Chm 1316 Honors Freshman Chemistry II
Spring 1999     Sample Exam 2

  1. From water’s heating curve shown in class (and Fig. 10.42), we estimated that a mole of ice at 0°C and a mole of steam at 100°C would, in thermal contact, come to equilibrium as two moles of water at about 80°C.  I don’t believe that.  So let’s do the DH conservation thing to find out what equilibrium would really prevail. Water’s DHfus = 6.02 kJ/mol, DHvap = 40.7 kJ/mol, CP = 4.18 J/gm°C.

    Every joule the cold mole gains, the hot mole must lose. Just looking at the relative magnitudes of the two phase change enthalpies, it’s clear that steam is going to win; that is, only a fraction (6.02/40.7 = 0.148) of the steam needs to condense to supply all the ice with its fusion enthalpy. That leaves us with 1 mole of water at 0°C, 0.148 moles of water at 100°C and (1-0.148 = 0.852) moles of steam at 100°C. Thanks to Hess, we can continue heating and cooling in any order that makes sense to us as long as we end up with all of the phases at the same temperature. I choose to warm up the cold water by condensing more steam because I know that the water-steam equilibrium will never fall below 100°C until all of the steam condenses. I have 0.852 mol × 40.1 kJ/mol = 34.2 kJ latent heat of vaporization left to “spend” in that pursuit, but I don’t know if I need it all or more in heating the ice water.

    Normally, you’d set that heat = n CP DT and solve for DT; that gives

    DT = 34,200 J/mol / [ 4.184 J/gm°C × (18 g/mol) ] = 454 °C

    which is ludicrous. The ice water can’t rise 354°C above the hot water! So we need to solve the problem the other way; what would it take to warm the ice water to 100°C?

    DH = n CP DT = [ 4.184 J/gm°C (18 g/mol) ] (100°C) = 7,530 J/mol = 7.53 kJ/mol

    Meaning that to heat all the ice water to 100°C, we only need 7.53 kJ out of the 34.2 kJ still available from the latent heat of vaporization! That means that only another small fraction of the steam, (7.53/40.1 = 0.188) needs condense for a total condensation of 0.148+0.188 = 0.336 moles of steam condensed, leaving 0.664 moles of steam and 1.336 moles of water all at 100°C as the equilibrium state!

     

  2. Mt. Everest at only 8,840 m would get lost in the Marianas Trench (10,430 m below sea level) off the Philippines.  Yet at such depths one might find a “black smoker,” a volcanic vent spewing superheated water.  Hot as it is, the water doesn’t boil due to the enormous pressure at the bottom of the ocean.  Estimate the boiling point of seawater there, assuming it was pure.

    Since water is relatively incompressible (the hallmark of “condensed phases”), we presume that it is 1.00 g/cc density throughout (ignoring also density changes with temperature). Every 0.760 m × (13.6/1.00) = 10.34 m of water is worth 1 standard atmosphere. So the pressure at the bottom of the Marianas Trench must be 10,430 m / 10.34 m/atm = 1,009 atm! (And life still populates such ecological niches!) Clausius-Clapeyron can be rearranged to tell us what T will boil water at such enormous pressures knowing that 373 K suffices at 1 atm.

    ln(P / 1 atm) = - (DHvap/R) [ (1 / T) - (1 / 373 K) ]
    or
    1 / T = (1 / 373 K) - ln(P / 1 atm) / (DHvap/R)
    1 / T = 2.68×10-3 K-1 - ln(1,009)/(40,100 J/mol / 8.314 J/mol K) = 1.25×10-3 K-1

    T = 803 K

     


  3. The seawater in problem #2 isn’t pure; it’s 0.6 M NaCl.  That has two effects.  First, seawater is slightly more dense than pure water since it’s holding the weight of that salt.  Second, the salt is a solute which changes water’s boiling point via Kb = 0.51 °C kg/mol.  Which effect dominates the change in that boiling point in the Marianas Trench?  (Assume 0.6 m NaCl.)

    I’m willing to bet that the extra pressure due to the increased density will change T more than does the boiling point elevation colligative property, but let’s see.

    DT = Kb m = 0.51°C m-1 × 0.6 m = 0.31°C (nothin’)

    but 0.6 M NaCl adds 23.99+35.45 = 59.44 g × 0.6 = 35.7 g to each (1 kg) liter, a 3.57 % increase raising the density of seawater to 1.0357 g/cc (not counting the stuff in seawater beyond NaCl, like Ca2+ , etc.). We’re also presuming that the organization of the water around the ions doesn’t contribute to a density change, but with the mole ratio of water to ions of 55.5/1.2, I’m not worried. So using the new density, the new pressure must be 1,009 atm (1.036) = 1045 atm, and the new boiling point becomes:

    1 / T = 2.68x10-3 K-1 - ln(1045) / (40,100 J/mol / 8.314 J/mol K) = 1.239×10-3 K-1

    T = 807 K


    Since the 1.0 density water gave 803 K, DT from this change is 4 K not 0.3 K.

     


  4. Imagine some metal can be found solidified in all three cubic forms.  If its atomic radius is the same size in each, what are the ratios of their densities?

    Density is the ratio of mass to volume. So we need here the volume of each unit cell (in terms of the atomic radius) and the mass therein (in terms of the number of atoms).

    SIMPLE CUBIC holds (1/8) of 8 atoms at the corners of a cube for a total of one atomic mass. The atoms “touch” along the unit cell’s edge; so the length of the cube is 2r where r is the atomic radius. That makes the volume 8r3 and the density proportional to one atomic mass per 8r3 volume or simply 1/8 = 0.125.

    FACE CENTERED CUBIC holds (1/8) of 8 corner atoms and (1/2) of 6 face atoms for a total of 4 atoms. The atoms “touch” along a face diagonal, and there are r+2r+r = 4r radii which span the 2½ s or s = 4r/2½. That makes V = s3 = 64r3/23/2 = 32r3/2½. Thus the density is proportional to 4(2)½/32 = 2½/8 = 0.177 or 2½ greater than simple cubic.

    BODY CENTERED CUBIC holds (1/8) each of 8 corner atoms and an additional central atom for a total of 2 atomic masses. The atoms “touch” along the body diagonal, spanning that 3½s length with r+2r+r = 4r. So s = 4r/3½ and V = 64r3/33/2. That makes its density proportional to 2(3)3/2/64 = 33/2/32 = 0.162 making BCC of intermediate density.

    So the density ratios SC:BCC:FCC are 1:1.30:1.42

     


  5. Rationalize the progression down the Periodic Table of the physical states of the dihalogens at standard conditions in terms of their intermolecular forces.

    Di-anything hasn’t a prayer of being polar since both atoms have exactly the same electronegativity. So the dihalogens must hold themselves together in condensed phases via the weak London forces. Those forces are apparently not strong enough to contain F2 or Cl2, both of which are gases at standard conditions (1 atm at 25°C). But as we go down the Periodic Table, all atoms become larger; their valence electrons lie further from the nuclei. That means that they are less strongly held as reflected in the decrease in electronegativity in the dihalogens going down the table.

    But electrons that are less strongly held are more polarizable (easily pushed around by external electric fields), and since London forces work by induced polarization, those forces should become stronger as we progress down the table. Thus, we aren’t surprised to find Br2 a liquid and I2 a solid, indicating progressively increasing London attractions.


  6. Use freezing point depression to explain why a mixture of two compounds has a melting range while either compound alone has a melting point.

    Pure compound phase changes always occur at unique temperatures (given fixed pressure) which makes them such good temperature standard points. (Actually, triple points are the best temperature reference since they can occur only at one T and one P.)

    But when you begin solidifying a compound from a mixture, it is the one with the higher melting point which solifies first, reducing its mole fraction in the liquid phase! The change in mole fractions in the liquid make for a different melting point lowering (same as freezing point depression), and the melting point continues to lower as that mole fraction changes. The same holds true in reverse for melting a mixture of the solids; the melting range is the same as the freezing range for the same mixture.

     


  7. For any pure compound, what are the limiting temperatures between which a liquid can always be found at some appropriate pressure?

    No liquid can be found above the critical point. For normal liquids (solid density higher than liquid density), no liquid can be found at a temperature below the triple point. While for water and other liquids more dense than their solids, liquids persist for temperaures below Ttp for high enough pressures, it is guaranteed that they’ll exist above Ttp for all compounds.

    Thus, liquids can always be found at temperatures between Ttp and Tcp.

    (No fair mentioning compounds so unstable that they detonate before melting!)

     


  8. Homemade ice cream (cream, eggs, sugar, flavoring) cannot be made without a cold bath made of ice and rock salt.  Why not?  What would happen without the salt?  Why?

    By freezing point lowering, the ice-water equilibrium occurs a T much below 0°C when the salt is used. That’s necessary since you’re trying freeze a mixture (milk, eggs, sugar, etc.) that is already has its freezing point depressed below water’s freezing point.

     


  9. At 20°C, the density of acetic acid maximizes at 1.0700 g/cc when its weight percent in water is 79%.  What is the molality and molarity of this solution?  (Ignore acid dissociation.)

    In 1 kg of solution is 790 g acetic acid (HC2H3O2) and 210 g water (H2O). Since acetic acid’s molecular weight is 2(12)+2(16)+4(1) = 60, that kg contains 790/60 = 13.17 moles acetic acid and 210/18 = 11.67 moles water for a total of 24.84 moles. Problem didn’t ask for the mole fractions, but they’re Xacetic = 13.17/24.84 = 0.5302 and Xwater = 0.4698. Problem did ask for molality, or moles acetic acid per 1 kg water, or 13.17 mol acetic (1000/210) = 62.67 m.

    However, 1 L of solution doesn’t weigh 1 kg; it weighs 1070.0 kg and contains (1070.0)×(0.79) = 845.3 g acetic acid or (845.3/60) = 14.08 moles acetic acid for a molarity of 14.03 M. Note the big difference when concentrations are high; low concentrations have the same numerical values for molality and molarity.

     


  10. 10. Benzene (C6H6) and naphthalene (C10H8) form ideal solutions.  What is Xbenzene in a mixture that has the same (initial) boiling point as water?  The normal boiling points and DHvap for benzene and naphthalene are (80.2°C, 30.8 kJ/mol) and (217.9°C, 40.4 kJ/mol) respectively.

    That (initial) boiling point would be 100°C, of course. We need P° for the pure components at this temperature; so we need to solve the Clausius-Clapeyron equation for both of them.

    P° = 1 atm e(-DHvap/R) [ (1/373 K) - (1/Tbp)]

    P°benzene = 1 atm e-(30,800 J/mol / 8.314 J/mol K) [(1/373 K) - (1/353 K)] = 1.76 atm

    P°naphthalene = e-(40,400 J/mol / 8.314 J/mol K) [(1/373 K) - (1/491 K)] = 0.044 atm


    Looks like naphthalene could almost be considered nonvolatile here. But we’ll treat it as volatile nonetheless. As Xnaphthalene varies from 0 to 1, the ideal solution vapor pressure (Raoult’s Law) varies linearly from 1.76 down to 0.04 atm. So if the solution loses 1.72 atm over 1.0 mole fraction, over what mole fraction will it lose only 0.76 (to become 1.00 atm)? 0.76 / 1.72 = Xnaphthalene = 0.44.

     

Comments to Chris Parr  Return to CHM 1316 HomePage  Last modified 14 March 1999.