Equilibrium By Dr. Srimala

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

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

5.0 Spontaneous Reaction Direction

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.

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

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

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

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

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.

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)

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

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.

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.

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

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

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

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 435.14 kJ/mol. Calculate its normal boiling point ?

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).

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 10.047 g/cm3 respectively. The heat of fusion of Bi is 11300 J/mole. Calculate the change in pressure for an increase of 1K of the melting point. The atomic wt of Bi is 208.98.

Solution 1.2: ΔV=mass/volume =208.98(1/9.72 – 1/10.047) =0.6998 cm3 =0.6998 x 10-6 m3

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:

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

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.

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

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

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

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)

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

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.

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 = 0.126 bar Pressure at triple point = 0.126 bar

2) Solution y = m x + c ln Ps = 16.8 – 7312 (1/ T) ΔHs / R = 7312 ΔHs = 7312x 8.314 = 60791.968 J/ mole Heat of sublimation = 60791.968 J/ mole ΔHv / R = 5452 ΔHv = 5452 x 8.314 ΔHv = 45327.928 J/ mole ΔHs = ΔHv + ΔHm ΔHm = ΔHs - ΔHv =60791.968 J/ mole - 45327.928 J/ mole =15464.04J/mole Heat of fusion = 15464.04J/mole

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