Thermodynamic of equilibria

Dr.P.GOVINDARAJ
Associate Professor & Head , Department of Chemistry
SAIVA BHANU KSHATRIYA COLLEGE
ARUPPUKOTTAI - 626101
Virudhunagar District, Tamil Nadu, India
THERMODYNAMICS OF EQUILIBRIA

• For an equilibrium reaction
A + B ⇌ C + D
K1
K2
• The rate of forward reaction
r1 = k1 [A][B]
• The rate of backward reaction
r2 = k2 [C][D]
• Since at equilibrium
r1 = r2
k1 [A][B] = k2 [C][D]
Equilibrium constant

k1
k2
=
[C] [D]
[A] [B]
KC =
[C][D]
[A][B]
Where KC & KP are known as equilibrium constant and PC , PD , PA , PB are
partial pressure of product and reactant
• Rearranging the above equation
• In terms of partial pressure of the reactant and product the above equation becomes
KP =
𝑃 𝐶 𝑃 𝐷
𝑃 𝐴
𝑃 𝐵
• i.e., The equilibrium constant is the ratio between the concentrations (or) partial pressures
of the products and reactants of an equilibrium reaction

Thermodynamics interpretation of law of mass action
• Consider two equilibrium in the box I and box II in which components A, B, C and D
are in equilibrium
• Where ‘C’ terms represent concentration and ‘p’ terms represent Partial pressure
A + B ⇌ C + D
• Statement
Law of mass action state that the ratio of the concentrations (or) partial pressures of
the products and reactants for an equilibrium reaction is constant

• Let one mole of ‘A’ is removed from box I by moving the piston infinitesimally slowly
outwards and the work done (W1) for the same in according to first law of
thermodynamics is
W1 = RT ln
𝑃 𝐴
𝑃′
𝐴
------(1)
• Similarly the work done (W2) in transferring one mole of ‘B’ from box I to box II is
W2 = RT ln
𝑃 𝐵
𝑃′
𝐵
------(2)
• Since the concentration of ‘A’ and ‘B’ remains constant in equilibrium, proportionate
amounts of ‘C’ and ‘D’ will react to form ‘A’ and ‘B’ in order to maintain the equilibrium
• So, one mole of ‘C’ is removed from box II to box I, then the work done (W3) is
W3 = RT ln
𝑃′ 𝐶
𝑃 𝐶
------(3)

• Similarly, the work done (W4 ) in transferring one mole of ‘D’ from box II to box I is
W4 = RT ln
𝑃′ 𝐷
𝑃 𝐷
------(4)
• Since this process is an isothermal cyclic process, the total amount of work done is zero
i.e., W1 + W2 + W3 + W4 = 0
i.e., RT ln
𝑃 𝐴
𝑃′
𝐴
+ RT ln
𝑃 𝐵
𝑃′
𝐵
+ RT ln
𝑃′ 𝐶
𝑃 𝐶
+ RT ln
𝑃′ 𝐷
𝑃 𝐷
= 0
ln
𝑃 𝐴
𝑃′
𝐴
+ ln
𝑃 𝐵
𝑃′
𝐵
= ln
𝑃 𝐶
𝑃′ 𝐶
+ ln
𝑃 𝐷
𝑃′ 𝐷
ln
𝑃 𝐴 𝑋 𝑃𝐵
𝑃′
𝐴
𝑋 𝑃′
𝐵
= ln
𝑃 𝐶 𝑋 𝑃𝐷
𝑃′
𝐶
𝑋 𝑃′
𝐷

𝑃′
𝐴
𝑋 𝑃′
𝐵
=
𝑃′
𝐶
𝑋 𝑃′
𝐷
𝑃′
𝐶 𝑋 𝑃′
𝐷
𝑃′
𝐴 𝑋 𝑃′
𝐵
=
---------(1)
• In terms of concentration terms the equation (1) becomes
𝐶′
𝐶 𝑋 𝐶′
𝐷
𝐶′
𝐴 𝑋 𝐶′
𝐵
=
𝐶 𝐶 𝑋 𝐶𝐷
𝐶 𝐴 𝑋 𝐶𝐵
---------(2)
• Equation (1) and (2) are mathematical statements for law of mass action
• On removing the ‘ln’ terms in both side of the above equation, we get
• On rearranging the above equation, we get
i.e., According to Law of mass action the ratio of the concentrations (or) partial
pressures of the products and reactants for an equilibrium reaction is constant

Relation between Equilibrium constant & Free energy
• Consider a reversible reaction
a A + b B ⇌ l L + m M
∆G = (G) products – (G) reactants -------(1)
• Since [(G) T,P,N = n1μ1 + n2μ2 ….] , equation (1) becomes
∆G = (lμL + mμM ) – (aμA + bμB ) -------(2)
where μA , μB , μL , μM are chemical potential of A, B , L and M
• We know that the chemical potential for a ‘i’ th gas in a gaseous mixture is
μi = μ𝑖
𝑜
+ RT ln pi -------(3)
• Change in free energy for the reversible reaction is

• Substituting chemical potential of A, B, L, and M in equation (2)
∆G = {l (μ 𝐿
𝑜
+ RT ln pL ) + m(μ 𝑀
𝑜
+ RT ln pM ) } – {a (μ 𝐴
𝑜
+ RT ln pA ) + b(μ 𝐵
𝑜
+ RT ln pB ) }
• Rearranging , we get
∆G = {(lμ 𝐿
𝑜
+ mμ 𝑀
𝑜
) – (aμ 𝐴
𝑜
+ bμ 𝐵
𝑜
) } + RT ln
pL
𝑙
pM
𝑚
pA
𝑎
pB
𝑏
∆G = ∆G0 + RT ln J -------(4)
where J =
pL
𝑙
pM
𝑚
pA
𝑎
pB
𝑏
• At equilibrium , ∆G = 0 and J becomes equilibrium constant KP then the equation (4,)
becomes
0 = ∆G0 + RT ln KP
∆G0 = - RT ln KP -------(5)
• Equation (5) is the relation between equilibrium constant and free energy and also called
as Van’t Hoff reaction isotherm

• Consider an equilibrium reaction
Derivation of Van’t Hoff reaction isotherm
A + B ⇌ C+ D
∆G = (G) product – (G) reactant
∆G = (GC + GD) – (GA + GB) -------(1)
∆G = (μ C + μ D) – (μ A + μ B) -------(2)
where μA , μB, μC, μD are the chemical potential of A, B, C and D
• Since μ = μ0 + RT ln p, the equation (2) becomes
∆G = (μ 𝑐
𝑜+ RT ln pC + μ 𝐷
𝑜 + RT ln pD) – (μ 𝐴
𝑜 + RT ln pA + μ 𝐵
𝑜 + RT ln pB)
∆G = (μ 𝑐
𝑜
+ μ 𝐷
𝑜
) – (μ 𝐴
𝑜
+ μ 𝐵
𝑜
) + RT ln pC + RT ln pD – RT ln pA – RT ln pB
• Change in free energy for the equilibrium reaction is
• Since [(G) T,P,N = n1μ1 + n2μ2 ….] , equation (1) becomes

∆G = (G0)products – (G0)reactants + RT ln
pC pD
pApB
∆G = ∆G0 + RT ln
PC PD
PA PB
--------(3)
• At equilibrium , ∆G = 0, KP =
pC pD
pApB
(Equilibrium constant), equation (3) becomes
∆G0 + RT ln KP = 0
∆G0 = - RT ln KP --------(4)
• Substituting (4) in (3) , we get
∆G = - RT ln KP + RT ln
pC pD
pA pB
-∆G = RT ln KP - RT ln
pC pD
pA pB
--------(5)
• Since [(G0) T,P,N = n1 μ1
𝑜 + n2 μ2
𝑜….] , equation (2) becomes

• For general reaction
n1A + n2B + …. ⇌ n3C + n4D ….
• Equation (4) becomes
-∆G = RT ln KP - RT ln
(pC)n3(pD)n4
(pA)n1(pB) n2
-∆G = RT ln KP – RT 𝑛 𝑙𝑛 𝑝 --------(6)
• Equation (4) & (6) are known as Van’t Hoff reaction isotherm
∆G0 = - RT ln KP & -∆G = RT ln KP – RT 𝑛 𝑙𝑛 𝑝

• The Van’t Hoff reaction isotherm is
Derivation of Van’t Hoff reaction isochore
-∆G = RT ln KP – RT 𝑛 𝑙𝑛 𝑝 -------(1)
• In terms of concentration, equation (1) becomes
-∆G = RT ln Kc – RT 𝑛 𝑙𝑛 𝐶 -------(2)
• At constant volume equation (2) modifies to
-∆A = RT ln Kc – RT 𝑛 𝑙𝑛 𝐶
• Differentiating w.r.t temperature at constant volume
𝜕(−∆A)
𝜕𝑇 V = RT
𝑑(ln Kc)
𝑑𝑇
+ R ln Kc – R𝑇
𝑑
𝑑𝑇
( 𝑛𝑙𝑛 𝐶 ) – R 𝑛𝑙𝑛 𝐶 -------(3)

• Since concentration ‘C’ is not a function of temperature . The term R𝑇
𝑑
𝑑𝑇
( 𝑛𝑙𝑛 𝐶 )= 0,
equation (3) becomes
𝜕(−∆A)
𝜕𝑇 V = RT
𝑑(ln Kc)
𝑑𝑇
+ R ln Kc – R 𝑛𝑙𝑛 𝐶
-
𝜕(∆A)
𝜕𝑇 𝑉 = RT
𝑑
𝑑𝑇
(ln Kc) + RT ln Kc – R 𝑛𝑙𝑛 𝐶
- T
𝜕(∆A)
𝜕𝑇 V = RT2 𝑑
𝑑𝑇
(ln Kc) + RT ln Kc – RT 𝑛𝑙𝑛 𝐶 -------(4)
- T
𝜕(∆A)
𝜕𝑇 V = RT2 𝑑
𝑑𝑇
(ln Kc) – ∆A -------(5)
• Since – ∆A = RT ln Kc – RT 𝑛𝑙𝑛 𝐶 , equation (4) becomes

∆ A = ∆ E + T
𝜕(∆A)
𝜕𝑇 V
• Gibb’s Helmholtz equation is
- T
𝜕(∆A)
𝜕𝑇 V = ∆ E − ∆ A --------(6)
• Comparing equation (5) and (6) we get
• Since ∆ E is the heat absorbed (qv) at constant volume i.e., ∆ E = qv , equation(7) becomes
∆ E − ∆ A = RT2 𝑑
𝑑𝑇
(ln Kc) – ∆A
∆ E = RT2 𝑑
𝑑𝑇
(ln Kc)
∆ E
RT2 =
𝑑
𝑑𝑇
(ln Kc) --------(7)
qv
RT2 =
𝑑
𝑑𝑇
(ln Kc)
d (ln Kc) =
qv
RT2 dT -------(8)

• Integrate the equation (7) with in the limits
𝐾𝑐1
𝐾𝑐2
d (ln Kc) = 𝑇1
𝑇2 ∆E
RT2 dT
[ln 𝐾 𝑐 ] 𝐾𝑐1
𝐾𝑐2 =
∆E
R
−
1
𝑇 𝑇1
𝑇2
ln
𝐾𝑐2
𝐾𝑐1
=
∆E
R
−
1
𝑇2
+
1
𝑇1
2.303 log
𝐾𝑐2
𝐾𝑐1
=
∆E
R
𝑇2
−𝑇1
𝑇2
𝑇1
--------(9)
• Equation (9) & (8) are Van’t Hoff reaction isochore
where Kc1 and Kc2 are equilibrium constants at the temperatures T1 and T2

Thermodynamic interpretation for Le - Chatelier’s Principle
Le – Chatelier’s Principle
It is stated that when a change is applied on an equilibrium system , the equilibrium
shifts in such a direction in order to nullify the effect of change.
Example
Increase of external pressure will cause the equilibrium to shift in the direction which
will bring about a lowering of pressure.
PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g)
i.e., Increase of pressure on the above equilibrium system causes the equilibrium
to shift from right to left

• Consider an equilibrium reaction
where ‘ε’ is the extent of the reaction
Thermodynamic treatment for Le - Chatelier’s Principle
aA + bB ⇌ mM + nN
• ∆G of the reaction is the function of T, P and ε
i.e., ∆G = f (T, P, ε ) ------(1)
d(∆G) =
𝜕(∆G)
𝜕𝑇 P, ε dT +
𝜕(∆G)
𝜕𝑃 T, ε dP +
𝜕(∆G)
𝜕𝜀 T,P dε ------(2)
Put
𝜕G
𝜕𝜀 P,T = ∆ G in equation (2) we get
d(∆G) =
𝜕
𝜕𝑇
𝜕G
𝜕𝜀 T,P P, ε dT +
𝜕
𝜕𝑃
𝜕G
𝜕𝜀 T, P T,ε dP +
𝜕
𝜕𝜀
𝜕G
𝜕𝜀 T,P T,P dε
• On doing partial differentiation, equation (1) becomes

d(∆G) =
𝜕
𝜕𝜀
𝜕G
𝜕𝑇 P, ε T,P dT +
𝜕
𝜕𝜀
𝜕G
𝜕𝑃 T, ε T,P dP +
𝜕2
G
𝜕𝜀2 T,P dε -------(3)
• Substitute
𝜕G
𝜕𝑇 P = - S and
𝜕G
𝜕𝑃 T = V in equation (3), we get
d(∆G) = -
𝜕𝑆
𝜕𝜀 T,P dT +
𝜕𝑉
𝜕𝜀 T,P dP +
𝜕2
G
𝜕𝜀2 T,P dε -------(4)
d(∆G) = - ∆S dT + ∆V dP +
𝜕2
G
𝜕𝜀2 T,P dε -------(5)
• Substitute -
𝜕𝑆
𝜕𝜀 T,P = ∆S and
𝜕𝑉
𝜕𝜀 T,P = ∆V in equation (4), we get
• At equilibrium ∆G = 0 and ∆S =
∆ 𝐻
𝑇
as per the equation ∆G = ∆ H - T∆S,
so that the equation (5) becomes
-
∆ 𝐻
𝑇
dT + ∆V dP + 𝐺′′ dε = 0 -------(6)
where
𝜕2
G
𝜕𝜀2 T,P = 𝐺′′

• At constant Temperature, dT = 0 and the equation (6) becomes
∆V dP + 𝐺′′ dε = 0 -------(7)
• In the above equilibrium reaction ∆V is positive i.e., [V product > V reactant ] ,then the
RHS in equation (8) would be positive, so that {
𝜕𝜀
𝜕𝑃
}T would be negative
• i.e., Increase in pressure (dP > 0 ) in the above equilibrium will decrease ε (dε < 0), so that
equilibrium shifted towards the reactant side. On the other hand decrease in pressure (dP < 0 )
will increase ε (dε > 0), so that the equilibrium shifted towards the product side
• On rearranging the equation (7), we get
∆V = - 𝐺′′ [
𝜕𝜀
𝜕𝑃
]T -------(8)

Conclusion
An increase in pressure would shift the equilibrium towards the low volume side
of the reaction (i.e., reactant side in the above equilibrium), where as a decrease in
pressure would shift the equilibrium towards the high volume side(i.e., product
side in the above equilibrium). This is one of the statements of Le - Chatelier’s
Principle

Thermodynamic of equilibria

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