The Factors that Determine the Equilibrium State
The equilibrium state (or the ratio of products to reactants) is determined by two factors:
1. Energy
Systems tend to move toward a state of minimum potential energy (enthalpy) to create
products that are more stable. Therefore enthalpy changes favour the exothermic
direction of a reaction since these involve products with lower potential energy.
A
+
B
C + D
+ kinetic energy
2. Degree of Randomness
Systems tend to move spontaneously toward a state of maximum randomness or
disorder; toward maximum entropy. This means that the entropy state favours the side
of the reaction which produces more particles since this creates greater disorder.
In addition, the states of matter have different degrees of disorder:
Most disorder
High entropy
Least disorder
Low entropy
Reactions that produce the most gas molecules will be favoured.
AB (s)
A (g)
+
B (g)
3. The Compromise
The composition of the equilibrium state is a compromise between these two factors: i)
minimum potential energy (enthalpy) and ii) maximum randomness (entropy).
If both factors favour the products, reactions are usually considered spontaneous.
If both factors favour reactants, they are considered non-spontaneous.
If these factors oppose each other, a mixture of products and reactants can exist at
equilibrium. At very low temperatures, enthalpy tends to be the more important factor.
Under such conditions, equilibrium favors the exothermic direction. At very high
temperatures, randomness becomes more important. Under such conditions,
equilibrium favors the direction that produces the most random distribution of particles
without regard to enthalpy.
Consider these reactions:
1)
H2O2 (l)
H2O(l) + O2(g)
∆H = -98.2 kJ/mol
Lower Enthalpy Favours: _____________
Higher Entropy Favours: ______________
2)
N2 (g) +
2 O2 (g) + 67.7 kJ
2 NO2 (g)
Lower Enthalpy Favours: _____________
Higher Entropy Favours: ______________
3)
NaCl (s)
NaCl (aq)
∆H = +3.88 kJ/mol
Lower Enthalpy Favours: _____________
Higher Entropy Favours: ______________
4)
CO (g) + 2 H2 (g)
CH3OH (g)
+ 209 kJ
Lower Enthalpy Favours: _____________
Higher Entropy Favours: _____________
5)
CO (g) + ½ O2 (g)
CO (g)
∆H = -282.5 kJ/mol
Lower Enthalpy Favours: _____________
Higher Entropy Favours: _____________
6)
H2SO4 (aq)
H2SO4 (l)
∆H = +79.4 kJ/mol
Lower Enthalpy Favours: _____________
Higher Entropy Favours: _____________
7)
Pb2+ (aq) + 2I- (aq)
PbI2 (s)
∆H = -585 kJ/mol
Lower Enthalpy Favours: _____________
Higher Entropy Favours: _____________
8)
C3H8 (l) + 349.5 kJ
C3H8 (g)
Lower Enthalpy Favours: ____________
Higher Entropy Favours: _____________
9)
CS2 (g)
+
3 O2 (g)
CO2 (g) + 2 SO2 (g)
Lower Enthalpy Favours: ____________
Higher Entropy Favours: _____________
+ 1107 kJ
The Factors Affecting Equilibrium
The composition of the equilibrium state (i.e. the proportion of reactants and
products) is determined by:
1) The Tendency Towards Minimum Potential Energy (Enthalpy or H)

H = Hproducts - Hreactants
H < 0: The forward reaction is exothermic and favoured.
H > 0: The forward reaction is endothermic and the reverse reaction is
favoured.
Most reactions we study are exothermic because these tend to be spontaneous.
2) The Tendency Towards Maximum Randomness (Entropy or S)

S = Sproducts - Sreactants
S > 0: Products are more random and the forward reaction is favoured.
S < 0: Products are less random and the reverse reaction is favoured.
Reactions favour the direction that produces maximum randomness.
At low temperatures, enthalpy change has the greatest influence and exothermic
reactions are generally spontaneous.
At high temperatures, the random motion of molecules is increased and the entropy
factor (S) has more influence on the equilibrium state.
Gibb’s Free Energy Change - G
Both enthalpy and entropy factors are incorporated into Gibbs Free
Energy Change. This concept allows us to predict whether the forward
or reverse reaction is favoured. Reactions that favour the products are
considered spontaneous.
G  H  TS
If G < 0 (  )
The forward reaction is spontaneous.
If G > 0 ( + )
The forward reaction is not spontaneous.
If G = 0
Equilibrium is reached.
Sample Problem:
e.g. Calculate the normal boiling point of mercury at 101.3 kPa.
⇌
H = + 61.4 kJ/mol
S = + 0.09907 kJ/K
* At the boiling point, the two phases are at equilibrium ( G = 0)
Hg (l)
Hg (g)
The Factors Affecting Equilibrium
1. For each of these reactions, state whether the entropy increases or decreases in
the forward direction:
a)
b)
c)
d)
e)
f)
g)
h)
i)
2 H2 (g) + O2 (g)
2 H2O (g)
2 SO3 (g)
2 SO2 (g) + O2 (g)
MgCO3 (s) + 2 H3O+ (aq)
Mg2+ (aq) + 3 H2O (l) + CO2 (g)
Ag+ (aq) + Cl- (aq)
AgCl (s)
Cl2 (g)
2 Cl (g)
H2O (l)
H2O (g)
Mg (s) + 2 H3O+ (aq)
Mg2+ (aq) + H2 (g) + 2 H2O (l)
2 C2H2 (g) + 5 O2 (g)
4 CO2 (g) + 2 H2O (g)
NH3 (g) + HCl (g)
NH4Cl (s)
2. For each reaction, calculate the value of Gibb’s Free Energy (∆G) and predict
whether the reaction is spontaneous or not spontaneous in the forward direction.
∆H (kJ)
T∆S (kJ)
a) H2O (l)
H2O (g)
284.2
41.8
b) 2 Fe(s) + ½ N2
Fe2N (s)
-3.76
-14.63
c) C6H14 (g)
6 C(s) + 7 H2 (g)
166.9
167.4
+
d) HCl (g) + H2O (l)
H3O (aq) + Cl (aq) -75.2
-39.3
3. For each of the following, calculate the value of ∆G and then predict whether
there will be more reactants or products at equilibrium at room temperature (298
K).
∆H
∆S
a) N2 (g) + 3H2 (g)
⇌
-91.96 kJ
-0.1982 kJ/K
-106 kJ
0.058 kJ/K
c) N2 (g) + 3 F2 (g) ⇌ 2 NF3 (g)
-249 kJ
-0.278 kJ/K
d) N2 (g) + 3 Cl2 (g) ⇌ 2 NCl3 (g)
+460 kJ
-0.275 kJ/K
e) N2F4 (g) ⇌ 2 NF2 (g)
+85 kJ
+0.198 kJ/K
f) 2 H2O (l) ⇌ 2 H2 (g) + O2 (g)
+572 kJ
+0.329 kJ/K
b) H2O2 (g) ⇌
2 NH3 (g)
H2O (g) + ½ O2 (g)