lecture 4. phase equilibrium phd falfushynska h

Lecture 4. Phase Equilibrium

PhD Falfushynska H.

Sulfur phase diagram triple point

Gibb’s Phase Rule: a tool to define the number of phases and/or degrees of phase changes that

can be found in a system at equilibrium

• For any system under study the rule determines if the system is at equilibrium

• For a given system, we can use it to predict how many phases can be expected

• Using this rule, for a given phase field, we can predict how many independent parameters (degrees of freedom) we can specify

• Typically, N = 1 in most condensed systems – pressure is fixed!

where:

F is # degrees of freedom of the system (independent parameters)

C is # components (elements) in system

P is # phases at equil.

N is # "noncompostional" parameters in system (temp &/or Press

F C P N

ure)

A system involving one pure chemical is an example of a one-component system.

A typical phase diagram. The solid green line applies to most substances; the dotted green line gives the anomalous behavior of water. The green lines mark the freezing point and the blue line the boiling point, showing how they vary with pressure.

Clausius–Clapeyron relationis a way of characterizing a discontinuous phase transition between two phases of matter of a single constituent.

dP/dT is the slope of tangent to the coexistence curve at any point

Gibbs–Duhem relation

For a liquid-gas transition,  L  is the specific latent heat (or specific enthalpy) of vaporization, whereas for a solid-gas transition, L  is the specific latent heat of sublimation

Application of Clausius–Clapeyron relation

Dependence of melting on solid solution composition

Dependence of crystallization temperature on fusion

The iron–iron carbide (Fe–Fe3C) phase diagram. The percentage of carbon present and the temperature define the phase of the iron carbon alloy and therefore its physical characteristics and mechanical properties. The percentage of carbon determines the type of the ferrous alloy: iron, steel or cast iron

Eutectic system

• A eutectic system is a mixture of chemical compounds or elements that has a single chemical composition that solidifies at a lower temperature than any other composition made up of the same ingredients.

Lidocaine and prilocaine, both solids at room temperature, form a eutectic that is an oil with a 16 °C (61 °F) melting point, used in eutectic mixture of local anesthetic (EMLA) preparations.

the intersection of the eutectic temperature and the eutectic composition gives the eutectic point

Ternary igneous phase diagrams are triangular diagrams that show melting relationships involving three chemical components. The diagrams may involve eutectic, peritectic and cotectic relationships.

Metamorphic T-X Phase DiagramsFor some rock compositions, metamorphic assemblages vary greatly depending on the composition of the fluid present. Metamorphic fluids are generally dominated by H2O and CO2, and the ratio H2O:CO2 can control mineral stability. So, phase equilibria are plotted on T-X diagrams instead of P-T diagrams. (X refers to the mole fraction of 2or H2O in the metamorphic fluid that is present.)

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L +

liquidus

solidusCu-Niphase

diagram

Phase Diagrams:

• Rule 1: If we know T and Co then we know the # and types of all phases present.

• Examples:A(1100°C, 60):

1 phase:

B(1250°C, 35): 2 phases: L +

B (1

250°

C,35

) A(1100°C,60)

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L +

Cu-Ni system

Phase Diagrams:

• Rule 2: If we know T and Co we know the composition of each phase

• Examples:TA A

35C o

32C L

At T A = 1320°C: Only Liquid (L)

C L = C o ( = 35 wt% Ni)

At T B = 1250°C: Both and L C L = C liquidus ( = 32 wt% Ni here) C = C solidus ( = 43 wt% Ni here)

At T D = 1190°C: Only Solid ( ) C = C o ( = 35 wt% Ni )

C o = 35 wt% Ni

adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM

International, Materials Park, OH, 1991.

BT B

DT D

tie line

4C3

• Rule 3: If we know T and Co then we know the amount of each phase (given in wt%)

• Examples:

At T A : Only Liquid (L) W L = 100 wt%, W = 0

At T D : Only Solid ( ) W L = 0, W = 100 wt%

C o = 35 wt% Ni

Therefore we define

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L +

Cu-Ni system

TA A

35C o

32C L

BT B

DT D

tie line

4C3

R S

At T B : Both and L

% 733243

3543wt

= 27 wt%

WL S

R + S

W R

R + SNotice: as in a lever “the opposite leg” controls with a balance (fulcrum) at the ‘base composition’ and R+S = tie line length =

difference in composition limiting phase boundary, at the temp of interest

wt% Ni20

120 0

130 0

30 40 50110 0

L (liquid)

(solid)

L +

L +

T(°C)

A

35C o

L: 35wt%Ni

Cu-Nisystem

• Phase diagram: Cu-Ni system.• System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; phase field extends from 0 to 100 wt% Ni.

Adapted from Fig. 9.4, Callister 7e.

• Consider Co = 35 wt%Ni.

Ex: Cooling in a Cu-Ni Binary

46354332

: 43 wt% Ni

L: 32 wt% Ni

L: 24 wt% Ni

: 36 wt% Ni

B: 46 wt% NiL: 35 wt% Ni

C

D

E

24 36

Partition Coefficient Kp (Distribution Coefficient Kd)

When a compound is shaken in a separatory funnel with two immiscible solvents, the compound will distribute itself between the two solvents.

Normally one solvent is water and the other solvent is a water-immiscible organic solvent.

Most organic compounds are more soluble in organic solvents, while some organic compounds are more soluble in water.

Here is the universal rule:

At a certain temperature, the ratio of concentrations of a solute in each solvent is always constant. ﾊ And this ratio is called the distribution coefficient, K.

(when solvent1 and solvent2 are immiscible liquids

For example,Suppose the compound has a distribution coefficient K = 2 between solvent1 and solvent2

By convention the organic solvent is (1) and waater is (2)

(1) If there are 30 particlesof compound , these aredistributed between equalvolumes of solvent1 and

solvent2..

(2) If there are 300 particles of compound , the same distribution ratio is observed in solvents 1 and 2

(3) When you double the volume of solvent2 (i.e., 200 mL of solvent2 and 100 mL of solvent1),the 300 particles of compound distribute as shownIf you use a larger amount of extraction solvent, more solute is extracted

Separatory Funnel Extraction Procedure

1. Support the separatory funnel in a ring on a ringstand. Make sure stopcock is closed

2. Pour in liquid to be extracted

3. Add extraction solvent

4. Add ground glassStopper (well greased)

Separatory Funnel Extraction Procedure

Pick up the separatory funnel with the stopper in palce and the stopcock closed, and rock it once gently.

Then, point the stem up and slowly open the stopcock to release excess pressure. Close the stopcock. Repeat this procedure until only a small amount of pressure is released when it is vented

Shake the separatory funnel.

Separatory Funnel Extraction Procedure

Shake the separatory funnel vigorously.

Now, shake the funnel vigorously for a few seconds. Release the pressure, then again shake vigorously. About 30 sec total vigorous shaking is usually sufficient to allow solutes to come to equilibrium between the two solvents.

Vent frequently to prevent pressure buildup, which can cause the stopcock and perhaps hazardous chemicals from blowing out. Take special care when washing acidic solutions with bicarbonate or carbonate since this produces a large volume of CO2 gas

Separatory Funnel Extraction Procedure

Separate the layers.

Let the funnel rest undisturbed until the layers are clearly separated

While waiting, remove the stopper and place a beaker or flask under the sep funnel.

Carefully open the stopcock and allow the lower layer to drain into the flask. Drain just to the point that the upper liquid barely reaches the stopcock

Chlorophyll extraction