CHM 3411 Physical Chemistry I Fall 1998

FINAL EXAM 11 December 1998

Work any 8. Identify the 2 not to be graded!

  1. Although ( H/ p)T could be derived tortuously from heat capacity expressions, there’s a much, much easier way if you use Maxwell’s equations! In fact, all you’ll need to know is the integral, G = H – TS, and total differential, dG = Vdp – SdT, versions of the Gibb’s Free Energy. Use those to derive the simple expression for ( H/ p)T, the isothermal Joule-Thompson Coefficient, m T.

    H = G + TS
    (H/p)T = (G/p)T + T(S/p)T
    dG = Vdp - SdT means that (G/p)T = V
    and Maxwell's equation from this total differential is
    (S/p)T = - (V/T)p by the equivalence of the cross-derivatives. So
    mT = (H/p)T = V - T(V/T)p

  2. The expression you’re looking for in (1) is mT = V – T(V/T)p, and we’ve already shown in class that it must be zero for ideal gases, pV=RT (V presumed molar). But it’s not zero for a Van der Waals gas. The major feature that makes m T non-zero is VdW’s attraction term not its molecular volume term. So use p = (RT/V) – a/V2 where a = 1.408 atm L2/mol2 for N2, and obtain m T at 1 atm and 25° C. Experimentally, it’s – 0.072 L. (You may need the quadratic formula x = [–b±(b2–4ac)½]/2a , but only the + sign works.)
  3. p = (RT / V) - a / V2
    0 = (p/T)p = R / V - (RT / V2)(V/T)p - (-2a / V3)(V/T)p
    (V/T)p = (R / V) / [ (RT / V2) - (2a / V3) ]
          = R / [ (RT / V) - (a / V2) - (a / V2) ] = R / [ p - (a / V2) ]
    T(V/T)p = RT / [ p - (a / V2) ]

    p = 1 atm = (RT / V) - (a / V2) rearrange to solve for V (/mol presumed)
    pV2 = RTV - a
    p V2 - RT V + a = 0
    V = [ + RT + { (RT)2 - 4 pa }½ ] / 2p
    RT = 0.08206 atm L mol-1 K-1 (298 K) = 24.45 atm L (surprise)2
    V = [(24.45 atm L)+{(24.45 atm L)2-4(1 atm)(1.408 atm L2)}½]/2 atm
       = 24.39 L not quite 24.45 due to the attractions

    mT = V - T(V/T)p = V - { RT / [p - (a / V2)] }
        = 24.39 L - { 24.45 atm L / [1.0 atm - 1.408 atm L2 / (24.39 L)2] }
        = - 0.118 L (not bad for an estimate; only 64% high)

  4. Using the partial VdW equation of question (2), p = (RT/V) – a/V2, find the temperature at which N2’s fugacity is 9.8 atm when p = 10 atm. (Hint: when you integrate, substitute the ideal gas equation to simplify! But don’t do that before you integrate!) a is as before.

    f / p = f = 9.8 atm / 10.0 atm = 0.98
    lnf = integral0p (Z-1)/p dp = ln(0.98) = - 0.202
    Z = pV / RT = 1 - a / (RTV)
    Z-1 = -a / RTV
    lnf = (-a / RT) integral0p (1/pV) dp (substitute pV~RT)
        ~ (-a / RT) (1/RT) integral0p dp = [-a / (RT)2] p
    T2 = - a p / (R2 lnf)
    T = [- a p / (lnf)]½ / R
        = [-1.408 atm L2 × 10 atm / (-0.0202) ]½ / (0.08206 atm L K-1)
        = 322 K

  5. Nitrogen’s normal boiling point is 77.34 K. But it doesn’t go critical until 130 K. If the critical pressure is 34 atm, what is nitrogen’s average D vapH over that temperature range? (Hint: at its normal boiling point, D vapH is 5.586 kJ/mol, but you don’t need that. It’s just a confirmation.)

    p = p0 e-(DvapH / R)×[(1/T) - (1/T0)] or
    -(DvapH / R) = ln(p/p0) / [(1/T) - (1/T0)] or
    DvapH = R ln(p/p0) / [(1/T0 - (1/T)]
        = 8.314 J/mol K-1 ln(34 atm/1 atm) / [(1/77.34) - (1/130)] K-1
        = 5598 J/mol = 5.598 kJ/mol

  6. At 136.7° C, chlorobenzene’s vapor pressure is 863 torr while bromobenzene’s is only 453 torr. Assume that they form an ideal mixture. (a) What’s the mole fraction of chlorobenzene in such a mixture that has a vapor pressure of 760 torr at that temperature? (b) Suppose you prepared such a mixture, raised its temperature to 136.7° C, and it wasn’t boiling! What would that tell you about the relative (non-ideal) intermolecular forces?

    Ideal mixture implies a linear relation between both partial pressures and the corresponding liquid mole fraction. That implies a linear relation of the total pressure (if the vapor's ideal too).

    1. p = p0fBr + XfCl(p0fCl - p0fBr)
      p = 453 + (863-453)XfCl = 453 + 410 XfCl all pressures in torr
      XfCl = (p-453)/410 = (760-453)/410 = 0.749 hence
      XfBr = 1 - 0.749 = 0.251

    2. Since ambient pressure is 760 torr, we expect that mixture to be boiling. If it's not, the London and dipole-dipole forces binding the chlorobenzene to the bromobenzene must be stronger than those binding each molecule to its own kind, making the solution a more attractive phase, and lowering the vapor pressures.

    1. Derive the pH expression for the stoichiometric point in the titration of a weak base, pKb, by a strong acid in terms of the concentration of the "salt," [BH+] = S.

    2. Hydrazine is a weak base with pKb of 5.77. If a 0.01 molar solution of it is titrated with 0.01 M HCl, at what pH should we stop the titration?

    1.  
      BH+ à BH(aq)+ H+       Kconj acid = Kw / Kb = Ka
      Ka = [BH][H+]/[BH+] = x2/(S-x) ~ x2/S = [H+]2/S
      pKa = 2 pH + log10S
      pH = ½pKa - ½log S = ½pKw - ½pKb - ½log S

    2. S = [hydrazine H+] = 0.01 M (½) (since the volume doubles with the addition of the acid)
      pH = ½(14) - ½(5.77) - ½log(0.005) = 5.27
  7. Nitrate ion can oxidize mercury(I) to mercury(II) in acid solution. What is the 25° C equilibrium constant for that reaction? To how many significant figures are you entitled?

    NO3- + 4 H+ + 3 e- à NO(aq) + 2 H2O     E° = + 0.96 V

    2 Hg2+ + 2 e- à Hg22+                         E° = + 0.92 V

    E° = + 0.96 - (+ 0.92) = + 0.04 V (one significant figure!)
    represents the reaction (2× the first minus 3× the second):

    2 NO3- + 8 H+ + 6 Hg2+ à 2 NO(aq) + 3 Hg22+ + 4 H2O       n = 6

    ln K = - DG° / RT = + (nF / RT)E° = 9.34
    K = eln K = e+9.34 = 1.14×104 ~ 104 (only 1 digit good)

  8. The limiting ionic conductivities (in mS m2 / mol) for the hydrogen and hydroxyl ions are 34.96 and 19.91, respectively. What is the resistance (R in ohms = S- 1) of pure water at 25° C in a vessel 1 m long with cross-sectional area of 0.1 m2? (Hint: R = 1/G and G = k A/L where k is the conductivity and L and A are the length and cross-sectional area.)

    k = 34.96 [H+] + 19.91 [OH - ] mS m2 /mol
        = (34.96 + 19.91)×10-7 mol/L mS m2 /mol
        = 5.487×10-9 S m2 / 10-3 m3 = 5.487×10-6 S/m

    R(W) = L / (kA) = 1 m / (5.487×10-6 S m-1 × 0.1 m2) = 1.822×106 S-1
        = 1.822 MW

  9. The proposed mechanism for the decomposition of acetaldehyde to form methane and CO involves the methyl radical as follows:

    CH3CHO à ·CH3 + ·CHO                         k1

    CH3CHO + ·CH3 à CH4 + CO + ·CH3         k2

    2 ·CH3 à C2H6                                         k3

    Ignoring the fate of the ·CHO but not knowing which rate constant is rate-limiting, you can still obtain the overall rate of appearance of methane by presuming that the methyl radical is in steady-state, e.g., its steady-state rate is zero. Notice that methyl radical catalyzes the production of methane in reaction 2, so · CH3’s rate of change is not effected by reaction 2. (Hint: there are no equilibrium reactions in this scheme. Don’t get anxious about non-integer orders; happens all the time.)

    d[CH4]/dt = k2 [CH3CHO][·CH3]

    0 = d[·CH3]/dt = k1 [CH3CHO] - 2 k3 [·CH3]2
    [·CH3]2 = ½(k1/k3) [CH3CHO]
    [·CH3] = (k1/2k3)½ [CH3CHO]½

    d[CH4]/dt = k2(k1/2k3)½ [CH3CHO]3/2 (if we believe the mechanism)

  10. Collision theory bimolecular rate constants result in k2 µ T½ exp(- Ea/RT). How should we correct the usual method of obtaining Ea from plots of ln k2 versus 1/T?

    Normally, we'd equate Ea/R with the negative slope of ln(k2) vs (1/T), but now there's another T-dependent term.

    ln k2 = ½ ln T - Ea/RT + ln C not T-dependent
    ln k2 = - ½ ln(1/T) - (Ea/R)(1/T) + ln C
    d ln k2/d(1/T) = - ½ (1/T)-1 - Ea/R = - ½T - Ea/R

    Ea = - R×slope - ½RT

    So while the ln k2 vs (1/T) plot won't be linear, we can still estimate the slope at various temperatures then subtract ½RT from the usual Ea calculation to make the modest correction.


Last modified 14 December 1998.