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A. K. M. B. RashidProfessor, Department of MMEBUET, Dhaka

Lecture 6

Application of Thermodynamicsin Phase Diagrams

The phase diagrams and its applications

The structure of phase diagrams

Construction of unary phase diagram

The free energy – composition (G - X) diagrams

Problem solving

Today’s Topics

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The study of phase transformations, like fusion, vaporisation, sublimation,

and allotropic transformations is extremely important in materials science.

A phase is a homogeneous system of which the intensive properties are

uniform or, at most, vary continuously throughout the system.

Phase diagrams are the primary thinking tool in materials science which

provide the basis for predicting or interpreting the changes in internal

structure of a material that accompany its processing or subsequent

service.

What are Phase Diagrams?

(a) Phase diagram for pure metal copper

(b) Phase diagram for unary ceramic compound SiO2

Fig. 7.1

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(c) Binary phase diagram of Pb-Sn metallic system

(d) Binary phase diagram of SiO2-Al2O3 ceramic system

Fig. 7.1

(e) Ternary phase diagram of CaO-SiO2-Al2O3 ceramic system

Fig. 7.1

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Phase diagram is a map that shows the domains of stability of phases and their combinations.

A point in this diagram represents astate of the system and lies within a specific domain on the map.

Reading the phase diagram will then tell you, at that state, when it comes to equilibrium,

1. what phases are present,2. the state of those phases, and3. the relative quantities of each phase.

Equilibrium in Multi-component System

Analogy similar to the unary heterogeneous systems can be applied to determine the conditions for equilibrium for multi-component, multi-phase systems.

The limits of stability of phases are indicated by the lines, called phase boundaries, on these diagrams and they are defined by the conditions under which pairs of phases may coexist at equilibrium.

T

P

For a system containing C component and P phases, the change in entropy for any process would be

dS’sys = dU’a + dV’a – mka dnk

a1Ta

Pa

Ta

1Ta

P

a=1C

k=1

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Conditions for equilibrium for multi-component, multi-phase systems:

Ta = Tb = Tg =...... = TP

Pa = Pb = Pg =...... = PP

. . . . . . . . . . . . .

. . . . . . . . . . . . .

m1a = m1

b = m1g = …. = m1

P

m2a = m2

b = m2g = …. = m2

P

mCa = mC

b = mCg = …. = mC

P

A crucial aspect in phase diagrams is the number of variables that need to

be specified in order to determine a thermodynamic state, indicated by its

degrees of freedom.

It is the smallest number of intensive variables, such as temperature,

pressure, concentrations, and so on, that the experimenter can vary

independently without changing the number of phase at equilibrium.

The Gibbs Phase Rule

This can be calculated from the difference in the number of total number

of variables (both dependent and independent) required to identify the

system and the number of available relations corresponding to the

conditions for equilibrium in the system.

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The state of any phase is completely determined by its pressure, temperature and composition.

If a phase contains C number of components, then (C -1) variables must be specified for the compositions, and 2 variables for P and T.

This makes a total of (C +1) variables for each phase.

So, for a system containing P phases, the total number of variables

m = P (C +1)

Equations on the equality of the temperatures and pressures yield 2(P -1) relations

While the equality of the chemical potentials of each species for the P phases yield C (P -1) additional relations.

However, if the system is in equilibrium, all of these variables are not independent: they are related by the conditions for equilibrium.

So the total number of relations is

n = 2 (P -1) + C (P -1)

n = (C + 2) (P – 1)

Ta = Tb = Tg =...... = TP

Pa = Pb = Pg =...... = PP

. . . . . . . . . . . .

m1a = m1

b = m1g = …. = m1

P

m2a = m2

b = m2g = …. = m2

P

mCa = mC

b = mCg = …. = mC

P

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Accordingly, for a system with C components and P phases, the number of degrees of freedom (i.e. the number of independent variables) is:

This equation is known as the Gibbs phase rule.

F = number of total variables (m) – number of relations (n)

F = P (C +1) – (C +2)(P -1)

F = C – P + 2

In a single-phase region, two variables need to be specified

(e.g., T and P).

In a two-phase region, one variable needs to be specified

Other variables change automatically in order to maintain the existing two-phase equilibria.

If T and P both are changed, the two-phase equilibriawould be disturbed by disappearing one of the two phases.

Self-Assessment Question 7.1What is the degree of freedom of the liquid phase of Fe-C-Si alloys?

Example : One-component system

S

L

G

P

T

triple point

F = C – P + 2

In a three-phase region, no variables need to be specified

There is only one point (the triple point) which occurs at a determined temperature and pressure (zero degree of freedom).

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The Structure of Phase Diagrams

Representation of structure of phase diagram using three types of

co-ordinate systems:

1. All the axes are thermodynamic potentials (T, P or m)

2. One axis is a potential and the others are not

3. All axes are not potentials.

(a) Both axes are potential (b) One axis is potential (c) Neither axis is potential

Additional Benefits of Using Non-Potential Axes

For diagrams constructed using all-potential-axes are simple and easy to understand.

However, they do not give information about the relative amounts of the phases presents in two- or three-phase equilibria.

Similarly, three-phase equilibria are presented as area when diagrams are constructed using all non-potential-axis.

This makes easy to calculate the relative amount of co-existing phases by constructing tie lines and using lever rules.

For diagrams constructed using one-potential-axis, two-phase equilibria are presented as areas.

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Number of Axes Required

Using Gibbs Phase Rule, the single phase regions of a component

have the highest number of degrees of freedom:

F = C - P + 2 = C + 1That is, single-phase regions require the largest number of variables, i.e., (C+1) for their specification (e.g., 2 for unary, 3 for binary, 4 for ternary etc.).

The graphical space in which the phase diagram is constructed

must have (C+1) independent co-ordinates.

Unary systems is two-dimensional [in (T, P) space]

Binary systems is three-dimensional [ in (T, P, a2) or (T, P, X2) space]

Ternary systems is four-dimensional [in (T, P, X2, X3) space]

Because of its limitation in presentation, most multi-component phase diagrams are represented as sections.

This is obtained by fixing a value for one (for binary systems) or two (for ternary systems) of the independent variables.

Fig. 7.3:A binary phase diagram plotted on(a) thermodynamic potentials (P, T, a2) space.(b) a more complex (P, T, X2) space.(c) a section taken at constant P.(d) the familiar (T, X2) space.

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Construction of Unary Phase Diagram

Generally plotted using (T, P) co-ordinate

systems.

The lines, indicating two-phase equilibria,

appeared on the diagram have the

general form of equation P = P (T).

During construction of the diagram:

Chemical potential surface of each phase is produced first.

The curve of intersection of any two chemical potential surface is then traced down to (P, T) space to obtain the relationship P = P (T).

P

T

S

L

G

The Chemical Potential

Using Gibbs free energy

dG = - SdT + VdP + mdn

the chemical potential can more suitably be defined as

m = G’n T, P

For unary system, G = nG:

m = G’n T, P

= (nG)n T, P

= G

Then the dependence of the chemical potential upon temperature and pressure in a unary system can be written as

dm = dG = - S dT + V dP

m = U’n S’, V’

The chemical potential expressed earlier in terms of internal energy

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If a phase is taken through an arbitrary change in state,

dma = - Sa dTa + Va dPa

This can be integrated over the limits of T and P to yield the chemical potential surface of a phase with the function ma = ma (T, P)

A chemical potential surface for b phase represented by the function mb = mb (T, P) can also be constructed in similar way.

m

T

P

ma = ma (P, T)

When the curves of both phases are superimposed into a single plot, the two surfaces intersect along a space curve AB.

At any point on that space curve, the temperatures, pressures, and chemical potentials of the two phases are identical.

Ta = Tb Pa = Pb ma = mb

Thus the curve of intersection of the two chemical potential surfaces, AB, is the locus of points for which the a and b phases are in equilibrium.

A

B

A’

B’

When the a = b equilibrium curve AB traced down to the P-T space, the line for P = P (T) relation, indicated by the curve A’B’, is obtained.

m

As we have seen earlier that, these three conditions must be met precisely in order for a and b phases to coexist in equilibrium.

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The Clapeyron Equation

We have stated earlier that, for unary systems, the two phase equilibriacurves in (P, T) space may each be described mathematically by the function P = P (T).

The Clapeyron equation is a differential form of this equation.

For any pair of coexisting phases in the unary system, integration of the Clapeyron equation yields a mathematical expression for the corresponding phase boundary on the phase diagram.

Repeated application to all the pairs of phases that may exist in the system yields all possible two-phase domains. Intersections of the two-phase curves produce triple point where three phases coexist.

Thus, Clapeyron equation is the only relation required for calculating a unary phase diagram.

Consider a b Equilibrium

Change in chemical potential of a phase when taken through any arbitrary

change in its state

dma = Va dPa – Sa dTa

Similarly, for any arbitrary change state of b phase, the change in

chemical potential

For equilibrium, dma = dmb, dPa = dPb, and dTa = dTb, so that

Va dP – Sa dT = Vb dP – Sb dT

dP/dT = (Sb – Sa)/(Vb – Va)

dP/dT = S/V

dmb = Vb dPb – Sb dTb

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In experiments, change in S is not measure directly.

Calorimetric measurements at constant pressure (i.e., QP) results values for heat of transformations (HF, HV, etc.).

Since, for a b equilibrium, G = 0 = H – TS, we have S = H/T.

Thus, expression for Clapeyron equation becomes

HT V

dPdT

=

Check units:

LHS:dP/dT = atm/deg

RHS:H = cal/mol, V = cc/molH/TV = cal/cc-deg

1 atm = 41.293 cal/cc

dP/dT = S/V

1. Predicting the effects of P on Tt (e.g., Tm, Tb)

describing the relationship between P and T at which a and b phases can exist in equilibrium

2. Calculating precisely Ht (e.g., HF, HG)

knowing the derivative (dP/dT), and V of the transformation

3. Obtaining an equation for a given equilibrium, P = P(T)

knowing any reference point to integrate dP/dT(e.g., normal melting point, normal boiling point)

Applications of Clapeyron Equation

HT V

dPdT

=

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Integration of Clapeyron Equation

Using appropriate reference points, the Clapeyron equation can be integrated for a two-phase equilibrium to yield the phase boundary P = P(T) for the equilibrium.

While integrating, the dependency of H and V on P and T must, however, be considered first.

HT V

dPdT

=

To obtain H = H (P, T) for a b equilibrium:

d (H) = dHb - dHa

Now, for the function H = H (P, T):

dH = CPdT + V(1 – Ta) dP

For all practical purposes, dependency of H on P can be ignored. Thus, dHP = CPdT, and

d (H) CP dT

where CP = CPb – CP

a = a + bT + cT-2

(7.21)

Eq (7.21) is often known as the Kirchhoff’s equation.

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To obtain V = V (P, T) for a b equilibrium:

d (V) = dVb - dVa

If b is an ideal gas phase and a is either a solid or liquid phase, then Vb >>Va , and

At STP, molar volume of:Gas = 22400 ccSolid/Liquid ≈ 10 cc

V = VG – Va VG = RT/P (for ideal gas)

If both b and a phases are condensed phases (solid or liquid),

For an approximate calculation (where P < a few tens of atm), V can be treated as constant.

For a precise calculation, dependency of V on P and T must be considered according to the familiar formula

V = V (P,T): dV = Va dT - Vb dP

Consider the equilibrium between the condensed phase and its vapour

phase, i.e., (a G) equilibrium

If CPG = CP

a, then H is independent of T. Then

dP / dT = H / TV

dP / dT HG / TVG = PHG / RT2

dP / P = (HG / RT2) dT

Eq. (7.25) is known as the Calusius - Clapeyron equation.

This is an approximate equation, derived by assuming that

1. HG is constant2. VG >>Va so that V VG.3. Vapour phase behaves ideally.

(7.25)

The Clausius - Clapeyron Equation

d ln P = dTHG

T2

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Integration of Clausius - Clapeyron Equation

d ln P = (HG / RT2) dT

Integrating between the limits (P1, T1) and (P2, T2):

Integrating indefinitely:

ln P = – (HG / RT) + C

P = A exp (– HG / RT)

where A = ln C

C = ln ASlope = - H/R

ln P

1/T

P2

P1

1T1

1T2

ln = – ln –HG

R

The Condensed Phase Equilibria

The Clausius – Clapeyron equation is applied only when there is a gas

phase involved in the equilibra.

Example: S G equilirbium, L G equilirbium, etc.

For all other equilibria involving condensed phases only (e.g., a b

equilirbium, S L equilirbium, etc), the Clausius-Clapeyron equation

cannot be applied.

In such cases, the original Clapeyron equation should be used.

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For a precise calculation, T and P dependency of S and V must be considered

while integrating the Clapeyron equation.

SV

dPdT

=

An approximate calculation of the phase boundaries between two

condensed phases can be made by ignoring T and P dependency of H and V.

In such cases, considering S and V as constant, integration of the

Clapeyron equation becomes straightforward:

P2 - P1 = ( T2 - T1 )SV

H = latent heat of fusion, etc.= ln (T2 / T1)

HV

P2 - P1

The gradient of P = P (T) line (i.e. S/V) for S L equilibrium for most substances is positive, which means that V for the transformation is positive (since S is always positive), i.e., the volume expands.

The only exception is water, for which the slope is negative, since ice contracts when it becomes water.

SL

G

P

T

ICE

WATER

STEAM

P

T

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Richards’ rule

The entropy of fusion of most elements is constant.

SF = 9 J/mol-KHF

TF

Trouton’s rule

The entropy of vapourisation of most elements is constant.

SG = 21 cal/deg-molHG

Tb

The Triple Point

At the triple point, S L, L G, and S G equilibrium lines meet.

P – T diagram for pure ironAt the triple point, Ga = Gb = Gg

For most elements and compounds, the triple point pressure is well below the atmospheric pressure.

The only exception is CO2, for which Ptp = 0.006 atm.

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Using the equations for L G and S G equilibrium lines, the triple point temperature and pressure is calculated as follows:

PG = AG exp (– HG / RT)

PS = AS exp (– HS / RT)

At the triple point (Ptp, Ttp), these two lines intersect. Thus

Ptp = AG exp (– HG / RTtp) = AS exp (– HS / RTtp)

Ttp = HS – HG

R ln (AS/AG)Ptp = AS exp

HG

HG – HS

The triple point of S, L and G phases can be determined if any two of the three equilibrium (S L, L G, S G) lines are known.

;

The most useful tool to obtain connections between the phase diagrams and their underlying thermodynamics principles.

Free Energy – Composition Diagrams

The fundamental principles of G-X diagrams are:

For a system of definite composition at constant P and T,

The stable phase has the lowest free energy, G.

The free energy is the same for coexisting phases.

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G-X Diagrams for Ideal Solutions

The molar free energy of ideal binary solutions:

GS = G1 + GM (1)

where, GM, the molar free energy of mixing of component 1 and 2, is:

GM = HM – TSM = RT (X1 ln X1 + X2 ln X2) (2)

and G1, the molar free energy of unmixed solution is:

G1 = X1 G01 + X2 G0

2 = G01 + (G0

2 – G01)X2 (3)

Combining eq.(1) and eq.(3), we get,

GS = G01 + (G0

2 – G01) X2 + GM (4)

X21 2

Low T

High T

GM

HM = 0

-TSM = GM

GM

G02

G01

G01 + (G0

2 – G01)X2

GS

X21 2

The GM – X plot shows that:

The curve is symmetrical at X2=0.5 and has a vertical slope at X1=1 and X2=1.

The curve has a minimum value of –RT ln2 at X2=0.5. This magnitude increases linearly with T.

Since, throughout the curve, GM is –veand GS is less than G1, components 1 and 2 prefer to form a solution.

GS = G01 + (G0

2 – G01) X2 + GM

GM = RT (X1 ln X1 + X2 ln X2)

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Thus, for a given phase of the solution, G-X diagram is a plot of the molar Gibbs free energy of mixing, GM, versus the mole fraction of component B, XB, at a fixed P and T.

Each existing phase in a system has its own G-X curve.

The competition for domains of stability of the phases and interactions that produce two and three phase fields that separate them can be visualised by comparing the G-X curves for all the phases in the system.

For such a comparison of free energies of mixing to be valid, it is absolutely essential that the energies of each component in all the phases be referred to the same reference state.

G-X Diagrams for Non-ideal Solutions

The form of G-X curve for non-ideal or real solution is given by the equation

GM = RT (X1 ln a1 + X2 ln a2)

GM = Gex + RT (X1 lnX1 + X2 lnX2)

where the excess free energy of mixing, Gex, can be positive or negative depending on the type of deviation from ideality.

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Ga

GM

A BXB

GM

ab

c

d

e

To produce G-X curves for non-ideal binary solutions of components A and B, the following steps can be followed:

Let the stable forms of pure A and B at the given T and P be a (fcc) and b (bcc), respectively.

The molar free energies of fcc A and bcc B are shown as point a and b in the figure.

To draw the free energy curve of the fcc aphase, convert the stable bcc arrangement of B atoms into an unstable fcc arrangement. This requires an increase in free energy, bc.

Construct the free energy curve for the aphase now be by mixing fcc A and fcc B as shown in the figure. The distance de will give GM for such solution of composition XB.

A similar procedure produces the molar free energy curve for the b phase.

In the following figure, G-X diagrams for both a and b phases are constructed in a single plot for comparison.

Ga

A BXB

GM

ab

cd

Gb

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Comparing Free Energy of Solutions with that of its Unmixed Components

B

GM

XB A

a

bc

d

G-X curves “convexing downwards”

Consider mixing of two separate solutions of A and B, marked by the points a and b.

The free energy of unmixed solutions is given by the point c.

The free energy of mixed homogeneous solution will be represented by the point d, which is lower than that of point c.

Thus, the resultant single-phase solution is stable relative to any two unmixed portions.

This is true for any single phase region in which the G-X curve is “convex downwards.”

GM

XBA B

x

y

z

p

q

m

n

G-X curves “convexing upwards”

The two separated solutions represented by point p and q have a free energy corresponding to point y, which is lower than that for the single homogeneous solution represented by point x.

The configuration with the lowest free energy for this composition is obviously the point z on the line mn, where two separate solutions of compositions m and n are in equilibrium with each other.

Thus, the stable system with compositions

from XB=0 to XB=m are composed of a single solution

from XB=m to XB=n consists of a mixture of two solutions of compositions m and n

from XB=n to XB=1 the stable system is again composed of a single solution.

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GM

XBA B

x

y

z

p

q

m

n

Single phase solutions from XB=m to XB=n are metastable with respect to the un-mixed two-phase system.

This situation is typical for systems exhibiting a miscibility gap in the phase diagram and is associated with sufficiently great positive deviation from the ideal behaviour.

System Consisting Two or More Separate Phases

Similar kind of miscibility may occur.

The A-rich alloys will have the lowest free energy as a homogeneous a phase and B-rich alloys as b phase.

For alloys with compositions near the cross-over in the G curves, the total free energy can be minimised by the atoms separating into two phases.

Ga

A BX0

Gb

G1

G0a

G0b

G1a

G1b

a1 b1

A BX0

Ge

mBa = mB

b

mAa = mA

b

G0a

G0b

Gea Ge

b

ae be

Construction of a tangent line common to two (G-X) curves for a pair of phases identifies the compositions of those two phases that coexist in equilibrium at the

indicated temperature and pressure.

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Examples of Some Common G-X Diagrams

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7.13 The vapour pressure of liquid zinc as a function of temperature is given as:

log P (mm Hg) = – 6620/T – 1.255 logT + 12.34.

Calculate the heat of vaporisation of zinc at its boiling point 907 C. If heat of sublimation of zinc at the boiling temperature is 30 kcal/mol, what will be the heat of fusion of zinc at its boiling temperature?

7.15 Mercury boils at 375 C with HG=14130 cal/mol. The heat capacity of liquid mercury is 6.61 cal/mol-K; that of mercury vapour, 4.97 cal/mol-K. What is the vapour pressure over liquid mercury at 25 C? At 100 C?

Problem Solving

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7.18 Below the triple point (-56.2 C) the vapour pressure of solid CO2 is given as

ln P (atm) = –3116 / T + 16.01.

The molar heat of melting of is 8330 J.

Calculate the vapour pressure exerted by liquid CO2 at 25 C, and illustrate why solid CO2, sitting on the laboratory bench, evaporates rather than melts.

7.19 The triple point of iodine I2 occurs at 112.9 C and 11.57 kPa. The heat of fusion at the triple point is 15.27 kJ/mol, and the following vapour pressure data are available for solid iodine:

Vapour pressure, kPa 2.67 5.33 8.00Temperature, C 84.7 97.5 105.4

Estimate the normal boiling point of molecular iodine.

7.22 Estimate the change in the equilibrium melting point of copper caused by a change of pressure of 10 kbar.

The molar volume of copper is 8.0x10–6 m3 for the liquid, and 7.6x10–6 m3 for the solid phase. The latent heat of fusion of copper is 13.05 kJ/mol. The melting point is 1085 C.

7.25 Carbon has two allotropes, graphite and diamond. At 25 C and 1 atm, graphite is the stable form.

Calculate the pressure that must be applied to graphite at 25 C in order to bring about its transformation to diamond.

Given data:S298 (graphite) = 5.73, S298 (diamond) = 2.43 J/mol/K,H298 (graphite) – H298 (diamond) = -1900 J/mol,@1 atm and 25 C, D(graphite) = 2.22 and D(diamond) = 3.515 g/cc

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Lecture 7

Thermodynamics of Reactive Systems

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