1. Lecture 4 Chemical potential and phase equilibrium

2. What are the mole amounts of n1 , n2 ,… and nm at equilibrium?
Chemical system
Chemical system
T constant temperature [K]
p constant total pressure [Pa]
ni the number of moles of each species (compound or element)
in the system [mol]
xi mole fractions of each species (compound or element)
in the system [-]
x1 + x2 +… xm = 1
T, p
n1, n2,... nm
(x1, x2....xm)
GOAL
What are the mole amounts of n1 , n2 ,… and nm at equilibrium?

3. The first law of thermodynamics
Signs for work and heat
W = Wout – Win
Q = Qin - Qout
-Win
Qin
Energy balance
U(B) - U(A) = Q – W or
U(B) - U(A) = (Qin – Qout) + (Win – Wout)
Work done by the system can be divided into two parts
Expansion work Wex = p[V(B) - V(A)]
Technical work Wtec
Total work W = Wtec + Wex
Closed system
UA(p,v,T)
UB(p,v,T)
Wout
-Qout
W work [J]
Q heat [J]
U internal energy [J]
For a heat engine that undergoues a cycle process based on Carnot’s idea about
the heat engine: Wout = Qin – Qout and Win = Qout – Qin , where
Wout and Win represent the technical work

4. Condition for spontaneous changes and thermodynamic equilibrium at constant T and p
The change of U from the initial state A to the final state B at constant p and T
U(B) - U(A) = Q – W = (Qin – Qout) – (Wout – Win)
Work done by the system is only the expansion work (Wtec = 0) =>
W = p[V(B) - V(A)] => Q = [(U(B) + pV(B)] - [U(A) + pV(A)]
According to the second law
T is constant =>
Q  T [S(B) - S(A)]
=>
[U(B) + pV(B)] - [U(A) + pV(A)]  T[S(B) - S(A)] => [U(B) + pV(B)] - TS(B)  [U(A) + pV(A)] - TS(A)]
where G = U + pV – TS = H-TS Gibbs Energy =>
G(B)  G(A)
Note that the temperature and pressure
of the system are constant

5. What does the result G(B)  G(A) mean?
For spontaneous changes at constant pressure and temperature, where the only form of work is expansion work, the change of Gibbs Energy is always negative => G = G(B) – G(A)  0
For chemical reactions, the condition G = G(B) – G(A)  0 means that spontaneous chemical reactions at constant T and p are possible only when the Gibbs Energy decreases.
If G(B) = G(A) no changes occur in the system which means that the system is at
thermodynamic equilibrium.
If G(B) = G(A) the Gibbs Energy of the system has reached its minimum and the general
condition for thermodynamic equilibrium is Gsystem = min
The equilibrium theory does not pay any attention to how much time it takes to reach
thermodynamic equilibrium (seconds, hours, years?). This depends on the system and
factors like reaction kinetics.

6. Chemical equilibrium of the system
Chemical system
T constant temperature [K]
p constant total pressure [Pa]
ni the number of moles of each species (compound or element)
in the system [mol]
xi mole fractions of each species (compound or element)
in the system [-]
x1 + x2 +… xm = 1
T, p
n1, n2,... nm
(x1, x2....xm)
At equilibrium G(T,p,n1,n2...nm) = min

7. The Gibbs energy of formation for compounds
The Gibbs energy of formation for compounds at the standard state is defined as follows
Gf°=Hf°-TSf°
For elements at the reference state (the most stable state)
Hf° = Sf° = 0 => Gf°= 0
What does the Gibbs energy of formation for compounds mean?
If Gf°  0, its value describes the minimum work needed to “build” the compound
from its elements at the standard state.
If Gf°  0, its absolute value describes the maximum work that is released when
the compound is built from its elements at the standard state.
The Gibbs free energy of formation for compounds is analogous the enthalpy and
entropy of formation for compounds.

8. Calculating Gf° What is Gf° for CO2 at 700K? C(s) + O2(g)  CO2(g)
From Janaf tables
Hf°(700K) = Hf°(CO2) - Hf°(O2) - Hf°(C) = {0 + 0} = kJ/mol
Sf°(700K) = so(CO2) - so(C) - so(O2) = = J/(molK)
Gf°(700K) = Hf°(CO2) - TSf°(O2) = 2.031 = kJ/mol
You find the same value in JANAF Tables.
Note that Gf° > Hf°. This means that it might be possible to get more work than heat out of the system when the oxidation reaction occurs isothermally. Fuel cells are based on this idea and their operation principles are not based on Carnot’s idea about an ideal heat engine. We will return to this topic later.

9. Gf° calculated in Janaf tables

10. Chemical potential of a species
Chemical system
T constant temperature [K]
p constant total pressure [Pa]
n the number of moles of each species (compound or element)
in the system [mol]
x mole fractions of each species (compound or element)
in the system [-]
T, p
n1, n2,... nm
(x1, x2....xm)
The Gibbs Energy for the system is
G = G(T,p,n1,…nm) = H(T,p,n1…nm) -TS(T,p,n1,…nm)
The chemical potential i (J/mol) is defined as follows
=> i = hi – Tsi
=>
Like the enthalpy and entropy
the chemical potential is also a total differential

11. The change of the chemical potential of a species with respect to T and p
The change of the chemical potential with respect to T (p is constant)
, molar entropy of a species [J/(molK)]
The change of the chemical potential with respect to p (T is constant)
, molar volume of a species [m3/mol]
i = hi –Tsi
Derivate of a product d(fg) = f’g + fg’

12. Calculating the chemical potential  for a pure substance from a given reference point T0 and p0 (no phase transition)
The value of  from the intial state (T0, p0) to the final state (T, p)
= h –Ts
For the change of enthalpy may be written
For the change of entropy may be written
= Volumetric thermal expansion coefficient

13. Calculating the chemical potential  for a pure substance from a given reference point T0 and p0
When p0 = po (= pressure at the standard state, 1bar) =>
where
h (T0, po) = Hfo at 25oC and 1bar
s (T0, po) is usually the absolute entropy at 25oC and 1bar (sometimes the entropy
of formation at 25oC and 1bar may also be used is some sources)

14. Calculating the chemical potential  for a pure substance from a given reference point T0 and p0
For a pure substance the chemical potential at T and p may now be written
Using the following notation
The chemical potential for a pure substance at T and p becomes
o (T) is the chemical potential of a pure substance at the standard state,
where p is usually 1bar.

15. Chemical potential for an ideal gas, solids and liquids at T and p
v =RT/p =>
For pure solids and liquids v(T,p)  v(T) =>
o(T) represents the chemical potential at the standard state. po is usually 1bar.

16. Chemical potential at the standard state for a pure substance in JANAF Tables
At the standard state ( p = 1bar)
The equation above may also be written as follows
Values for the enthalpy and entropy are found in JANAF Tables
Hfo is usually defined at 25oC in practical calculations
and so(T) is the absolute entropy.

17. Thermodynamic values In JANAF Tables
Cp specific heat capacity [J/molK]
So absolute entropy [J/molK]
- [Go – Ho (Tr)]/T = gef [J/molK]
Ho – Ho(Tr) the change of sensible
enthalpy [kJ/mol]
fHo the enthalpy of formation at the
temperature of T [kJ/mol]
fGo = fHo -T fSo = Gibbs energy of
formation at the temperature of T [kJ/mol]
Kf Equilibrium constant
- The reference/zero point for the
enthalpy is 25oC ( = Tr)
The pressure is 1 bar (the standard state)

18. Example
Calculate the chemical potential o(T) for CH4 at the standard state at the temperature
of 1500K . Use tabulated values.
=> °(1500K) = · · · = - 416·103 J/mol
What is the chemical potential of CH4 at 1500 K and 15bar?
o(T) = ho- Tso

19. How to calculate o(T) using gef?
Note that the gef has been tabulated as –gef in JANAF tables. This means that you must multiple the gef value in Janaf tables by -1 to get the correct chemical potential from the equation above.
Solution of the previous example when gef is used
= · ·( ) = - 416·103 J/mol
Definition of the gef

20. Phase equilibrium between pure liquid and vapor
At equlibrium
Vapor
dT = dp =0 and the total mole amount (ntot) of vapor and liquid
is constant =>
nl + ng = ntot => d(nl + ng) = 0 => dnl = -dng
Liquid
=>
T and p are constant
in the system
=>
l = g

21. Example of phase equilibrium between saturated water and vapor
Calculate chemical potentials for saturated water and vapor at 12 bar.
At 12 bar, the temperature of saturated water and vapor is 188oC (461.15K)
For saturated water at 12 bar: h’ = kJ/mol s’ = kJ/(molK) (from the steam table)
For saturated vapor at 12 bar: h’’ = kJ/mol s’’ = kJ/(molK) (from the steam table)
’ (12bar) = h’ – Ts’ = –  = kJ/mol
’’ (12bar) = h’’ – Ts’’ = 0.117 = kJ/mol

22. Phase equilibrium between CaCO3(s) and CaO(s)
When the temperature is high enough, CaCO3(s) may dissolve into CaO(s) + CO2(g) according to the following equation: CaCO3(s) = CaO(s) + CO2(g).
For example, lime kilns at pulp mills combust typically oil or natural gas to dissolve CaCO3(s) into CaO(s) + CO2(g).
What must the partial pressure of CO2 be at 1100K that CaCO3(s) dissolves into CaO(s) + CO2(g) when the total pressure is 1 bar?
At equilibrium, all species exist and we can write
m[CaCO3] = m[CaO] + m[CO2].
Lime kiln (meesauuni) is a component which belongs to the
chemical recovery process at a pulp mill.

23. Phase equilibrium between CaCO3(s) and CaO(s)
Chemical potentials for each species at the standard state are calculated as follows:
°CaO(T) = · · · = -707·103 J/mol (From tables)
°CO2(T) = · · · = -657·103J/mol (From tables)
°CaCO3(T) = ·103 J/mol
°CaCO3(T) has been calculated using the average cp between K and 1100K
(26.1 cal /molK) and the following equation
This causes some inaccuracy for the calculation.

24. Phase equilibrium between CaCO3(s) and CaO(s)
At the equilibrium
-1385·103 = - 707·103 + [(- 657·103) + RTln(pCO2/po)] =>
RTln(pCO2/po)= ( ) · 103 = -21·103 J/mol =>
pCO2/po= exp[-21·103/(8.314·1100)] = =>
pCO2 = 0.1bar
If pCO2 < 0.1bar CaO(s) exists only
Íf pCO2 > 0.1bar CaCO3 (s) exists only
=> pCO2 < 0.1bar at 1100K
m[CO2] = mo[CO2] + RTln(pCO2/po)
m[CaCO3] = mo[CaCO3] + v(ptot - po)
m[CaO] = mo[CaO] + v(ptot - po)
For CaCO3 and CaO v(ptot - po) = 0, because ptot = pó =1bar

25. Derivation of Clayperon equation
For both phases
p = p(T) =>
=>
Considering
and

26. Derivation of Clayperon equation
=>
=>
For pure substances () = () => h() –Ts() = h() –Ts()
=>
Which is a Clayperon equation

27. Clausius-Clayperon equation
Equilibrium between vapor ( = g) and liquid ( = l)
Clayperon equation
In most cases, v(g) >> v(l)
If vapor can be treated as an ideal gas =>
h() – h() = l l = vaporization heat
=>
=>
Which is Clausius-Clayperon equation
Note
For example

28. Example At what pressure does water freeze at -5oC?
= liquid (water) and  = solid (ice)
Clayperon equation
, where r (T) is the fusion heat of water (J/mol)
It is reasonable to assume that v(l), v(s) and r(T) are independent on the temperature =>
=>
where

29. Example Initial values
Tr = K T = K r(273.15K) = 6000J/mol pr = 1bar
vliquid (275.15K) = 1.8*10-5 m3/mol vice (273.15K) = 1.998*10-5 m3/mol
Substituting initial values in the calculation equation the pressure becomes
p = 555bar
For example, the pressure under the skate may be several hundred bars.