CHM 5414 Thermodynamics

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Exam #3 Solutions

5 December 1996

  1. Choose the tool(s) from this list:

    1. DIPPR
    2. NIST
    3. STANJAN
    4. Stull's "Chemical Thermodynamics of Organic Compounds" book
    5. None of the above

    most appropriate to the tasks below:

    1. a study of the influence of geometrical isomerism on bond energies
    2. a search for optimal propellants
    3. the heat capacities of a series of industrial organic solvents near their boiling points
    4. lambda max for a homologous series of ketones which are liquid below 100°C
    5. optimization of yield in a near thermoneutral reaction within limited pressure and temperature ranges.

  1. All isomers are easily found with the molecular formula input of NIST.
  2. The database and P,T manipulations of STANJAN make it the obvious choice.
  3. Industrial chemicals and a vast variety of chemical/physical data regarding them is the strong point of DIPPR.
  4. lambda max betokens spectroscopic data which is offered by none of the above
  5. Reactions yields require at least equilibrium data which is STANJAN's long suite. Of course if the reactions involve organic compounds, you'll need Stull's book.

    1. Confirm that water refuses to "spread" on mercury.


    • This will be difficult to do since it does spread on mercury!

    Your instructor transposed some digits before entering them in his calculator. Worse still, it is because the water-mercury surface tension is so much lower than the mercury-air one, that part (b) finds no surface energy penalty to immersion of mercury in water!

    So the question should have read Confirm that water "spreads" on mercury.

    Nobody loses credit on this one (except Dr. Parr).

    Here's what proves the point:

    S = gammaHg-air - ( gammaHg-water + gammawater-air )

    is the spreading coefficient measuring the energy cost of leaving the Hg in the air relative to having water grow its areal contact with both the Hg and the air as it spreads. If that's negative, we leave well enough alone. If it's positive, you'd lower your energy by covering up the Hg even at the expense of creating additional water surface area!

    S = 485 - (375 + 73) = + 37 mN/m > 0

    meaning that water does spread on mercury! Maybe I'll learn to work a calculator by next semester.

  • It might seem that the reverse (spreading of Hg on water) is meaningless because of the density difference: 13.6 g/cc Hg sinks through 1.0 g/cc water. But wait! Didn't Galileo float an iron needle on water, using its surface tension to refute Aristotle's "all things find their level?" If so, is it possible to find a macroscopic (don't riddle me "atoms" here) Hg droplet small enough to be supported by water's surface tension? Prove your answer!

    • If that were possible, it would mean countering gravity's energy lowering with -dh by an energy rise in turning Hg-air + water-air surface into water-Hg instead. Imagine and element of surface dA changing hands in that way; the surface energy change would be:

    dE = ( gammaHg-water - gammaHg-air - gammawater-air ) dA
    or
    dE = (375 - 485 - 73) dA = - 183 dA < 0 (for dA>0)

    In other words, immersing the Hg bears no energetic penalty but rather an energy lowering in the same direction as the gravity lowering. There is no gravity/surface-tension competition; they cooperate to sink the mercury regardless of droplet size.

  • The following data were observed for the adsorption of N2 on a mica surface at 90°K. V is measured (as always) as if it were at 1 atm and 273°K. The isotherm of this process suggests that only a monolayer is important under these conditions. Find the area covered by a single N2 molecule if the density of liquid N2 is 810 kg m-3, and estimate the surface area of the sample (per gram of mica).

    P (atm)2.83.44.0 6.09.417.123.5
    V (10-6 m3 / g)12.013.4 15.117.023.928.230.8

    • If only monolayers are important, the Langmuir isotherm equations should suffice:

    P/V = P/Vm + 1/aVm

    tells us to plot P/V against P and take the slope to be 1/Vm, the (inverse of the) monolayer volume per gram of sample (normalized to STP).

    Open this Graph!

    So that gets us Vm = 1/(25,300 g m-3) = 3.95×10 -5 m3 N2 / g (of mica).

    At STP, that's worth 3.95×10 -5 m3 ÷ (22.4 L/mol × 10 -3 m3/L) = 1.76×10 -3 mol N2 adsorbed as a monolayer on 1 gram of mica (at 90°K). What surface area does that many molecules occupy?

    We presume each N2 occupies a small cube for simplicity. Knowing the MW and density of N2, we can calculate the volume of each molecule, Vmolecule=s3 where s is the length of the side of that molecular cube. It's area must then be s2=V2/3, and the total area occupied by a monolayer (per gram of mica) would be that times the number of N2 found above...well, actually, the number of moles we found above times Avogadro's Number.

    So each molecule's "area" of contact with the mica is:

    a = [ MW / ( rho ×NAv) ]2/3
    a = [ 0.028 kg/mol / (810 kg/m3×6.022×1023) ]2/3 = 1.49×10-19 m2

    And the area occupied by a monolayer of N2 molecules per gram of mica is

    Amono = 1.49×10-19 m3/molecule × 1.76×10-3 moles per gram mica × 6.022×1023 molecule/mole
    Amono = 158 m3 per gram of mica

    ...not a whole lot for a powder; a square about 6 yards on a side is all. Whoopee.

  • A macromolecule M2 is in a form of a dimer in equlibrium with its monomer, M,

    M2 arrow right 2 M

    Given a reaction progress parameter, x, such as we use in Chem. I, it is clear that a number-averaged (osmotic pressure) molecular weight would vary with x, being 2m when x=0 but only m when x=1.

    1. Derive the number-averaged molecular weight as a function of x.


    • Call it <#> for short. The weights used in this average are just the number (molecules, moles, or even mole fractions...any of them will do the trick since they are just scalings of one another) of compounds with different masses. Here we have only two types, and we can represent their number of moles, let us say, as nmono and ndimer.

    From the reaction equation, it's clear that when x=0, we have ndimer=1 and nmono=0, but when x=1, we have ndimer=0 and nmono=2. So ndimer=(1-x) and nmono=2x.

    <#> = Summation nimi / Summation ni = [ (1-x)2m + 2x m ] / [ (1-x) + 2x ] = 2m / (1+x)

    which fits our expectations perfectly, since <#>=2m when x=0 but only m when x=1.

  • But if the molecule weight is determined by light scattering, it will be a mass-averaged molecular weight. Derive that molecular weight as a function of x.

    • Now the weighting factor for each compound becomes nimi instead of just ni, and

    <mass> = Summation nimi2 / Summation nimi

    or

    <mass> = [ (1-x)(2m)2 + 2x(m)2 ] / [ (1-x)2m + 2x m ]
    <mass> = (2-x)m

    also fitting our expectations when x=0 or 1 but definitely not the same function of x as <#>.

  • When x is either 0 or 1, only one kind of molecule exists, and the polydispersity ratio will be 1. But that's not the case for the general x between 0 and 1. Derive the polydispersity ratio as a function of x.

    • Since our <mass> and <#> were the same at 0 and 1, their ratio must be 1.000 as required for only one compound present. But in general,

    PDR = polydispersity ratio = <mass> / <#>
    PDR = (2-x)m(1+x)/2m = (2+x-x2)/2

    which satisfies at the 0 and 1 limits as expected.

  • What value of x maximizes the polydispersity ratio for this mixture?

    • A function of x is extremal when its first derivative is zero (remember that far back?).

    Partial [(2+x-x2)/2]/ Partial x = (1/2) [ 1-2x ] = 0

    which occurs when x=1/2. Not surprising.

  • What, therefore, is the maximal polydispersity ratio to be found in a dimeric/monomeric mixture?

    • At last...a simple substitution:

    PDRmax = PDR(x=½) = [ 2 + ½ - (½)2 ]/2 = 1.125

    not a whole lot different from 1.000 for such simple mixtures.
    Last modified 12 December 1996.