CHM 3312 Exam #3 Solutions 30 April 1997

Instructor: Chris Parr BE3.506 883-2485 parr@utdallas.edu
                       http://wwwpub.utdallas.edu/~parr/


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  1. Use Trouton's Rule (p. 134) to estimate the ratio of the weights of (dominant) configurations available to the system of a liquid and its vapor at its normal boiling point, W(g)/W(l).

Trouton's Rule is that Delta Svap~85 J/mol K.

Delta Svap = S(g, 1 atm) - S(liq) = R ln[W(g, 1 atm)] - R ln[W(liq)], if we believe Boltzmann.

Thus, Delta Svap = R ln[W(g, 1 atm)/W(liq)] or W(g, 1 atm)/W(liq) = e Delta Svap/R.

So W(g, 1 atm)/W(liq) = e[85 J/mol K]/8.314 J/mol K = e10.2 = 28,000.

Either W is inconceivable large but that for the vapor is 28,000 times larger.

What is the ratio of those weights for the universe while the liquid vaporizes?

Phase changes are the perfect example of reversible processes; an absolutely trivial change in the temperature stops the boiling and starts the condensation. That means that the normal boiling point has Delta Gvap = 0 but Delta G = -T Delta Suniverse. So Delta Svap(universe) = 0.

By Boltzmann's expression Wuniverse(g, 1 atm) = Wuniverse(liquid) at the normal boiling point.

  • Imagine a system of 3 harmonic oscillators with a total of 3 quanta (the levels are all one quanta apart). Find pi, the probability of occupancy of level i, for all occupied levels of the dominant configuration and then again for all configurations. Which is the better model for a Boltzmann distribution for this N=3 case?

    • Configuration, j I II III
    Level 3, p3[j] X, 1/3 _, 0 _, 0
    Level 2, p2[j] _, 0 X, 1/3 _, 0
    Level 1, p1[j] _, 0 X, 1/3 XXX, 1
    Level 0, p0[j] XX, 2/3 X, 1/3 _, 0
    Wj 3!/(1!0!0!2!) = 3 3!/(0!1!1!1!) = 6 3!/(0!0!3!0!) = 1
    Wj/ Summation Wj 0.3 0.6 (dominant) 0.1
    The dominant configuration, II, gives {pi} = 1/3, 1/3, 1/3, 0 or all occupied levels having the same occupancy. That doesn't happen in the Boltzmann world until the temperature becomes infinite! Bad model. Bad, bad model.

    On the other hand, I gives {pi} = 2/3, 0, 0, 1/3 not much better.
    Or III gives {pi} = 0, 1, 0, 0 completely silly.

    However, if we average all those together, giving each the fractional weight of their configuration, we get

    p0 = 0.3×(2/3) + 0.6×(1/3) + 0.1×(0) = 0.4
    p1 = 0.3×(0) + 0.6×(1/3) + 0.1×(1) = 0.3
    p2 = 0.3×(0) + 0.6×(1/3) + 0.1×(0) = 0.2
    p3 = 0.3×(1/3) + 0.6×(0) + 0.1×(0) = 0.1

    Ooooo...a monotonically descending occupancy scheme: {pi} = 0.4, 0.3, 0.2, 0.1 smacks of Boltzmann to me!

  • Oddly enough*, you only need nu (H2) = 4400 cm-1 (and the mass ratios) in order to calculate the equilibrium constant for

    H2 + T2 arrow right 2 HT

    at 1000°K (where so many rotational states are turned on, even for H2, that nuclear statistics are irrelevant). Remember to include e Delta D0/RT.

    • * since they all have the same bond distance and force constant

    Ooooo... Delta nu = 0.

    So multiplicative constants will cancel numerator and denominator out of the partition function expression for

    K = [HT]2 / ( [H2] [T2] )

    K = { [q(HT)]2 / [q(H2) q(T2)] } e- Delta epsilon 0/RT

    q = qtr qrot qvib qel

    In order to find qvib and epsilon 0, we'll need nu proportional to µ-½ and µ = m1m2/(m1+m2).

    µ(H2) = 0.5 amu; µ(T2) = 1.5 amu; µ(HT) = 0.75 amu.

    That makes nu (T2) = (0.5/1.5)½×4400 cm-1 = 2540 cm-1
    and nu (HT) = (0.5/0.75)½×4400 cm-1 = 3593 cm-1.

    Since epsilon 0 = ½h nu , Delta epsilon 0 = ½ [ 2(3593) - 4400 - 2540 ] = 123 cm-1 (trivial)
    so e- Delta epsilon 0/RT = e-123 cm-1×1.4388 cm K / 1000°K = 0.838 (not impressive).

    And since qvib(T2) = 1/(1-e-2540 cm-1× 1.4388 cm K/1000°) = 1.027, the higher frequency HT and H2 will give even less impressive qvib which we can safely ignore.

    qrot = kT/ sigma B proportional to µ/ sigma since B=hc/2I and I=µr2. We can leave all those other items out since, being the same for all these molecules, they'll cancel! Haaa ha ha ha ha ha.

    OK... sigma for H2 and T2 are both 2 but sigma for HT is 1.

    Thus, Krot = [qrot(HT)]2 / [ qrot(H2) qrot(T2) ]
    = µ(HT)2/[ µ(H2) µ(T2) ] × [ sigma (H2) sigma (T2)] / [ sigma (HT)]2
    = (0.75)2/[(0.5)(1.5)] × (2)(2)/(1)2 = 3 (at all T high enough for nuclear statistics of ortho and para hydrogen to be ignored).

    The easy one is saved for dessert:
    Ktr = {[M(HT)]2/[M(H2) M(T2)]}1.5 = [ 42 / (2×6) ]1.5 = 1.54

    No, there was an easier one: Kel = 12/(1×1) = 1.

    So K = 1.54×3×1.027-1×1×0.838 = 3.77 ~ 4 which is what we would've guessed!

  • Compared to an isotopically pure crystal, what's the molar residual entropy of natural abundance B(s), 19.87% 10B + 80.22% 11B?

    • The entropy of an isotopically pure crystal (of 11B, say) would be zero. But the mixing of 0.1987 10B into that causes the residual entropy to be the entropy of mixing, viz.,

    Sresidual = - R Summation pi ln(pi)
    = - 8.314 [ 0.1987 ln(0.1987) + 0.8022 ln(0.8022) ] = 4.139 J/mol K

    Lake Eyring

  • The potential energy surfaces we've seen in Chapter 27 all involve a single activated complex located at the lone mountain pass along the reaction coordinate. But some reactions involve a metastable intermediate with a deep lake between the entrance and (now) exit passes. If the reaction is thermoneutral, the lake and passes are symmetrical. For thermal collisions, what would you expect the transmission coefficient, kappa, to be in this case? Why would the dynamics yield that expectation?

    • The figure is from Eyring, Walter, and Kimball's Quantum Chemistry and purports to show the reaction potential surface for H+H2 going to H2+H (say an ortho-para conversion). The target feature is a metastable H3 which, unfortunately, doesn't exist. This potential is wrong, but other examples of metastable complexes can be found; this one has the advantage of rigorous thermoneutrality.

    Thermal collisions implies that the energy in excess of the activation barrier (that pinched area in the contour map) is only about kT, not very much compared to the barrier itself. That means that once you fall into the lake, while you do have energy enough to get out again, you have to have near perfect aim at the barrier to do that.

    That metastable intermediate is rather like a real molecule then. It will vibrate about its equilibrium position (the bottom of the lake) in a manner unlike that of the activated complex which sits on the top of a barrier. The latter has no choice but to fall apart along its imaginary asymmetric stretch. The metastable lake dweller has real vibrations in all 4 modes (symmetric, asymmetric and a pair of bends...it's linear after all).

    Hirschfelder's trajectory on Eyring's Surface (This figure is also from EWK. It shows the world's first chemical trajectory done by Hirshfelder on an adding machine! The story goes that Hirshfelder only got this far, months of work on and adding machine, no doubt, before threatening to quit graduate school! Eyring hastily put him on another project, and the world's first chemical trajectory was never finished. Someone claimed to have extended it on a computer; it was unreactive.)

    Its vibrations cause its trajectory to wander around the lake until, by chance, it finds either the entrance or the exit pass, at which point it leaves. The likelihood of finding the exit pass (leading to reaction) first is, not surprisingly, 50%. That means that 50% of the time it will find the entrance pass first and unreact.

    That makes the transmission coefficient 0.5.
    Last modified 5 May 1997.