1 / 34

PHASE EQUILIBRIA

PHASE EQUILIBRIA. INTRODUCTION. At sea level. At top of a mountain. Water boils at < 100 C. Water boils at 100 C. b.P = f(P). P = 1.0133 bar. bp = 100 C. elevation = 8848 m. P = 0.26 bar. bp = 69 C. P = 2 atm , b.p = 124 C

dea
Download Presentation

PHASE EQUILIBRIA

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. PHASE EQUILIBRIA

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

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

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

  7. Phase = region of a substance that is: • Uniform in chemical composition • physically distinct • mechanically separable • 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 (N2, O2, Ar, CO2) • 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 1 2  1 Water (l) diamond Ice cube in liquid water (slush) > 1 White gold (Au-Ag-Ni) CCl4 – H2O

  9. ONE-COMPONENT SYSTEM (N = 1) s = l coexistence curve  = 2 Critical point (374C, 218 atm) P (atm) Liquid  = 1 l = v coexistence curve  = 2 1 4.58 mm Hg Solid  = 1 Gas  = 1 s = v coexistence curve  = 2 Triple point  = 3 0 0.01 100 T (C) Phase diagram of water

  10. POLYMORPHS Different atomic arrangement at constant composition B : 95,5C and 0,51 Pa C : 115C and 2,4 Pa E : 151C and 1,31108 Pa Phase diagram of sulfur

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

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

  13. TWO-COMPONENT SYSTEM (N = 2) There are tree variables that can affect the phase equilibria of a binary system: T, P, and C P P T c T

  14. Two-component phase diagram 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. BINARY SYSTEM– TYPE 1 • Complete solubility in solid and liquid states • Change of state (s = l) • Metal (Hume-RotheryRule) • Similar crystal structure • Similar atomic volumes • Small values of electronegativity For interstitial solid solutions, the Hume-Rothery rules are: Solute atoms must be smaller than the pores in the solvent lattice. The solute and solvent should have similar electronegativity.

  16. For substitutional solid solutions, the Hume-Rothery rules are: The atomic radii of the solute and solvent atoms must differ by no more than 15%: The crystal structures of solute and solvent must match. 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. 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.

  17. T Liquid  = 1 Liquidus Liquid + solid  = 2 Solidus Solid  = 1 A c B

  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. Similar  Solid solution of naphthalene - -naphthol

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

  21. BINARY SYSTEM– TYPE 2 • Partial or limited solubility  miscibility gap • No change of state  always liquid or always solid T • l • = 1 • s l1 + l2 Coexistence curve  = 2  +  A c B

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

  23. lower critical solution temperature (lower consolute temperature)

  24. Partial or limited solubility • Change of state BINARY SYSTEM– TYPE 3 EUTECTIC DIAGRAM T Liquid solution l  = 1 l +  l +   = 2 Solid solution (B rich)  = 2 Solid solution (A rich)    = 1  = 1   = 2 Eutectic l A c B

  25. 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

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

  27. Phase diagram for system NaCl – H2O

  28. ENTHALPY-COMPOSITION DIAGRAM

  29. Solution A having composition of xAand enthalpy of HA is mixed adiabatically with solution A having composition of xBand enthalpy of HB. The composition and the enthalpy of the mixture is calculated using material balance: Total balance: Component balance: Combining both equations yields:

  30. 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:

  31. 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 MgSO4 by mass at 110F is cooled to 70F. Evaporation and radiation losses may be neglected. SOLUTION (a) Initial solution (A) xA = 0.3 HA = - 31 Btu/lb Cooled system (B) xB = 0.3 HB = - 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 MgSO4.7H2O is in equilibrium with solution.

  32. The crystal MgSO4.7H2O contains: xC = 0.49 From the graph: xA = 0.26 The MgSO4.7H2O crystal yield is 869.6 lb

More Related