n initial H2(g) + I2(g) 2HI(g)

0.200 0.200 0
-x +2x
n -x 2x
nat equilibrium 0.2 – 2x 0.2-x
HI  2x
[ ]at equilibrium H2   0.2  x I2   0.2  x
10 10 10

74

SOLUTION :

KC  49.5 H2  0.2  0.1558
10
49.5  HI2
H2 I2  •4.42 x 103M

 2x 2 I2  0.2  0.1558
 10  10
49.5   0.2  x 2

 10   4.42 x 103M

49.5  2x HI  2(0.1558)
0.2  x
10
7.0356  2x  0.031M
0.2  x

1.4071  7.0325x  2x

2x  7.0325x  1.4071

x  0.1558

75

4. A pressure of 0.464 atm of CO2 is introduced
into a closed vessel that contain graphite at
1000 K. The reaction occurs as follow :

CO2 (g) + C (s) 2CO (g)

When the system reached equilibrium, the
pressure in the vessel was found to be 0.746
atm. Calculate:

a) the partial pressure of each species at
equilibrium

b) the equilibrium constant, Kp

76

SOLUTION : + C (s) 2CO (g)

CO2 (g) - 0
- +x
P initial 0.464 - +x
P -x

P at equilibrium 0.464-x

77

Answer

a) the partial pressure of each species
at equilibrium

a) PT  0.746 atm

PT  PCO2  PCO

 0.464  x  2x 0.746 b) the equilibrium constant, Kp

x 0.282 b)  KP PCO 2
PCO2
PCO2  0.464  x
 0.464 0.282  0.5642
 0.182 atm
0.182
PCO  2 x
 2 x 0.282  1.75
 0.564 atm

78

5. Nitric oxide, NO decomposes according to the
following reaction, for which Kc = 83.3 at 4000oC.

2NO (g) N2 (g) + O2 (g)

If 4.0 g of NO (g) is placed in a 2.0 L vessel and
allowed to come to equilibrium, calculate :

i. the equilibrium concentrations of NO, N2 and
O2.

ii. the equilibrium partial pressures for each gas.

iii. the mass of O2 in gram which is formed at
equilibrium.

79

SOLUTION : N2(g) + O2 (g)

2NO (g) 0 0
+x +x
n initial 0.1333 x x

n -2x
nat equilibrium 0.1333 – 2x

 [ ]at equilibrium NO  0.1333  2x N2   x O2   x
2 2
2

80

SOLUTION :

a) nNO   4g N2  0.0632
(14 16)gmol1 2

 0.1333 mol

KC  83.3  3.16 x 102M
83.3 
N2 O2  0.0632
2
NO2 O2 

 x 2  3.16 x 102M
2
83.3   0.1333  2x 2

2 HI  0.1333  2(0.0632)

83.3  x 2
 3.45 x 103 M
0.1333  2x

9.1269  x

0.1333  2x

1.2166  18.2538x  x

x  18.2538x  1.2166

x  0.0632 mol

81

SOLUTION :

b) nO2 (at equilibriu m)  nN2 (at equilibriu m)  0.0632 mol

PN2 V  nN2RT

PN2 (2)  0.0632 (0.08206)(4273.15)

PN2  0.0632 (0.08206 ) ( 4273 .15)
2

PO2  PN2 11.08 atm

nNO(at equilibriu m)  6.9x103mol

PNOV  nNORT

PNO (2)  6.9x103 (0.08206)(4273.15)

PN2  6.9x 10 3 (0.08206 ) ( 4273 .15)
2

PNO 1.21atm

82

SOLUTION :

c) Massof O2  0.0632 x 16
 2.02 g

83

6. A certain amount of C1a0C0O0o3Cundinergaoes0.t6h5e4rmaLl
decomposition at

container until the system reached

equilibrium. The reaction established as

follow:

CaCO3 (s) CaO (s) + CO2 (g)

I3f.9thx1e0-e2q, ucilaiblcruiulamtecothnestamnat,ssKP(inat 1000oC is
g) of CaO

produced at equilibrium.

84

SOLUTION :

nNO   4g l1
(14 16)gmo

 0.1333 mol

KP  PCO2  3.9 x 102

nCO2  PV
RT
3.9 x102 0.654
 0.08206 1273 

 2.44 x104 mol

mass CaO  2.44 x104 56

 0.0137g

85

LEARNING OUTCOMES 6.2
• At the end of this topic, students should be able

to :

(e) define and determine the degree of
dissociation, α .

(f) deduce the expression for reaction quotient,
Q and predict the direction of nett reaction by
comparing the values of Q and Keq.

86

Degree of dissociation, α

• Dissociation reaction : a molecule is broken
down into smaller molecules, atoms or ions.

• Degree of dissociation, α : fraction of molecules
that dissociates.

α= Amount dissociates x
Initial amount
=c

• Percentage of dissociation : percentage of

molecules that dissociates. 87
% α = Amount dissociates x 100%

Initial amount

• If α = 1 or 100% : complete dissociation
occurs

• If α less than 1 or 100% : incomplete or
partial dissociation occur.

88

Exercise 1 :

If 3.00 moles of A are placed in 1.00 L container at

298 K, calculate the degree of dissociation, α, of A
for the following reaction :

A (g) 2B (g)

The equilibrium constant, Kc at 298 K is 0.10.

89

SOLUTION :

A (g) 2B(g)

[ ] initial 3 0
[] +2x
-x 2x

[ ] at equilibrium 3- x

KC  B2
A

0.10   2x2
3  x 

x  0.2616 M

  0.2616
3

 0.0872

90

Exercise 2 :

At a pressure of 4.4 x 105 Pa and a
temperature of 150 oC, phosphorus

pentachloride, PCl5 is 25% dissociated.
Calculate the partial pressure equilibrium

constant for this reaction.

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

91

SOLUTION : PCl3 (g) + Cl2 (g)

PCl5 (g)

n initial 1 00
n -0.25
n at equilibrium +0.25 +0.25
1- 0.25
0.25 0.25

92

Answer   b)
KP  PPCl3 PCl2
a) PT  0.746 atm PCl5

nT  nPCl5  nPCl5  nCl2  
 1 0.25  0.25  0.25 8.8 x104 2
1.25 mol 2.64 x104

PPCl5  XPCl5PT  2.9 x 104
 0.75 x 4.4 x105
1.25
 2.64 x105 Pa

PPCl5  XPCl3PT
 0.25 x 4.4 x105
1.25
 8.8 x104 Pa

PCl2  PPCl3  8.8 x104 Pa

93

Exercise 3 :

At 700K, the equilibrium constant, Kp for this
reaction:

CCl4 (g) C (s) + 2Cl2 (g)

is 0.76. A pressure of 2.0 atm of CCl4 is introduced
into a container and allowed to come to equilibrium.

a) What is the value of Kc at 700K ?
b) What is the percentage of CCl4 that is been

dissociated at equilibrium?
c) What are the equilibrium partial pressure for

CCl4 and Cl2 ?

94

Exercise 4

The degree of dissociation of dinitrogen tetroxide at
250°C and 1 atm is 0.15. Calculate the degree of
dissociation at 250°C and 5 atm.

N2O4 (g) 2NO2 (g)

95

SOLUTION : N2O4 (g) 2NO2 (g)

n initial 1 0
n +2x
n at equilibrium -x 2x

1- x

96

Answer

Let the degree of dissociation at 1 atm = x

  0.15  x  KPPNO2 2
PN2O4
PNO2  XNO2PT
 0.3 x 1  0.2612
1.15
 0.261 atm 0.739

PPCl5  XPCl5PT  9.21x 10-2 atm
 0.85 x 1
1.15
 0.739 atm

97

SOLUTION :

Let the degree of dissociation at 5 atm = b

N2O4 (g) 2NO2 (g)

n initial 1 0
n +2b
n at equilibrium -b 2b

1- b

98

  0.15 b

Answer

PNO2  X NO2PT

1 2b 2b x 5
b

PN2O4  X PCl5PT
11bb2b x 5

1 2b 2b x 52
b
K 
P 1b
1b  2b x 5

 9.21 x 102

b  0.068

99

Equilibrium Calculations When Kc is Very Small
• The concentration change, x can often be
neglected if Kc is very small.

• By assuming x is very small than [A]initial ,

∴ [A]equilibrium = [A]initial - x

≈ [A]initial

• Note : Check the assumption is valid or not :

100