Chemical engineering thermodynamics II
Topic 1: Solution Thermodynamics Theory
-
Applications are to gas mixtures and liquid solutions
-
Determination of thermodynamic properties for the above systems
-
Also applies to processes that involve composition changes such as mixing, separation,
mass transfer operations and chemical reactions
Fundamental property relations
First law of thermodynamics: 𝑑(𝑛𝑈) = 𝑑𝑊 + 𝑑𝑄
(1)
But 𝑑𝑄 = 𝑇𝑑(𝑛𝑆) and 𝑑𝑊 = −𝑃𝑑(𝑛𝑉)
This implies that: 𝑑(𝑛𝑈) = 𝑇𝑑(𝑛𝑆) − 𝑃𝑑(𝑛𝑉)
(2)
By definition: (𝑛𝐻) = (𝑛𝑈) + 𝑃(𝑛𝑉) and (𝑛𝐺) = (𝑛𝐻) − 𝑇(𝑛𝑆)
This means: (𝑛𝐺) = (𝑛𝑈) + 𝑃(𝑛𝑉) − 𝑇(𝑛𝑆)
(3)
Differentiating each term in Equation 3, applying product rule, and substituting for 𝑑(𝑛𝑈), we
obtain:
𝑑(𝑛𝐺) = (𝑛𝑉)𝑑𝑃 − (𝑛𝑆)𝑑𝑇
(4)
This implies that for a closed system: (𝑛𝐺) = 𝑔(𝑃, 𝑇)
For an open system with N components: (𝑛𝐺) = 𝑔(𝑃, 𝑇, 𝑛1 , 𝑛2 , 𝑛3 , … . , 𝑛𝑁−1 , 𝑛𝑁 )
Therefore: 𝑑(𝑛𝐺) = (
𝜕(𝑛𝐺)
𝜕𝑃
)
𝜕(𝑛𝐺)
𝑇,𝑛
𝑑𝑃 + (
𝜕𝑇
𝜕(𝑛𝐺)
The last derivative in Equation 5, (
𝜕𝑛𝑖
)
𝜕(𝑛𝐺)
𝑃,𝑛
)
𝑑𝑇 + ∑𝑖 (
𝜕𝑛𝑖
)
𝑃,𝑇,𝑛𝑗
𝑑𝑛𝑖
(5)
is called the chemical potential of species i in the
𝑃,𝑇,𝑛𝑗
mixture or solution with symbol 𝜇𝑖
Comparing Equation 5 with Equation 4 and substituting for 𝜇𝑖 , we obtain:
𝑑(𝑛𝐺) = (𝑛𝑉)𝑑𝑃 − (𝑛𝑆)𝑑𝑇 + ∑𝑖 𝜇𝑖 𝑑𝑛𝑖
(6)
𝜕(𝑛𝐺)
Where (𝑛𝑉) = (
𝜕𝑃
)
𝜕(𝑛𝐺)
𝑇,𝑛
and (𝑛𝑆) = − (
𝜕𝑇
)
𝑃,𝑛
Equation 6 is the fundamental equation for solution thermodynamics which involves phase
equilibria and chemical reaction equilibria.
Partial properties
-
In a solution, each species contributes to the overall property of the system
-
Addition of a differential amount of species i to a finite amount of solution at constant
temperature and pressure causes a change to the system properties
𝜕(𝑛𝑀)
̅𝑖 = (
The partial molar property of species i in the mixture or solution is given by: 𝑀
𝜕𝑛𝑖
)
𝑃,𝑇,𝑛𝑗
Where 𝑀 is the solution property eg V, H, S, G
̅𝑖 is the partial property of species i in the mixture or solution
𝑀
𝑀𝑖 is the pure-species property of species i at the solution temperature and pressure
If 𝑀 = 𝐺, we have 𝐺̅𝑖 and this is equal to 𝜇𝑖
Chemical potential and phase equilibria
-
When two or more phases are in equilibrium, temperature and pressure are uniform
throughout the whole system
-
Equation 6 applies to species i in all phases in question
Dealing with two phases, 𝛼 and 𝛽, each extensive property of the whole system is a sum of it in
the 𝛼 phase and the 𝛽 phase.
This is represented as: (𝑛𝑀) = (𝑛𝑀)𝛼 + (𝑛𝑀)𝛽
(7)
For the 𝛼 phase: 𝑑(𝑛𝐺)𝛼 = (𝑛𝑉)𝛼 𝑑𝑃 − (𝑛𝑆)𝛼 𝑑𝑇 + ∑𝑖 𝜇𝑖 𝛼 𝑑𝑛𝑖 𝛼
For the 𝛽 phase: 𝑑(𝑛𝐺)𝛽 = (𝑛𝑉)𝛽 𝑑𝑃 − (𝑛𝑆)𝛽 𝑑𝑇 + ∑𝑖 𝜇𝑖 𝛽 𝑑𝑛𝑖 𝛽
Finding total differential change in Gibbs Free Energy according to Equation 7:
𝑑(𝑛𝐺) = (𝑛𝑉)𝑑𝑃 − (𝑛𝑆)𝑑𝑇 + ∑ 𝜇𝑖 𝛼 𝑑𝑛𝑖 𝛼 + ∑ 𝜇𝑖 𝛽 𝑑𝑛𝑖 𝛽
𝑖
𝑖
If the system is a closed system: ∑𝑖 𝜇𝑖 𝛼 𝑑𝑛𝑖 𝛼 + ∑𝑖 𝜇𝑖 𝛽 𝑑𝑛𝑖 𝛽 = 0
(8)
If the system is not yet at equilibrium, there will be mass transfer of species i between the phases
and this is represented by 𝑑𝑛𝑖 𝛼 and 𝑑𝑛𝑖 𝛽 in Equation 8. Therefore, any increase in the amount of
species i in the 𝛼 phase is equal to the corresponding decrease in the amount of species i in the 𝛽
phase until equilibrium is established as shown in Figure 1.
α phase
Exchange of species i
due to mass transfer
β phase
Figure 1
This means: 𝑑𝑛𝑖 𝛼 = −𝑑𝑛𝑖 𝛽
∑𝑖(𝜇𝑖 𝛼 − 𝜇𝑖 𝛽 )𝑑𝑛𝑖 𝛼 = 0
This implies that:
Since 𝑑𝑛𝑖 𝛼 ≠ 0, this implies that 𝜇𝑖 𝛼 − 𝜇𝑖 𝛽 = 0
Therefore: 𝜇𝑖 𝛼 = 𝜇𝑖 𝛽 = ⋯ = 𝜇𝑖 𝜋 for i = 1, 2, 3, ….., N and there are 𝜋 phases
Conclusion: multiple phases at the same temperature and pressure are in equilibrium when the
chemical potential of each species is the same in all phases.
Gibbs/Duhem equation
Thermodynamic properties of a homogenous phase: (𝑛𝑀) = 𝑔(𝑃, 𝑇, 𝑛1 , 𝑛2 , 𝑛3 , … . , 𝑛𝑁−1 , 𝑛𝑁 )
Therefore: 𝑑(𝑛𝑀) = (
𝜕(𝑛𝑀)
𝜕𝑃
)
𝑇,𝑛
𝑑𝑃 + (
𝜕(𝑛𝑀)
𝜕𝑇
)
𝑃,𝑛
𝜕(𝑛𝑀)
𝑑𝑇 + ∑𝑖 (
𝜕𝑛𝑖
)
𝑃,𝑇,𝑛𝑗
𝑑𝑛𝑖
Since the total number of moles in the system, n, is constant:
𝜕𝑀
⟹ 𝑑(𝑛𝑀) = 𝑛 ( 𝜕𝑃 )
𝜕𝑀
𝑇,𝑥
𝑑𝑃 + 𝑛 ( 𝜕𝑇 )
𝑃,𝑥
̅𝑖 𝑑𝑛𝑖
𝑑𝑇 + ∑𝑖 𝑀
(9)
But 𝑛𝑖 = 𝑛𝑥𝑖 , meaning 𝑑𝑛𝑖 = 𝑥𝑖 𝑑𝑛 + 𝑛𝑑𝑥𝑖 . Also 𝑑(𝑛𝑀) = 𝑛𝑑𝑀 + 𝑀𝑑𝑛
Substituting into Equation 9 and simplifying gives:
[𝑑𝑀 − (
𝜕𝑀
𝜕𝑀
̅𝑖 𝑑𝑥𝑖 ] 𝑛 + [𝑀 − ∑ 𝑥𝑖 𝑀
̅𝑖 ] 𝑑𝑛 = 0
) 𝑑𝑃 − ( ) 𝑑𝑇 − ∑ 𝑀
𝜕𝑃 𝑇,𝑥
𝜕𝑇 𝑃,𝑥
𝑖
𝑖
The size of the system, n, and the differential change, dn, can be arbitrarily chosen and are therefore
not equal to zero, it means the terms in the square brackets must be zero.
𝜕𝑀
Therefore: 𝑑𝑀 = ( 𝜕𝑃 )
𝜕𝑀
𝑇,𝑥
𝑑𝑃 + ( 𝜕𝑇 )
𝑃,𝑥
̅𝑖 𝑑𝑥𝑖
𝑑𝑇 + ∑𝑖 𝑀
̅𝑖
And: 𝑀 = ∑𝑖 𝑥𝑖 𝑀
(10)
(11)
̅𝑖
Multiplying Equation 11 by n gives (𝑛𝑀) = ∑𝑖 𝑛𝑖 𝑀
̅ 𝑖 + ∑𝑖 𝑀
̅𝑖 𝑑𝑥𝑖
Differentiating Equation 11: 𝑑𝑀 = ∑𝑖 𝑥𝑖 𝑑𝑀
Substituting this result into Equation 10 and simplifying gives:
𝜕𝑀
( 𝜕𝑃 )
𝜕𝑀
𝑇,𝑥
𝑑𝑃 + ( 𝜕𝑇 )
𝑃,𝑥
̅𝑖 = 0
𝑑𝑇 − ∑𝑖 𝑥𝑖 𝑑𝑀
(12)
Equation 12 is the Gibbs/Duhem equation.
At constant temperature and pressure, the Gibbs/Duhem equation reduces to:
̅𝑖 = 0
∑𝑖 𝑥𝑖 𝑑𝑀
(13)
Equations 11 and 13 show that each species in a mixture or solution cannot have its own private
properties. This is due to molecular interactions within the system. However, partial properties
have all the characteristics of individual species properties as they exist in solution.
̅1 𝑥1 + 𝑀
̅2 𝑥2
Applying Equation 11 to a binary system: 𝑀 = 𝑀
̅1 𝑑𝑥1 + 𝑥1 𝑑𝑀
̅1 + 𝑀
̅2 𝑑𝑥2 + 𝑥2 𝑑𝑀
̅2
Differentiating gives: 𝑑𝑀 = 𝑀
̅1 + 𝑥2 𝑑𝑀
̅2 = 0 (from Gibbs/Duhem equation)
But 𝑥1 𝑑𝑀
̅1 𝑑𝑥1 + 𝑀
̅2 𝑑𝑥2
Therefore: 𝑑𝑀 = 𝑀
(14)
Since 𝑥1 + 𝑥2 = 1, this means 𝑑𝑥1 = −𝑑𝑥2
⟹
𝒅𝑴
̅𝟏 −𝑴
̅𝟐
= 𝑴
𝒅𝒙𝟏
̅1 𝑥1 + 𝑀
̅2 𝑥2 and using 𝑥1 + 𝑥2 = 1
Back to Equation 14: 𝑀 = 𝑀
Simplifying yields the following equations:
̅2 = 𝑀 − 𝑥1 𝑑𝑀
𝑀
𝑑𝑥
(15)
̅1 = 𝑀 + 𝑥2 𝑑𝑀
𝑀
𝑑𝑥
(16)
1
1
̅𝑖 = 𝑀𝑖
To calculate pure-species property of species i: lim 𝑀 = lim 𝑀
𝑥𝑖 →1
𝑥𝑖 →1
̅𝑖 = 𝑀
̅𝑖 ∞
To calculate property when species i approaches infinite dilution: lim 𝑀
𝑥𝑖 →0
Relations among partial properties
𝜕𝐺̅𝑖
𝜕 𝜕(𝑛𝐺)
𝜕 𝜕(𝑛𝐺)
𝜕
𝜕(𝑛𝑉)
(𝑛𝑉) = (
( ) =
(
)
=
(
)=
)
= 𝑉̅𝑖
𝜕𝑃 𝑇,𝑥 𝜕𝑃 𝜕𝑛𝑖 𝑃,𝑇,𝑛
𝜕𝑛𝑖
𝜕𝑃
𝜕𝑛𝑖
𝜕𝑛𝑖 𝑃,𝑇,𝑛
𝑗
𝑗
𝜕𝐺̅𝑖
𝜕 𝜕(𝑛𝐺)
𝜕 𝜕(𝑛𝐺)
𝜕
𝜕(𝑛𝑆)
(−𝑛𝑆) = (−
( ) =
(
)
=
(
)=
)
= −𝑆𝑖̅
𝜕𝑇 𝑃,𝑥 𝜕𝑇 𝜕𝑛𝑖 𝑃,𝑇,𝑛
𝜕𝑛𝑖
𝜕𝑇
𝜕𝑛𝑖
𝜕𝑛𝑖 𝑃,𝑇,𝑛
𝑗
𝑗
Therefore: 𝑑𝐺̅𝑖 = 𝑉̅𝑖 𝑑𝑃 − 𝑆𝑖̅ 𝑑𝑇
Conclusion: the same relations on molar properties apply to partial properties
Ideal gas model
𝑃𝑉 𝑖𝑔 = 𝑅𝑇
⟹ 𝑉 𝑖𝑔 = 𝑅𝑇⁄𝑃
This shows that all ideal gases (be it a single component or mixture) have the same molar volume
when at the same temperature and pressure
Determining partial molar volume
𝜕(𝑛𝑉 𝑖𝑔 )
𝜕(𝑛 𝑅𝑇⁄𝑃)
𝑅𝑇 𝜕𝑛
𝑖𝑔
( )
𝑉̅𝑖 = (
)
= (
)
=
𝜕𝑛𝑖
𝜕𝑛
𝑃
𝜕𝑛𝑖 𝑛
𝑖
𝑇,𝑃,𝑛
𝑇,𝑃,𝑛
𝑗
𝜕𝑛
But 𝑛 = 𝑛𝑖 + ∑𝑗 𝑛𝑗
𝑖𝑔
This means: 𝑉̅𝑖 =
⟹ (𝜕𝑛 )
𝑖
𝑗
𝑗
=1
𝑛𝑗
𝑅𝑇
𝑃
𝑖𝑔
Therefore: 𝑉̅𝑖 = 𝑉 𝑖𝑔 = 𝑉𝑖 𝑖𝑔
(17)
Determining partial pressure
𝑃𝑖 = 𝑦𝑖 𝑃 =
𝑦𝑖 𝑅𝑇
𝑉𝑖
𝑖𝑔
=
𝑦𝑖 𝑅𝑇
𝑉 𝑖𝑔
Conclusion: since the ideal gas model assumes molecules with zero volume (volume of molecules
is negligible compared to volume of container), which do not interact (no intermolecular forces,
elastic collisions between molecules), other thermodynamic properties other than molar volume of
the constituent species or components are independent of one another. Each species has its own
set of private properties
Gibbs theorem
A partial molar property (other than volume) of a constituent species in an ideal gas mixture is
equal to the corresponding molar property of the species as a pure ideal gas at the mixture
temperature but at a pressure equal to its partial pressure in the mixture.
̅𝑖 𝑖𝑔 (𝑇, 𝑃) = 𝑀𝑖 𝑖𝑔 (𝑇, 𝑃𝑖 ) where 𝑀 ≠ 𝑉
This means: 𝑀
(18)
Determining partial molar enthalpy
The enthalpy of an ideal gas depends on temperature only. This means it is independent of pressure.
̅𝑖 𝑖𝑔 (𝑇, 𝑃) = 𝐻𝑖 𝑖𝑔 (𝑇, 𝑃𝑖 ) = 𝐻𝑖 𝑖𝑔 (𝑇, 𝑃)
∴ 𝐻
̅𝑖 𝑖𝑔 = 𝐻𝑖 𝑖𝑔
⟹ 𝐻
(19)
Determining partial molar entropy
From the 1st law: 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊
Substituting: 𝑑𝑈 = 𝐶𝑉 𝑑𝑇, 𝑑𝑊 = −𝑃𝑑𝑉, 𝑃 =
We obtain: 𝑑𝑆𝑖 𝑖𝑔 =
𝐶𝑉
𝑇
𝑑𝑇 +
𝑅
𝑉
𝑅𝑇
𝑉 𝑖𝑔
, 𝑑𝑆𝑖 𝑖𝑔 =
𝑑𝑄
𝑇
𝑑𝑉
At constant temperature: 𝑑𝑇 = 0
𝑉
1
Also: 𝑉 𝑑𝑉 = 𝑑 ln 𝑉 = ln 𝑉𝑓𝑖𝑛𝑎𝑙 − ln 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = ln 𝑉 𝑓𝑖𝑛𝑎𝑙
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑉𝑓𝑖𝑛𝑎𝑙
From ideal gas law: 𝑉
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
1
=
𝑃𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑉
𝑃𝑓𝑖𝑛𝑎𝑙
𝑃𝑓𝑖𝑛𝑎𝑙 −1
= (𝑃
𝑃
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
−1
Therefore: 𝑉 𝑑𝑉 = ln 𝑉 𝑓𝑖𝑛𝑎𝑙 = ln (𝑃 𝑓𝑖𝑛𝑎𝑙 )
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
)
𝑃
= − ln 𝑃 𝑓𝑖𝑛𝑎𝑙 = −𝑑 ln 𝑃
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
This means: 𝑑𝑆𝑖 𝑖𝑔 = −𝑑 ln 𝑃
𝑃
Integrating from Pi to P: 𝑆𝑖 𝑖𝑔 (𝑇, 𝑃) − 𝑆𝑖 𝑖𝑔 (𝑇, 𝑃𝑖 ) = −𝑅 ln 𝑦 𝑃 = 𝑅 ln 𝑦𝑖
𝑖
⟹ 𝑆𝑖 𝑖𝑔 (𝑇, 𝑃𝑖 ) = 𝑆𝑖 𝑖𝑔 (𝑇, 𝑃) − 𝑅 ln 𝑦𝑖
𝑖𝑔
Applying Equation 17: 𝑆𝑖̅ (𝑇, 𝑃) = 𝑆𝑖 𝑖𝑔 (𝑇, 𝑃𝑖 )
𝑖𝑔
Therefore: 𝑆𝑖̅ (𝑇, 𝑃) = 𝑆𝑖 𝑖𝑔 (𝑇, 𝑃) − 𝑅 ln 𝑦𝑖
∴ 𝑆𝑖̅
𝑖𝑔
= 𝑆𝑖 𝑖𝑔 − 𝑅 𝑙𝑛 𝑦𝑖
(20)
Determining partial Gibbs Free Energy
𝑖𝑔
̅𝑖 𝑖𝑔 − 𝑇𝑆𝑖̅ 𝑖𝑔
Since 𝐺 = 𝐻 − 𝑇𝑆, this means: 𝐺̅𝑖 = 𝐻
𝑖𝑔
⟹ 𝐺̅𝑖 = 𝜇𝑖 𝑖𝑔 = 𝐻𝑖 𝑖𝑔 − 𝑇(𝑆𝑖 𝑖𝑔 − 𝑅 ln 𝑦𝑖 ) = (𝐻𝑖 𝑖𝑔 − 𝑇𝑆𝑖 𝑖𝑔 ) + 𝑅𝑇 ln 𝑦𝑖
∴ 𝜇𝑖 𝑖𝑔 = 𝐺𝑖 𝑖𝑔 + 𝑅𝑇 𝑙𝑛 𝑦𝑖
Applying Equation 11 to Equations 19, 20 and 21 yields the following results:
(21)
𝐻 𝑖𝑔 − ∑𝑖 𝑦𝑖 𝐻𝑖 𝑖𝑔 = 0
(22)
𝑆 𝑖𝑔 − ∑𝑖 𝑦𝑖 𝑆𝑖 𝑖𝑔 = −𝑅 ∑𝑖 𝑦𝑖 𝑙𝑛 𝑦𝑖
(23)
𝐺 𝑖𝑔 − ∑𝑖 𝑦𝑖 𝐺𝑖 𝑖𝑔 = 𝑅𝑇 ∑𝑖 𝑦𝑖 𝑙𝑛 𝑦𝑖
(24)
The terms on the LHS of Equations 22, 23 and 24 represent property changes of mixing of ideal
gases, as illustrated in the next topic.
Alternative expression for chemical potential of species i in an ideal gas mixture
𝑑𝐺𝑖 𝑖𝑔 = 𝑉𝑖 𝑖𝑔 𝑑𝑃 − 𝑆𝑖 𝑖𝑔 𝑑𝑇
𝑅𝑇
At constant temperature and using 𝑉𝑖 𝑖𝑔 = 𝑅𝑇⁄𝑃 : 𝑑𝐺𝑖 𝑖𝑔 = 𝑃 𝑑𝑃 = 𝑅𝑇𝑑 ln 𝑃
Integrating both sides: 𝐺𝑖 𝑖𝑔 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑃
(25)
where Γ𝑖 (𝑇) is an integration constant that depends on the species and is a function of temperature
only.
Substituting into Equation 21: 𝜇𝑖 𝑖𝑔 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑃 + 𝑅𝑇 𝑙𝑛 𝑦𝑖 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛(𝑦𝑖 𝑃)
(26)
Fugacity and fugacity coefficient: Pure species
Fugacity is defined as the effective partial pressure of a real gas.
For a real gas: 𝐺𝑖 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑓𝑖
(27)
Comparing Equation 27 with Equation 25: 𝑃 = 𝑓𝑖 𝑖𝑔
𝑓
Subtracting Equation 25 from Equation 27: 𝐺𝑖 − 𝐺𝑖 𝑖𝑔 = 𝐺𝑖 𝑅 = 𝑅𝑇 𝑙𝑛 ( 𝑃𝑖 )
(28)
The term in the brackets in Equation 28 is called the fugacity coefficient of pure species i which
is: 𝜙𝑖 =
𝑓𝑖
𝑃
For an ideal gas: 𝑓𝑖 = 𝑃, meaning 𝜙𝑖 = 1 and also 𝐺𝑖 𝑅 = 0
Vapour/Liquid equilibrium for pure species
For saturated vapour: 𝐺𝑖 𝑣 = Γ𝑖 (𝑇) + 𝑅𝑇 ln 𝑓𝑖 𝑣
For saturated liquid: 𝐺𝑖 𝑙 = Γ𝑖 (𝑇) + 𝑅𝑇 ln 𝑓𝑖 𝑙
Change of phase from saturated liquid to saturated vapour or vice versa occurs at T and Pisat
At equilibrium: 𝑑𝐺 = 𝐺𝑖 𝑣 − 𝐺𝑖 𝑙 = 0
But 𝐺𝑖 𝑣 − 𝐺𝑖 𝑙 = 𝑅𝑇 ln (
𝑓𝑖 𝑣
𝑓𝑖 𝑙
)
This means at equilibrium: 𝑓𝑖 𝑣 = 𝑓𝑖 𝑙 = 𝑓𝑖 𝑠𝑎𝑡
Therefore: 𝜙𝑖 𝑠𝑎𝑡 =
𝑓𝑖 𝑠𝑎𝑡
⁄ 𝑠𝑎𝑡
𝑃𝑖
Fugacity and fugacity coefficient: Species in solution
-
Applies when species is in a mixture of real gases or a solution of liquids
𝜇𝑖 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑓̂𝑖
(29)
For phase equilibrium the chemical potential of species i in all phase must be equal. This means
𝛼
𝛽
𝜋
that: 𝑓̂𝑖 = 𝑓̂𝑖 = ⋯ = 𝑓̂𝑖
This is the criterion for equilibrium that is used to solve phase equilibrium problems.
𝑣
𝑙
For vapour/liquid equilibrium: 𝑓̂𝑖 = 𝑓̂𝑖
This is the criterion for equilibrium used to solve vapour/liquid equilibrium problems.
̂
𝑅
𝑓
Subtracting Equation 26 from Equation 29: 𝜇𝑖 − 𝜇𝑖 𝑖𝑔 = 𝐺̅𝑖 = 𝑅𝑇 𝑙𝑛 (𝑦 𝑖𝑃)
𝑖
(30)
The term in the brackets in Equation 30 is called the fugacity coefficient of species i in solution
̂
𝑓
which is: 𝜙̂𝑖 = 𝑦 𝑖𝑃
𝑖
For an ideal gas: 𝑓̂𝑖 = 𝑦𝑖 𝑃, meaning 𝜙̂𝑖 = 1 and also 𝐺̅𝑖
𝑅
=0
Ideal solution model
The ideal solution model equation is analogous to Equation 21.
𝑖𝑑
𝜇𝑖 𝑖𝑑 = 𝐺̅𝑖 = 𝐺𝑖 (𝑇, 𝑃) + 𝑅𝑇 𝑙𝑛 𝑥𝑖
(31)
Determining partial molar volume for species i in an ideal solution
It is given by: 𝑉̅𝑖
𝑖𝑑
= (
𝑖𝑑
𝜕𝐺̅𝑖
𝜕𝑃
𝜕𝐺
)
𝑇,𝑥
= ( 𝜕𝑃𝑖 ) = 𝑉𝑖
(32)
𝑇
Determining partial molar entropy for species i in an ideal solution
𝑆𝑖̅
𝑖𝑑
= −(
𝑖𝑑
𝜕𝐺̅𝑖
𝜕𝑇
)
𝑃,𝑥
𝜕𝐺
= − ( 𝜕𝑇𝑖 ) − 𝑅 𝑙𝑛 𝑥𝑖 = 𝑆𝑖 − 𝑅 𝑙𝑛 𝑥𝑖
(33)
𝑃
Determining partial molar enthalpy for species i in an ideal solution
̅𝑖 𝑖𝑑 = 𝐺̅𝑖 𝑖𝑑 + 𝑇𝑆𝑖̅ 𝑖𝑑
𝐻
̅𝑖 𝑖𝑑 = 𝐺𝑖 + 𝑅𝑇 𝑙𝑛 𝑥𝑖 + 𝑇𝑆𝑖 − 𝑅𝑇 𝑙𝑛 𝑥𝑖 = 𝐻𝑖
⟹ 𝐻
(34)
Applying Equation 11 to Equations 31, 32, 33 and 34 yields the following results:
𝑉 𝑖𝑑 = ∑𝑖 𝑥𝑖 𝑉𝑖
(35)
𝐻 𝑖𝑑 − ∑𝑖 𝑥𝑖 𝐻𝑖
(36)
𝑆 𝑖𝑑 = ∑𝑖 𝑥𝑖 𝑆𝑖 − 𝑅 ∑𝑖 𝑥𝑖 𝑙𝑛 𝑥𝑖
(37)
𝐺 𝑖𝑑 = ∑𝑖 𝑥𝑖 𝐺𝑖 + 𝑅𝑇 ∑𝑖 𝑥𝑖 𝑙𝑛 𝑥𝑖
(38)
Lewis/Randall Rule
𝑓̂
Subtracting Equation 27 from Equation 29: 𝜇𝑖 = 𝐺𝑖 + 𝑅𝑇 𝑙𝑛 (𝑓𝑖 )
(39)
𝑖
Applying Equation 39 to an ideal solution: 𝜇𝑖
𝑖𝑑
Comparing Equation 40 with Equation 31: 𝑥𝑖 =
̂ 𝑖𝑑
𝑖𝑑
𝑓
= 𝐺̅𝑖 = 𝐺𝑖 + 𝑅𝑇 𝑙𝑛 ( 𝑓𝑖 )
𝑖
𝑖𝑑
𝑓̂𝑖
𝑓𝑖
𝑖𝑑
⟹ 𝑓̂𝑖 = 𝑥𝑖 𝑓𝑖
(40)
(41)
Equation 41 is called the Lewis/Randall Rule. It applies to each species in an ideal solution at all
conditions of temperature, pressure and composition. 𝑓𝑖 is determined when pure species i is in the
same physical state as the solution and at the same temperature and pressure. The Lewis/Randall
Rule is analogous to Raoult’s Law.
Dividing Equation 41 by 𝑃𝑥𝑖 :
𝑖𝑑
𝑓̂𝑖
𝑃𝑥𝑖
=
𝑓𝑖
𝑃
𝑖𝑑
⟹ 𝜙̂𝑖 = 𝜙𝑖
(42)
Excess properties
Liquid solution thermodynamic properties are dealt with easily by their departure from ideal
solution behaviour.
At the same temperature, pressure and composition: 𝑀𝐸 = 𝑀 − 𝑀𝑖𝑑
The same applies for partial properties
𝜇𝑖 = 𝐺̅𝑖 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑓̂𝑖
(29)
𝑖𝑑
Applying Equation 29 to an ideal solution: 𝜇𝑖 𝑖𝑑 = 𝐺̅𝑖 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑥𝑖 𝑓𝑖
(43)
̂
𝑖𝑑
𝐸
𝑓
Subtracting Equation 43 from Equation 29: 𝐺̅𝑖 − 𝐺̅𝑖 = 𝐺̅𝑖 = 𝑅𝑇 𝑙𝑛 (𝑥 𝑓𝑖 )
(44)
𝑖 𝑖
𝑓̂
The term in the brackets in Equation 44 is called activity coefficient: 𝛾𝑖 = 𝑥 𝑓𝑖
𝑖 𝑖
The activity coefficient accounts for liquid phase non-idealities. Summability (Equation 11) and
Gibbs/Duhem equation (Equation 13) can be applied to Equation 44.
Topic 2: Solution Thermodynamics Applications
We will focus on property changes of mixing
Δ𝑀 = 𝑀 − ∑ 𝑥𝑖 𝑀𝑖
𝑖
Recall equation for excess properties: 𝑀𝐸 = 𝑀 − 𝑀𝑖𝑑
Excess molar Gibbs Free: 𝐺 𝐸 = 𝐺 − 𝐺 𝑖𝑑 = 𝐺 − ∑𝑖 𝑥𝑖 Gi − 𝑅𝑇 ∑𝑖 𝑥𝑖 ln 𝑥𝑖 = Δ𝐺 − 𝑅𝑇 ∑𝑖 𝑥𝑖 ln 𝑥𝑖
Excess molar entropy: 𝑆 𝐸 = 𝑆 − 𝑆 𝑖𝑑 = 𝑆 − ∑𝑖 𝑥𝑖 Si + 𝑅 ∑𝑖 𝑥𝑖 ln 𝑥𝑖 = Δ𝑆 + 𝑅 ∑𝑖 𝑥𝑖 ln 𝑥𝑖
Excess molar enthalpy: 𝐻 𝐸 = 𝐻 − 𝐻 𝑖𝑑 = 𝐻 − ∑𝑖 𝑥𝑖 Hi = Δ𝐻
Excess molar volume: 𝑉 𝐸 = 𝑉 − 𝑉 𝑖𝑑 = 𝑉 − ∑𝑖 𝑥𝑖 Vi = Δ𝑉
Heat effects of mixing processes
The enthalpy change of mixing is also called heat of mixing. This is analogous to the heat of
reaction though it is relatively smaller in magnitude.
Heats of solution
Heat effect when solids or gases are dissolved in liquids. It is based on 1 mole of solute and has
̃ . The enthalpy change of mixing, Δ𝐻 is the heat effect per mole of solution (mol of
the symbol Δ𝐻
solution = mol of solute + mol of solvent)
𝑚𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
If species 1 is the solute: 𝑥1 = 𝑚𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Therefore, ratio of mol of solvent per mol of solute is: 𝑛̃ =
̃ =
Hence: Δ𝐻
∆𝐻
𝑥1
=
1−𝑥1
1
⟹ 𝑥1 = 1+𝑛̃
𝑥1
∆𝐻
1+𝑛̃
Solution processes are largely physical changes and can be represented as:
For an anhydrous salt: 𝐿𝑖𝐶𝑙(𝑠) + 12𝐻2 𝑂(𝑙) → 𝐿𝑖𝐶𝑙(12𝐻2 𝑂)
For a hydrated salt: 𝐿𝑖𝐶𝑙 ∙ 5𝐻2 𝑂(𝑠) + 7𝐻2 𝑂(𝑙) → 𝐿𝑖𝐶𝑙(12𝐻2 𝑂)
̃ =⋯
Δ𝐻
̃ =⋯
Δ𝐻
Heat of formation of solute in solvent, for example of LiCl in 12 mol of H2O is represented as:
1
𝐿𝑖 + 𝐶𝑙2 + 12𝐻2 𝑂(𝑙) → 𝐿𝑖𝐶𝑙(12𝐻2 𝑂)
2
This is a result of adding the enthalpy change or heat of formation of LiCl and the heat of solution
of 1 mol of LiCl and 12 mol of H2O
Heat of formation of a hydrated salt, for example LiCl·5H2O is the sum of the heat of formation of
LiCl and the heat of formation of the water of hydration and is given by:
1
5
𝐿𝑖 + 𝐶𝑙2 + 5𝐻2 + 𝑂2 → 𝐿𝑖𝐶𝑙 ∙ 5𝐻2 𝑂(𝑠)
2
2
The heats of solution can be obtained by combination of the above reactions or graphically from a
̃ vs 𝑛̃ diagram. Figure 2 shows heats of solution for HCl(g) and LiCl(s).
Δ𝐻
Figure 2
An enthalpy/concentration diagram, H vs x, makes it easier to determine the enthalpy of a solution
with a certain concentration at a particular temperature. Figure 3 shows the variation of enthalpies
of NaOH solutions of different concentrations with temperature.
Figure 3
For adiabatic mixing of two solutions of different concentrations, the final solution lies on a
straight line that joins the two initial solutions.
Topic 3: Vapour/Liquid Equilibria
Phase rule and Duhem theory
Phase rule: 𝐹 = 2 − 𝜋 + 𝑁
where F is the number of intensive variables that can be independently fixed
π is the number of phases in the system
N is the number of chemical species or components in the system
Duhem theory: for any closed system formed initially from given masses of prescribed chemical
species, the equilibrium state is completely determined when any two independent variables are
fixed. They may be intensive or extensive.
Phase diagrams
-
P vs x,y phase diagram at constant T
-
T vs x,y phase diagram at constant P
You should know the different regions, curves and important points.
•
Dew point curve is the P or T vs y curve and represents all saturated vapour mixtures
(vapour at boiling point).
•
Bubble point curve is the P or T vs x curve and represents all saturated liquid mixtures
(liquid at boiling point).
•
Mixtures boil over a range of temperatures hence the bubble point curve is different from
the dew point curve. Between the bubble point curve and the dew point curve exist two
phases, saturated vapour and saturated liquid that are at equilibrium. The equilibrium
compositions are determined by tie line (horizontal lines drawn at a certain temperature or
pressure). The overall composition of the mixture does not change in this region and can
be determined using the lever rule. For a binary mixture, both curves begin and terminate
at saturation points of the pure species.
•
Superheated vapour mixtures exist at temperatures above dew point temperature of the
mixture or at pressures below the dew point pressure of the mixture.
•
Subcooled liquid mixtures exist at temperatures below bubble point temperature of the
mixture or at pressures above the bubble point pressure of the mixture.
Simple models for Vapour/Liquid Equilibria (VLE)
-
Raoult’s law
-
Henry’s law
-
Modified Raoult’s law
Raoult’s law
-
Applicable at low to moderate pressures
-
The major assumptions are: 1) the vapour phase is an ideal gas
2) the liquid phase is an ideal solution. An ideal solution is formed by species that are
chemically similar (in terms of magnitude and type of intermolecular forces). Therefore,
the species must not be too different in size and must be of the same chemical nature such
as adjacent members of a homologous series
-
Applies to species of known vapour pressure which requires the species to be subcritical,
that is at a temperature below critical temperature, Tc
The equation for Raoult’s law is: 𝑃𝑖 = 𝑦𝑖 𝑃 = 𝑥𝑖 𝑃𝑖 𝑠𝑎𝑡
𝑃𝑖 𝑠𝑎𝑡 is the vapour pressure of pure species i at the system temperature
This shows that the graph of 𝑃𝑖 vs 𝑥𝑖 is a straight line, where 𝑃𝑖 = 0 when 𝑥𝑖 = 0 and 𝑃𝑖 = 𝑃𝑖 𝑠𝑎𝑡
when 𝑥𝑖 = 1.
For a binary system: when 𝑥1 = 0, 𝑥2 = 1, 𝑃1 = 0, 𝑃2 = 𝑃2 𝑠𝑎𝑡 and 𝑃 = 𝑃2 𝑠𝑎𝑡
When 𝑥1 = 1, 𝑥2 = 0, 𝑃2 = 0, 𝑃1 = 𝑃1 𝑠𝑎𝑡 and 𝑃 = 𝑃1 𝑠𝑎𝑡
Therefore the graph of P vs x1: 𝑃 = 𝑥1 𝑃1 𝑠𝑎𝑡 + 𝑥2 𝑃2 𝑠𝑎𝑡 = 𝑃2 𝑠𝑎𝑡 + 𝑥1 (𝑃1 𝑠𝑎𝑡 − 𝑃2 𝑠𝑎𝑡 ) is a straight
line since 𝑃1 𝑠𝑎𝑡 and 𝑃2 𝑠𝑎𝑡 are constants at a particular temperature.
Deviations from Raoult’s law
There are two types of deviations from Raoult’s law:
-
Negative deviation
-
Positive deviation
Negative deviation
When the bubble point pressure curve lies below the Raoult’s law line. This means Pactual < Pideal.
If the deviations are sufficiently large for a binary mixture of liquids 1 and 2, the 1—2
intermolecular forces will be stronger than those of both pure liquids 1—1 and 2—2 and a
minimum point is formed on the bubble point pressure curve. The dew point pressure curve also
exhibits a minimum at the same point such that 𝑥1 = 𝑦1 . A boiling liquid (saturated liquid) of this
composition produces a saturated vapour of exactly the same composition, behaving like a pure
liquid hence no separation by distillation is possible for this constant boiling solution. This solution
is called an azeotrope and for a solution that shows negative deviation from Raoult’s law, it is
called a minimum pressure azeotrope or a maximum boiling azeotrope.
Positive deviation
When the bubble point pressure curve lies above the Raoult’s law line. This means Pactual > Pideal.
If the deviations are sufficiently large for a binary mixture of liquids 1 and 2, the 1—2
intermolecular forces will be weaker than those of both pure liquids 1—1 and 2—2 and a maximum
point is formed on the bubble point pressure curve. The dew point pressure curve also exhibits a
maximum at the same point such that 𝑥1 = 𝑦1. This solution is called a maximum pressure
azeotrope or a minimum boiling azeotrope. If 1—1 and 2—2 are so strong such that complete
miscibility of the liquids is prevented, the system forms two separate liquid phases over a range of
compositions, which is the basis of liquid-liquid equilibrium.
Solving VLE problems using Raoult’s law
𝐵
𝑖
The Antoine equation gives the variation of 𝑃𝑖 𝑠𝑎𝑡 with T: ln 𝑃𝑖 𝑠𝑎𝑡 (𝑘𝑃𝑎) = 𝐴𝑖 − 𝑇(℃)+𝐶
𝑖
There are four classes of problems:
1) BUBL P: Calculate yi and P given xi and T
2) DEW P: Calculate xi and P given yi and T
3) BUBL T: Calculate yi and T given xi and P
4) DEW T: Calculate xi and T given yi and P
Solving 1st class of problems
Step 1: At the given T, calculate Pisat for each species
Step 2: Since xi is also given, P is calculated as follows:
𝑦𝑖 𝑃 = 𝑥𝑖 𝑃𝑖 𝑠𝑎𝑡
∑𝑖 𝑦𝑖 𝑃 = 𝑃 ∑𝑖 𝑦𝑖 = ∑𝑖 𝑥𝑖 𝑃𝑖 𝑠𝑎𝑡
Summing both sides over all i:
But ∑𝑖 𝑦𝑖 = 1, therefore: 𝑃 = ∑𝑖 𝑥𝑖 𝑃𝑖 𝑠𝑎𝑡
(45)
Step 3: Calculate yi of the required species using Raoult’s law
Solving 2nd class of problems
Step 1: At the given T, calculate Pisat for each species
Step 2: Since yi is also given, P is calculated as follows:
𝑦𝑖 𝑃 = 𝑥𝑖 𝑃𝑖 𝑠𝑎𝑡
𝑦𝑃
⟹ 𝑥𝑖 = 𝑃 𝑖𝑠𝑎𝑡
𝑖
𝑦𝑃
But ∑𝑖 𝑥𝑖 = 1, therefore: 𝑃 =
𝑦
𝑖
𝑖
∑𝑖 𝑥𝑖 = ∑𝑖 𝑠𝑎𝑡
= 𝑃 ∑𝑖 𝑃 𝑠𝑎𝑡
𝑃
Summing both sides over all i:
𝑖
𝑖
1
(46)
𝑦
𝑖
∑𝑖 𝑠𝑎𝑡
𝑃
𝑖
Step 3: calculate xi of the required species using Raoult’s law
Solving 3rd class of problems
Making T the subject of the formula in the Antoine equation: 𝑇𝑖 𝑠𝑎𝑡 (℃) =
For a binary system, Equation 45 can be written as: 𝑃2 𝑠𝑎𝑡 =
where 𝛼 =
𝑃1 𝑠𝑎𝑡
𝑃2
𝑠𝑎𝑡
𝐵𝑖
𝐴𝑖 −𝑙𝑛 𝑃 (𝑘𝑃𝑎)
− 𝐶𝑖
𝑃
(47)
(48)
𝑥1 𝛼+𝑥2
𝑦 ⁄𝑥
which is the relative volatility using Raoult’s law, that is, 𝛼12 = 𝑦1⁄𝑥1
2
2
𝐵
𝐵
1
2
⟹ 𝑙𝑛 𝛼 = 𝑙𝑛 𝑃1 𝑠𝑎𝑡 − 𝑙𝑛 𝑃2 𝑠𝑎𝑡 = 𝐴1 − 𝑇+𝐶
− (𝐴2 − 𝑇+𝐶
)
1
2
(49)
The iterative steps for T are as follows:
Step 1: Calculate T1sat and T2sat using Equation 47
Step 2: Select any temperature between T1sat and T2sat and calculate P1sat and P2sat using the Antoine
equation at that temperature
Step 3: Calculate α0 using the P1sat and P2sat determined in Step 2
Step 4: Calculate P2sat using α and Equation 48
Step 5: Calculate T2sat using P2sat as P in Equation 47
Step 6: Calculate new α using T2sat as T in Equation 49
Step 7: Repeat steps 4 to 6 until T2sat converges. This is the required temperature, T.
Step 8: Calculate P1sat using T in step 7 and the Antoine equation
Step 9: Calculate y1 using Raoult’s law
Solving 4th class of problems
For a binary system, Equation 46 can be written as: 𝑃1 𝑠𝑎𝑡 = 𝑃(𝑦1 + 𝑦2 𝛼)
where 𝛼 =
(50)
𝑃1 𝑠𝑎𝑡
𝑃2 𝑠𝑎𝑡
The iterative steps for T are as follows:
Step 1: Calculate T1sat and T2sat using Equation 47
Step 2: Select any temperature between T1sat and T2sat and calculate P1sat and P2sat using the Antoine
equation at that temperature
Step 3: Calculate α0 using the P1sat and P2sat determined in Step 2
Step 4: Calculate P1sat using α and Equation 50
Step 5: Calculate T1sat using P1sat as P in Equation 47
Step 6: Calculate new α using T1sat as T in Equation 49
Step 7: Repeat steps 4 to 6 until T1sat converges. This is the required temperature, T.
Step 8: Calculate P1sat using T in step 7 and the Antoine equation
Step 9: Calculate x1 using Raoult’s law
Henry’s law
-
Applies to species whose Tc is less than operating temperature (these species do not have
vapour pressure at the operating temperature).
-
Pressure is low enough such that the vapour phase is assumed to be an ideal gas
-
The species is also present as a very dilute solute in the liquid or approaches infinite dilution
Henry’s law is given by: 𝑦𝑖 𝑃 = 𝑥𝑖 𝐻𝑖
Hi is the Henry’s law constant of species i. The larger the Henry’s law constant of the species, the
less soluble it is in the liquid phase.
Modified Raoult’s law
The liquid phase is not an ideal solution. The law is given by: 𝑦𝑖 𝑃 = 𝑥𝑖 𝛾𝑖 𝑃𝑖 𝑠𝑎𝑡
The activity coefficient, 𝛾𝑖 can be modelled by various correlations from Solution
Thermodynamics Applications as a function of composition and one of them is:
ln 𝛾1 = 𝐴𝑥2 2 and
ln 𝛾2 = 𝐴𝑥1 2 where 𝐴 = 𝐴(𝑇)
We will focus on three classes of problems:
1) BUBL P: Calculate yi and P given xi and T
2) DEW P: Calculate xi and P given yi and T
3) Azeotropic composition and pressure at a given T
Solving 1st class of problems
Step 1: At the given T, calculate Pisat for each species and A if A is given as a function of T.
Step 2: At the given xi, calculate 𝛾𝑖 for each species
Step 3: Since xi is given, P is calculated as follows:
𝑦𝑖 𝑃 = 𝑥𝑖 𝛾𝑖 𝑃𝑖 𝑠𝑎𝑡
∑𝑖 𝑦𝑖 𝑃 = 𝑃 ∑𝑖 𝑦𝑖 = ∑𝑖 𝑥𝑖 𝛾𝑖 𝑃𝑖 𝑠𝑎𝑡
Summing both sides over all i:
But ∑𝑖 𝑦𝑖 = 1, therefore: 𝑃 = ∑𝑖 𝑥𝑖 𝛾𝑖 𝑃𝑖 𝑠𝑎𝑡
(51)
Step 4: Calculate yi of the required species using the modified Raoult’s law
Solving 2nd class of problems
Step 1: At the given T, calculate Pisat for each species and A when A is given as a function of T.
Step 2: Since xi is not given, it is not possible to calculate the activity coefficient of each species.
An iterative procedure is required where the initial values of each activity coefficient are
determined by assuming an ideal solution, that is each 𝛾𝑖 = 1
Step 3: Since yi is also given, P is calculated as follows:
𝑦𝑃
𝑦𝑖 𝑃 = 𝑥𝑖 𝛾𝑖 𝑃𝑖 𝑠𝑎𝑡
⟹ 𝑥𝑖 = 𝛾 𝑃𝑖 𝑠𝑎𝑡
𝑖 𝑖
𝑦𝑃
But ∑𝑖 𝑥𝑖 = 1, therefore: 𝑃 =
𝑦
∑𝑖 𝑥𝑖 = ∑𝑖 𝑖 𝑠𝑎𝑡 = 𝑃 ∑𝑖 𝑖𝑠𝑎𝑡
𝛾𝑃
𝛾𝑃
Summing both sides over all i:
𝑖 𝑖
𝑖 𝑖
1
∑𝑖
𝑦𝑖
𝛾𝑖 𝑃𝑖 𝑠𝑎𝑡
Step 4: Calculate xi of each species using the modified Raoult’s law
Step 5: Calculate the new activity coefficient of each species using the given correlation.
Step 6: Repeat steps 3 to 5 until P converges. From this value of P, determine the required xi
(52)
Solving 3rd class of problems
𝑦 ⁄𝑥
An azeotrope exists when 𝑥𝑖 = 𝑦𝑖 . This means: 𝛼12 = 𝑦1⁄𝑥1 = 1
2
2
𝛾 𝑃 𝑠𝑎𝑡
From modified Raoult’s law: 𝛼12 = 𝛾1 𝑃1𝑠𝑎𝑡
2 2
First of all, a check must be done on whether an azeotrope exists. 𝛼12 is a continuous function of
x1 therefore an azeotrope exists when for 0 ≤ 𝑥1 ≤ 1, 𝛼12 at one limit is less than 1 and 𝛼12 at the
other limit is greater than 1. This ensure that the graph of 𝛼12 vs x1 definitely passes through 1.
Otherwise, an azeotrope does not exist.
𝛾 𝑃 𝑠𝑎𝑡
Since for an azeotrope 𝛾1 𝑃1𝑠𝑎𝑡 = 1, each activity coefficient is given by a correlation as a function
2 2
of composition and each 𝑃𝑖 𝑠𝑎𝑡 can be determined from the Antoine equation at the given
temperature, the azeotropic composition can be calculated.
The azeotropic pressure is then calculated using the modified Raoult’s law at the azeotropic
composition.
Topic 4: Phase equilibria
Our focus will be on osmotic equilibrium and osmotic pressure
The equilibrium criterion is always: (𝑑𝐺 𝑡 ) 𝑇,𝑃 = 0
𝑑𝑓̂
For stability: (𝑑𝐺 𝑡 ) 𝑇,𝑃 < 0. This leads to another stability criterion which is: 𝑑𝑥𝑖 > 0 at constant
𝑖
T and P.
Figure 4
Species 1 is the solute and species 2 is the solvent. The membrane is permeable to species 2
(solvent) only.
If 𝑃′ = 𝑃, the stability criterion applied to species 2 is applicable:
⟹ 𝑓̂2 (𝑇, 𝑃′ = 𝑃, 𝑥2 < 1) < 𝑓̂2 (𝑇, 𝑃, 𝑥2 = 1)
𝑑𝑓̂2
𝑑𝑥2
>0
but for the RHS, since 𝑥2 = 1, 𝑓̂2 = 𝑓2
⟹ 𝑓̂2 (𝑇, 𝑃′ = 𝑃, 𝑥2 < 1) < 𝑓2 (𝑇, 𝑃)
This difference in fugacities represents a driving force for mass transfer, where the solvent diffuses
from right to left through the semi-permeable membrane in Figure 4. This is called osmosis. This
mass transfer increases pressure on the left side until equilibrium is established at a new pressure
P*.
Therefore: 𝑓̂2 (𝑇, 𝑃′ = 𝑃∗ , 𝑥2 < 1) = 𝑓2 (𝑇, 𝑃)
(equilibrium criterion)
The pressure difference 𝑃∗ − 𝑃 is called osmotic pressure of the solution with symbol Π.
Since T is constant: 𝑓̂2 (𝑃 + Π, 𝑥2 ) = 𝑓2 (𝑃)
Taking the LHS: 𝑓̂2 (𝑃 + Π, 𝑥2 ) = 𝑓2 (𝑃) ∙
Looking at the first ratio:
𝑓̂2 (𝑃,𝑥2 )
𝑓2 (𝑃)
Looking at the second ratio:
𝑓̂2 (𝑃,𝑥2 ) 𝑓̂2 (𝑃+Π,𝑥2 )
∙ 𝑓̂ (𝑃,𝑥 )
𝑓2 (𝑃)
2
2
= 𝑥2 𝛾2
𝑓̂2 (𝑃+Π,𝑥2 )
𝑓̂2 (𝑃,𝑥2 )
(from the definition of 𝛾𝑖 )
is called the Poynting factor
Recall: 𝜇𝑖 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑓̂𝑖
(29)
̂
𝜕𝜇
Differentiating with respect to P: ( 𝜕𝑃𝑖 )
𝑇,𝑥
𝜕 ln 𝑓
= 𝑉̅𝑖 = 𝑅𝑇 ( 𝜕𝑃 𝑖 )
𝑇,𝑥
𝑉̅2
𝜕 ln 𝑓̂2
=
𝑅𝑇
𝜕𝑃
⟹
Separating variables and integrating from 𝑃 to 𝑃 + Π
𝑃+Π,𝑥2
⟹ |ln 𝑓̂2 |𝑃,𝑥
= ∫
2
𝑃+Π
𝑃
𝑉̅2
𝑑𝑃
𝑅𝑇
𝑃+Π ̅
𝑓̂2 (𝑃 + Π, 𝑥2 )
𝑉2
⟹
= 𝑒𝑥𝑝 ∫
𝑑𝑃
̂
𝑅𝑇
𝑓2 (𝑃, 𝑥2 )
𝑃
But 𝑓̂2 (𝑃, 𝑥2 ) = 𝑥2 𝛾2 𝑓2 (𝑃) from first ratio and 𝑓̂2 (𝑃 + Π, 𝑥2 ) = 𝑓2 (𝑃) from equilibrium
criterion.
̅2
𝑃+Π 𝑉
Substituting them results in: 𝑒𝑥𝑝 ∫𝑃
𝑅𝑇
𝑑𝑃 =
̅2
𝑃+Π 𝑉
Taking natural logarithm both sides: ∫𝑃
𝑅𝑇
1
𝑥2 𝛾2
𝑑𝑃 = − ln(𝑥2 𝛾2 )
Looking at the LHS: If 𝑉̅2 is a weak function of P (common in liquids), it means that:
̅2
𝑃+Π 𝑉
∫𝑃
⟹
𝑅𝑇
̅2
Π𝑉
𝑅𝑇
𝑑𝑃 =
̅2
𝑉
𝑅𝑇
(𝑃 + Π − P) =
= − ln(𝑥2 𝛾2 )
̅2
Π𝑉
𝑅𝑇
𝑅𝑇
⟹ Π = − 𝑉̅ ln(𝑥2 𝛾2 )
2
If the solution is very dilute in the solute (species 1): 𝑥1 → 0, 𝑥2 → 1, 𝑉̅2 ≈ 𝑉2 and 𝛾2 ≈ 1
⟹ ln(𝑥2 𝛾2 ) = ln(𝑥2 ) = ln(1 − 𝑥1 )
Applying Maclaurin series and neglecting x12 and higher powers of x1: ln(1 − 𝑥1 ) = −𝑥1
Therefore: 𝚷 =
𝒙𝟏 𝑹𝑻
𝑽𝟐
this is called the van’t Hoff equation.
The van’t Hoff equation applies where species 1 is a non-electrolyte.
If 𝑃′ > 𝑃 + Π then: 𝑓̂2 (𝑃′ , 𝑥2 ) > 𝑓2 (𝑃)
This difference in fugacities is a driving force where solvent molecules are transferred from left to
right in Figure 4. This process is called reverse osmosis, which is a pressure-driven process.
Topic 5: Chemical reaction equilibria
A chemical reaction is represented as: |𝑣1 |𝐴1 + |𝑣2 |𝐴2 + ⋯ → |𝑣3 |𝐴3 + |𝑣4 |𝐴4 + ⋯
Where |𝑣𝑖 | is the stoichiometric coefficient and 𝐴𝑖 is the chemical formula.
𝑣𝑖 itself is called the stoichiometric number and is positive for products and negative for reactants.
For an inert species (a species that does not take part in the reaction), it is zero.
As the chemical reaction progresses, the reactants are consumed whilst products are formed. The
changes in the number of moles (positive for products and negative for reactants) of each species
are proportional according to the stoichiometry of the reaction.
⟹
⟹ 𝑑𝑛𝑖 = 𝑣𝑖 𝑑𝜀
𝑑𝑛1
𝑑𝑛2
𝑑𝑛3
𝑑𝑛4
𝑑𝑛𝑖
=
=
=
=
= 𝑑𝜀
𝑣1
𝑣2
𝑣3
𝑣4
𝑣𝑖
where 𝜀 is called the reaction coordinate which characterizes the extent of
reaction or degree of completion
𝑛
𝜀
Integrating both sides: ∫𝑛 𝑖 𝑑𝑛𝑖 = 𝑣𝑖 ∫0 𝑑𝜀
⟹ 𝑛𝑖 = 𝑛𝑖0 + 𝑣𝑖 𝜀
𝑖0
Let 𝑛 = ∑𝑖 𝑛𝑖 , 𝑛0 = ∑𝑖 𝑛𝑖0 , 𝑣 = ∑𝑖 𝑣𝑖 :
Therefore: 𝒚𝒊 =
𝒏𝒊
𝒏
=
⟹ 𝑛 = 𝑛0 + 𝑣𝜀
𝒏𝒊𝟎 +𝒗𝒊 𝜺
𝒏𝟎 +𝒗𝜺
Equilibrium of chemical reactions
The equilibrium criterion: (𝑑𝐺 𝑡 ) 𝑇,𝑃 = 0
Recall: 𝑑(𝑛𝐺) = (𝑛𝑉)𝑑𝑃 − (𝑛𝑆)𝑑𝑇 + ∑𝑖 𝜇𝑖 𝑑𝑛𝑖
But 𝑑𝑛𝑖 = 𝑣𝑖 𝑑𝜀, therefore 𝑑(𝑛𝐺) = (𝑛𝑉)𝑑𝑃 − (𝑛𝑆)𝑑𝑇 + ∑𝑖 𝑣𝑖 𝜇𝑖 𝑑𝜀
Since 𝑛𝐺 or 𝐺 𝑡 is a state function, the RHS is an exact differential expression, meaning:
𝜕(𝑛𝐺)
𝜕𝐺 𝑡
∑ 𝑣𝑖 𝜇𝑖 = (
) = (
)
𝜕𝜀 𝑇,𝑃
𝜕𝜀 𝑇,𝑃
𝑖
(6)
𝜕𝐺 𝑡
At equilibrium: ( 𝜕𝜀 )
𝑇,𝑃
= ∑𝑖 𝑣𝑖 𝜇𝑖 = 0
Recall: 𝜇𝑖 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑓̂𝑖
(29)
For pure species i in its standard state (usually P = 1 bar or 1 atm) at the same temperature, Equation
27 can be written as: 𝐺𝑖 𝑜 = 𝛤𝑖 (𝑇) + 𝑅𝑇 𝑙𝑛 𝑓𝑖 𝑜
(53)
𝑓̂
Subtracting Equation 53 from Equation 29: 𝜇𝑖 = 𝐺𝑖 𝑜 + 𝑅𝑇 ln (𝑓 𝑖𝑜 )
𝑖
𝑓̂𝑖
⟹ ∑ 𝑣𝑖 [𝐺𝑖 𝑜 + 𝑅𝑇 ln ( 𝑜 ) ] = 0
𝑓𝑖
𝑖
𝑣𝑖
𝑓̂𝑖
⟹ ∑ 𝑣𝑖 𝐺𝑖 + 𝑅𝑇 ∑ ln ( 𝑜 ) = 0
𝑓𝑖
𝑜
𝑖
𝑖
𝒗𝒊
∑𝒊 𝒗𝒊 𝑮𝒊 𝒐
𝒇̂𝒊
⟹ 𝐥𝐧 ∏ ( 𝒐 ) = −
𝑹𝑻
𝒇𝒊
𝒊
𝑓̂
𝑣𝑖
But ∏𝑖 (𝑓 𝑖𝑜 ) = 𝐾 which is the equilibrium constant at constant pressure and ∑𝑖 𝑣𝑖 𝐺𝑖 𝑜 = ∆𝐺 𝑜
𝑖
which is the standard Gibbs Free Energy change of reaction
∆𝑮𝒐
⟹ 𝐥𝐧 𝑲 = −
𝑹𝑻
Since pressure is fixed at standard state conditions, ∆𝐺 𝑜 is a function of T only.
𝑓̂
Fugacity ratios 𝑓 𝑖𝑜 give the connection between the equilibrium state of interest and standard state
𝑖
of each species.
The general formula for standard property changes of reaction is: ∆𝑀𝑜 = ∑𝑖 𝑣𝑖 𝑀𝑖 𝑜 which are also
functions of T only
𝜕(𝐺 ⁄𝑅𝑇 )
From Gibbs Free Energy as a generating function: (
𝜕𝑇
𝐻
) = − 𝑅𝑇 2
𝑃
Applying this: 𝐻𝑖 𝑜 = −𝑅𝑇 2
𝑑(𝐺𝑖 𝑜 ⁄𝑅𝑇)
𝑑𝑇
∑𝑖 𝑣𝑖 𝐻𝑖 𝑜 = −𝑅𝑇 2
Multiplying by 𝑣𝑖 and summing over all i:
⟹
𝑑(∆𝐺 𝑜 ⁄𝑅𝑇 )
𝑑𝑇
∆𝐻 𝑜
= − 𝑅𝑇 2
⟹
𝒅 𝐥𝐧 𝑲
𝒅𝑻
=
𝑑(∑𝑖 𝑣𝑖 𝐺𝑖 𝑜 ⁄𝑅𝑇 )
𝑑𝑇
∆𝑯𝒐
𝑹𝑻𝟐
This shows that for an exothermic reaction (∆𝐻 𝑜 is negative), K decreases as T is increased. The
converse is true for an endothermic reaction (∆𝐻 𝑜 is positive)
If ∆𝐻 𝑜 is assumed to be independent of temperature, integrating both sides yields:
𝐾
∫𝐾′ 𝑑 ln 𝐾 =
∆𝐻 𝑜
𝑅
𝑇 1
𝐾
∫𝑇 ′ 𝑇 2 𝑑𝑇
⟹ ln (𝐾′ ) = −
1
∆𝐻 𝑜 1
𝑅
1
(𝑇 − 𝑇 ′ )
Therefore a plot of ln 𝐾 vs 𝑇 gives a straight line as shown in Figure 5.
Figure 5
Relation of equilibrium constant to composition: Gas-phase reactions
𝑓̂
Recall that: 𝐾 = ∏𝑖 (𝑓 𝑖𝑜 )
𝑣𝑖
𝑖
The standard state of a gas is the ideal gas state of the pure gas at the standard-state pressure (𝑃𝑜 )
of 1 bar or 1 atm.
But 𝑓𝑖 𝑖𝑔 = 𝑃, therefore this implies that 𝑓𝑖 𝑜 = 𝑃𝑜
𝑣𝑖
𝑓̂𝑖
⟹ 𝐾 = ∏ ( 𝑜)
𝑃
𝑖
𝑣𝑖
̂
𝜙𝑦𝑃
But 𝑓̂𝑖 = 𝜙̂𝑖 𝑦𝑖 𝑃. This implies that: 𝐾 = ∏𝑖 ( 𝑖𝑃𝑜𝑖 )
𝑃
𝑣𝑖
𝑃
𝑣
But ∏𝑖 (𝑃𝑜 ) = (𝑃𝑜 ) where 𝑣 = ∑𝑖 𝑣𝑖
𝑃 𝑣
𝑣𝑖
⟹ 𝐾 = ( 𝑜 ) ∏(𝜙̂𝑖 𝑦𝑖 )
𝑃
𝑖
𝑃 −𝑣
𝑣𝑖
̂
⟹ ∏(𝜙𝑖 𝑦𝑖 ) = 𝐾 ( 𝑜 )
𝑃
𝑖
If the equilibrium mixture is assumed to be an ideal solution of gases: 𝜙̂𝑖
⟹ ∏(𝝓𝒊 𝒚𝒊
)𝒗𝒊
𝒊
⟹ ∏(𝒚𝒊
𝒊
= 𝜙𝑖
(42)
𝑷 −𝒗
= 𝑲 ( 𝒐)
𝑷
If the equilibrium mixture is assumed to be an ideal gas: 𝜙̂𝑖
)𝒗𝒊
𝑖𝑑
𝑖𝑔
= 1
𝑷 −𝒗
= 𝑲 ( 𝒐)
𝑷
This shows that whilst K is not a function of P, ∏𝑖 (𝑦𝑖 )𝑣𝑖 is a function of P.
-
An increase in K at constant pressure increases the product ∏𝑖(𝑦𝑖 )𝑣𝑖 which implies a shift
of equilibrium to the right (more products are formed)
-
If 𝑣 is negative, an increase in P (equilibrium mixture pressure) at constant temperature
increases the product ∏𝑖 (𝑦𝑖 )𝑣𝑖 but has no effect on K. If 𝑣 = 0, P has no effect on the
product ∏𝑖 (𝑦𝑖 )𝑣𝑖
-
Addition of inert gases does not affect K (since 𝑣𝑖𝑛𝑒𝑟𝑡 = 0). But it does affect the
equilibrium composition. Since 𝑦𝑖 =
𝑛𝑖
𝑛
, adding inerts increase n.
𝑛𝑖 𝑣𝑖
𝑃 −𝑣
⟹ ∏ ( ) = 𝐾 ( 𝑜)
𝑛
𝑃
𝑖
1 𝑣𝑖
1 𝑣
𝑃
−𝑣
But ∏𝑖 (𝑛) = (𝑛) . This therefore means: ∏𝑖 (𝑛𝑖 )𝑣𝑖 = 𝐾 (𝑛𝑃𝑜 )
•
If 𝑣 is negative, an increase in n at constant temperature and pressure decreases the
product ∏𝑖(𝑛𝑖 )𝑣𝑖 which implies a shift of equilibrium to the left (more reactants are
formed). If 𝑣 = 0, addition of inerts has no effect on the product ∏𝑖(𝑛𝑖 )𝑣𝑖
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The factor by which the reaction is multiplied is the power by which K is raised.
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The larger the K, the more complete the forward reaction
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If the chemical reaction is reversed, new K is the reciprocal of the old K
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If the overall reaction is a sum of different individual reactions, K for the overall reaction
is equal to the product of the K of each individual reaction