PChem I     Exam 4 Solutions     3 December 1999

Answer any 5 of the 6.

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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.

    n = pV/RT = 1 atm × 4.8 L / [(0.08206 atm L /mol K) × (273+37)] = 0.19 mol
    N = n NAv = 0.19 × 6.022×1023 = 1.14×1023 molecules
    A = pR2 = ¼pD2 = ¼p(2.5×10-2 m)2 = 4.9×10-4 m2
    dN/dt = Z×A
    Z = p / (2pmkT)½
    Z = 1.01×105 Pa / (2p × [29×10-3 / NAv] kg × 1.38×10-23 J/K × 310 K)½
    Z = 2.81×1027 molecules m-2 s-1
    dN/dt = 2.81×1027 m-2 s-1 × 4.9×10-4 m2 = 1.38×1024 molecules/s

    T ~ N / (dN/dt) = 1.14×1023 / 1.38×1024 = 0.083 s = 83 ms
    Not even a respectable cough.

  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.

    MW = 2(14.01)+2(16.00)+2(1.008) = 62.04 g/mol
    n0 = m/MW = 5.00×10-2 g / 62.04 g/mol = 8.06×10-4 mol
    ng = pV/RT
    ng = 1 bar (1 atm / 1.01 bar) 6.59×10-3 L / [(0.08206 atm L/mol K) × 288 K ]
    ng = 2.76×10-4 moles gas
    therefore nt = (8.06-2.76)×10-4 = 5.30×10-4 mol solid

    nt = n0 e-kt   or   k = - (1/t) ln(nt / n0) = [ ln(n0 / nt) ] / t
    k = ln(8.06×10-4 / 5.30×10-4) / 4200 s = 1.00×10-4 s-1

    T½ = 0.693/k = 6930 s = 115.5 min

  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

    ln k = ln(c) + n ln(T) - E/RT
    d(ln k)/dT = 0 + n/T + E/RT2
    T2 d(ln k)/dT = nT + E/R

    T ( °C )39.9 41.8543.8 45.4547.1
    k ( s-1 ) 4.67×10-67.22×10-6 10.0×10-6
    T(K)313.05315.00 316.95318.6320.25
    ln(k)-12.274-11.839-11.513
    Dln(k) +0.435+0.326
    DT 3.903.30
    T2 d(ln k)/dT 1107010030

    Two equations and two unknowns: Ti2 Di = nTi + E/R

    Taking the difference at the two midpoint temperatures, the E/R term cancels out, leaving

    n = [ T22D2 - T12D1 ] / [ T2 - T1 ]
    n = [10030 - 11070] / [318.6 - 315.0] = - 289 (!)

    Although we've manipulated it correctly, we've just applied a 2nd order formula to a first order process! The hydrolysis of ATP is first order. We should have applied it instead to a second order process since both the collision and activated complex theory k's are k2.

  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.

    - d[N2O5]/dt = - d[NO]/dt = k2 [NO] [NO3]

    0 ~ d[NO3]/dt = k1[N2O5] - k-1[NO2][NO3]ss - k2[NO][NO3]ss
    [NO3]ss ~ k1[N2O5] / ( k2[NO] + k-1[NO2] )

    -d[N2O5]/dt ~ k1k2[NO][N2O5] / ( k2[NO] + k-1[NO2] )

  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.]

    At the activated complex, the exothermicity has not yet been felt. So a "reactant-like" complex, A-BC, feels the enormous forces of the exothermic falloff as B is jumping toward A; that's AB vibration. If the complex were a "product-like" one, AB-C, those same forces would be felt as C jumps away from AB; that's translation of the separating products instead.

    In terms of the "bent gutter" view of the ABC potential surface, A-BC has not yet "turned the bend," so the exothermic falloff pulls the system forcibly in the -RAB direction. But AB-C has already "turned the bend," and those falloff forces push on +RBC instead.

    So high vibration in highly exothermic reactions is consistent with Hammond's Postulate that such reactions are associated with "early" activated complexes (or "reactant-like").

  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?

    Aside from the fact that we're mixing apples and oranges since the hard sphere s deals with long-range van der Waals forces which deflect gas molecules even during near misses, and BET's s deals with liquid packing dimensions, the (sBET / sHS )½ value of 0.61 appears to represent the ratio of diameters vaguely consistent with the notion that N2 "thickness" (across the N atom) is perhaps 2/3 of its "length" (along the internuclear direction). That would suggest that N2 may be standing up on the surface rather than "lying down" with both atoms at the same average distance from the surface. Such a posture would maximize the number of N2 molecules (and N atoms) in contact with the surface.

    (There are molecules know not to stand up; H2 is catalytically decomposed on many metal surfaces, resulting in twice as many H atoms skittering around.)


Last modified 2 December 1999