CHM 3411   Physical Chemistry I   Fall 1999

FINAL EXAM SOLUTIONS   8 December 1999

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

  1. In last year's final, students were asked to derive an expression for the isothermal Joule-Thompson Coefficient that turned out to be very simple:

    mT = (H/p)T = V - T(V/T)p     which you may assume for the purposes of this problem.

    Using the full expression for the Van der Waals gas, show that an excellent approximation to mT is given by

    mT = nb - 2an/RT

    or, for a single mole,

    mT = b - (2a/RT)

    (The approximations that reduce the final expression to this are that V/(V-nb) ~ 1 and 2an2 << nRTV, both of which are excellent. And remember the trick to set p = stuff so that partials with p fixed will set the partials of stuff to zero.)

    (p + an2/V2) (V - nb) = nRT
    p = nRT/(V-nb) - an2/V2
    0 = (p/T)p = nR/(V-nb) - [nRT/(V-nb)2](V/T)p + [2an2/V3](V/T)p
    (V/T)p = { nR/(V-nb) } / { [nRT/(V-nb)2] - 2an2/V3 }
    T(V/T)p = { nRT/(V-nb) } / { [nRT/(V-nb)2] - 2an2/V3 }
    V = nb + (V-nb)     so
    V - T(V/T)p = nb + (V-nb) - T(V/T)p
    = nb + { [(V-nb)nRT/(V-nb)2 - 2an2(V-nb)/V3] - nRT/(V-nb) } / { [nRT/(V-nb)2] - 2an2/V3 }
    = nb - 2an2(V-nb)/V3 / [ nRT/(V-nb)2 - 2an2/V3 ]
    = nb - 2an2(V-nb)3/V3 / [ nRT - 2an2(V-nb)2/V3 ]
    ~ nb - 2an2 / [ nRT - 2an2/V ] ~ nb - 2an2/nRT = nb - 2an/RT = mT

  2. Using that last expression for mT, obtain DH for the isothermal compression of a mole of ethane at 298 K from 1 atm to 100 atm. For ethane, a = 5.562 atm L2 mol-1 and b = 0.06380 L mol-1.

    DH ~ Dp × (H/p)T = Dp × m T = Dp × (b - 2a/RT)
    DH ~ 99 atm × (0.06380 L - [2×5.562 atm L2 / {0.08206 atm L / mol K × 298 K}]
    DH ~ - 38.9 atm L (8.314 J/0.08206 atm L) = - 3.94 kJ

  3. The conductivity of a 0.0312 M solution of a weak base is 1.53×10-4 S cm-1. If the sum of the limiting ionic conductances of BH+ and OH- is 237.0 S cm2 / mol, what is the value of the base constant, Kb?

    k = L° c
    L° = n+l+ + n-l- = 237.0 S cm2 / mol
    c = k / L° = 1.53×10-4 S cm-1 / 237.0 S cm2 mol-1 = 6.45×10-7 mol/mL
    c = 6.45×10-4 M = [BH+] = [OH-]
    [B] = 0.0312 - c = 0.0306

    Kb = [BH+] [OH-] / [B] = (6.45×10-4)2 / 0.0306 = 1.36×10-5

  4. One liter of a 0.1 M solution of substance A mixes ideally with 3 liters of a 0.05 M solution of substance B at 25°C. What is DmixS? And DmixH?

    5. Water mixing with water represents no entropy change, but moles of A mixing with moles of B surely does. And we have 0.1 moles of A and 0.15 moles of B for a total of 0.25 moles of solute. That's n. And the final mole fraction of A is XA = 0.1/0.25 = 0.4 meaning that XB = 0.6.

    DmixS = - n R (XAlnXA + XBlnXB) = - ¼ mol 8.314 J/mol K (0.4ln0.4 + 0.6ln0.6)
    DmixS = + 1.40 J/K

    DmixH = 0.00 by definition of ideal mixture.

  5. In the older literature, pressures are in mm Hg and logarithms are to the base 10. Fortunately, T is still in Kelvins! In such an article, you find the vapor pressure curve of some substance listed as

    log10(p) = 7.9741 - (1318.9 / T)

    Obtain its normal boiling point temperature and DvapH there.

    First thing first; we ought to convert to natural logarithm via ln(x) = 2.3026 log(x), so

    ln(p) = 18.361 - (3036.9/T)

    Even without that, the normal boiling point is easy; p = 760 mm Hg. So

    ln(760) = 18.361 - (3036.9/T)
    T = 3036.9 / [ 18.361 - ln(760) ] = 258.95 K = - 14.20°C

    Now oddly enough, we don't have to change to bars or atmospheres to proceed to the enthalpy of vaporization. That's because a conversion factor inside the logarithm just becomes an additive term, altering the magnitude of the 18.361 but not changing the slope with respect to 1/T, which, by the Clausius-Clapeyron equation, is DvapH.

    DvapH/R = - d(lnp)/d(1/T) = 3036.9
    DvapH = 8.3145 J/mol K × 3036.9 = 25.250 kJ/mol

  6. A biological cell at 37°C needs to run the hypothetical reaction:
    A + B <===> Z       DG° = + 23.8 kJ/mol
    and gets it to be spontaneous by using an enzyme that couples that reaction with
    ATP <===> ADP + phosphate       DG° = - 31.9 kJ/mol
    Find the equilibrium constant the cell manages for
    A + B + ATP <===> Z + ADP + phosphate

    (All values pertain to 37°C.)

    DoverallG° = + 23.8 - 31.9 = - 8.1 kJ/mol

    K = e- DoverallG° / RT = e8100 J/mol / ( 8.314 J/mol K × 310 K ) = 23.2 > 1

  7. The following data are for isopropanol (I) in benzene (B) at 25°C.

    XI 0 0.059 0.521 0.924 1
    PI(torr) 0 12.9 30.5 42.2 44.0
    PB(torr) 94.4 86.6 75.3 24.2 0

    (a) Does this solution exhibit positive, negative, or no deviation from ideality?

    Were the solution ideal, we'd expect p0.5 to be ½ way from 0 to p*, but 30.5 > 22.0 and 75.3 > 47.7 for XI~0.5, hence the deviations are quite positive. We wouldn't have expected otherwise from the mixture of an organic (no hydrogen bonding) and an alcohol (lots of hydrogen bonding).

    (b) Estimate KHenry for both substituents. Are they consistent with your answer to (a)?

    psolute ~ KH Xsolute
    KH ~ psolute / Xsolute

    KH(I) ~ 12.9 torr / 0.059 = 219 torr >> 44 torr (positive deviation)

    KH(B) ~ 24.2 torr / (1-0.924) = 318 torr >> 94.4 (positive deviation)

  8. The reaction
    cis-Cr(en)2(OH)2+ <===> trans-Cr(en)2(OH)2+
    is first order in both directions. At 25°C, the equilibrium constant is 0.16 and the rate constant k1 = 3.3×10-4 s-1. In an experiment starting from the pure cis form, how long would it take for half the equilibrium amount of the trans isomer to be formed?

    (Of course [cis]+[trans] = A0, the original concentration of cis. So there's really only one variable to solve the rate equation for, [trans]=x, as a function of time.)

    Let x=[trans] and A0 be [cis]0, the initial cis concentration, then

    dx/dt = k1[cis] - k-1[trans] = k1(A0-x) - k-1x = k1A0 - (k1+k-1)x
    dx / [k1A0 - (k1+k-1)x] = dt which integrates to
    - ln[k1A0 - (k1+k-1)x] |0x / (k1+k-1) = t |0t
    ln{ [k1A0 - (k1+k-1)x] / [k1A0] } = - (k1+k-1)t
    [k1A0 - (k1+k-1)x] = k1A0 e- (k1+k-1) t

    or

    x = A0 [ k1 / (k1+k-1) ] (1 - e- (k1+k-1) t ]

    Although we know that k1[cis]eq = k-1[trans]eq, or rather K = [trans] / [cis] = k1/k-1, we can actually arrive at that by setting t to infinity (where equilibrium has been established). Then
    x = [trans]eq = A0 [ k1 / (k1+k-1) ]
    [cis]eq = A0 - x = A0 { 1 - k1 / (k1+k-1) } = A0 [ k-1 / (k1+k-1) ]
    hence [trans]eq / [cis]eq = k1 / k-1 = K deja vu

    So k-1 = k1 / K = 3.3×10-4 s-1 / 0.16 = 2.1× 10-3 s-1 and k1+k-1 = 2.39×10-3 s-1

    That's all we need to establish t½ where x = ½A0 [k1 / (k1+k-1)]

    t½ = - [1 / (k1+k-1)] ln { [ k1A0 - ½A0(k1+k-1)k1 / (k1+k-1) ] / k1A0 }
    t½ = - [1 / (k1+k-1)] ln (½) = 0.693 / 2.39×10-3 s-1 = 290 s < 5 min.

  9. The following mechanism has been proposed for the thermal decomposition of pure ozone in the gas phase:
    2 O3 ---> O3 + O + O2       k1
    O + O3 ---> 2 O2       k2
    Derive the rate equation.

    - d[O3]/dt = k1 [O3]2 + k2 [O] [O3]

    d[O]/dt = k1 [O3]2 - k2 [O]ss [O3] ~ 0
    [O]ss = k1 [O3]2 / ( k2 [O3] ) = (k1 /k2) [O3]

    - d[O3]/dt = k1 [O3]2 + k2 (k1 /k2) [O3]2 = 2 k1 [O3]2
    since steady state in this mechanism means that every reaction 1 is followed immediately by reaction 2 at the same rate, and each consumes one O3.

  10. In Alaska, you put heating elements on your car's engine at night not only to keep the block from cracking in the intense cold but also, perhaps, to keep the battery warm. The battery ads that tout "cold cranking amps" do so to encourage buyers in cold climates. What's the deal? How does a lead acid battery's voltage change between 25°C and, say, -25°C? Bear in mind that a "battery" means a collection of cells in series to add their voltages. In an automobile "battery," there are 6 cells. (Which is why you have to top off the distilled water in each of six little holes. Have you remembered to do that lately?)

    In case you don't remember it, an auto battery reacts Pb and PbO2 to produce PbSO4 from each in sulfuric acid solution. For our purposes, we'll let the acid activity be 1, but in real batteries it's more concentrated.

    Species DfG° (kJ/mol) DfH° (kJ/mol) S° (J/mol K)
    Pb0064.81
    PbO2(s) -217.33-277.468.6
    PbSO4(s) -811.24-981.39147
    H+(aq) 000
    SO42-(aq) -744.53-909.2720.1
    H2O(aq) -237.13-285.8369.91

    - nFE° = DG°
    E° = - DG° / nF
    DG° = DH° - TDS°
    dE°/dT = - (dDG°/dT) / nF = + DS° / nF

    Pb + PbO2 + 4 H+ + 2 SO42- ---> 2 PbSO4 + 2 H2O       n = 2
    DS° = 2 (147) + 2(69.91) - 64.81 - 68.6 - 4 (0) - 2 (20.1) = 260 J/mol K

    dE°/dT = DS° / nF = 260 J/mol K / [2 × 96,450 C/mol] = + 1.35 mV/K

    DE° ~ (dE°/dT) DT = 1.35 mV/K (- 50 K) = - 0.067 V per cell
    DE° ~ 6(0.067) = - 0.40 V per battery

Last modified 8 December 1999.