1. Equilibrium By
Dr. Srimala

2. Content 1.0 Introduction 2.0 Phase Equilibrium
2.01 Liquid-Vapor Phase Equilibrium
3.0 Gibbs Phase Rule
4.0 P-T Phase Diagrams & Clausius Clapeyron Equation
4.1 The Clausius-Clapeyron Equation
4.2 Liquid-Vapour (Vaporization) Equlibrium
5.0 Triple Point
5.1 Calculation solid-liquid-gas triple point

3. 1.0 Introduction When is a system at equilibrium?
For a system to be at equilibrium there can be no spontaneous processes occurring within the system !
1. Temperature
It must be at the same temperature as the surroundings
It must have a uniform temperature
Steady state is not the same as equilibrium T2 ! T1

4. 5.0 Spontaneous Reaction Direction

5. Steady state Equilibrium
At steady state different temperatures can exist at different points around the system, but the system does not change with time.
At equilibrium the temperature is the same throughout the system, and the system does not change with time.

6. 2. Energy
No moving parts
This means that a system at equilibrium can not have moving parts, because in real systems motion leads to friction – which is irreversible
Example:
Mechanical Energy can be converted completely to some other form of mechanical energy
It can also be converted completely to heat by a frictional process
Heat can not be converted completely to energy by a frictional process

7. 3.Constant Pressure
In the absence of restraining gravity, spring, electrostatic, magnetic, osmotic, or surface forces, at equilibrium the system must be at uniform pressure
If it’s not, the pressure difference causes motion
4.No flow of electricity
Electricity flowing through a resistor causes the wire to heat up – the current is changed into heat, which is an irreversible process

8. 2.0 Phase Equilibrium Typical phase equilibria:
vaporization (liquid  vapor)
sublimation (solid  vapor)
melting/freezing ( solid  liquid)
allotropy (solid phase I  solid phase II)

9. 2.01 Liquid-Vapor Phase Equilibrium
Consider the liquid –vapor equilibrium
To be at equilibrium, the rate of water molecules leaving the liquid must be the same as the rate of molecules returning to the liquid
Evaporation = condensation
Vapor pressure of the liquid = pressure of the vapor

10. If the system is not at equilibrium
The liquid either spontaneously boils to transfer mass into the vapor phase until equilibrium is attained, or…
The vapor condenses until the gas pressure equals the vapor pressure of the liquid.

11. 3.0 Gibbs Phase Rule For a system at equilibrium, F = C - P + 2
C = the number of components (1 so far in this chapter)
P = the number of phases present
F = the number of degrees of freedom (the number of intensive variables such as temp, pres, or mol frac that can be changed without disturbing the number of phases in equilibrium)

12. Cont …… 2 1 3 For a unary system
For a one-component system the phase rule becomes
F = 3 - P
When only one phase is present both p and T are independently variable. (An area on the p-T diagram)
When two phases are present there is only one p possible for a given T. (A line on the p-T diagram)
Three phases can be present (triple point) but there is no variation allowed in p or T.
For a unary system
Phases
Degrees of Freedom
Components 2 1 3

13. 4.0 P-T Phase Diagrams & Clausius Clapeyron Equation
For any of the phase boundary lines , the slope is given by
dP/dT = DS/DV = DH/TDV
For the transitions s  l, l  g, and s  g, DS > 0
For l  g, and s  g, DV > 0
while for s  l, DV is almost always > 0
This explains the positive slopes.

14. 4.1 The Clausius-Clapeyron Equation
This equation is of great importance for calculating the effect of change of pressure (P) on the equilibrium transformation temperature (T) of a pure substance and represented as
(the Clapeyron equation)
The Clausius-Clapeyron equation is applicable to any phase change-fusion, vaporization, sublimation, allotropic transformation, etc.

15. 4.2 Liquid-Vapour (Vaporization) Equlibrium
Applying Eq 1 to a liquid-vapour equlibrium, we have
Where Hv is the molar heat of vaporization/ latent heat of evaporation Vvap is the molar volume of vapour
Vliq is the molar volume of liquid
Since molar volume of vapour is very large than the molar volume of liquid, Vliq is negligible and hence

16. Cont....
Assuming that the vapour behaves as an ideal gas, the volume Vvap related to
Substituting the value of Vvap in Eq (3),
The equation can be rearranged in the most generally used differential form

17. Cont...
The equation can be integrated between the limits P1 and P2 corresponding to temperatures T1 and T2 respectively.

18. Exercise 1.1
Application of the Clasius Clayperon equation to liquid –vapour equlibrium
The vapour pressure of liquid titanium at 2227oC is 200 N/m2. The heat of vaporization at the normal boiling of titanium is kJ/mol. Calculate its normal boiling point ?

19. Solution 1.1:
Let’s assume that the normal boiling point of is Tb. At this temperature, the vapour pressure is equal to 1 atm (101,325 N/m2).

20. Exercise 1.2
Application of the Clasius Clayperon equation to solid-liquid equlibrium
The melting point of Bi is 544.5oK. The densities of solid and liquid Bi are 9.72 and g/cm3 respectively. The heat of fusion of Bi is J/mole. Calculate the change in pressure for an increase of 1K of the melting point. The atomic wt of Bi is

21. Solution 1.2: ΔV=mass/volume =208.98(1/9.72 – 1/10.047) =0.6998 cm3
= x m3

22. 5.0 Triple Point
At the triple point the pressure and temperature for boiling and sublimation are the same:
Solving for T and back substituting to find P:

23. 5.1 Calculation solid-liquid-gas triple point
Vaporization and sublimation (log P vs 1/T) are linear in the low pressure range where this triple point appears
Draw appropriate axes (Fig 1)-log P vs 1/T
Plot the melting point and boiling point on the 1 atm line.
Tm  Tt
Use equation to compute the
vapour pressure at this temperature Pt

24. vapour pressure at chosen temperature. Finally plot a straight line.
5. Draw a straight line from the boiling point to (Pt, Tt) -this is the (L+G) two phase line. Also draw from melting point (+L)
6. Sublimation curve also a straight line on this plot. Compute the slope of the sublimation line (ΔHs) from the heat of melting and vaporization
7. Use to compute As and calculate
vapour pressure at chosen temperature. Finally plot a straight line.

25. Cont… Fig 1 100 log P 10-20 1/T (Tm, 1 atm) (Tb, 1 atm) Slope =
-ΔHV /2.303R
-(ΔHV + ΔHm ) /2.303R
(Tt , Pt)
100
10-20
Fig 1

26. Example : Phase diagram for silicon
It’s given Tb=2750K, Tm = 1683K, ΔHV =297 kJ; ΔHm = 46.4 kJ P=1 atm
The vapour pressure curve is given by

27. Set the triple point temperature equal to the melting point Tt = Tm = 1683K. Substitute this value into the vapor pressure equation to obtain pressure at triple point
Plot the triple point (Pt, Tt) as the point O and construct the lines MO and BO. Apply equation heat of sublimation.
The triple point (Pt, Tt) also lies on the sublimation curve. Substitute this info together with the heat of sublimation to evaluate AS

28. The sublimation curve is given by
Let’s choose T=300K
This point is labeled P. A straight line from OP is the sublimation curve. The calculate triple point occurs at (2.65 x 10-4 atm, 1683K)

29. Illustration of the calculation for silicon
log P
1/600
M B O
100
10-30
1/300
1/T (K-1)
10-20
10-10 L G S P

30. Question
Near the triple point of iodine, the equilibrium vapor pressure of solid and the liquid are well represented by:
ln Ps = 16.8 – 7312 / T
ln Pl = 12.0 – 5452 / T
Where the temperature and pressure are express in Kelvin and bar units
a) Calculate the temperature Ttp and pressure Ptp of the triple point.
b) Consider the functional form of the integrated Clausius-Clapeyron equation (assume ΔH is constant) to state the molar heat of sublimation and the molar of fusion for iodine.

31. 1) Solution: At triple point, Ps = Pl
T = 387.5K
Temperature at triple point = 387.5K
ln Ps = 16.8 – 7312 / T
= 16.8 – 7312 / 387.5K
Ps = bar
Pressure at triple point = bar

32. 2) Solution y = m x + c ln Ps = 16.8 – 7312 (1/ T) ΔHs / R = 7312
ΔHs = 7312x 8.314
= J/ mole
Heat of sublimation = J/ mole
ΔHv / R = 5452
ΔHv = 5452 x 8.314
ΔHv = J/ mole
ΔHs = ΔHv + ΔHm
ΔHm = ΔHs - ΔHv
= J/ mole J/ mole
= J/mole
Heat of fusion = J/mole

33. Replacement Class 3rd Sept 2007 (1 hours class)
Tuesday (28th Aug)-11/2hours
8am-9.00am
Venue:BSK1 Materials and Mineral Resources Engineering