1. CHEM171 – Lecture Series Two : 2012/01 PHASE CHEMISTRY AND COLLIGATIVE PROPERTIES  Phase Diagrams  Solutions  Solution Concentrations  Colligative Properties Brown et al., Chapter 10, 385 – 394, Chapter 11, 423-437

2. When a gas is cooled – condenses to liquid phase – temperature called the boiling point temperature, T b Boiling point temperature is pressure dependent. TbTb P PHASE DIAGRAMS CHEM171 – Lecture Series Two : 2012/02

3. Common plot of pressure vs temperature – PHASE DIAGRAM P T LIQUID VAPOUR 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. CHEM171 – Lecture Series Two : 2012/03

4. 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. CHEM171 – Lecture Series Two : 2012/04

5. P T SOLID LIQUID T f = freezing point temperature T P SOLIDLIQUID VAPOUR CHEM171 – Lecture Series Two : 2012/05

6. Transition of a solid directly into a gas is called sublimation. Opposite is called deposition SOLID LIQUID VAPOUR P T X CHEM171 – Lecture Series Two : 2012/06

7. 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: 150T P 0.05 0 SOLID LIQUID VAPOUR CHEM171 – Lecture Series Two : 2012/07

8. 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 CHEM171 – Lecture Series Two : 2012/08 When the liquid and the vapour are at equilibrium then the pressure within the container is known as the VAPOUR PRESSURE of the liquid.

9. 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. CHEM171 – Lecture Series Two : 2012/9

10. Various types of solutions CHEM171 – Lecture Series Two : 2012/10

11. Amount of solute present in a specified amount of solution or solvent Expressed as molarity (M) molality (m) mole fraction (x) SOLUTION CONCENTRATIONS CHEM171 – Lecture Series Two : 2012/11

12. m = number of moles of solute (mol) mass of solvent (Kg) Molality is useful for colligative properties. 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? CHEM171 – Lecture Series Two : 2012/12

13. SOLUTION The molar concentration is 4.27 mol dm -3 For 1.00 dm 3 of solution, mass = ρ × V = 1.25 g mL -1 × 1000 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 × 98.12 g mol -1 = 419 g  mass of H 2 O = mass of solution – mass of acid = 1250 g – 419 g = 831 g = 0.831 kg The molality of the solution is therefore m = 4.27 mol/0.831 Kg = 5.14 m CHEM171 – Lecture Series Two : 2012/13

14. EXERCISE FOR THE IDLE MIND Practice Exercise p 425; Exercise 11.29, 11.30 and 11.31 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 CHEM171 – Lecture Series Two : 2012/14

15. EXAMPLE An aqueous solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in the solution. SOLUTION 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/(0.99 + 3.6) = 0.22 CHEM171 – Lecture Series Two : 2012/15

16. 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. SOLUTION (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 = 10.86 m CHEM171 – Lecture Series Two : 2012/16

17. (b) Moles of water = 500 g/18.02 g mol -1 = 27.75 mol x glycerol = 5.43 mol/(5.43 + 27.75) mol = 0.164 (c) Volume of solution = 1000 g/1.10 g cm -3 = 909.1 cm 3 Molarity = moles/volume of solution = 5.43 mol/0.9091 dm 3 = 5.97 M EXERCISE FOR THE IDLE MIND Exercise 11.32, 11.33 and 11.37 p 444. CHEM171 – Lecture Series Two : 2012/17

18. 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 CHEM171 – Lecture Series Two : 2012/18

19. 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. CHEM171 – Lecture Series Two : 2012/19

20. The lowering of the vapour pressure can be quantified by Raoult’s Law. Vapour pressure of a liquid is a measure of the position of equilibrium between the rate of evaporation and the rate of condensation. 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. CHEM171 – Lecture Series Two : 2012/20

21. 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 1.00 × 10 2 g of water. SOLUTION P * water = 1.00 atm at 100°C Number of moles of water = 100 g/18.02 g mol -1 = 5.55 mol Mole fraction of water 5.55/(5.55 + 2.92 × 10 -2 ) = 0.995 Number of moles of sucrose = 10.0 g/342.3 g mol -1 = 2.92 ×10 -2 mol P solution = x solvent P * solvent = 0.995 × 1.00 atm = 0.995 atm CHEM171 – Lecture Series Two : 2012/21

22. EXAMPLE SOLUTION 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. 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 = 1.731 mol Number of moles of C 10 H 8 = 40.3 g/128 g mol -1 = 0.315 mol x solvent = 1.731/(1.731 + 0.315) = 1.731/2.046 = 0.846 CHEM171 – Lecture Series Two : 2012/22

23. P solution = (0.846)(74.6 Torr) = 63.1 Torr Therefore, V.P. of C 6 H 6 changes by 11.5 Torr VP T solvent solution P T T b solution T b solvent 1 atm Therefore the boiling point temperature of a solvent is elevated upon addition of a non-volatile solute CHEM171 – Lecture Series Two : 2012/23

24. The degree of change in the boiling point temperature can also be quantified by the following equation Δ T b = K b × m where m = molality K b = molal boiling point elevation constant CHEM171 – Lecture Series Two : 2012/24

25. EXAMPLE What is the normal boiling point temperature of a 1.45 mol dm -3 aqueous solution of sucrose? SOLUTION Δ T b = 0.512°C kg mol -1 × 1.45 mol Kg -1 Δ T b = K b × m = 0.742°C The boiling point temperature of this solution is: 100°C + 0.742°C = 100.742°C CHEM171 – Lecture Series Two : 2012/25

26. EXAMPLE What mass of naphthalene, C 10 H 8, must be dissolved in 422 g of nitrobenzene to produce a solution which boils at 213.76°C at 1.00 atm? SOLUTION The normal boiling point of nitrobenzene is 210.88°C and the molal boiling point elevation constant is 5.24°C Kg mol -1. From the equation: Δ T b = K b × m We can calculate Δ T b : Δ T b = (213.76 – 210.88)°C = 2.88°C m = 2.88°C/(5.24°C Kg mol -1 )(1.00) = 0.5496 mol Kg -1 CHEM171 – Lecture Series Two : 2012/26

27. Therefore for every Kg of nitrobenzene we add 0.5496 mol of naphthalene For 422 g of nitrobenzene must add: (0.5496 mol)(0.422 Kg)/1 Kg = 0.232 mol of naphthalene mass of naphthalene = 0.232 mol × 128.16 g mol -1 = 29.7 g CHEM171 – Lecture Series Two : 2012/27

28. 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  T f is always positive  T f = T f (solvent) – T f (solution) Note: CHEM171 – Lecture Series Two : 2012/28

29. EXAMPLE Calculate the normal freezing point temperature of a 1.74 m aqueous solution of sucrose. SOLUTION  T f = 1.86°C Kg mol -1  1.74 mol Kg -1 = 3.24°C.  the freezing point of the solution is (0 – 3.24°C) = -3.24°C CHEM171 – Lecture Series Two : 2012/29

30. 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. SOLUTION  T f = K f (benzene) × m, K f = 5.12°C kg mol -1 T f of pure benzene = 5.48°C Thus, m = (5.48 – 4.92)°C/5.12 Kg mol -1 = 0.109 m  number of moles of solute = (0.109 mol)(0.050 Kg)/1 Kg = 5.469  10 -3 mol Molar mass = 1.20 g/5.469 × 10 -3 mol = 219 g mol -1 CHEM171 – Lecture Series Two : 2012/30

31. 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, Π V = nRT where Π is the osmotic pressure in Pa and R is the gas constant (8.3143 J K -1 mol -1 ). CHEM171 – Lecture Series Two : 2012/31

32. 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 11.12 and 11.13 p436 EXERCISE FOR THE IDLE MIND Practice Exercise p429, 432 (2 questions), 435 and 437. Exercise 11.44, 11.45, 11.51 and 11.54 p 445. CHEM171 – Lecture Series Two : 2012/32