PHASE CHEMISTRY AND COLLIGATIVE PROPERTIES

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1 PHASE CHEMISTRY AND COLLIGATIVE PROPERTIES Phase Diagrams Solutions Solution Concentrations Colligative Properties Brown et al., Chapter 10, , Chapter 11, CHEM120 Lecture Series Two : 2011/01

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3 Phase Diagrams Solid Liquid Freezing/Melting Solid Vapour Sublimation/Deposition Liquid Vapour Boiling/Condensing Triple point (O) - 3 phases in equilibrium Critical Point (C) - Above T C & P C supercritical fluid Ref: Brown et al., chapter 10.6, pages

4 PHASE DIAGRAMS When a gas is cooled condensed or liquid phase temperature called the boiling point temperature, T b Boiling point temperature is pressure dependent. T b P CHEM120 Lecture Series Two : 2011/02

5 Common plot of pressure vs temperature PHASE DIAGRAM P LIQUID VAPOUR T T b = boiling point temperature On the line, both phases are in equilibrium. Gases have 3 degrees of freedom of movement; rotate, vibrate and translate. CHEM120 Lecture Series Two : 2011/03

6 Molecules lose a degree of freedom (rotation) when they move from the gaseous state to the liquid state. When liquid solidifies, the molecules lose their translational motion, they are in fixed positions within the solid structure. When we cool a liquid, the molecules become part of a rigid structure, a solid and the temperature at which this takes place is called the freezing point temperature. Freezing point temperature is dependent on the pressure, if the pressure is increased, the denser phase forms readily. As the pressure increases, the liquid freezes sooner and the freezing point temperature increases. CHEM120 Lecture Series Two : 2011/04

7 P SOLID LIQUID T f = freezing point temperature P SOLID LIQUID VAPOUR T T CHEM120 Lecture Series Two : 2011/05

8 Transition of a solid directly into a gas is called sublimation. Opposite is called deposition P SOLID LIQUID VAPOUR X T CHEM120 Lecture Series Two : 2011/06

9 The solid/vapour and liquid/vapour lines are very important when understanding the concept of the vapour pressure of a substance. Let s take an imaginary substance with the following phase diagram: P 0.05 SOLID LIQUID 0 VAPOUR 150 T CHEM120 Lecture Series Two : 2011/07

10 If we place some of our substance in the liquid phase into an evacuated container (i.e. P ~ 0) at 150 C, on the phase diagram we are in the region where our substance wants to be a vapour. Therefore our substance will start to vapourise and molecules will go from the liquid phase to the vapour phase. As this happens, the pressure inside our container will increase as more gas molecules will collide with the sides of the container. This cannot go on indefinitely, in fact, both the liquid and vapour phases will be present at equilibrium when P = 0.05 and T = 150 C CHEM120 Lecture Series Two : 2011/08

11 When the liquid and the vapour are at equilibrium then the pressure within the container is known as the VAPOUR PRESSURE of the liquid. CHEM120 Lecture Series Two : 2011/09

12 SOLUTIONS A solution is a homogeneous mixture of a solute in a solvent. A homogeneous mixture is one in which the composition is the same throughout the solution. There is also only one phase throughout the mixture An aqueous solution is one in which water the solvent and the dissolved substance, the solute. CHEM120 Lecture Series Two : 2011/10

13 Various types of solutions CHEM120 Lecture Series Two : 2011/11

14 SOLUTION CONCENTRATIONS Amount of solute present in a specified amount of solution or solvent Expressed as molarity (M) molality (m) mole fraction (x) CHEM120 Lecture Series Two : 2011/12

15 Solution Concentrations Mole Fraction Ratio of the moles of one component to the total number of moles present X A = Moles of A Total Number of Moles Molality Ratio of the moles of the solute per mass of SOLVENT /kg m = Moles of Solute (mol) Mass of Solvent (kg) 15

16 Molality is useful for colligative properties. m = number of moles of solute (mol) mass of solvent (kg) EXAMPLE The acid that is used in car batteries is 4.27 mol dm -3 aqueous sulfuric acid, which has a density of 1.25 g per millilitre. What is the molality of the acid? CHEM120 Lecture Series Two : 2011/13

17 ANSWER The molar concentration is 4.27 mol dm -3 For 1.00 dm 3 of solution, mass = ρ V = 1.25 g ml ml = 1250 g We know this solution contains 4.27 mol of H 2 SO 4 mass of H 2 SO 4 = 4.27 mol g mol -1 = 419 g mass of H 2 O = mass of solution mass of acid = 1250 g 419 g = 831 g = kg The molality of the solution is therefore m = 4.27 mol/0.831 Kg = 5.14 m CHEM120 Lecture Series Two : 2011/14

18 EXERCISE FOR THE IDLE MIND Practice Exercise p 425; Exercise 11.29, and p 444. Mole fraction is the number of moles of individual component divided by total number of moles Mole fraction (x A ) = moles A total moles in solution CHEM120 Lecture Series Two : 2011/15

19 EXAMPLE An aqueous solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in the solution. ANSWER In 100 g of acid, we have 36 g HCl and 64 g H 2 O Moles of HCl = (36 g/36.5 g mol -1 ) = 0.99 mol Moles of H 2 O = (64 g/18 g mol -1 ) = 3.6 mol x HCl = moles HCl/total moles = 0.99/( ) = 0.22 CHEM120 Lecture Series Two : 2011/16

20 EXAMPLE (Practice Exercise p 427) A solution containing equal masses of glycerol (C 3 H 8 O 3 ) and water has a density of 1.10 g cm -3. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol and (c) the molarity of glycerol in the solution. ANSWER (a) For a 1000 g solution, have 500 g of glycerol and 500 g of water. Moles of glycerol = 500 g/92.08 g mol -1 = 5.43 mol Molality = 5.43 mol/0.5 Kg = m CHEM120 Lecture Series Two : 2011/17

21 (b) Moles of water = 500 g/18.02 g mol -1 = mol x glycerol = 5.43 mol/( ) mol = (c) Volume of solution = 1000 g/1.10 g cm -3 = cm 3 Molarity = moles/volume of solution = 5.43 mol/ dm 3 = 5.97 M EXERCISE FOR THE IDLE MIND Exercise 11.32, and p 444. CHEM120 Lecture Series Two : 2011/18

22 COLLIGATIVE PROPERTIES Physical properties of solutions that depend primarily on the number of particles present and not on their nature, e.g. vapour pressure lowering boiling point elevation freezing point depression osmosis CHEM120 Lecture Series Two : 2011/19

23 Colligative Properties - Summary COLLIGATIVE PROPERTIES Depend on number of particles present in solution Molality Vapour Pressures Phase Changes van't Hoff factor Osmotic Pressure Raoult's Law P = X P o (Daltons law & X = 1) Boiling Point Elevation T b = K b m Freezing Point Depression T f = K f m V = nrt

24 Vapour Pressure The pressure exerted by a vapour when it is in dynamic equilibrium with its liquid at a fixed temperature. Vapour vs Gas Vapour can be liquified by increasing the pressure at constant temperature but a gas cannot. 24

25 A liquid in a closed container will establish an equilibrium with its vapour; pressure exerted by vapour in equilibrium is vapour pressure. Substance with a vapour pressure is volatile. Substance with no measurable vapour pressure is nonvolatile. CHEM120 Lecture Series Two : 2011/20

26 QUESTION Consider the following data and determine which is the more volatile; ethanol or methanol. Explain your answer. C 2 H 5 OH CH 3 OH T ( C) VP (Torr) T ( C) VP (Torr) CHEM120 Lecture Series Two : 2011/21

27 Vapour Pressure Raoult s Law The vapour pressure of a component in a solution is directly proportional to the mole fraction of that component in the solution. P A = X A x P o A Where P A vapour pressure of A above the soln X A the mole fraction of component A in the soln P o A the vapour pressure of pure component A 27

28 Vapour pressure of a liquid is a measure of the position of equilibrium between the rate of evaporation and the rate of condensation. The lowering of the vapour pressure can be quantified by Raoult s Law. Raoult s Law: The vapour pressure of the solvent in a solution containing a non-volatile solute is directly proportional to the mole fraction of the solvent in the solution. P solution = x solvent P* solvent P* is the vapour pressure of the pure solvent. CHEM120 Lecture Series Two : 2011/22

29 Raoult s Law - Assumption Raoult s law assumes no intermolecular forces of either attraction or repulsion between the molecules present. Vapour Pressure 29

30 EXAMPLE Calculate the vapour pressure (atm) of an aqueous solution at 100 C which contains 10.0 g of sucrose, C 12 H 22 O 11, in g of water. ANSWER P * water = 1.00 atm at 100 C Number of moles of water = 100 g / g mol -1 = 5.55 mol Number of moles of sucrose = 10.0 g / g mol -1 = mol Mole fraction of water 5.55 / ( ) = = x solvent P * solvent = atm = atm P solution CHEM120 Lecture Series Two : 2011/23

31 Vapour Pressure Raoult s Law - 2 Volatile components(a & B) The total pressure above the solution is the sum of the individual partial pressures as calculated using Raoult s law. Dalton s law of Partial pressures P total = P A + P B Raoult s Law P A = X A x P o A & P B = X B x P o B where X B = 1 - X A 31

32 EXAMPLE Calculate by how much the vapour pressure of the solvent changes when 40.3 g of naphthalene, C 10 H 8, is added to 135 g of benzene, C 6 H 6, at 20 C. The vapour pressure of benzene at 20 C is 74.6 Torr. Consider naphthalene to be non-volatile for this problem. ANSWER P solution = x solvent P * solvent (P * solvent = 74.6 Torr) Number of moles of C 6 H 6 = 135 g/78 g mol -1 = mol Number of moles of C 10 H 8 = 40.3 g/128 g mol -1 = mol x solvent = / ( ) = 1.731/2.046 = CHEM120 Lecture Series Two : 2011/24

33 P solution = (0.846)(74.6 Torr) = 63.1 Torr Therefore, V.P. of C 6 H 6 changes by 11.5 Torr VP solvent P 1 atm solution T T b solvent T b solution T Therefore the boiling point temperature of a solvent is elevated upon addition of a non-volatile solute CHEM120 Lecture Series Two : 2011/25

34 Boiling Point Elevation Quantification ΔT b = K b x m where ΔT b m K b = T b (solution) T b (solvent) = molality (moles solute per mass solvent) = molal boiling point elevation constant or ebullioscopic constant Note: K b is a constant for the solvent involved 34

35 The degree of change in the boiling point temperature can also be quantified by the following equation ΔT b = K b m where m = molality K b = molal boiling point elevation constant CHEM120 Lecture Series Two : 2011/26

36 EXAMPLE What is the normal boiling point temperature of a 1.45 mol dm -3 aqueous solution of sucrose? ANSWER ΔT b = K b m ΔT b = C kg mol mol Kg -1 = C The boiling point temperature of this solution is: 100 C C = C CHEM120 Lecture Series Two : 2011/27

37 EXAMPLE What mass of naphthalene, C 10 H 8, must be dissolved in 422 g of nitrobenzene to produce a solution which boils at C at 1.00 atm? ANSWER The normal boiling point of nitrobenzene is C and the molal boiling point elevation constant is 5.24 C Kg mol -1. From the equation: ΔT b = K b m We can calculate ΔT b : ΔT b = ( ) C = 2.88 C m = 2.88 C / (5.24 C Kg mol -1 )(1.00) = mol Kg -1 CHEM120 Lecture Series Two : 2011/28

38 Therefore for every Kg of nitrobenzene we add mol of naphthalene For 422 g of nitrobenzene must add: ( mol)(0.422 Kg)/1 Kg = mol of naphthalene mass of naphthalene = mol g mol -1 = 29.7 g CHEM120 Lecture Series Two : 2011/29

39 Freezing Point Depression Quantification ΔT f = K f x m P Solution Pure solvent where ΔT f m K f = T f (solution) T f (solvent) = molality (moles solute per mass solvent) = molal freezing point depression constant or cryoscopic constant Note: K f is a constant for the solvent involved T 39

40 The addition of a non-volatile solute will decrease or depress the freezing point temperature of a solvent. The effect is quantified by the equation: T f = K f m Note: T f is always positive T f = T f (solvent) T f (solution) CHEM120 Lecture Series Two : 2011/30

41 EXAMPLE Calculate the normal freezing point temperature of a 1.74 m aqueous solution of sucrose. ANSWER T f = 1.86 C Kg mol mol Kg -1 = 3.24 C. the freezing point of the solution is ( C) = C CHEM120 Lecture Series Two : 2011/31

42 EXAMPLE 1.20 g of an unknown organic compound was dissolved in 50.0 g of benzene. The solution had a T f of 4.92 C. Calculate the molar mass of the organic solute. ANSWER T f = K f (benzene) m, K f = 5.12 C kg mol -1 T f of pure benzene = 5.48 C Thus, m = ( ) C/5.12 Kg mol -1 = m number of moles of solute = (0.109 mol)(0.050 Kg)/1 Kg = mol Molar mass = 1.20 g/ mol = 219 g mol -1 CHEM120 Lecture Series Two : 2011/32

43 Freezing Point Depression & Boiling Point Elevation 43

44 Osmotic Pressure Osmotic Pressure The pressure that would be applied to a solution to stop the passage through a semipermeable membrane of solvent molecules from the pure solvent. ΠV = nrt OR where Π = osmotic pressure V = volume n = moles R = Universal Gas Constant ( L atm K -1 mol -1 or J K -1 mol -1 ) T = temperature (K) Π = n x RT = M x RT V 44

45 Osmosis - tendency of solvent molecules to pass through a semipermeable membrane from a more dilute to a more concentrated solution Osmotic pressure - the excess hydrostatic pressure on the solution compared to pure solvent. Reverse Osmosis - if a pressure greater than the osmotic pressure is exerted on the solution, the solvent passes back through the membrane to the dilute side. When a solution containing n moles of solute in a volume V m 3 is in contact with the pure solvent at a temperature T K, ΠV = nrt where Π is the osmotic pressure in Pa and R is the gas constant ( J K -1 mol -1 ). CHEM120 Lecture Series Two : 2011/33

46 The colligative properties of solutions provide a useful means of experimentally determining molar mass. Any of the four properties can be used. Refer to Sample Exercises and p436 EXERCISE FOR THE IDLE MIND Practice Exercise p429, 432 (2 questions), 435 and 437. Exercise 11.44, 11.45, and p 445. CHEM120 Lecture Series Two : 2011/34

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