Chemical Thermodynamics

II. Thermodynamics of soution formation

A. Demonstration of solvent and soute mixing
1. separated solute and sovent in containers
2. spread out solute and solvent
    (breaking intermol. attractions)
3. mixing of solute and solvent
    (forming new intermol. attractions)
 
B. Translation to thermodynamic quantities
        e.g. making pickling brine: NaCl (s) + H₂O -> NaCl (aq)
1. breaking up salt lattice (-LE)
2. solvating gaseous ions (DH_hyd = breaking up water-water interactions and forming NaCl-water interactions)
3. DH_soln = DH_hyd - LE_NaCl
4. DHsoln is small and positive;
        Why is NaCl so soluble in water? Because of big DS value.
        Therefore DGsoln is negative and the solubility of NaCl varies with temperature.

A. 'Like dissolves like'
1. Types of interactions: Ion-ion to Van der Waals
2. Example: salad dressing; mix. of oil and water
    form a temporary emulsion by shaking before pouring
a. oil: nonpolar; hydrophobic: weak intermol. interactions
b. water: polar; hydrophilic: strong intermol. interactions
3. Trends
a. Polar molecules and ions will be soluble in polar solvents.
b. Nonpolar molecules will be soluble in nonpolar solvents.
c. Trends in terms of thermodynamics:
DHsoln large and positive when trying to mix polar/nonpolar bec. of needing to break up water-water interactions w/o forming new strong intermol. interactions.

B. Effect of pressure on solubility
1. Henry's Law: P = kC
    P = partial pressure of solute
    k = constant unique to solution
    C = [solute] (M)
2. Example: club soda bottle
a. new bottle: bottled under pressure
=> closed system in equilibrium
b. open new bottle: release pressure
=> lots of fizz (P decreases; C decreases)
c. old bottle: has gone 'flat'
=> hit equilibrium of P and C where P is ambient press.
 
C. Effect of Temperature on solubility
    -remember the Gibbs free energy equation
1. Salts dissolving: depends on DS
a. DS positive: increase solubility w/increased T
b. DS negative: decrease solubility w/increased T
2. Gases dissolving: DS is always negative; solubility always decreases w/increased T in an open system.
3. Special example: formation of CaCO₃ from boiled water (boiler scale). Use acetic acid to remove CaCO₃.
    Colligative Properties: properties of solvents that are altered by the presence of solutes (Pvap, osmotic pressure, bp, mp).

A. Raoult's Law
1. Psoln = CsolventP?solvent
          P = partial pressure
         C = mole fraction of solvent
    Think of it as due to the fact that there are less solvent molecules at the liquid/gas interface, so less solvent molecules will have the opportunity to vaporize at any time.
2. Essentially Psoln is proportional to Csolvent
Plot general trend, stress as ideal behavior.
 
B. Applications of Raoult's Law
1. Measurement of MW of solute (nonionizing, nonvolatile)
    Experimentally determine Psoln for know solution.
a. weigh solute
b. weigh solvent
c. combine in a closed system
    (no continual loss of solvent)
d. let equilibrate
e. measure Psoln
    Example: 50.0g of solid solute was dissolved in 600.0g water at 25?C.
        After equilibration, the measured Psoln was 23.6 torr. The standard Pwater at 25?C is 23.8 torr.
            Cwater = (Psoln)?(P?water) = (23.6 torr)?(23.8 torr) = 0.992
             nwater = (600g)?(18.015 g/mole) = 33.30 moles water
            Cwater = 0.992 = (nwater)?(nwater + nsolute)
            = (33.30)?(33.30 + nsolute)
                    0.992(33.30 + nsolute) = 33.30 moles
            nsolute = (33.30)?(0.992) - 33.30 = 0.269 moles solute
            MW solute = (50.0g)?(0.269 moles) = 186 g/mole
2. Formation of large crystals of solute: vapor diffusion.
    Requirements:
a. solvent1 and solvent2 are miscible
b. solute is soluble in solvent1
c. solute is not soluble in solvent2
d. Pvap of solvent1 is much lower or much higher than Pvap of solvent2
3. Binary mixtures of volatile liquids.
a. When two volatile liquids are mixed, both will have vapor pressures - what is the overall vapor pressure of the solution?
    Component A: Psoln = XAP?A
    Component B: Psoln = XBP?B
    But the true Psoln is made up of both PA and PB: PT = XAP?A + XBP?B
    Dalton's Law of Partial Pressures + Rauolt's Law!
    Example: A mixture of 50.0g acetone (CH3C(O)CH3) and 50.0g methanol (CH3OH) is prepared. What is the vapor pressure of the solution at 25?C? At 25?C, the vapor pressures of acetone and methanol are 271 and 143 torr, respectively.
    Need to know: to use the combined Raoult's Law: PT = XAP?A + XBP?B
        also: XA = nA/(nA + nB) ; XB = nB/(nA + nB)
    MW acetone: 58.08 g/mol ; MW methanol: 32.04 g/mol
    Need to look up: P? methanol = 143 torr; P? acetone = 271 torr
    Define Variables: A = acetone, B = methanol
        P?A = 271 torr P?B = 143 torr
        XA: nA = 50.0g/58.08 g/mol = 0.8609 mol acetone
        nB = 50.0g/32.04 g/mol = 1.560 mol methanol
        XA = 0.8609/(0.8609+1.560) = 0.3555
        XB: XB = 1.560/(0.8609+1.560) = 0.6445
        PT = XAP?A + XBP?B
            PT = (0.3555)(271 torr) + (0.6445)(143 torr)
                 = 96.35 + 92.16 torr = 188.5 torr = 189 torr (3 s.f.)
b. What if the actual pressure of the solution is 180 torr? What does this mean?
    Ideal Binary Mixture: The two components form an ideal solution, and follow Raoult's Law;
        A::::A and B::::B = A::::B; DHsoln = 0
    Negative Deviation: The vapor pressure of the new solution is less than the predicted ideal vapor pressure;
        the components have stronger intermolecular interactions with each other than exist in the pure liquids
        (A::::A and B::::B < A::::B); DHsoln < 0; it is harder to vaporize molecules in the solution.
    Positive Deviation: The vapor pressure of the new solution is more than the predicted ideal vapor pressure;
        the components have weaker intermolecular interactions with each other than exist in the pure liquids
        (A::::A and B::::B > A::::B); DHsoln > 0; it is easier to vaporize molecules in the solution.