Advanced Thermodynamics   Exam #1

(Open handouts & note sheet)

  1. Use thermodynamical arguments to find which of the gaseous saturated hydrocarbons (methane through butane) burns with the hottest flame (stoichiometrically in pure oxygen) and explain why.

    You may not have time to calculate all the flame temperatures; so you're going to have to reason this one out. As a hint, think about how enthalpy is extensive but flame temperature is not. The answer will have a lot to do with stoichiometry...say per O2 molecule consumed, for example.

    Species (gaseous)CH4C2H6 C3H8C4H10 CO2H2OO2
    Delta Hf° (kJ/mol) -74.4-83.8-104.7-125.6-393.51-241.8260.00
    ~ CP (J/mol K) 35.352.6??37.133.629.4

OK, so what is the combustion stoichiometry on a per O2 basis?

½CH4 + O2 arrow right ½CO2 + H2O
(2/7) C2H6 + O2 arrow right (4/7) CO2 + (6/7) H2O
(1/5) C3H8 + O2 arrow right (3/5) CO2 + (4/5) H2O
(2/13) C4H10 + O2 arrow right (8/13) CO2 + (10/13) H2O

which shows a 1 : 2 mix of products (CO2 : H2) approaching 1 : 1 as the hydrocarbon grows. But that richer proportion of CO2 carries with it CO2's 66% more negative enthalpy of formation (relative to water's), yielding more combustion energy per O2 consumed.

But wait! To get flame temperatures, the overall exothermicity has to be divided by the sum of product heat capacities (scaled stoichiometrically as well). And CP for CO2 is higher than that for water. But not 66% higher; in fact, only about 10% higher!

Thus the numerator in - Delta Ho/ Summation nu i CP (products only) will pull ahead of the denominator, raising the flame temperature as the hydrocarbons grow.

If you had the time to take that ratio, you would've found the adiabatic flame temperatures (K) of methane through butane to be 8000, 8460, 8600, and 8700. (The big jump going to ethane comes from methane's absence of a carbon-carbon bond.)

  • Give the KP at ordinary temperatures for the ff reaction to one significant figure. (All the electronic ground states are singlets.)
    D2 + CH4 arrow right HD + CDH3
    Use your chemical intuition about the molecular parameters! These molecules should be old friends by now. Of course D = 2H.

    • Right away we're not interested in KPelectronic since it's going to be 1. And we can boot KPvibrational too since although the deuterated molecules will have frequencies 2½ smaller than the hydrogenated ones (frequency is inversely proportional to the square root of mass), they'll still be so high as to give zvib of at most 1.05, unlikely to influence a one significant figure answer!

    More important, but to be ignored, would be Delta Do which will have non-negligible terms since the zero-point differences for H2 and D2 will ½[1-(½)½](4400 cm-1) or about 660 cm-1, important at ordinary temperatures where kT~200 cm-1. However, the effect is countered (not exactly) by the change in the C-D vs. C-H zero-points. While D2 goes up to HD, CH goes down to CD, and the effect ought to be nearly a washout. So we'll guess the zero-point correction to be 1.0, uninteresting.

    OK, that leaves KPtranslation at [MHDMCDH3/MD2MCH4] 3/2 = [3×17/(4×16)] 3/2 = 0.71 and KProtation. Ah, the very different sigma symmetry factors will make the largest contribution! Let's do them first.

    HD has sigma =1 since no rotation (other than the identity) makes it look like itself. In D2, on the other hand, sigma =2 since a C2 will exchange identical nuclei. DCH3 is easy as well; it works just like NH3 (but with a more interesting "lone pair") and must have sigma =3. The hard one is CH4. We might remember it from class, or recall the mental rotation drill (grab any one of 4 H atoms and spin about that C-H axis to 3 new symmetrical orientations for a sigma of 12) or you could run the scheme, find CH4 to be Td and count the 12 rotations (including the identity) in that group.

    Since the sigma terms are in zrot's denominator, KP will have a factor like
    sigma D2 sigma CH4/ sigma HD sigma CDH3 = (2×12)/(1×3) = 8.
    I'm impressed.

    Thus far, KP = 0.71×8 = 6 (to one significant figure after all).

    But we're not through worrying about KProtation. We need to consider the effects of B and (ABC)½. They are going to counter one another too...just like the zero point energies did, but B depends on inverse mass and not its square root, so the effect is likely to be bigger.

    In fact, B= hbar 2/2I where I=µr2. The neat thing about isotopic substitutions is that changing the masses has no effect upon the electrons, so molecular geometries and energies (save zero point) are unchanged! So the only important part of B is 1/µ. And µ is "reduced mass" for the diatomic, viz., m1m2/(m1+m2) = 1 for D2 but 2/3 for HD! So B's for those two are proprotional to 1 and 3/2.

    Just the diatomics are going to contribute a factor of BD2/BHD = 2/3 to KProtation. Not negligible but about to be modified by the polyatomics!

    Similarly, the ABC of CH4 depend only on the H mass in motion (each proportional to 1/3 since the C-H bond about which you spin leaves that H at rest). And the very high symmetry of CH4 makes A=B=C, a "spherical top" molecule. Not so with DCH3; when you hold D and spin, you get 1/3 as before for A (proportional), but B=C involve spinning D as well and would be closer to 1/4 [1/(1+1+2)] instead.

    So the A½ factors will be identical between CH4 and DCH3 and will thus cancel out of KProtation. But those (BC)½ terms will be different, and will figure into KProtation as follows:
    (BCCH4/BCDCH3)½ = BCH4/BDCH3 = 4/3

    which makes the rest of KProtation (beyond symmetry factors)
    (2/3)×(3/4) = ½ (!)

    Thus (still ignoring the small zero point energy differences), KP = 0.71(8)½ = 3

  • Nash thoughtfully provided statistical mechanical expressions for many thermodynamic variables of interest, such as

    E = kT2[d ln(Z)/dT]V      S = k ln(Z) + E/T      A = - kT ln(Z)

    but he left out such an expression for the variable most often used by chemists, enthalpy, H = E + PV. Its dependence (on Z, T, and V) turns out to have a satisfyingly symmetrical expression. Find that expression, using the thermodynamic pressure, P = -[dA/dV]T.

    This result will not be confined to the ideal gas! It will serve for any gas. Remember where that expression for P came from: it resulted from the total differential, dA = -SdT-PdV.

    • Much too simple a plug-in. I should've made you find P=-[dA/dV]T instead of giving it to you in the hint. Oh well, maybe it'll relieve you of Calcaphobia.

    H = E + PV
    H = E + V[d(-A)/dV]T
    H = E + V[d(kTlnZ)/dV]T
    H = E + VkT[d(lnZ)/dV]T
    H = kT2[d(lnZ)/dT]V + kTV[d(lnZ)/dV]T
    H = kT × { T[d(lnZ)/dT]V + V[d(lnZ)/dV]T }

  • What's the entropy of 3 harmonic oscillators sharing 4 quanta?

    • S = k lnW and W = (Q+N-1)!/[Q!(N-1)!] = 6!/(4!×2!) = 15

    S = k ln(15) = 2.71 k

    Alternately, you could construct all of the microstates:

    5    Pi
    4o    .067
    3 o   .133
    2  oo o .2
    1 o  oo .267
    0oo o o  .333
    wi363 3 W=15


    Gets S = k ln(15) again. What you cannot do is use the entropy of mixing formula with the Pi level probabilities, since that formula was derived using Boltzmann populations over large numbers of oscillators, not just 3. However, S = -k Summation pi ln(pi) isn't wrong; entropy is, even in this case, the entropy of mixing, but it's mixing of microstates! So since there are 15 microstates, each one equally likely, pi=1/15 for each, and S = k ln(15) yet again!
    Last modified 4 November 1997.