1. PHASE EQUILIBRIUM (GROUP SECOND) 1) Muhammad Shafique 2) Noor Yussuf 3) Noor Nadia Syahira 4) Norfaizah 5)Nur Alya 6) Nur Azahariah 7) Wan Muhammad Hakimi

2. TWO COMPONENT : (LIQUID – LIQUID SYSTEM)

3. Definition & Application of Roult Law Raoult’s law - vapour pressure of the solute-containing solution (P A) is equal to the mole fraction of the solvent (XA) times the vapour pressure of the pure solvent (P*A). P A =X A P* A P SOLUTION =X SOLVENT P* SOLVENT

4. Phase Diagram for Water Pressure/atm Solution Solvent Temp. /K

5. Phase Diagram for Water Pressure/atm Temp. /K ΔTbΔTb Freezing point of the solution ΔTfΔTf Vapour pressure lowering ΔPΔP Boiling point of the solution

6. (a) Vapour-pressure lowering If a nonvolatile solute (one that has no tendency to escape from a solution) is dissolved in a liquid solvent, the vapour pressure of the solvent is lowered.

7. P solution = X solven t P o solvent If the solution contains only one solute: X A + X B =1 P A = (1-X B )P o A P A = P o A - X B P o A (P o A – P A ) = X B P o A (Raoult’s law ) ∆P = X B P o A ∆P A = vapour-pressure lowering X B = the mole fraction of solute P o A = vapour-pressure of the pure solvent

8. Example : The vapour pressure of pure water at 20 ○ C is 25.21 mmHg. If a nonvolatile solute,sucrose is added to a mole fraction of 20%,what is the resulting vapour pressure of the solution at 20 ○ C? Solution P A =X A P* A 0.80 x 17.5mmHg 14 mmHg

9. (b) Boiling point elevation The boiling point of the solution is higher than the boiling point of the pure solvent. ∆T b = boiling point elevation. For dilute solutions, the increase in boiling point depends on the molality of the solute in the solution. ∆T b = k b m ∆T b = boiling point of solution – boiling point of solvent k b = molal boiling point elevation constant

10. (b) Boiling point elevation  T b = T b – T b o T b - boiling point of the pure solvent T b - boiling point of the solution T b > T b o  T b > 0  T b = K b m m - molality of the solution K b - molal boiling-point elevation constant ( o C/m)

11. (c) Freezing point depression The freezing point of the solution is lower than the freezing point of the pure solvent. The freezing point depression = ∆T f. For dilute solutions, the decrease in freezing point depends on the molality of the solute in the solution.

12. Freezing-Point Depression  T f = T f – T f T f - freezing point of the pure solvent T f - freezing point of the solution T f > T f o  T f > 0  T f = K f m m - molality of the solution K f - molal freezing-point depression constant ( 0 C/m)

13. Example : What are the freezing point and boiling point of a solution containing 6.50 g of ethylene glycol (C 2 H 6 O 2 ), commonly used as an automotive antifreeze, in 200 g water?

14. Solution To determine ∆T f and ∆T b. The molality of the solution has to be calculated. The number of moles of C 2 H 6 O 2 is, 6.50g= 0.105 mol 62.1gmol -1 The mass of solvent in kg is, 200 kg = 0.200 kg H 2 O 1000

15. The molality is therefore, = 0.525 m C 2 H 6 O 2 For H 2 O, K f = 1.86 ○ C m -1 K b = 0.51 ○ C m -1

16. The changes in the freezing point and boiling point: ∆T f = (1.86 ○ C/m)(0.525m) = 0.967 ○ C ∆T b = (0.51 ○ C/m)(0.525m) = 0.270 ○ C T f = 0 o C - ∆T f = 0 o C – 0.967 o C The freezing point of the solution (T f ) =  0.976 ○ C ∆T b = boiling point of solution – boiling point of solvent (100 o C) The boiling point of solution = 100.27 ○ C  T f = T f – T f

17. What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.  T f = K f m K f water = 1.86 0 C/m m =m = moles of solute mass of solvent (kg) = 3.202 kg solvent 478 g x 1 mol 62.01 g = 2.41 m  T f = K f m = 1.86 0 C/m x 2.41 m = 4.48 0 C  T f = T f – T f T f = T f –  T f = 0.00 0 C – 4.48 0 C = -4.48 0 C

18. DEFINITION:  Mixture of two or more liquid substances in such a ratio that the composition of the mixture is not changed by simple distillation. AZEOTROPIC MIXTURE

19. POSITIVE AZEOTROPES -minimum boiling mixtures. -Boiling point of an azeotrope is lower than that of any of its constituents.

21. NEGATIVE AZEOTROPES - maximum boiling mixtures. -Boiling point higher than that of any of its constituents.

23. HETEROAZEOTROPE. Two liquids may be nearly immiscible but, when their mixture is boiled, their vapor consists of a fixed ratio of the two and coexists with the two liquid phases. AZEOTROPIC Combinations of solvents that do not form an azeotrope when mixed in any proportion.

24. Phase diagram of a heteroazeotrope. Vertical axis is temperature, horizontal axis is composition. The dotted vertical line indicates the composition of the combined layers of the distillate whenever both layers are present in the original mixture

25. NON-IDEAL SOLUTION & IDEAL SOLUTION

26. 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 : AABB AB = WEAKER THAN SO,the molecules in the solution have lower tendency to escape into vapour phase. Therefore the process is EXOTHERMIC

27. 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. 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. 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. Notice the much bigger difference this time due to the presence of the new ionic interactions Azeotrop e

28. USING THE DIAGRAM Distilling dilute nitric acid Start with a dilute solution of nitric acid with a composition of C 1 and trace through what happens. 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. 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. 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.

29. Distilling nitric acid more concentrated than 68% by mass This time you are starting with a concentration C 2 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. 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. 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.

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

31. 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.

32. 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. It is known as a constant boiling mixture or an azeotropic mixture or an azeotrope. AZEOTROPE SUMMARY 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.

33. Calculation to determine the solution is +ve or –ve deviation If PT (actual) > P (calculated) -ve deviation If PT (actual) = P (calculated) Ideal solution If PT (actual) < P (calculated) +ve deviation EXAMPLE 1: Given that P⁰A is 50 kPa and P⁰B is 25 kPa.XA is 0.3 A solution of A and B has a vapour pressure of 60 kPa. Determine whether the solution is ideal or not? SOLUTION: PA=XA (P⁰A) PB =XB(P⁰B) PT (calculated)=PA+PB =(0.3)(50) =(1-0.3)(25) =15kPa+17.5kPa =15 kPa =(0.7)(25) =32.5 kPa =17.5 kPa PT(actual) > PT(calculated) Therefore, it is a non-ideal solution and has a +ve deviation (:

34. EXAMPLE 2: Given that P⁰A is 32 kPa and P⁰B is 60 kPa A solution has a vapour pressure 50 kPa contains 12 mol of A and 15 mol of B. Does this solution behave ideally? SOLUTION: PA=XA(P⁰A) PB=XB(P⁰B) PT(calculated)=14.22+33.60 =[12÷(12+15)] [32] =[15(12+15)] [60] =47.8 kPa =14.22 kPa =33.60 kPa As PT(actual) > PT(calculated) Therefore, the solution is non-ideal solution. The answer is, NO,the solution doesn’t behave ideally.

35. Principle in simple distillation & fractional distillation of a Binary Liquid Mixture Prepared By : Noor Yussuf b. Othman Nur Alya bt. Razak

36. What is Simple Distillation?? > The process of heating and condensing to purify substances

37. SIMPLE DISTILLATION Is used to separate a mixture of liquid with difference in boiling point It is less effective than fractional distillation as the distillate is still a mixture of both components The principle is to conduct multiple, little distillations. Each vaporisation or condensation cycle will result in the new vapour phase being enriched in the lower boiling point (higher vapour pressure component) By using more of these cycles, two components with different boiling points can be separated.

38. Simple distillation from pure water from salt water.

39. 1. The impure liquid is heated in distillation flask. Since water vaporise at much lower temperature than salt, pure water vapour is formed. 2. Water vapour is cooled in condenser and condenses back into liquid water which is collected in another flask. The collected water is pure water and does not contain any salt. 3. The liquid obtained through condensation of vapour in the distillation process is known as distillate. 4. The salt behind in distillation flask is known as residue.

40. Fractional distillation Definition : separation of a mixture into its component parts or fractions

41. Fractional distillation Separates chemical compounds by their boiling points by heating them to a temperature at which several fractions of the compound will evaporate Generally the components parts boil at less than 25’C from each other under one atmospheric pressure

42. A fractionating column is inserted between the distillation flask and the distillation head It provides a large surface area in which mixture can be continuously vaporized and condensed

43. Principle of fractional distillation As the vapours ascend the column from the boiling mixture below, the high boiling components are condensed and returned to the flask, the ascending column of vapour thus being steadily ‘scrubbed’ by the descending column of liquid condensate The ascending column of the vapour becomes therefore steadily richer in the lowest boiling component, and the descending column of condensate steadily richer in the highest boiling component

45. Boiling point – composition diagram WXY TwTw TxTx TBTB TATA

46. Boiling point-composition diagram A has a lower boiling point so A is more volatile than B. Assume we have a liquid mixture with composition W. This mixture will boil at T w and give a vapour of composition X, which is richer in A. If this vapour of composition of X is condensed, it will form a liquid which also has composition X.

47. Boiling point-composition diagram This new liquid will boil at temperature T x (which is lower than T w ) to give a vapour of composition Y, which condenses to give a liquid of composition Y. This liquid is even richer in A The coiling and condensation cycle can be repeated until a vapour consisting of pure A is abtained which can be condensed into a liquid of pure A. Pure A is the distillate. The less volatile liquid remaining in the flask is pure B and is known as the residue.

48. Vapour pressure-composition diagram

49. Assume you have an ideal mixture of two liquids, A and B. Each of A and B is making its own contribution to the overall vapour pressure of the mixture. Suppose you double the mole fraction of A in the mixture. According to Raoult’s Law, you will double its partial vapour pressure. If you triple the mole fraction, its partial vapour pressure will triple, and so on.

50. Vapour pressure-composition diagram In other words, the partial vapour pressure of A at a particular temperature is proportional to its mole fraction. If you plot a graph of the partial vapour pressure of A against its mole fraction, you will get a straight line sloping up. Now we will plot the line for B. The mole fraction of B falls as A increases so the line will slope down. As the mole fraction of B falls, its vapour pressure will fall at the same rate.

51. Vapour pressure-composition diagram Notice that the vapour pressure of pure B is way higher than A. That means that molecules must break away more easily form the sureface of B than A. B is more volatile than A. To get the total vapour pressure, you need to add the values of A and B together at each composition. The net effect of that is to give you a straight line as shown in the diagram.