1. Chemical Equilibrium
Chapter 6

2. 6.0 Chemical Reaction Chemical reaction may occur in two ways:-
a) a non-reversible reaction
- occurs in one direction
b) a reversible reaction
- occurs in both direction
i.e (forward & reverse reaction)

3. 6.1.1 A non-reversible reaction
Example 1
NaCl + AgNO3 AgCl + NO3
One/both reactants will be completely consumed and amount of products formed depends on the limiting reactant.
A single arrow (→) is used to represent reaction.

4. 6.1.2 Reversible Reaction Example 2 H2(g) + I2(g) ⇌ 2HI(g)
Occurs in both direction.
⇌ shows that reactants react to form product, and product reacts to form reactants.
Amount of products formed does not depend on the amount of reactant used.
Reaction will reach equilibrium when the concentration of products & reactants remained constant.
H2(g) + I2(g) ⇌ 2HI(g)

5. 6.2 Equilibrium System
A system is at equilibrium when there is no observable change occurs.
Equilibrium system can be observed in:
a. Physical Equilibrium
b. Chemical Equilibrium

6. ⇌ 6.2.1 Physical Equilibrium Example 3
vaporisation in a closed container.
Involves physical change of substance.
Level of H2O in the container remains constant at equilibrium because the rate of evaporation equals the rate of condensation.
H2O (l)
H2O (g) ⇌

7. 6.2.2 Chemical Equilibrium Example 4 2NO2(g) ⇌N2O4(g)
Initially NO2 molecules combine to form N2O4 (brown gas appears)
As soon as N2O4 is formed, it undergoes reverse reaction to form NO2.
At equilibrium, the system contains a constant amount of NO2 & N2O4.

8. 6.3 Dynamic Equilibrium A system reaches a dynamic equilibrium when
the rate of forward reaction equals the rate of reverse reaction.
the concentrations of both reactants and products remain constant.
no observable change occurs but the conversions of reactants to products and products to reactants continue.

9. After time t1, both concentrations remain constant.
[N2O4] at equilibrium
[NO2 ]at equilibrium
The reaction does not stop but the rate of forward reaction equals the rate of reverse reaction.

10. ⇌ 6.4 Equilibrium constant, K
Since concentrations at equilibrium remain constant, the equilibrium can be expressed by a constant.
Consider:
Kc is known as equilibrium constant Concentrations of species are expressed in molar.
aA (g) + bB (g) cC (g) + dD (g) ⇌
Kc =
[C]c[D]d
[A]a[B]b

11. Table 6.1
constant
N2O4 (g) NO2 (g) ⇌
K =
[NO2]2
[N2O4]
= 4.63 x 10-3

12. 6.5 Equilibrium Law: Law of Mass Action
When a system has reached equilibrium, the ratio of multiplied concentrations of products to the multiplied concentrations of reactants (each raised to the power of the respective stoichiometric coefficient) is a constant at constant temperature. OR
Value of equilibrium constant, Kc is a constant at constant temperature.

13. 6.6 Types of System in Chemical Equilibrium
a) Homogeneous equilibrium
Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
b) Heterogeneous equilibrium
A system in which the reactants and products are not in the same phase.

14. 6.7 Homogeneous Equilibrium System
6.7.1 Liquid phase
Important variable: concentration
Equilibrium constant: Kc
Example 5
CH3COOH (l) + CH3OH (l) ⇌ CH3COOCH3 (l) + H2O (l)
Kc=

15. Example 6
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
Kc=

16. 6.7.2 Gas phase
The quantitative aspects to be considered are the concentration and pressure.
We can use Kcand Kp to represent the equilibrium constant.
Kp =
(PA)a
(PB)b
(PC)c (PD)d
the pressure of gases are expressed in atm or other units of pressure

17. Example 7 Write the equilibrium law in the form of Kp
and Kc for the following reactions:
a) N2(g) + 3H2(g) NH3(g)
b) 2NO(g) + O2(g) N2O4(g)
Answer:
Kp =
Kc =

18. Expression of Kc Example 8
Expression of Kc depends on the equilibrium equation given
Example 8
Kc = [SO2]2 [O2]
[SO3]2
2SO3 (g) ⇌ 2SO2 (g) + O2 (g)
1. SO3 (g) ⇌ SO2 (g) + ½ O2 (g)
Kc1 = [SO2] [O2]1/2
[SO3]
2. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
Kc2 = [SO3]2
[SO2]2 [O2]

19. Example 9
Find the relationship between Kc1 and Kc2 for the following equilibrium equations.
1. SO3 (g) ⇌ SO2 (g) + ½ O2 (g)
Kc1 = [SO2] [O2]1/2
[SO3]
2. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
Kc2 = [SO3]2
[SO2]2 [O2]

20. Example 10
Consider the following reaction:
A(g) + B(g) ⇌ 2C(g)
If 5 moles of A are allowed to mix and react with 3 moles of B in 1dm3 container, 2 moles of C are produced at equilibrium. What is the value of Kc for this reaction?
(Ans: 0.5 )

21. Example 11
Consider the reaction:
CO(g) + 2H2(g) ⇌ CH3OH(g)
The following equilibrium concentrations are achieved at 400K:
[CO] = 1.03M
[H2] = 0.332M
[CH3OH] = 1.56M
Determine the equilibrium constant at 400K
(Ans: M-2)

22. Example 12
Nitrogen gas reacts with oxygen gas at high temperature to form nitrogen monoxide gas, NO. The equilibrium constant for the reaction is 4.1 x 10-4 at 200oC.
If the concentration for both N2(g) and O2(g) at equilibrium are 0.40 moldm-3 and 1.3 moldm-3 respectively, calculate the concentration of NO gas at equilibrium?
(Ans: M)

23. Example 13
The equilibrium constant Kp for the reaction;
2NO2 (g) ⇌ 2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if the P NO2 = atm and PNO = atm?

24. Answer; PNO PO Kp = PNO PNO PO = Kp PO = 158 x (0.400)2 (0.270)2
= 347 atm

25. 6.7.3 Relationship between Kp and Kc
Consider
aA (g) + bB (g) cC (g) + dD (g) ⇌
Kc =
[C]c[D]d
[A]a[B]b
Since P = nRT V
Thus,
Kc =
{PC /(RT)}c {PD / (RT)}d
{PA /(RT)}a {PB /(RT)}b
[ ] = P RT

26. ⇌ For the reaction aA (g) + bB (g) cC (g) + dD (g)
(RT) (a + b) – (c+d)
Kc =
PCc PDd
PAa PBb
(RT) (a + b) – (c+d)
Kc = Kp
(RT) (c + d) – (a+b)
Kp = Kc
(RT)Δn
Kp = Kc

27. The equilibrium concentrations for the reaction between carbon monoxide and chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. 1.
CO (g) + Cl2 (g) COCl2 (g) ⇌

28. Answer ; [COCl2] [CO][Cl2] = 0.14 0.012 x 0.054 Kc = = 220 M -1
Kp = Kc(RT)n
n = 1 – 2 = -1
Kp = 220 x ( x 347)–1
= 7.7 atm–1

29. 6.8 Heterogeneous Equilibrium system
Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
Kc and Kp can be used to represent the equilibrium constant.
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

30. ⇌ CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) 1. Kc =
[CH3COO-][H3O+]
[CH3COOH][H2O]
H2O a pure liquid, has a constant concentration and thus is not included in the expression.
Therefore, the equilibrium constant is written as,
Kc =
[CH3COO-][H3O+]
[CH3COOH]
Unit : M

31. ⇌ CH3COOH (aq) + C2H5OH (aq) CH3COOC2H5 (aq) + H2O (l) 2. Kc =
Carboxylic
acid
Alcohol
Ester
Kc =
[CH3COOC2H5]
[CH3COOH]
[C2H5OH]

32. ⇌ ‘ 3. CaCO3 (s) CaO (s) + CO2 (g) [CaO][CO2] Kc = [CaCO3]
[CaCO3] = constant
[CaO] = constant
Kc = [CO2]
Kp = PCO 2

33. ⇌ CaCO3 (s) CaO (s) + CO2 (g) PCO = Kp PCO
does not depend on the amount of CaCO3 or CaO

34. ⇌ Consider the following equilibrium at 295 K:
The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g) ⇌ 5.

35. Answer; Kp = P P = 0.265 atm x 0.265 atm = 0.0702 atm2 Kp = Kc(RT)n
NH3
H2S P
= atm x atm
= atm2
Kp = Kc(RT)n
n = 2 – 0 = 2
T = 295 K
Kc =
atm2
( atm L mol–1 K–1 x 295 K )2
= x 10–4 mol2 L–2
= x 10–4 M2

36. 6.
The value of Kc for the following equation is 1.0 x 10–3 M at 200°C.
2NOBr (g) ⇌ 2NO (g) + Br2 (g)
Calculate Kc’ for the following equation at 200°C.
NO (g) + ½ Br2 (g) ⇌ NOBr (g)

37. Answer;
Kc = [NO]2 [Br2]
[NOBr]2
Kc’ = [NOBr]
[Br2]1/2 [NO]
[NO]2 [Br2]
[NOBr]2 1 = 1 Kc
[NO]2 [Br2]
[NOBr]2 = 1 Kc
[NO] [Br2]
[NOBr] =
= Kc’

38. Kc’ = 1 Kc
Kc’ =
1.0 x 10–3 M
= 31.6 M-½
= 31.6 dm3/2 mol–½

39. ⇌ 7. At 440C, the equilibrium constant Kc for reaction, H2(g) + I2(g)
2HI(g) ⇌
has a value of If mole of H2 and 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?

40. ⇌ [ ]0 / M Δ[ ] / M [ ]⇌ / M Answer; [H2]0 = 0.200 /10.0 = 0.0200 M
[I2]0 = /10.0 = M
H2 (g) I2 (g) HI (g) ⇌
[ ]0 / M
0.0200
0.0200
Δ[ ] / M -x -x
+2x
[ ]⇌ / M x x
+2x

41. Kc = [HI]2
[H2] [I2]
49.5 = (2x)2
( – x) ( – x)
7.036 = x x
x =

42. ⇌
H2 (g) I2 (g) HI (g)
[ ]0 / M
0.0200
0.0200
Δ[ ] / M -x -x
+2x
[ ]⇌ / M x x
+2x = = =
Therefore concentration of each substance at equilibrium:
[H2] = M
[I2] = M
[HI] = M

43. 8.
A L flask is filled with mol of H2 and mol of I2 at 448°C. The value of the equilibrium constant, Kp for the reaction,
H2 (g) + I2 (g) ⇌ 2HI (g)
at 448°C is What are the partial pressures of H2 , I2 and HI in the flask at equilibrium.

44. Initial pressures of H2 and I2.
Answer;
Initial pressures of H2 and I2.
PH = 2
n RT H2 V
= (1.000)( )(721)
1.000
= atm
PI = 2
n RT I2 V
= (2.000)( )(721)
1.000
= atm

45. ⇌ H2 (g) + I2 (g) 2HI (g) P0 59.19 118.4 ΔP -x -x +2x P⇌ 59.19 - xΔP -x -x
+2x P⇌ x x
+2x
P P H2 I2
Kp = PHI 2
50.5 = (2x)2
(59.19 – x) (118.4 – x) ⇒
46.5x2 – 8969x = 0
Partial pressure of gases at equilibrium: H2 P
= 3.89 atm ;
PHI
= atm I2 P
= 63.1 atm ;

46. 6.2.6 Degree of Dissociation, α
The fraction of a molecule dissociated. α ΔC C0 =
ΔC = changes in concentration
C0 = initial concentration
A complete dissociation occurs if α = 1

47. AB ⇌ A B– a
C0 / M
ΔC / M -x +x +x
C⇌ / M
a-x +x +x α x a =

48. 9. The concentration of H+ ion measured for HCOOH
0.5 M is M. What is the degree of dissociation
of the compound?
a = 0.5
x =
HCOOH ⇌ H HCOO–
C0 / M
0.5
ΔC / M
C⇌ / M
0.0089
0.0089 = α x a =
0.0089
0.5 =
= = 1.78%

49. 6.2.7 Predicting the direction of reaction
For any reversible reaction, we can determine the direction of a reaction (whether is moving forward or reverse) by comparing the value of Q with Kp or Kc.
Q is the reaction quotient
Q has the same expression of Kc and Kp but the numerical value gained is NOT at equilibrium.

50. ⇌ Predicting the direction of reaction:
aA (g) + bB (g) cC (g) + dD (g) ⇌
Qc =
[C]c[D]d
[A]a[B]b
If Q = K
The system is at equilibrium
If Q < K
Reaction is moving forward
- more reactants are present in the mixture at t
If Q > K
Reaction is reversed
- more products are present in the mixture at t

51. 10.
At 12800C the equilibrium constant (Kc) for the reaction
is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.
Br2 (g) Br (g) ⇌
Br2 (g) Br (g) ⇌
C0/M
0.063
0.012
[Br]2
[Br2]
Qc =
= (0.012)2
0.063
= x 10–3
Qc > Kc
Equilibrium position shifts
from right to the left.

52. ⇌ C0/M 0.063 0.012 ΔC/M C⇌/M 0.063 + x 0.012 - 2x Br2 (g) 2Br (g) x +
Kc =
(0.012 – 2x)2 x =
1.1x10-3

53. 4x2 – 0.048x x10-4 = 6.93x x10-3x
4x2 – x x10-5 = 0
x =
x = 1.78x10-3 C⇌
[Br] = – 2x
[Br2] = x
If x =
= M
= M
= M
If x = 1.78x10-3
= M

54. 11. The equilibrium constant Kc for the reaction
I2 (g) + H2(g) ⇌ 2HI (g)
is 54.0 at certain temperature. At equilibrium, the conc. of I2, H2 and HI are M, M and 1.47 M respectively. If 0.500M of HI is added, what are the conc. when equilibrium is re-established?
Qc =
[HI]2
[H2] [I2]
Since Qc > Kc, the system is not at equilibrium
The net reaction will proceed from RIGHT to LEFT until equilibrium is re-established. =
( )2
(0.200) (0.200)
= 97.0

55. H2 (g) I2 (g) HI (g) ⇌
C0/M
0.200
0.200
1.47
Kc = 54
Qc = 97 +x +x
-2x
ΔC/M
C⇌/M x x
1.97 – 2x = =
= 1.863
Kc = [HI]2
[H2] [I2]
54.0 =
( x) ( x)
(1.97 – 2x)2
7.35 = x
1.97 – 2x
x = M

56. 6.3 Factors affecting the equilibrium position
The position of equilibrium in a system is affected by:
a) concentration
b) pressure
c) temperature
The effects can be predicted & explained by the :
Le Chatelier’s Principal

57. 6.3.1 Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

58. A) Effect of Concentration:
Changing the concentration of the reactants or products causes the equilibrium to shift to the direction that will reduce the effect .
Example:
Increasing the concentration of reactant/s causes the equilibrium position to shift forward in order to reduce the disturbance (to eliminate the additional concentration of reactant/s)
more products will be formed until the equilibrium is reached again (i.e: composition of equilibrium reaches Kc value which remains the same at a fixed temperature).

59. Example ; N2 (g) + 3H2 (g) 2NH3 (g) ⇌ Change
Equilibrium Shift
Increase concentration of product(s)
To the left
Decrease concentration of product(s)
To the right
Increase concentration of reactant(s)
To the right
To the left
Decrease concentration of reactant(s)

60. Example : a) Iron (III) or Fe+3 b) Sodium hydroxide, NaOH
Consider the reaction below:
Fe 3+(aq) SCN-(aq) Fe(SCN)2+(aq)
(Yellowish) (Blood red)
Discuss the effect of adding the following substances to the equilibrium mixture.
a) Iron (III) or Fe+3
b) Sodium hydroxide, NaOH

61. Answer; Consider the reaction below:
Fe 3+(aq) SCN-(aq) Fe(SCN)2+(aq)
(Yellowish) (Blood red)
a) When Fe3+ is added to the equilibrium,
the concentration of Fe3+ in the system increases.
the position of equilibrium shifts to right in order to eliminate the effect of the increasing Fe3+ ions.
more products will be formed and the solution turns blood red.

62. b) When NaOH is added to the system:
Consider the reaction below:
Fe 3+(aq) + SCN-(aq) Fe(SCN)2+(aq)
(Yellowish) (Blood red)
It reacts with Fe3+ to form Fe(OH)3; thus the concentration of Fe3+ decreases.
In order to achieve equilibrium, the position of equilibrium moves to the left to replace the Fe3+ that has reacted.
More Fe3+ ions will be formed and the solution becomes yellowish brown.

63. B) Effect of Temperature:
The effect of temperature on an equilibrium depends on the type of reactions, whether it is:
i) an endothermic reaction or
ii) an exothermic reaction
Temperature is the only factor that affects the value of Kc

64. i) Endothermic reaction
An endothermic reaction is a reaction that absorbs heat to form products.
Example:
Consider the following system:
N2O4(g) NO2(g) ΔH = +57 kJ
‘+’ sign indicates that for the forward reaction absorbs heat
If forward reaction absorbs heat, then the reverse reaction releases heat.

65. N2O4(g) 2NO2(g) ΔH = +57 kJ Increasing the temperature:
Means that heat is added to the system
Reaction will move forward (forward reaction absorbs heat) so as to reduce the added heat.
More NO2 gas will be released.
Since the amount of product increases, value of Kc increases 65

66. Lowering the temperature: Heat is removed from the system
N2O4(g) NO2(g) ΔH = +57 kJ
Lowering the temperature:
Heat is removed from the system
The position of equilibrium will move from right to left to replace the heat lost. (since reverse reaction is exothermic)
More N2O4 will be formed.
Since more product is formed, value of Kc decreases
Concentration of N2O4 increases, thus Kc value decreases

67. Example: Consider the reaction:
Co(H2O)62+(aq)+ 4HCl(aq) CoCl42+(aq) + H2O
ΔH = +ve
pink
blue
To overcome the low temperature reaction should release heat
System at equilibrium
Disturbance:
Low temperature
High temperature

68. C) Effect of Pressure The pressure of a system may be changed by:
i) changing the moles of reactant or product
ii) changing the volume
iii) adding inert gas to the system
Affects only the reversible reaction involving gaseous reactants/products

69. i) Changing moles of reactant or product
Adding or removing the gaseous reactant or
product at constant volume has the same effect as
changing the concentration
Example:
Consider the following system at equlibrium
2SO2(g) + O2(g) SO3(g)
When SO2 gas is added to the system, the partial pressure of SO2 increases (which means [SO2] increases)
Equilibrium shifts to the right to reduce the effect
of high concentration of SO2.

70. ii. Changing the volume of the container:
Decreasing the volume:
Causes the pressure of the system to increase
The position of equilibrium will shift in a direction that will lower down the pressure
The pressure can be brought down by having reaction that produces less number of molecules

71. Example N2(g) + 3H2(g) 2NH3(g)
Consider the above system at equilibrium:
When the Vol. of the container decreases, the P of the system increases.
position of equilibrium will shift in the direction that brings the pressure down by reducing the number of moles present in the system
The equilibrium shifts to the right to produce less number of moles of gases in the system
More ammonia is formed
Value of Kc remains the same

72. ii. Changing the volume of the container
Increasing the volume:
Causes the pressure in the system to decrease
The equilibrium position will shift in the direction that will raise the pressure up.
It moves to the direction that produces more number of molecules (higher number of moles)

73. Example
Consider the reaction:
when the volume of the container increases, the pressure of the system decreases
Equilibrium position will shift to the direction that produces more moles
The equilibrium shifts to the left.
2SO2(g) + O2(g) SO3(g)

74. iii) Changing pressure by adding inert gas
a) Adding inert gas at constant volume:
Adding inert gas causes the total pressure in the container to increase.
However the partial pressure of each gas reacting in the reversible reaction remains the same.
Since the partial pressure of the gases remain the same & Kp is constant at constant temperature, the equilibrium position does not change.

75. b) Adding inert gas at constant pressure:
Adding inert gas to the system increases the total pressure.
In order to keep pressure constant, the volume of the container has to expand.
This causes the partial pressures of the gases to drop.
System will move in the direction that produces higher number of moles

76. The value of Kc for the reaction is 1x10–3 at high
Example:
For the following chemical equation,
N2 (g) + O2 (g) ⇌ 2NO (g)
The value of Kc for the reaction is 1x10–3 at high
temperature and 4.8x10–31 at 25°C. Is the forward reaction exothermic or endothermic?
Kc = [NO]2
[N2] [O2]
[NO]↑ T↑
Kc↑
Equilibrium position
shifts from LEFT to the RIGHT
Forward reaction is ENDOTHERMIC.

77. D) The effect of Catalyst
Catalysts are substances that increase the rate of a chemical reaction but are not used up in the reaction.
The rate of forward reaction and reverse reaction increase.
A catalysts allows a system to reach equilibrium faster but does not shift the position of an equilibrium system.

78. Le Châtelier’s Principle
Change Equilibrium
Constant, K
Change
Equilibrium shift
Concentration
yes no
Pressure
yes no
Volume
yes no
Temperature
yes
yes
Catalyst no no

79. Example :
Consider the chemical reaction at equilibrium given below
CaCO3 (s) ⇌ CaO (s) + CO2 (g) ΔH =+ve
(a) Calculate the values of Kp and Kc for the system at 525oC, with the equilibrium pressure of CO2 is atm.
(b) Predict the equilibrium position when the following changes are made to the system
(i) some of the CaO is removed from the system
(ii) the pressure of the system is increased
(iii) the temperature of the system is raised
(iv) Helium gas is added at constant volume
(v) Helium gas is added at constant pressure

80. = 0.220 atm Kp = Kc(RT)Δn Δn = 1 – 0 = 1 Kp= Kc(RT) = 3.36 x 10–3 M
Kp = P
CO2
= atm
Kp = Kc(RT)Δn
Δn = 1 – 0 = 1
Kp= Kc(RT)
Kc = Kp
(RT)
= atm
atm L mol –1 K –1 x 798 K
= 3.36 x 10–3 M

81. (b)
(i) Does not change
(the partial pressure of CO2 is not affected by the removal of SOLID CaO from the system).
(ii) Shifts to the LEFT.
(P↑, ; to the side with less number of moles of gas).
(iii) Shifts to the RIGHT.
(Endothermic, T↑, shifts to the right to reduce heat).
(iv) Does not change
(Addition of inert gas at constant V does not change the partial pressure of gas).
(v) Shifts to the RIGHT.
(Addition of inert gas at constant P, V increase, therefore equilibrium position shifts to the side with greater number of moles of gas).

82. HABER PROCESS A process that produces ammonia from H2 and N2
The process is introduced by German scientists Fritz Haber and Karl Bosch. (sometimes called Haber-Bosch) in
Main objectives of the industry are to obtain a high yield of the product while keeping the cost down

83. Consider
N2(g) + 3H(g) NH3(g)
ΔH = kJ
A higher yield of NH3 can be obtained by carrying out reaction under high pressures and at the lowest temperature possible.
However the rate of production is low at a lower temperature. Commercially, faster production is preferable.
So combination of high-pressure, high temperature and proper catalyst is the most efficient way to produce ammonia on a large scale.
Temperature used = 500oC
Pressure = 500 – 1000 atm
Catalyst = Osmium and Ruthenium