Intermolecular Forces and Liquids and Solids Chapter 12

Intermolecular Forces and Liquids and Solids Chapter 12

Midterm II • Any conflicts with March 20? If yes, let me know ASAP. The original date was March 22.

Phase Diagram of Water • Note the high critical temperature and critical pressure: – These are due to the strong van der Waals forces between water molecules. • The slope of the solid–liquid line is negative. – This means that increasing the pressure above 1 atm will raise the boiling point and lower the melting point. – Lower the melting point?

Phase Diagram of Carbon Dioxide Carbon dioxide cannot exist in the liquid state at pressures below 5. 11 atm; CO 2 sublimes at normal pressures.

Phase Diagram of Carbon Dioxide Carbon dioxide cannot exist in the liquid state at pressures below 5. 11 atm; CO 2 sublimes at normal pressures. At 1 atm, solid CO 2 does not melt at any temperature. Instead, it sublimes to form CO 2 vapor. Why might it be useful as a refrigerant?

Phase Diagram of Carbon Dioxide Carbon dioxide cannot exist in the liquid state at pressures below 5. 11 atm; CO 2 sublimes at normal pressures. If you want to send something frozen across the country, you can pack it in dry ice. It will be frozen when it reaches its destination, and there will be no messy liquid left over like you would have with normal ice.

The slope of the curve between solid and liquid is positive for CO 2 as well as almost all other substances. Why does water differ?

Freeze-drying Normal (right) and freeze-dried (left) spaghetti • Completely remove water from some material, such as food, while leaving the basic structure and composition of the material intact • Two reasons – – • Keeps food from spoiling for a long period of time Significantly reduces the total weight of the food How? – – – Freeze the material Lower the pressure (<0. 006 atm) Increase the temperature slightly

Freeze-drying Normal (right) and freeze-dried (left) spaghetti • How? – Freeze the material – Lower the pressure – Increase the temperature slightly

Physical Properties of Solutions Chapter 13

A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount 13. 1

A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 13. 1

Solutions The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.

How Does a Solution Form? As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

How Does a Solution Form If an ionic salt is soluble in water, it is because the iondipole interactions are strong enough to overcome the lattice energy of the salt crystal.

Energy Changes in Solution • Simply, three processes affect the energetics of the process: Ø Separation of solute particles Ø Separation of solvent particles Ø New interactions between solute and solvent

Energy Changes in Solution The enthalpy change of the overall process depends on H for each of these steps.

Three types of interactions in the solution process: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction Hsoln = H 1 + H 2 + H 3 13. 2

“like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. • non-polar molecules are soluble in non-polar solvents CCl 4 in C 6 H 6 • polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O • ionic compounds are more soluble in polar solvents Na. Cl in H 2 O or NH 3 (l) 13. 2

Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass mass of solute x 100% % by mass = mass of solute + mass of solvent mass of solute x 100% = mass of solution Mole Fraction (X) moles of A XA = sum of moles of all components 13. 3

Concentration Units Continued Molarity (M) M = moles of solute liters of solution Molality (m) m = moles of solute mass of solvent (kg) 13. 3

What is the molality of a 5. 86 M ethanol (C 2 H 5 OH) solution whose density is 0. 927 g/m. L? m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Strategy: Find mass of solvent Know mass of solute + mass of solvent = mass of solution If mass of solution and mass of solute known, can calculate mass of solvent Can calculate mass of solute from moles of solute Can calculate mass of solution from density and volume of the solution Solve 13. 3

What is the molality of a 5. 86 M ethanol (C 2 H 5 OH) solution whose density is 0. 927 g/m. L? m = moles of solute mass of solvent (kg) M = moles of solute liters of solution 0. 586 moles of solute per 1 L of solution: 5. 86 moles ethanol = 270 g ethanol 927 g of solution (1000 m. L x 0. 927 g/m. L) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0. 657 kg m = moles of solute mass of solvent (kg) = 5. 86 moles C 2 H 5 OH 0. 657 kg solvent = 8. 92 m 13. 3

Temperature and Solubility Solid solubility and temperature No clear correlation between ΔHsoln and the variation of solubility with temperature solubility decreases increases with increasing temperature 13. 4

Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO 3 contaminated with 10 g Na. Cl. Fractional crystallization: 1. Dissolve sample in 100 m. L of water at 600 C 2. Cool solution to 00 C 3. All Na. Cl will stay in solution (s = 34. 2 g/100 g) 4. 78 g of PURE KNO 3 will precipitate (s = 12 g/100 g). 90 g – 12 g = 78 g 13. 4

Temperature and Solubility Gas solubility and temperature solubility usually decreases with increasing temperature 13. 4

Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c = k. P c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution k is a constant (mol/L • atm) that depends only on temperature low P high P low c high c 13. 5

Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 0 1 Raoult’s law P 10 = vapor pressure of pure solvent X 1 = mole fraction of the solvent If the solution contains only one solute: X 1 = 1 – X 2 P 10 - P 1 = P = X 2 P 10 X 2 = mole fraction of the solute 13. 6

Ideal Solution PA = XA P 0 A PB = XB P 0 B PT = PA + PB PT = XA P 0 A + XB P 0 B 13. 6

PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law Force < A-A & B-B A-B Force > A-A & B-B A-B 13. 6

Fractional Distillation Apparatus 13. 6

Boiling-Point Elevation Tb = Tb – T b 0 is the boiling point of the pure solvent T b is the boiling point of the solution Tb > T b 0 Tb > 0 Tb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0 C/m) 13. 6

Freezing-Point Depression Tf = T 0 f – Tf T 0 Tf f is the freezing point of the pure solvent is the freezing point of the solution T 0 f > Tf > 0 Tf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0 C/m) 13. 6

13. 6

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. Tf = Kf m m = Kf water = 1. 86 0 C/m moles of solute mass of solvent (kg) 478 g x = 1 mol 62. 01 g 3. 202 kg solvent = 2. 41 m Tf = Kf m = 1. 86 0 C/m x 2. 41 m = 4. 48 0 C Tf = T 0 f – Tf = 0. 00 0 C – 4. 48 0 C = -4. 48 0 C 13. 6

Osmotic Pressure (p) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis. dilute more concentrated 13. 6

Osmotic Pressure (p) High P Low P p = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K) 13. 6

A cell in an: isotonic solution hypertonic solution 13. 6

Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 10 Boiling-Point Elevation Tb = Kb m Freezing-Point Depression Tf = Kf m Osmotic Pressure (p) p = MRT 13. 6

Colligative Properties of Electrolyte Solutions 0. 1 m Na. Cl solution 0. 1 m Na+ ions & 0. 1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0. 1 m Na. Cl solution van’t Hoff factor (i) = 0. 2 m ions in solution actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be nonelectrolytes Na. Cl Ca. Cl 2 1 2 3 13. 7

Colligative Properties of Electrolyte Solutions Boiling-Point Elevation Tb = i Kb m Freezing-Point Depression Tf = i Kf m Osmotic Pressure (p) p = i. MRT 13. 7

A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution • collodial particles are much larger than solute molecules • collodial suspension is not as homogeneous as a solution 13. 8

The Cleansing Action of Soap 13. 8

Chemistry In Action: Desalination