Exam 4 Solutions

4 December 1998

Your browser must support super and subscripts and Greek symbols (abg).
  1. Atkins made an error which got compounded in the homework to Ch. 27. So let's correct the development and use the proper version here! His activated complex rate expression was correct to
    k2 = B eD#S/R e-D#H/RT       (OK)
    and
    Ea = D#H + 2 RT       (OK)
    but when he put those together, he got the sign of the 2RT wrong and wrote
    k2 = B e-2 eD#S/R e-Ea/RT       (wrong)
    leading some of you to wonder why his homework solutions could say
    D#S = R [ ln(A/B) - 2 ]       (OK)
    which is inconsistent. That's because
    k2 = B e+2 eD#S/R e-Ea/RT       (OK)
    is correct instead. So his homework solution was OK, and mine (trusting the erroneous text equation) was wrong (explaining that screwy positive D#S)!

    Let's presume Ea = 50 kJ/mol and A = 1012 L mol-1 s-1 for some reaction at 25°C. Find D#H, D#S, and D#G for that reaction at that temperature.

    D#H = Ea - 2 RT = 50 kJ/mol - 2(8.314×10-3 kJ/mol K-1)(298 K)
           = 45 kJ/mol

    B = (kT/h)(RT/pq)
       = [(1.38×10-23 J/K)(298 K) / (6.63×10-34 J s)] ×
            [0.0821 L atm/(mol K) × (298 K) / (0.987 atm)]
       = 1.54×1014 L mol-1 s-1

    D#S = R [ ln(A/B) - 2 ]
          = 8.314 J/mol K-1 [ ln(1012 / 1.54×1014) - 2 ] = - 58.5 J/mol K-1

    D#G = D#H - TD#S = 45 kJ/mol - (298 K)(-58.5×10-3 kJ/mol K-1
           = + 62 kJ/mol

  2. Diffusion in one dimension is said to result in a Gaussian concentration distribution that flattens out to uniformity at infinite time while normalizing its area to the original number of moles. With the correct normalization, that key result is
    c(x,t) = [ n0 / A(pDt)1/2 ] e-x2/4Dt
    Show that it is a valid solution to the diffusion equation,
    dc/dt = D d2c/dx2

    Let [...] stand for [ n0 / A(pDt)1/2 ] .

    dc/dx = [ ... ] e-x2/4Dt (-1/4Dt) (2x)
    d2c/dx2 = (-2/4Dt) [...]e-x2/4Dt + 4x2 (1/4Dt)2 [...] e-x2/4Dt
    D d2c/dx2 = [...]e-x2/4Dt { (x2/4Dt2) - 1/2t }

    dc/dt = (-x2/4D)(-1/t2) [...]e-x2/4Dt + e-x2/4Dt [...]t1/2 (-1/2t3/2)
          = [...]e-x2/4Dt { (x2/4Dt2) - 1/2t } = D d2c/dx2

    Of course the typo on the exam didn't help. That diffusion equation was for a spherically symmetric problem and the spherical polar second differential not d2/dx2. Full credit if you got dc/dt = [...]e-x2/4Dt { (x2/4Dt2) - 3/2t } and concluded that the diffusion equation or its result must be wrong!

  3. While a He balloon makes an unpleasant bang if you burst it, the pressure differential between the inside and outside is negligibly different from zero! If a 10 cm radius He balloon deflates in 24 hours, what fraction of its surface area represents effusion openings? Use initial rate loss (presumed constant) so you can count on the initial volume, (4/3)pr3, and surface area, 4pr2. Use room T and p. Ignore air diffusing in.

    The original He pressure was 1 atm. It falls to zero over 24 hours. Hence
    dp/dt = - 1 atm/day (1 day/24 hr)(1 hr/3600 s) = - 1.16×10-5 atm / s
    but
    dp/dt = - (kT/2pm)½ pA0/V

    A0 = - (dp/dt)(V/p)(2pm/kT)½
    V = (4/3)pr3 = (4/3)(3.14)(0.1 m)3 = 4.19×10-3 m3
    m = 4.003 amu (1.67×10-27 kg/amu) = 6.68×10-27 kg

    A0 = + 1.16×10-5 atm/s (4.19×10-3 m3 / 1 atm) ×
            { [2(3.14)(6.68×10-27 kg)] / [(1.38×10-23 J/K)(298 K)] }½
        = 1.55×10-10 m2
    S = 4pr2 = 4(3.14)(0.1 m)2 = 0.126 m2
    A0/S = 1.23×10-9

    Alternately, there's S/A0 = 810 million times as much balloon material as leak area. The collision frequency is very impressive to overcome that!

  4. Cyclopropane decomposes to propene at 500°C leaving it partial pressures at 0, 100, and 200 seconds in one experiment to be 400, 373, and 347 torr, respectively. What's the reaction order and rate constant? [Atkins P25.10]

    Fortunately, the time is linear so we can check the fall-off in equal times: 373/400 = 0.933 and 347/373 = 0.930 virtually the same! That means this is a first order reaction and
    ln([A]/[A]0) = - kt
    or
    k = (-1/t) ln([A]t/[A]0) = (-1/200 s) ln(347/400) = 7.1×10-4 s-1

  5. The redox reaction
    I - + OCl - arrow right Cl - + OI -
    is thought to proceed by the mechanism below:
    OCl - + H2O arrow double HOCl + OH -       K1 (rapid equilibrium)
    HOCl + I - arrow right HOI + Cl -       k2     (slow)
    HOI + OH - arrow double OI - + H2O         K3 (rapid equilibrium)
    Predict the overall rate expression for the appearance of Cl - .

    d[Cl - ]/dt = k2 [HOCl] [I - ]
    but that includes a reaction intermediate and so isn't a proper overall rate.

    K1 = [HOCl] [OH - ] / [OCl - ]
    excludes the water since it's in dilute aqueous solution where the water's activity isn't imagined to change.
    Thus [HOCl] = K1 [OCl - ] / [OH - ]
    so d[Cl - ]/dt = k2K1 [I - ] [OCl - ] / [OH - ]
    which is a proper overall expression, confirming that OH - inhibits the reaction (by driving rxn 1 toward reactants, minimizing free HOCl upon which the product depends).

Last modified 2 December 1999