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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:
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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; CO2 sublimes at normal pressures.
Phase Diagram of Carbon Dioxide Carbon dioxide cannot exist in the liquid state at pressures below 5.11 atm; CO2 sublimes at normal pressures. At 1 atm, solid CO2 does not melt at any temperature. Instead, it sublimes to form CO2 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; CO2 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 CO2 as well as almost all other substances. Why does water differ?
Freeze-drying • 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 Normal (right) and freeze-dried (left) spaghetti
Freeze-drying • How? • Freeze the material • Lower the pressure • Increase the temperature slightly Normal (right) and freeze-dried (left) spaghetti
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 ion-dipole 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: • Separation of solute particles • Separation of solvent particles • 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 DHsoln = DH1 + DH2 + DH3 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 • CCl4 in C6H6 • polar molecules are soluble in polar solvents • C2H5OH in H2O • ionic compounds are more soluble in polar solvents • NaCl in H2O or NH3 (l) 13.2
moles of A XA = sum of moles of all components x 100% mass of solute x 100% = mass of solution mass of solute mass of solute + mass of solvent Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = Mole Fraction(X) 13.3
moles of solute liters of solution moles of solute m = mass of solvent (kg) M = Concentration Units Continued Molarity(M) Molality(m) 13.3
moles of solute m= moles of solute M = mass of solvent (kg) liters of solution What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? 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
moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution 5.86 moles C2H5OH = 0.657 kg solvent What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? • 0.586 moles of solute per 1 L of solution: • 5.86 moles ethanol = 270 g ethanol • 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg = 8.92 m 13.3
solubility increases with increasing temperature solubility decreases with increasing temperature Temperature and Solubility Solid solubility and temperature No clear correlation between ΔHsoln and the variation of solubility with 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 KNO3 contaminated with 10 g NaCl. • Fractional crystallization: • Dissolve sample in 100 mL of water at 600C • Cool solution to 00C • All NaCl will stay in solution (s = 34.2g/100g) • 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g 13.4
Temperature and Solubility Gas solubility and temperature solubility usually decreases with increasing temperature 13.4
low P high P low c high c 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 is the concentration (M) of the dissolved gas c = kP P is the pressure of the gas over the solution k is a constant (mol/L•atm) that depends only on temperature 13.5
= vapor pressure of pure solvent 0 P1 = X1 P 1 0 0 0 P 1 P 1 P 1 - P1 = DP = X2 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 X1= mole fraction of the solvent Raoult’s law If the solution contains only one solute: X1 = 1 – X2 X2= mole fraction of the solute 13.6
0 0 0 0 PA = XA P A PB = XB P B PT = XA P A +XB P B Ideal Solution PT = PA + PB 13.6
Force A-A Force A-A Force B-B Force B-B Force A-B Force A-B < > & & PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law 13.6
0 DTb = Tb – T b 0 T b is the boiling point of the pure solvent 0 Tb > T b DTb = Kbm Boiling-Point Elevation T b is the boiling point of the solution DTb > 0 m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) 13.6
0 DTf = T f – Tf 0 T f is the freezing point of the pure solvent 0 T f > Tf DTf = Kfm Freezing-Point Depression T f is the freezing point of the solution DTf > 0 m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) 13.6
0 DTf = T f – Tf moles of solute m= mass of solvent (kg) = 3.202 kg solvent 1 mol 62.01 g 478 g x 0 Tf = T f – DTf 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. DTf = Kfm Kf water = 1.86 0C/m = 2.41 m DTf = Kfm = 1.86 0C/m x 2.41 m = 4.48 0C = 0.00 0C – 4.48 0C = -4.48 0C 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. more concentrated dilute 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 hypotonic solution hypertonic solution 13.6
Vapor-Pressure Lowering Boiling-Point Elevation DTb = Kbm 0 P1 = X1 P 1 Freezing-Point Depression DTf = Kfm p = MRT Osmotic Pressure (p) 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. 13.6
actual number of particles in soln after dissociation van’t Hoff factor (i) = number of formula units initially dissolved in soln Colligative Properties of Electrolyte Solutions 0.1 m NaCl 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 NaCl solution 0.2 m ions in solution i should be 1 nonelectrolytes 2 NaCl CaCl2 3 13.7
Boiling-Point Elevation DTb = iKbm Freezing-Point Depression DTf = i Kfm p = iMRT Osmotic Pressure (p) Colligative Properties of Electrolyte Solutions 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
Chemistry In Action: Desalination