PHASE EQUILIBRIUM

1. They provide us with the knowledge of phase composition and phase stability as a function of temperature (T), pressure (P) and composition (C). PHASE EQUILIBRIUM one of the most important sources of information concerning the behavior of elements, compounds and solutions.

2. Chapter outlines

3. DEFINITION COMPONENTS PHASE PHASE RULE: DEGREE OF FREEDOM

4. The mixture of ice and water = have two phase which is solid and liquid The mixture of oxygen gas and nitrogen gas = have one phase which is gas phase (the system is homogen) The mixture of oil and water = have 2 same phase(liquid). Oil and water are not homogen and have the boundaries to separate both phase CaCO3(s) CaO(s) + CO2(g) = 3 phase (2 solid,1 gas) • A chemically and structurally homogeneous portion of material • Separated with other parts of the system PHASE • Determines the number of independent variables needed • Is the correlation between the number of phase (P), components (C), and degree of freedom

5. OR

6. PHASE RULES F = C – P +2 Degree of freedom or the number of independent variables The number of phase 2 variables (temperature and pressure) Number of component

7. DEGREE OF FREEDOM (F) The number of variables that may be changed independently without causing the appearance of a new phase or disappearance of an existing phase UNIVARIANT TYPES BIVARIANT EXAMPLES Calculate the degree of freedom (F) CaCO3(s) CaO(s) + CO2(g) The number of components is not always easy to determine at first glance, and it may require careful examination of the pyhsical conditions of the system at equilibrium F = C – P +2 = 2 – 3 + 2 = 1 (univariant) Means: only one variable, either temperature or pressure can be changed independently

8. Standard phase diagram for water (H2O) ONE COMPONENT SYSTEM Standard phase diagram for carbon dioxide (CO2)

9. H2O Standard phase diagram for one component system CO2 Critical point ??? B What does it means by: 1)AO curve 2)OB curve 3)OC curve 4)AOB curve 5)BOC curve 6)AOC curve O A ?????

10. Standard phase diagram for water (H2O) TA curve = known as melting point or freezing point Special case !!!!! Represent the equilibrium between ice and liquid Has a negative slope Water as the liquid is denser than the solid (ice floats on water). That means that an increase of pressure favors the formation of liquid and that the melting point of water falls with increasing pressure. According to Le Chatelier’s principle, when pressure is applied, the reaction shifts in the direction that can release the stress and cause ice to melt This unique properties of water is due to the network of hydrogen bonding in ice is more extensive than in liquid

11. LEARNING CHECK !!!!! (d) 50 (c) (b) (a)

12. LEARNING CHECK !!!!! 50 (a) (b) (c) (d) (e)

13. The point O is the triple point for CO2 (at 5.1 atm, -57o C). So, CO2 solid can’t changed to liquid form at 1 atm. Standard phase diagram for carbon dioxide (CO2) Critical point Has a positive slope So, its shows that the increases of pressure, will increased the melting point for CO2 solid O sublimation

14. Raoult’s law – in explaining the effect of non-volatile solute on vapour pressure of solvent and its melting and boiling point Two completely miscible liquid – ideal, non-ideal, positive and negative deviation TWO COMPONENT SYSTEM Eutectic system and cooling curves Composition diagram vs boiling point composition for ideal, non-ideal, negative n positive deviation Fractional distillation and azeotropic system

15. VAPOUR PRESSURE

16. Vapour pressure of a liquid Vapour pressure increases with increasing temperature due to its KE When a liquid evaporates in a closed vessel, its gaseous molecules formed above the liquid have high KE and exert a vapour pressure. The molecules collide with the pinston and push the pinston upward sublimation Microscopic equilibrium between gas and liquid. Note that the rate of evaporation of the liquid is equal to the rate of condensation of the gas. Microscopic equilibrium between gas and solid. Note that the rate of evaporation of the solid is equal to the rate of condensation of the gas.

17. Vapour pressure of a volatile liquid Volatile liquid is a liquid that can easily evaporate at one atmospheric pressure and room temperature Molecules of volatile liq escape the liquid phase into gaseous phase.(KE) A volatile liquid has a strong tendency to vapourize or evaporate into vapour, creating high vapour pressure. On contrary a less volatile liquid has low vapour pressure because of lower tendency to vapourize

18. Factor that affect Vapour pressure 1 ethyl ether (C4H10O)Pvapor (25oC) = 520 torr • Types of Molecules: the types of molecules that make up a solid or liquid determine its vapor pressure. If the intermolecular forces between molecules are: • relatively strong, the vapor pressure will be relatively low. • relatively weak, the vapor pressure will be relatively high. ethyl alcohol (C2H6O)Pvapor (25oC) = 75 torr Low Temperature 2 Temperature: at a higher temperature, more molecules have enough energy to escape from the liquid or solid. At a lower temperature, fewer molecules have sufficient energy to escape from the liquid or solid. High Temperature

19. COMPLETELY MISCIBLE LIQUID RAOULT’S LAW

20. 3 types : Complete Miscible liquid Half miscible liquid Immiscible liquid Methanol and ethanol Liquid solution in liquid ether and water Oil and water Ideal solution- mostly involve the substance that have similar physicochemical properties. Ex: MeOH/EtOH, benzene/toluene, n-hexane/n-heptane A A A B Complete Miscible liquid • 2 types of complete miscible liquid which is ideal and non-ideal solution • An ideal solution is a solution that obeys Raoult’s law and non-ideal solution disobey. • A solution is a ideal solution when: • The intermolecular attractions between the mixture of same molecule with the the mixture of different molecule are equal. • The volume of the mixture are the total volume of both liquid (volume of liquid A add with volume of liquid B) • No heat changes (no endo-exothermic process) • Obeys Raoult’s law =

21. Raoult’s law – ideal solution • Relationship between vapour pressure of a solvent and its mole fraction • States the vapour pressure of the solute containing solution (PA) is equal to the mole fraction of the solvent (XA) times the vapour pressure of the pure solvent (PoA) PA = XA PoA Psolution= Xsolvent Posolvent PT = PA + P B @ Ptotal= XAPoA + XBPoB EXAMPLE !!!!! A solution is prepared by adding 2.0 mole of glucose in 15.0 mole of water at 25 oC. The vapour pressure of pure water at 25 oC is 23.76 mmHg. Calculate the vapour pressure of the solution at 25 oC Psolution= Xsolvent Posolvent The mole fraction: XA = nA / nt = 15.0 /17.0= 0.88 Therefore, Psolution= 0.88 x 23.76 mmHg = 20.96 mmHg

22. EXAMPLE !!!!! At 25 oC the vapour pressure of pure benzene and toluene are 93.4 mmHg and 26.9 mmHg. If the mixture contains 60.0g of benzene and 40.0 g of toluene, calculate the vapour pressure of this solution PT = PA + P B @ Ptotal= XAPoA + XBPoB Calculate the number of moles for benzene and toluene: n of benzene : PA + P B = 60/78 = 0.77 mole n of toluene : PA + P B = 40/92 = 0.43 mole Ptotal= XAPoA + XBPoB = 0.77 x 93.4 mmHg + 0.43 x 26.9 mmHg 0.77 + 0.43 0.77 + 0.43 = 69.54 mmHg

23. IDEAL SOLUTION DIAGRAMS (RAOULT’S LAW) VAPOUR PRESSURE/ COMPOSITION DIAGRAM BOILING POINT/ COMPOSITION DIAGRAM

24. Vapour pressure/composition diagram(ideal solution) • There is actually no such thing as an ideal mixture! However, some liquid mixtures get fairly close to being ideal. These are mixtures of two very closely similar substances. Commonly quoted examples include: • hexane and heptane • benzene and methylbenzene • propan-1-ol and propan-2-ol Pure mixture vapour pressure Notice that the vapour pressure of pure B is higher than that of pure A. That means that molecules B must break away more easily than of A. B is the more volatile liquid. Pure vapour pressure Total vapour pressure of the mixture

25. Boiling point/composition diagram We'll start with the boiling points of pure A and B.B has the higher vapour pressure. That means that it will have the lower boiling point. The diagram just shows what happens if you boil a particular mixture of A and B. Notice that the vapour over the top of the boiling liquid has a composition which is much richer in B - the more volatile component. VP BP

26. EXAMPLE !!!!! Which vapour sample rich at this point? composition

27. NON-IDEAL DIAGRAMS • NEGATIVE DEVIATION • POSITIVE DEVIATION

28. Vapour composition of a negative deviation solution • Involves the intermolecular forces between molecules in solution are stronger than those in pure liquid • Therefore, vapour pressure of the solution is lower than vapour pressure of its components or pure liquid. • Example : A A B B A B = WEAKER THAN • SO,the molecules in the solution have lower tendency to escape into vapour phase. • Therefore the process is EXOTHERMIC

29. Nitric acid and water form mixtures in which particles break away to form the vapour with much more difficulty than in either of the pure liquids. Azeotrope That means that mixtures of nitric acid and water can have boiling points higher than either of the pure liquids because it needs extra heat to break the stronger attractions in the mixture. In the case of mixtures of nitric acid and water, there is a maximum boiling point of 120.5°C when the mixture contains 68% by mass of nitric acid. That compares with the boiling point of pure nitric acid at 86°C, and water at 100°C. Notice the much bigger difference this time due to the presence of the new ionic interactions

30. USING THE DIAGRAM As the acid loses water, it becomes more concentrated. Its concentration gradually increases until it gets to 68% by mass of nitric acid. At that point, the vapour produced has exactly the same concentration as the liquid, because the two curves meet. Distilling dilute nitric acid Start with a dilute solution of nitric acid with a composition of C1and trace through what happens. You produce a constant boiling mixture (or azeotropic mixture or azeotrope). If you distil dilute nitric acid, that's what you will eventually be left with in the distillation flask. You can't produce pure nitric acid from the dilute acid by distilling it. The vapour produced is richer in water than the original acid. If you condense the vapour and reboil it, the new vapour is even richer in water. Fractional distillation of dilute nitric acid will enable you to collect pure water from the top of the fractionating column.

31. Distilling nitric acid more concentrated than 68% by mass This time you are starting with a concentration C2 to the right of the azeotropic mixture. The vapour formed is richer in nitric acid. If you condense and reboil this, you will get a still richer vapour. If you continue to do this all the way up the fractionating column, you can get pure nitric acid out of the top. As far as the liquid in the distillation flask is concerned, it is gradually losing nitric acid. Its concentration drifts down towards the azeotropic composition. Once it reaches that, there can't be any further change, because it then boils to give a vapour with the same composition as the liquid. Distilling a nitric acid / water mixture containing more than 68% by mass of nitric acid gives you pure nitric acid from the top of the fractionating column and the azeotropic mixture left in the distillation flask.

32. Vapour composition of a positive deviation solution Formed when the intermolecular forces between molecules in the mixture are weaker than those in pure liquids. A A B B A B = The solution has a greater tendency to evaporate or escape into vapour Vapour pressure of the solution is higher than expected STRONGER THAN The process is endothermic

33. A large positive deviation from Raoult's Law produces a vapour pressure curve with a maximum value at some composition other than pure A or B. If a mixture has a high vapour pressure it means that it will have a low boiling point The molecules are escaping easily and you won't have to heat the mixture much to overcome the intermolecular attractions completely. The implication of this is that the boiling point / composition curve will have a minimum value lower than the boiling points of either A or B.

34. USING THE DIAGRAM Suppose you are going to distil a mixture of ethanol and water with composition C1 as shown on the next diagram. It will boil at a temperature given by the liquid curve and produce a vapour with composition C2. When that vapour condenses it will, of course, still have the composition C2. If you reboil that, it will produce a new vapour with composition C3.

35. This particular mixture of ethanol and water boils as if it were a pure liquid. It has a constant boiling point, and the vapour composition is exactly the same as the liquid. AZEOTROPE It is known as a constant boiling mixture or an azeotropic mixture or an azeotrope. SUMMARISE • Distilling a mixture of ethanol containing less than 95.6% of ethanol by mass lets you collect: • A distillate containing 95.6% of ethanol in the collecting flask (provided you are careful with the temperature control, and the fractionating column is long enough • Pure water in the boiling flask.

36. FRACTIONAL DISTILLATION

37. Typical fractional distillation in the lab the thermometer bulb is placed exactly at the outlet from the fractionating column to give the maximum possible surface area for vapour to condense on Some fractionating columns have spikes of glass sticking out from the sides which serve the same purpose In some cases, where you are collecting a liquid with a very low boiling point, you may need to surround the collecting flask with a beaker of cold water or ice.

38. Relating what happens in the fractionating column to the phase diagram Boil a mixture with composition C1. Which compound that rich at this point? What phase? Which compound that rich at this point? What phase The vapour over the top of the boiling liquid will be richer in the more volatile component, and will have the composition C2.

39. Which compound that rich at this point? What phase? Each time the vapour condenses to a liquid, this liquid will start to trickle back down the column where it will be reboiled by up-coming hot vapour. Each time this happens the new vapour will be richer in the more volatile component. The aim is to balance the temperature of the column so that by the time vapour reaches the top after huge numbers of condensing and reboiling operations, it consists only of the more volatile component - in this case, B. The boiling points of the two liquids. The closer they are together, the longer the column has to be.

40. what is the point of the packing in the column? • To make the boiling-condensing-reboiling process as effective as possible, it has to happen over and over again. • By having a lot of surface area inside the column, you aim to have the maximum possible contact between the liquid trickling down and the hot vapour rising. • If you didn't have the packing, the liquid would all be on the sides of the condenser, while most of the vapour would be going up the middle and never come into contact with it.

41. BOILING POINT ELEVATION • FREEZING POINT DEPRESSION

42. BOILING POINT : is the temperature at which its vapour pressure equals the external pressure WHY?? Boiling point of pure solvent Boiling point solvent is higher than boiling point of solution ∆Tb = Tb - T ob Boiling point of solution Boiling point elevation ∆Tb is proportional to molality of solute in the solution, so…. ∆Tb = Kb m Molal (m) = mol solute Kg solvent Molar mass of solute = Kb (gram solute) (∆Tb ) (kg solvent) The molal boiling point elevation constant with unit oC/m or oC kg/mol

43. EXAMPLES: What is the boiling point elevation when 11.4 g of ammonia (NH3) is dissolved in 200. g of water? Kb for water is 0.52 °C/m. • Determine molality of 11.4 g of ammonia in 200. g of water: • 11.4 g / 17.031 g/mol = 0.6693676 mol • 0.6693676 mol / 0.200 kg = 3.3468 m 2) Determine bp elevation: Δt = Kbm Δt = (0.52 °C/m) (3.3468 m) Δt = 1.74 °C

44. EXAMPLES: Calculating Molecular Mass (Formula Weight) of Solute 1.15g of an unknown, nonvolatile compound raises the boiling point of 75.0g benzene (C6H6) by 0.275oC. Calculate the molecular mass (formula weight) of the unknown compound. Calculate the molality of solute particles:m = ΔTb ÷ KbΔTb = 0.275oC Kb = 2.53oCm-1 (from table above) m = 0.275 ÷ 2.53 = 0.109m Calculate the moles of solute present:molality = moles solute ÷ kg solvent n(solute) = m x kg solvent = 0.109 x 75.0 x 10-3 = 8.175 x 10-3 mol Calculate the molecular mass (formula weight) of the solute: n(solute) = mass(solute) ÷ MM(solute) MM(solute) = mass(solute) ÷ n(solute) = 1.15 ÷ 8.175 x 10-3 = 141 g/mol

45. FREEZING POINT DEPRESSION : is the temperature at which solid begins to appear in liquid or solution Boiling point of solution ∆Tf = T of - Tf T of > Tf Boiling point of pure solution ∆Tf is proportional to molality of solute in the solution, so…. ∆Tf = Kfm Molal (m) = mol solute Kg solvent Molar mass of solute = Kf (gram solute) (∆Tf ) (kg solvent)

46. the boiling point of the solvent in a solution is higher than that of the pure solvent; • the freezing point (melting point) of the solvent in a solution is lower than that of the pure solvent.

47. Learning check !!!!!! Automobile antifreeze is ethylene glycol, C2H6O2. It is a non-electrolyte. If a radiator contains 40.0% antifreeze and 60.0% water, by mass, what is the freezing point of the solution in the radiator? The normal freezing point for water is 0.0 °C and Kf is 1.86 °C mol/kg. Find the molality of the solution. The mass of the solvent is 0.0600 kg and the formula weight of the solute is 62.066 g/mol. Use the freezing point depression formula

48. It’s the BONUS TIME AND YOUR HAPPY HOUR. THIS IS YOUR HOMEWORKS TO FIND ABOUT EUTECTIC MIXTURE AND COOLING CURVES EUTECTIC MIXTURE AND COOLING CURVES YOU HAVE TO MAKE A SHORT NOTES ON THIS SUBTOPIC AND SUBMIT IT TO YOUR ‘KAWAII’ LECTURER GOOD LUCK GUYS