μ α =μ β = μ γ = =μ ω μ α =μ β =μ γ = =μ ω Thus for c components, the number of additional constraints is c(p 1) ( ) ( )

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1 Phase Diagrams 1 Gibbs Phase Rule The Gibbs phase rule describes the degrees of freedom available to describe a particular system with various phases and substances. To derive the phase rule, let us begin with a system that has c independent chemical species. - Factors in to account for any restrictions due to stoichiometry, equilibrium or mass or charge constraint - Thus for a system with acetic acid, H + and C 2 H 3 O 2 - are not independent of HC 2 H 3 O 2 (because [H + ] = [C 2 H 3 O 2 - ] and [C 2 H 3 O 2 - ] is related to [HC 2 H 3 O 2 ] via K a. We will assume no chemical reactions (which reduces the number of independent chemical species). Now let each chemical component exist in the maximum number of phases, p. When we include the two quantities, temperature and pressure, as degrees of freedom that can describe our system, the total degrees of freedom of the system (before considering thermodynamic constraints) is f = p c+ 2 However, the number of chemical components in each phase is not truly independent since the total of the each mole fraction must equal one. X1+ X2 + X3+ Xc = 1 This yields one constraint for each phase; therefore, the total degrees of freedom becomes ( ) f = p c+ 2 p = p c We are assuming equilibrium among all the components; therefore, each component has p 1 restrictions due to the equality of chemical potentials. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) μ α =μ β =μ γ = =μ ω μ α =μ β = μ γ = =μ ω ( ) ( ) ( ) ( ) μ α =μ β =μ γ = =μ ω c c c c Thus for c components, the number of additional constraints is c(p 1) ( ) ( ) f = p c c p 1 = pc p+ 2 pc+ c= c p+ 2 Thus we have derived the Gibbs phase rule: f = c p+ 2

2 Consider the degrees of freedom on a one-component phase diagram, that is, where c = 1 p 2 solid triple point phase boundary gas Within each phase region, p = 1; therefore, according to the Gibbs phase rule, the number of degrees of freedom that can be used to describe the system is 2. f = c p+ 2= = 2 In this case the degrees of freedom correspond to the temperature and pressure. As an example, consider a solid. Within the solid region, the temperature and pressure can change without changing the phase or composition of the solid: the solid has two degrees of freedom. Along a phase boundary, two phases are in equilibrium, that is, p = 2. Therefore, the degrees of freedom is reduced to one. f = c p+ 2= = 1 Consider the gas- boundary. When the gas and are in equilibrium, changing the temperature forces a change in the pressure (if the gas and are to remain in equilibrium!). Another perspective would be that boiling points are not independent of pressure or that the boiling points are a function of pressure. (If pressure is a degree of freedom, then temperature is not.) Finally consider the triple point, where p = 3. Then the degrees of freedom reduced to zero. f = c p+ 2= = 0 The triple point has no freedom. It has a specific pressure and a specific temperature. Changing either makes our system go away from the triple point. Can four phases exist together (as in two solid phases, a phase and the gaseous phase? No!! f = c p+ 2= = 1 Having -1 degrees of freedom is meaningless. No quartic points are possible. T

3 { Two-component Liquid-Vapor Phase Diagrams isopleth T dew point curve tie line vapor T(A) b 3 T(B) b vapor + bubble point curve 0 Z A 1 isopleth a line of constant composition (vertical line when x -axis is mole fraction) dew point curve boundary between the vapor phase and the vapor + phases bubble point curve boundary between the phase and the vapor + phases tie line horizontal line that connects the dew point curve and the bubble point curve with the vapor + phase region x A, x B mole fractions in phase y A, y B mole fractions in vapor phase z A, z B total mole fractions in both phases T vapor { { T(A) b T(B) b 0 1 z A z A x A

4 { Lever rule 4 Lever rule useful technique to that aids in the relative amount of vapor and and the composition of the vapor and the composition of the vapor T vapor { { T(A) b T(B) b 0 1 z A The ratio of vapor amount to amount is the same as the ratio of the lengths on the tie line. That is the length of the line segment to the right of z A is proportional to the amount of vapor and the length of the line segment to the left of z A is proportional to the amount of. z A x A n x z = n z y vap A A liq A A Alternately we have in terms of mole fraction of vapor and mole fraction of, X vap x = x A A z y A A za ya and Xliq = x y A A. The composition of the vapor is found where the tie line meets the bubble point line and the composition of the is found where the tie line meets the dew point line.

5 Brief derivation of lever rule 5 The total amount of component A, z A n, is the sum of the amount of A vap and A liq. zan = nliqxa + nvapya However, the total amount of both components is the sum of the amount of and the amount of vapor. n = n + n liq vap And the total amount of component A is z A n. zan = nliqza + nvapza Equating both expressions yields the lever rule. n z + n z = n x + n y n z n x = n y n z n z x = n y z liq A vap A liq A vap A liq A liq A vap A vap A ( ) ( ) liq A A vap A A n z x = n y z vap A A liq A A

6 Total vapor pressure of a mixture of two volatile s 6 Total vapor pressure as function of mole fraction Consider that the total vapor pressure above a solution comes from component A and component B (assuming that Raoult s law is valid for both components). (Remember x A is the mole fraction in the phase.) ( ) ( ) p= p + p = x p + x p = x p + 1 x p = p + x p p * * * * * * * A B A A B B A A A B B A A B Total vapor pressure as function of vapor mole fraction The mole fraction of the components in the vapor phase can be written in terms of the mole fractions in the phase, assuming Raoult s law. y p x p p = = = + pb yb = = 1 ya p * * A A A A A x * * * A * * p pb xa( pa pb) pb( 1 xa) + xapa To avoid issues like negative pressures and mole fractions let us, without loss of * * generality, assume that A is more volatile than B, that is, pb < pa. If so, then ya > yb * xapa * * * * ya = yapb + yaxa( pa pb) = x * * * ApA p + x p p ( ) B A A B ( ) y p = x p y p + p y * yapb xa = * * * p + y p p * * * * A B A A A A B A ( A A( B A) )

7 The total pressure in the vapor phase can be written in terms of the pure vapor pressures and the mole fraction of component A in the vapor phase. 7 ( ) ( ) p= x p + 1 x p = p + x p p * * * ( p p ) p y ( ) * * * * * A A A B B A A B * * * * * * * p B pa ( pa pb) y A + ( pa pb) pbya ( ) ( ) ( ) ( ) p= p + = * A B B A B * * * * * * pa pa pb ya pa pa pb ya p p p p y + p p y + p p y p p y pp = = * * * * * * * * * * * * B A A B A B B A A B A B B A B A * * * * * * pa pa pb ya pa pa pb ya

8 Distillation 8 Fractional distillation T T(A) b T(B) b 0 1 z A The goal of distillation is to purify a one-phase mixture of s. The purification is possible when the two pure s have different vapor pressures (that is different boiling points). When the mixture boils, the vapor has a greater mole fraction of the more volatile than the mole fraction of the, that is, y A > x A. The enriched vapor is condensed and boiled (at higher temperature) to further enrich the vapor with the more volatile component. Thus the goal of making a with a specific purity is a multi-step process which is called fractional distillation. The number of distillation steps (vaporization followed by condensation) needed to complete a separation of s is also called the number of theoretical plates. (The diagram above has 3 theoretical plates shown.) Assuming without loss of generally that p A * > p B *, the greater the ratio between p A * and p B *, the fewer theoretical plates are needed for a separation. Steam distillation Some s (such as essential oils and other fragrances) have decomposition temperatures below their boiling points. Thus fractional distillation is ineffective in purifying such s. Steam distillation starts differently than fractional distillation in that the mixture of s in immiscible (ideally) rather than miscible.

9 9 The boiling of the mixture occurs when the total vapor pressure equals the atmospheric pressure. The total vapor pressure can be considered to come from each * * component separately, i. e., p= pa + pb Since the total vapor pressure is greater than the vapor pressure of either component, the boiling will occur below the boiling point of either pure component. The condensate of steam distillation has the composition of the vapor. * pa ya = * * p + p Thus for oils of low volatility, the distillate has a very small mole fraction. A B Azeotropes T(azeo) b T(A) b T vapor T(B) b 0 Z A x aze o 1 In some mixtures, the bubble point line touches the dew point at a specific composition. When such a composition boils, the composition of the vapor is same as the composition of the, that is x A = y A. This composition is called the azeotropic composition and the mixture is said to have an azeotrope. Azeotropes often limit the degree of purification that a mixture can undergo via fractional distillation. For example, a mixture of water and ethanol has an azeotrope at 95% volume percent ethanol (95.6% weight%, mole fraction). Thus the maximum purity that a mixture of ethanol and water can achieve via distillation is 95% (or 190 proof). To make absolute ethanol, a drying agent (such as anhydrous calcium chloride) must be added to remove the water or another substance must be added to make a tertiary mixture without an azeotrope (such as benzene).

10 Liquid-vapor phase diagram with immiscible phase 10 The previous -vapor phase diagrams have had phases where the two s have been miscible in each other in all proportional. We need to consider the possibility that the two s have limited solubilities in each other. vapor immiscible A+B vapor+a vapor+b B A 0 1 X A In the diagram above, the two-phase portion (immiscible) connects with the dew point curve of the -vapor portion. Within the two-phase region, two solutions exist, one that is mostly A with a small amount of B impurity (labeled A ) and one that is mostly B with a small amount of A impurity (labeled B ) The composition of the solutions and the relative amount of each solution can be determined with a tie line and the lever rule. Use of a tie line and the lever rule can be used in any two phase region in the diagram (or any phase diagram!) such as the vapor + A region or the vapor + A region.

11 Two-component Liquid-Liquid Phase Diagrams 11 In the above phase diagram, the two-phase region had contact with the vapor region. In mixtures where the two regions do not make contact, the phase diagram can be studied separately from the -vapor phase diagram. T(K) 292 one phase T uc two phases X n itrob en zene In some mixtures, the s are miscible above a critical temperature. This critical temperature is called an upper consulate temperature. An example of such is mixture is nitrobenzene with hexane. T(K) 310 two phases T lc one phase X triethylamine In other mixtures, the s are miscible below a critical temperature. This critical temperature is called a lower consulate temperature. An example of such is mixture is triethylamine with water.

12 T(K) 483 one phase T uc 12 two phases 334 T lc one phase 0 1 X nicotine Rarely some mixtures have both an upper consulate temperature and a lower consulate temperature. An example is nicotine in water (in a pressurized system).

13 Two-component Liquid-Solid Phase Diagrams Simple diagram with eutectic point A simple two-component solid phase diagram has 4 regions, 2 lines and 1 significant point. 1. region one phase with a miscible mixture of the two-components 2. + A region two phases with solid A with a enriched with B 3. + B region two phases with solid B with a enriched with A 4. solid region two phases with solid A mixed with solid B T(B) m +B A+B us +A eutectic point solidus 0 X A 1 5. The line that indicates the lowest temperatures where only exists is called the us. 6. The line that indicates the highest temperature where only solid exists is called the solidus. The eutectic point (eutektos, Gr. easily melted) yields the temperature where all the in a mixture must freeze and composition where the composition of the melt always equals the composition of solid 1. The composition at the eutectic point is known as the eutectic composition. 2. The lowest melting point of solid mixture occurs at the eutectic composition. 3. Once the temperature of a mixture reaches the eutectic point, the temperature will remain constant until all of the mixture has solidified. 4. Applications of eutectics A. solder Sn/Pb (63/37) melts at 183 C B. fire ceiling sprinklers Wood s metal Bi/Pb/Sn/Cd (50.0/26.7/13.3/10) melts at 70 C C. emergency escape seals in gas cylinders Field s metal Bi/In/Sn (32.5/51.0/16.5) melts at 62 C, nontoxic alternative to Wood s metal D. mercury alternative Galinstan Ga/In/Sn (68.5/21.5/10.0), used in thermometers melts at -19 C E. pharmaceuticals lidocaine/prilocaine topical anesthetic (eutectic composition between drug and excipient are convenient to ensure consistency of formulation) 5. Not all solid mixtures have a eutectic point, the gold/silver -solid diagram looks like a simple -vapor diagram. 13 T(A) m

14 Cooling curves 14 A cooling curve is a plot temperature versus time plot for the cooling a particular composition of a two-component. The simplest useful cooling curve (corresponding) the simple -solid phase diagram has four features. 1. Liquid cooling top portion of the curve corresponds to the mixture losing thermal energy, the slope remains constant until solidification begins. 2. One-component solidification the slope of the curve decreases since now the cooling rate is partially cancelled by the energy released from solidification 3. Eutectic halt all of the remaining solidifies, all energy lost is from formation of bonding within the solid (metallic, dipole, dispersion, etc ). 4. Solid cooling once the eutectic halt is finished, the temperature continues to decrease (albeit with a larger slope, since the heat capacity of a solid is usually smaller that the. Liquid cooling Break in curve indicates us One-component solidification Eutectic halt Solid cooling The break in the curve between the cooling region and the one-component solidification region indicates the temperature of the us. The above curve is idealized primarily because the changing temperature affects the heat capacity of s and solids. Thus a cooling curve is not a collection of straight lines, but a messy collection of curves whose breaks in curvature are difficult to find.

15 15 By having a series of cooling curves with mixtures of varying compositions, a phase diagram can be reproduced T(A) m T(B) m +B +A A+B 0 1 X A X A = X A = X A = X A = X A = X A = Analysis of the cooling curves 1. Identifying the break between the cooling and the one-component solidification yields the temperature of the us. 2. The melting point of the pure substances are identified with halts. 3. The eutectic temperature is found in all of the mixture cooling curves. 4. Mixtures with more complicated phase diagrams would have more complicated cooling curves. Phase region boundaries are identified by finding breaks and halts.

16 Diagram with immiscible solid solutions 16 T(B) m T(A) m B+B +B immiscible A+B +A A+A 0 X A 1 The above diagram is a more realistic phase diagram for binary mixture. During cooling, as the temperature passes the us, the solid formed is not pure solid but a solid solution that is mostly A (A ) or a solid solution that is most B (A ). Because of the formation of impure solids, two new regions appear (A + A and B + B ) that are two-phase solid regions.

17 Kinetics versus thermodynamics in solidification 17 The diagram below traces the cooling of a mixture of two components from the onephase region to the two-phase solid region ( + A ). When total mole fraction is 0.60, the grain of solid that begins to freeze has a composition of X A = As the cooling continues, mole fraction of A in the grain decreases, say to X A = According to the thermodynamic properties of the mixture describe by the phase diagram, the composition of the entire solid should be X A = T(B) m B+B +B immiscible A+B X A +A X = 0.80 A+A T(A) m X = 0.90 However for the solid to have the same composition throughout the grain, the atoms in the solid would need to diffuse through the solid. The diffusion of atoms in a solid is very slow. Therefore, unless the cooling occurs over a very long time, the composition of the solid will not be in equilibrium. Therefore inner portion of the grain will have a different composition than the outer portion of the grain.

18 Diagram with compound formation (congruent melting) 18 T(AB) m 2 AB+ 2 T(A) m +A T(B) m +A B 2 B+ AB+A 2 B+A B X A Sometimes as a cools, the component combine chemically as solid is formed in a chemical reaction that can be written as 2 A (l) + B (l) A 2 B (s). The diagram shows that three solids can be formed (though not at the same time), the compound A 2 B, solid A and solid B. If the isopleth, X A = 0.67, is examined, the composition of solid is always the same as the composition of the. This circumstance is known as congruent melting. Note the congruent melting point occurs at a relative maximum on the phase diagram. (Impurties lower the melting points of a solid.)

19 Diagram with peritectic reaction (incongruent melting) 19 peritectic point +A B+ +A B 3 AB+A 3 B+A B 3 0 X A If a compound is formed, but the composition of the melt is always different from the composition of solid as the mixture cools; the substance undergoes incongruent melting and has a peritectic reaction. The reaction for the above diagram can be stated as 2 A(l) + B(l) + A(s) A 3 B (s). Note that reaction involves both the and the solid states. Generically the reaction can be written. Liquid + Solid Compound (s) A peritectic reaction occurs at the temperature of the peritectic point on the phase diagram. On the diagram above, note that if the isopleth, X A = 0.75 is examined; then as the melt cools and enters the + A region, solid is formed that is not A 3 B and remaining does not have X A = 0.75.