Exam 2 Solutions

19 October 1998

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  1. Trouton's rule, DvapS ~ 85 J K-1 mol-1, might be true simply because of the enormous DvapVm. Test that hypothesis by calculating DSm for expansion of water vapor (assumed ideal) from liquid to gaseous density at standard conditions. What fraction of DvapS does it represent?

    DexpS = R ln(Vf / Vi).
    Vi = MW/r = 18 g/mol / 1.0 g/cc = 18 cm3.
    Vf = (RT/p)q = 0.08206 atm L/mol K (298 K) / 1 atm = 24.45 L = 24,450 cm3
    DexpS = 8.315 J/mol K ln(24,450/18) = 60 J/mol K or 70% of Trouton's 85. So more than simple expansion is going on even without water's flakiness!

    Logically, we should have used T = 373 K not 298 K to estimate DS, in which case, we'd have obtained 61.8 J/mol K or 73% of Trouton. (Is that 730 mTrouton's?)


    Actually, water's hydrogen bonding makes a mockery of Trouton's rule, and it's DvapS at it's boiling point is 109 J K-1 mol-1.

  2. Estimate the correction to the chemical potential (J/mol) of Xe gas at STP due to its non-zero first virial coefficient, B' = -0.006861 atm-1. Recall that Z = 1 + B'p + C'p2 + ... but ignore C' and higher terms.

    m = m0 + RT ln(p/pq) + RT ln(F)
    ln(F) = [integral 0 to p] (Z-1)/p dp = B' [integral 0 to p] dp = B'p

    STP is 273 K and 1 atm for gases (ignoring "bars"),
    RT ln(F) = 8.315 J/mol K (273 K) (-0.006861 atm-1) 1 atm
    = - 15.57 J/mol

    but I'd take standard conditions (25°C and 1 atm) too . . .
    RT ln(F) = 8.315 J/mol K (298 K) (-0.006861 atm-1) 1 atm
    = -17 J/mol

    In either case, this terribly non-ideal noble gas doesn't have a very large correction to its chemical potential considering that RT is about 2.5 kJ/mol.

  3. Ether (1,1'-oxybis-ethane) is a terrible fire hazard (not to mention explosive when it peroxidizes with age). With its high flammability and low boiling point (only 34.5°C), it's the best argument for a flame-free laboratory! Demonstrate that by computing ether's vapor pressure at 25°C given DvapHq = 27.10 kJ/mol.

    p = p* e-(DvapH/R)×[(1/T)-(1/T*)]
       = 1 atm e-(27,100 J/mol/8.315 J/mol K)×[(1/298 K) - (1/307.7 K)]
       = 0.714 atm = 542 Torr

  4. Estimate the temperature at which DfGq(NH3) = 0 given that at 25°C, DfGq = - 16 kJ/mol and DfHq = - 45 kJ/mol.

    d(DG/T)dT = - DH/T2
    Integrating, D(DG/T) = + DH/T
    DG/T = DGq/Tq + DHq×[(1/T) - (1/Tq)]
    0 = DG = xDGq + DHq(1-x) where x = T/Tq
    x = DHq/(DHq - DGq) = -45/(-45+16) = 1.55
    T = 1.55×Tq = 1.55(298 K) = 462 K

    (You get the same answer if you set DG = DH - TDS = 0 and solve for T taking DH and DS at their standard conditions values.)

  5. For X~Y, ln(X/Y) = ln[ 1 - (Y/Y) + (X/Y) ] = ln[ 1 + (X-Y)/Y ] can be approximated by (X-Y)/Y. Use that approximation to estimate the triple point temperature of a compound for which you know all there is to know about its fusion and vaporization, viz., DfusV, Tfus, DfusH, and Tvap, DvapH. Your estimate will assume that the triple point pressure is nearly 1 atm and the triple point temperature is not very different from Tfus or Tvap. (We could use that first guess to refine the estimate via interation, but I don't want you to do that; just derive an expression for the first guess.)

    The triple point occurs where (p,T)liq,solid = (p,T)liq,gas.
    If the normal boiling point temperature Tvap=Tb and the normal freezing point temperature is Tfus=Tf and the triple point temperature is Ttp, then from the vaporization equation (Clausius-Clapeyron),

    ptp = 1 atm e-(DvapH/R)×[(1/Ttp)-(1/Tb)]
    ln(ptp/1 atm) = -(DvapH/R)×[(1/Ttp)-(1/Tb)]

    Apply the small ln estimate to p/1 atm and reduce the right hand side

    (ptp - 1 atm)/1 atm ~ -(DvapH/R)×[(Tb-Ttp)/(TbTtp)]
    ptp ~ 1 atm - (1 atm)(DvapH/R)(Tb-Ttp)/Tb2

    And from the melting equation (Clausius),

    ptp = p* + (DfusH/DfusV) ln(Ttp/Tf)
    where p* is the TOTAL pressure above the melt at the normal freezing point; so it too is 1 atm.
    ptp ~ 1 atm + (DfusH/DfusV)(Ttp-Tf)/Tf

    And equating the two version of ptp to solve for the common Ttp gives

    Ttp ~ [ - (DfusH/DfusV) + (1 atm)DvapH/RTv] /
        [(1 atm)(DvapH/RTb2) - (DfusH/TfDfusV)]

    You'd arrive at this same expression by using p = I + sT, where I is the intercept and s the slope of the phase change "lines" with constants evaluated at 1 atm & normal Ttrs.

Last modified 26 October 1998