2. At sea level At top of a mountain Water boils at 100  C Water boils at < 100  C b.P = f(P)

3. P = 1.0133 bar bp = 100  C

4. P = 0.26 bar bp = 69  C elevation = 8848 m

5. P = 2 atm, b.p = 124  C + EG 50/50 (v/v), b.p =129.5  C => b.p = f(P, c)

6. stability map stability of the state of aggregation Depending on the circumstances, molecules aggregate to form solid, liquid, or gas If I specify P, T, and c, what are the stable phases? Phase diagram

7. Examples of single-phase system (  = 1):  Pure water (l)  White gold (alloy of Au-Ag-Ni) => Ag and Ni substitute Au in its FCC lattice => homogeneous composition throughout => solution  Air (N 2, O 2, Ar, CO 2 ) Phase = region of a substance that is:  Uniform in chemical composition  physically distinct  mechanically separable Examples of dual-phase system (  = 2):  Ice cubes in liquid water  Milk (fat globules in aqueous solution)

8. Equilibrium =  the condition which represents the lowest energy level  the properties are invariant with time Component = the measure of chemical complexity N N  1 1 2 2 1 1 > 1 Water (l) diamond Water (l) diamond Ice cube in liquid water (slush) White gold (Au-Ag-Ni) White gold (Au-Ag-Ni) CCl 4 – H 2 O

9. Phase diagram of water P (atm) T (  C) 1 100 Solid  = 1 Gas  = 1 Liquid  = 1 l = v coexistence curve  = 2 s = v coexistence curve  = 2 s = l coexistence curve  = 2 00.01 Triple point  = 3 4.58 mm Hg Critical point (374  C, 218 atm)

10. POLYMORPHS Phase diagram of sulfur B : 95,5  C and 0,51 Pa C : 115  C and 2,4 Pa E : 151  C and 1,31  10 8 Pa Different atomic arrangement at constant composition

11. The transformation from one polymorph to another can be : 1.Reversible: two crystalline forms are said to be enantiotropic 2.Irreversibel: two crystalline forms are said to be monotropic

12. Pressure-temperature diagram for dimorphous substances: (a) enantiotropy, (b) monotropy

13. There are tree variables that can affect the phase equilibria of a binary system: T, P, and C T T P P c

14. Two-component phase diagram 3 dimension T-P-c 3 dimension T-P-c 2 dimension T-P P-c T-c Crystallization: liquid and solid The effect of P can be ignored T-c Diagram

15.  Complete solubility in solid and liquid states  Change of state (s = l) Metal (Hume-Rothery Rule) Similar crystal structure Similar atomic volumes Small values of electronegativity For interstitial solid solutions, the Hume-Rothery rules are:interstitial 1.Solute atoms must be smaller than the pores in the solvent lattice. 2.The solute and solvent should have similar electronegativity.electronegativity

16. 1.The atomic radii of the solute and solvent atoms must differ by no more than 15%:atomic radiisolvent For substitutional solid solutions, the Hume-Rothery rules are: 2.The crystal structures of solute and solvent must match.crystal structures 3.Maximum solubility occurs when the solvent and solute have the same valency. Metals with lower valency will tend to dissolve in metals with higher valency.solubilityvalency 4.The solute and solvent should have similar electronegativity. If the electronegativity difference is too great, the metals will tend to form intermetallic compounds instead of solid solutions. electronegativityintermetallic compounds

17. T cAB Liquid  = 1 Solid  = 1 Liquid + solid  = 2 Liquidus Solidus

18.  Liquidus: l  s + l : the lowest T for only liquid (at any c)  Solidus: s  s + l : the highest T for only solid (at any c)  The diagram represents the phase behavior of two components having many similarities (type of bonding, atomic size, crystal structure, etc), we call it ISOMORPHOUS DIAGRAM.  The shape of the diagram looks like a lens  lens-shape diagram / lenticular diagram.  Example: naphthalene -  -naphthol

19. Solid solution of naphthalene -  -naphthol Similar 

20. Solid solution of naphthalene -  -naphthylamine  >

21.  Partial or limited solubility  miscibility gap  No change of state  always liquid or always solid T cAB  = 2 l  = 1 s l 1 + l 2  +  Coexistence curve

22. Hexane - nitrobenzene upper critical solution temperature (upper consolute temperature)

23. lower critical solution temperature (lower consolute temperature)

25.  Partial or limited solubility  Change of state T cAB l l +  l +        Solid solution (A rich) Solid solution (B rich) Liquid solution  = 1  = 2 Eutectic     l EUTECTIC DIAGRAM

26.  Freezing point depression of both components (A and B)  Eutectic point is equilibrium of , , and l  Eutectic point is unique; it only happens at one T, P, and c

27. Phase diagram for the simple eutectic system naphthalene - benzene

28. Phase diagram for system NaCl – H 2 O

30. Solution A having composition of x A and enthalpy of H A is mixed adiabatically with solution A having composition of x B and enthalpy of H B. The composition and the enthalpy of the mixture is calculated using material balance: Total balance: Component balance: Combining both equations yields:

31. Similarly, if mixture A were to be removed adiabatically from mixture C, the enthalpy and composition of residue B can be located on the straight line through points A and C by means of the equation:

32. EXAMPLE Calculate (a) the quantity of heat to be removed and (b) the theoretical crystal yield when 5000 lb of a 30 per cent solution of MgSO 4 by mass at 110  F is cooled to 70  F. Evaporation and radiation losses may be neglected. SOLUTION (a)Initial solution (A)x A = 0.3H A = - 31 Btu/lb Cooled system (B)x B = 0.3H B = - 75 Btu/lb Enthalpy change  H = - 44 Btu/lb Heat to be removed= (- 44) (5000) = - 220000 Btu (b)The cooled system B is located in the region where MgSO 4.7H 2 O is in equilibrium with solution.

34. The crystal MgSO 4.7H 2 O contains: x C = 0.49 From the graph: x A = 0.26 The MgSO 4.7H 2 O crystal yield is 869.6 lb