Chm 1316 Honors Freshman Chemistry II Spring 2000
Exam 3 Solutions 5 April 2000

Open Book & Handouts. For any 8 of the 10. Neither Give Nor Receive Help!

  1. Ball lightning has almost an element of myth about it; it is so rare. However, qualified observers have witnessed these free floating electrical discharges, and even been "chased" by them. (When you run, the vortices creating in the air by your movement can push floating objects after you!)

    One such observer watched a ball lighting for a couple of minutes until it happened to float over to a metal bucket filled with 2 gallons of water. Its internal fields could not support themselves in the dead electrical short of the metal, and the ball exploded with a terrible bang. The observer reported that the water in the bucket, which had been at room temperature (25°C), was now boiling!

    What can we say about the minimum energy content (kJ) of that lightning ball? We can safely ignore the heat capacity of the bucket's metal, since it's swamped by water's 1.0 cal g-1. Water's density is 1.0 g cm-3.

    From the Periodic Table handed out in class, 1 gallon = 3.7854 L. So the volume of water raised to the boiling point was 7.57 L = 7.57×103 cc or (since water is the definition of unit specific gravity) 7.57×103 g.

    DE = C DT = 1.0 cal g-1 °C-1 (7.57×103 g) (75 °C) = 5.68×105 cal

    From the Periodic Table, 1 cal = 4.184 J, so

    DE = 5.68×105 cal (4.184 J/cal) = 2.38×103 kJ

    Clearly the ball lightning would have had more energy than that. However much water boiled (unknown) would have consumed its heat of vaporization, and the metal bucket itself had a heat capacity we've not taken into account. Still 2 MJ in a basketball sized glowing sphere is impressive!

  2. Lead melts at 327°C consuming 5.86 cal g-1 in the process. But before it can melt, it must warm from room temperature (25°) to its melting point, consuming about 0.0327 cal g-1 °C-1 in the process. How fast (m/s) would a lead bullet have to be going to melt when it strikes an immovable object?

    Watch your units! And it doesn't matter how much the bullet weighs, so pick a convenient weight. Ignore the immovable object's heat absorption; it happens so fast, the lead hasn't an opportunity to share the wealth. Now you know why most meteors (50,000 m/s) burn up even in the atmosphere.

    Choose a 1 gram bullet; it doesn't matter because a 2 gram bullet would have twice the intial kinetic energy at the same speed and require twice the energy to the problem is intensive rather than extensive.

    DE = C DT + DHfus = 0.0327 cal/°C (327-25)°C + 5.86 cal = 15.74 cal = 65.84 J

    ½ m v2 = ½ 10-3 kg v2 = 65.84 J = 65.84 kg m2 s-2
    v = [ 131,700 m2/s2 ]½ = 363 m/s = 804 mph > vsound (770 mph)

  3. The enthalpy of combustion of cinnamic aldehyde, C6H5-CH=CH-CHO, is found (via bomb calorimetry) to be -1,112.3 kcal/gm. Since the enthalpies of combustion of H2 and graphite are -285.9 kJ/mol and -393.5 kJ/mol, respectively, what's the standard enthalpy of formation of cinnamic aldehyde? (essence of cinnamon)

    C6H5-CH=CH-CHO + ½(21) O2(g)  ---> 9 CO2(g) + 4 H2O(liq)       DHcomb

    DHcomb = - 1,112.3 kcal/g (4.184 kJ/kcal) (132.15 g/mol) = - 6.15×105 kJ/mol (!?!)
    This is clearly way too high, but it's what you'd have to believe from the data given. So the data's wrong. It should have read "11,123 cal/g" instead. But we're stuck with the typo:

    9 C(graphite) + 9 O2(g)  ---> 9 CO2(g)       DH1 = 9(-393.5) kJ/mol
    4 H2(g) + 2 O2(g)  ---> 4 H2O(liq)       DH2 = 4 (-285.9) kJ/mol
    9 CO2(g) + 4 H2O(liq)  ---> C6H5CHCHCHO + ½ (21) O2(g)       DH = + 6.15×105 kJ/mol

    If we add the last 3 equations (and their enthalpies) together, we get

    9 C(graphite) + 4 H2(g) + ½ O2(g)  ---> C6H5CHCHCHO       DHf

    DHf = 6,150,000 - 3,541 - 1,144 kJ/mol = 6.14×105 kJ/mol

    Had we used the correct value for the combustion,

    DHf = 1,470 - 3,541 - 1,144 kJ/mol = - 3,215 kJ/mol (makes more sense)

  4. But enthalpy of combustion wasn't what was measured in the bomb calorimeter for cinnamic aldehyde!

    1. What was? Why?

      Energy of combustion because pressure wasn't fixed; volume was!
      DE = qV meaning heat transferred at fixed volume.

    2. What value (kcal/gm for comparison) for the quantity in (a) would have been obtained from the bomb?

      DE = DH - DPV = DH - RTDngas

      DE = 615,000 kJ/mol - 8.314×10-3 kJ/mol K (298 K) (9 - ½[21])
      DE = 615,004 kJ/mol

      Even with the corrected DHcomb = 1,470 kJ/mol, DE = 1,474 kJ/mol not different from DH within the significant figures.

  5. Iron(III) can be chelated by either one (hexadentate) EDTA or three (bidentate) ethylene diamines. Which reaction does entropy favor? Why? If the energetics were about the same, in what way would the complex equilibrium constants differ?

    3 ed + Fe3+  ---> Fe(ed)33+
    EDTA + Fe3+  ---> Fe(EDTA)3+
    (sure, EDTA complexes as an anion, EDTA-4, but we'll ignore that here)

    Clearly entropy favors the latter since only two free species get tied up in the complex rather than four! That means that the EDTA complex would have a larger stability (equilibrium) constant than the ethylene diamine one.

  6. When a mole of manganese(II) nitrate (anhydrate*) dissolves in 300 mol of water, the process is endothermic by 53.09 kJ at 25°C. Since all nitrates are highly soluble, that dissolution is spontaneous. That means that the entropy of solution can be no lower than what value (J mol-1 K-1)?

    * means no water of hydration in the solid

    Assuming the dissolution was just barely spontaneous, then DG ~ 0 and DS = DH/T or

    DSsolution = 53,090 J/mol / 298 K = 178 J/mol K

    Now DG might be much lower than 0, in which case, DS = DH/T - DG/T might be much larger than 178 J/mol K. So that value is a minimum.

  7. Acetobacter sours wine (oxidizes ethanol to acetic acid) in order to fuel its own existence. Like the rest of Life, acetobacter must make ATP from ADP, a reaction whose Gibbs Free Energy is +30.5 kJ/mol. If acetobacter were 100% efficient at its task, how many moles of ADP could it convert to ATP for every mole of ethanol it oxidizes?

    Molecule Ethanol, C2H5OH Acetic Acid, CH3CO2H H2O CO2
    DGf°, kJ/mol -174.8 -392 -237.2 -394.4

    C2H5OH(liq) + O2(g)  ---> CH3CO2H(liq) + H2O(liq)

    DG = DGf[C3CO2H] + DGf[H2O] - DGf[C2H5OH] - DGf[O2]
    DG = - 392 - 237.2 - (-174.8) - (0) = - 454 kJ/mol

    nATP = 454/30.5 = 14 and change

  8. The heat of fusion of sodium metal is 3.05 kJ/mol, and its entropy of fusion is 8.25 J/mol K. What is sodium's melting point (°C)? (Liquid sodium is used as a coolant in some nuclear reactors.)

    At the melting point, DGfus = 0 = DHfus - TDSfus

    So T = DHfus/DSfus = 3050 J/mol / 8.25 J/mol K = 370 K

  9. Consider crystals of benzene at zero degrees Kelvin.

    1. What is the absolute entropy of such crystals (per mole)?

      If they're perfect, S0 = 0.00 by definition. There's only one way, W, to build a perfect crystal, so R ln(W) = 0.

    2. If the benzene were monodeuterated (C6H5D), and there was no crystal positional preference for deuterium vs. hydrogen, what would the absolute entropy be (per mole)? [HINT: the position of the deuterium around the benzene ring makes this an entropy of mixing question.]

      Now there's 6 ways to find the molecule, so S0 = R ln(6) = 14.90 J/mol K. You could also arrive at that by the entropy of mixing formula:

      Smix = - R Si Xi ln(Xi)

      where the sum is over 6 terms each one having Xi = 1/6.
      Since ln(1/6) = - ln(6), the answers agree.

  10. Le Châtlier would have loved the (thankfully correct) version of

    ln K = - (DH°/R) (1 / T) + (DS°/R)

    because it vindicates what he said about temperature effects on equilibrium.

    1. What did he say?

      Well last semester anyway, he said, "Increasing temperature favors products of endothermic reactions but reactants of exothermic ones."

    2. How is this expression consistent with that?

      We want to see what sign we'd expect from the differential d ln(K) / dT. Fortunately, we know that dT -1/dT = - T -2 dT; so if DH° and DS° are sufficiently insensitive to T (over the range of dT of interest), then we can treat them both as constants, and

      d ln(K) = + (DH°/RT2) dT

      and if dT increases, a positive DH° will correlate with a positive d ln(K), favoring products. Vice versa for a negative (exothermic) DH°.

      [Of course, if ln(K) increases, then K must increase, favoring products. And if ln(K) decreases, K decreases, favoring reactants.]

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