PHASE CHEMISTRY AND COLLIGATIVE PROPERTIES Phase Diagrams Solutions Solution Concentrations Colligative Properties Brown et al., Chapter 10, 385 394, Chapter 11, 423-437 CHEM120 Lecture Series Two : 2013/01
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 : 2013/02
Common plot of pressure vs temperature PHASE DIAGRAM P LIQUID VAPOUR T T b = boiling p On the line, both phases are in equilibrium. Shows pressures and temperatures at which gaseous, liquid, and solid phases can exist. Allows us to predict phase of substance stable at any given T and P Plus melting, boiling pt, etc. Triple point: all 3 phases are in equilibrium Gases have 3 degrees of freedom of movement; rotate, vibrate and translate. CHEM120 Lecture Series Two : 2013/03
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: 2013/04
P SOLID LIQUID T f = freezing point temperature P SOLID LIQUID VAPOUR T T CO 2 H 2 O solid gas solid liquid gas CHEM120 Lecture Series Two : 2013/05
Transition of a solid directly into a gas is called sublimation. Opposite is called deposition P SOLID X LIQUID VAPOUR T CHEM120 Lecture Series Two : 2013/06
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 : 2013/07
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 : 2013/08
When the liquid and the vapour are at equilibrium then the pressure within the container is known as the VAPOUR PRESSURE of the liquid. CHEM120 Lecture Series Two : 2013/09
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 : 2013/10
Various types of solutions CHEM120 Lecture Series Two : 2013/11
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 : 2013/12
Mass percent Mass percent is represented as %(m/m) or %(w/w). It has the advantage of being unaffected by temperature. To calculate: mass of solute / mass of solution x 100 mass of solute + mass of solvent To use: g solute /100 g solution
exercise What is the mass percent of a solution made by dissolving 25.0 g of sodium phosphate in 200.0 g of water? SOLN: Mass percent of Soln = mass of solute / mass of solution x 100 = 25.0 g / 225.0 g x 100 = 11.11 %
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 : 2013/15
ANSWER 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 CHEM120 Lecture Series One : 2013/16
ADDITIONAL EXERCISE 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 CHEM120 Lecture Series One : 2013/17
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/(0.99 + 3.6) = 0.22 CHEM120 Lecture Series One : 2013/18
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 = 10.86 m CHEM120 Lecture Series One: 2013/19
(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 ADDITIONAL EXERCISE Exercise 11.32, 11.33 and 11.37 p 444. CHEM120 Lecture Series One : 2013/20
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 One : 2013/21
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 One: 2013/22
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) 17.7 20-6 20 34.9 100 34.9 200 63.5 400 49.9 400 78.4 760 64.7 760 CHEM120 Lecture Series One : 2013/23
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. P soln = vapour pressure of the solution x solvent = mole fraction of the solvent CHEM120 Lecture Series One : 2013/24
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. ANSWER P * water = 1.00 atm at 100 C Number of moles of water = 100 g / 18.02 g mol -1 = 5.55 mol Number of moles of sucrose = 10.0 g / 342.3 g mol -1 = 2.92 10-2 mol Mole fraction of water 5.55 / (5.55 + 2.92 10-2 ) = 0.995 P solution = x solvent P * solvent = 0.995 1.00 atm = 0.995 atm CHEM120 Lecture Series One : 2013/25
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 = 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 CHEM120 Lecture Series One : 2013/26
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 One : 2013/27
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 One : 2013/28
Ideal Solutions with two or more volatile components Consider a mixture of benzene (C 6 H 6 ) and toluene (C 7 H 8 ) containing 1.0 mol of benzene and 2.0 mol of toluene [x ben = 0.33 and x tol = 0.67). At 20 C the vapour pressure of the pure substances are P ben = 10.0 kpa and P tol = 2.90 kpa Soln: P ben = (0.33)(10.0 kpa) =3.3 kpa P tol = (0.67)(2.90 kpa) =1.9 kpa The total vapour pressure is: P total = (3.3 kpa)(1.9 kpa) =5.2 kpa The mole fraction of benzene in the vapour is given by X ben in vapour = P ben / P total = 3.3 kpa/ 5.2 kpa = 0.63 Although benzene constitutes only 33% of the molecules in the solution, it makes up 63 % of the molecules in the vapour. CHEM120 Lecture Series One : 2013/29
EXAMPLE What is the normal boiling point temperature of a 1.45 mol Kg -1 aqueous solution of sucrose? ANSWER ΔT b = K b m ΔT b = 0.512 C kg mol -1 1.45 mol Kg -1 = 0.742 C The boiling point temperature of this solution is: 100 C + 0.742 C = 100.742 C CHEM120 Lecture Series Two : 2013/30
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? ANSWER 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: ΔT b = K b m We can calculate ΔT b : ΔT 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 CHEM120 Lecture Series Two : 2013/31
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 CHEM120 Lecture Series Two : 2013/32
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 : 2013/33
EXAMPLE Calculate the normal freezing point temperature of a 1.74 m aqueous solution of sucrose. ANSWER 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 CHEM120 Lecture Series Two : 2013/34
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 = (5.48 4.92) C/5.12 Kg mol -1 = 0.109 m number of moles of solute = (0.109 m)(0.050 Kg)/1 Kg = 5.469 10-3 mol Molar mass = 1.20 g/5.469 10-3 mol = 219 g mol -1 CHEM120 Lecture Series Two : 2013/35
Osmosis - tendency of solvent molecules to pass through a semi-permeable 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 8.314 m 3.Pa.mol -1.K -1 = 8.314J K -1 mol -1 (1m 3 Pa = 1 J ). CHEM120 Lecture Series Two : 2013/36
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 ADDITIONAL EXERCISE Practice Exercise p429, 432 (2 questions), 435 and 437. Exercise 11.44, 11.45, 11.51 and 11.54 p 445. CHEM120 Lecture Series Two : 2013/37