Chm 1316 Honors Freshman Chemistry II
Spring 2001
Exam 4 due 11 AM on 23 April 2001

Open book; closed mouth. Solutions Email ME questions.
  1. Ozone is made photolytically in the stratosphere (around 20 km up) during the day, but at night, only the decay processes are at work. One such is:

    NO + O3 ® NO2 + O2

    The rate constant for that process is k = 1.8×10-12 cc molecules-1 s-1 e-1370/T where Ea/R = 1370 K. If you don't like molecules/cc as a concentration unit, 1 M = 6.023×1020 molecules/cc, but you'll find molecules/cc works just fine.

    The daytime concentration of the reactants are [NO]0~3×108 and [O3]0~2×1012 molecules/cc. If we ignore [O] and therefore the other half of the NOX catalytic cycle (O+NO2®O2+NO), we see that NO is the limiting reactant...so much so that [O3] will be effectively constant at [O3]0. (HINT: that makes k[O3] fixed!)

    The temperature at 20 km is about 220 K.

    How many minutes will it take for stratospheric O3 to eat up 99% of the stratospheric NO at 20 km in the dark?

  2. What would the reduction potential be for a silver half cell (Ag/Ag+ couple) where the Ag+(aq) source was Ag2CO3(s)?

  3. What would be the reduction potential for the Ag/Ag+ cell at the boiling point of water if [Ag+] = 1 M?

  4. The following table records initial rate of appearance of a product P as a function of initial concentrations of reactants A, B, and C.

    [A]0 (M) [B]0 (M) [C]0 (M) d[P]/dt (M/s)
    0.10.10.10.5
    0.20.20.22.0
    0.20.10.20.25
    0.20.20.42.0

    1. What is the rate expression for appearance of P?
    2. What is the rate constant (watch the units)?

  5. Your company wants to produce 1 ton (English) of electrolytically purified copper per week. How many amps will it be drawing from our electrical grid?

  6. There's an old wives' tale in kinetics that says reaction rates double in speed for every 10°C increase in temperature. Can't be true since it implies that all activation energies are the same. Besides 10°C is only 18°F, and that change in an oven's temperature doesn't cook meals twice as fast! So some kinetics are less sensitive than that to T. (And there aren't any home ovens calibrated to an accuracy of 18°F anyway.)

    1. Starting at 25°C, find the Eact implied by that "rule of thumb."
    2. Do reactions (like baking of bread, say) that are not that sensitive have higher or lower activation energies than that of (a)?

  7. Here below are two potential energy surfaces seen as energy contour maps. Their x-axes are the direction of approach of A onto BC, and their y-axes (even the bent one) is the direction of departure of AB from C, indicating a successful reactive collision.

    1. Late barrier; heavy central atom Indicate by the standard double dagger (‡) where the ABC‡ activated complex is.

    2. Since the activation energy hill is in the product valley, what kind of reactant energy would be required to pass over it? [HINT: Think like a bobsledder.]

    3. The diagram below differs from that above only in that it is squashed. (It turns out that this crushed diagram is appropriate for a very light mass B between much heavier A and C while the rectilinear diagram above is for a very heavy mass B between two much lighter A and C.)

      Be that as it may, the question posed by this new bobsled run is whether or not it is as demanding as the above in terms of the requirement of part (b)? As that product channel gets bent further and further down toward the reaction one, will the demand in (b) become more acute or less relevent to reaction? Why? [HINT: Think like a bobsledder.]

      Late barrier; light central atom
       

  8. The following standard reduction potentials are taken from Zumdahl's Table 17.1, p. 843.

    Cu+(aq) + e- ® Cu       E° = + 0.52V
    Cu2+(aq) + e- ® Cu+(aq)       E° = + 0.16V

    Is Cu+(aq) stable in solution? If so, why? If not, what happens to it?

  9. From the same table,

    Cu2+(aq) + 2 e- ® Cu       E° = + 0.34V

    Note that it is exactly the average between the two standard reduction potentials in (8). Why must that be so? [HINT: Think DG.]

  10. For a first order reaction rate, t½ = 0.693/k is the "half-life." But there's nothing magical about ½! Prove that by finding a similar expression for t0.1, the "tenth-life," where in each such period, the reaction falls to 1/10 of the value it had before the period started. So the time to fall from 100% to 10% is the same as that to fall from 10% to 1% for a first order reaction. In fact, however long it takes to fall in concentration by any factor f, it takes that long to fall by another factor of f.

    Deja vu.

Comments to Chris Parr Return to CHM 1316 HomePage Last modified 19 April 2001.