PChem I     Exam 4     3 December 1999

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    2001
  1. In the movie 2001, the astronaut took a deep breath and pursed his lips to hold it against the explosive decompression of his space pod. Not a prayer. Consider the chance you'd have of blowing a column of water up 30 feet (~1 atm), and you'll appreciate that the astronaut could not possibly prevail against the release of his 4.8 L lungfull of 37°C air (average MW=29 g/mol) into the void through his 2.5 cm diameter trachea. About how long would his lungs have taken to empty? (Assume the initial rate of loss is sustained...even though it isn't.)

    It would actually have taken just a tad longer due to the fact that human lungs branch down to aveoli sacs < 1 mm in diameter, but the simpler calculation puts the meaning of "explosive decompression" on a human scale.

  2. The first order decomposition of nitramide in the presence of bases,

    NH2NO2 ---> N2O(g) + H2O(liq)

    is conveniently analyzed by collecting the gas evolved during the reaction. During such an experiment, 50.0 mg of nitramide was allowed to decompose at 15°C. The volume of (dry) gas evolved after 70.0 min. was measured to be 6.59 cm3 at 1 bar pressure. Find the rate constant and the half-life for the nitramide decomposition.

  3. Although collision theory insists that
    k = a T ½ e-Ea / RT
    and activated complex theory demands that
    k = b T 2 e-D#H / RT
    practical kineticists fit their data instead to
    k = c T n e-E / RT
    where n isn't necessarily integer or even half integer. It's whatever the data says it is.

    Derive d(ln k)/dT for the last expression, and use estimates of its values at the midpoints between the temperatures in the table below to estimate n for this (hydrolysis of ATP) reaction.

    It'll be simplest to multiply both sides of your expression by T 2. Then if we call that derivative D, the expression you derive should read T 2 D = nT + E/R where n and E have been presumed constants. Solve that for n given the two midpoint estimates. Remember that T is in K.

    T ( °C )39.943.847.1
    k ( s-1 ) 4.67×10-67.22×10-6 10.0×10-6

  4. The overall reaction
    N2O5 + NO ---> 3 NO2

    is thought to proceed through the following mechanism:

    N2O5 ---> NO2 + NO3     k1
    NO2 + NO3 ---> N2O5     k-1
    NO + NO3 ---> 2 NO2     k2

    Derive the rate equation for the loss of N2O5 by assuming that the intermediate NO3 is in steady state.

  5. The nascent products of highly exothermic reactions are always highly vibrationally excited. Explain why this is consistent with Hammond's Postulate that their activated complexes look very "reactant-like" (as opposed to "product-like").

    In other words, why does the "reactant-like" A-BC as a complex geometry give higher AB vibration than does the "product-like" AB-C as a complex geometry (even if their reaction exothermicities were the same)?

    [HINT: use the shape of the potential energy surface for the A+BC ---> AB+C reaction.]

  6. Atoms are spherical. Molecules are not.

    The cross-section for N2 differs with the process considered for it. For example, its hard-sphere collision cross-section, s = 0.43 nm2, whereas its BET monolayer footprint cross-section, s = 0.16 nm2. What does that suggest to you about the orientation of N2 molecules in a surface monolayer?

Last modified 2 December 1999