CHM 3312 Final Solutions 7 May 1997

Instructor: Chris Parr BE3.506 883-2485 parr@utdallas.edu
                       http://wwwpub.utdallas.edu/~parr/


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  1. For the reaction, 2 NO2 arrow right N2O4# arrow right 2 NO + O2, Johnston et al. [JCP 25, 736 (1956)] have proposed an activated complex like:
    
               O   O   N
                \ / \ / \
                 N   O   O
    whose activation energy, Delta epsilon 0 = 111 kJ/mol. While # has 3(6)-6 or 12 vibrations, only those shown in the following table are important to the reaction at 500°K. To save you time, the rotational constants are give as the product ABC, which has units of (cm-1)3. Find the pre-exponential factor and bimolecular rate constant k2 (p. 942) at 500°K.

    Molecule gel sigma ROT ABC, (cm-1)3 (degeneracy) nu , cm-1
    NO2 2 2 1.53 1332, 751, 1616
    N2O4# 1 2 0.00745 (2) 1700, (2) 1300, (2) 700, 350, 250

k2 = (kT/h) K' where we've already swallowed our ignorance of the transmission coefficient (kappa) by setting it to unity. And K' is K# without the reaction coordinate imaginary vibration.

kT/h = (1.38×10-23 J/molec°K)(500°K)/(6.626×10-34 Js) = 1.04×1013 s-1.

K' = (RT/pØ) (qmØ/NAv)N2O4# / [(qmØ/NAv)NO2]2 × e- Delta epsilon 0 / RT
where we've retained the NAv denominators since we have a handy formula for qm which includes them.

RT/pØ = (0.08206 L atm / mol°K) (500°K)/(1 atm) = 41.03 L mol-1

e- Delta epsilon 0 / RT = e-(111,000 J/mol) / [(8.314 J/mol°K)(500°K)] = 2.53×10-12

Since qmØ/NAv = 2.561×10-2 T5/2 M3/2 for translations,

[qmØ/NAv]N2O4#] / ([qmØ/NAv]NO2)2
= [(MN2O4#)3/2/(MNO2)3] / [2.561×10-2 T5/2]
= [(92)3/2/(46)3] / [0.02561 (500)5/2]
= 6.33×10-8 for K'TRANS. It's not surprising since 2 molecules have become one!

Since qROT = 1.0270 T3/2 / [ sigma (ABC)½],

[qROT]N2O4# / ([qROT]NO2)2
= sigma 2NO2 ABCNO2 / [ sigma N2O4# 1.027 (500)3/2 (ABC)½N2O4# ]
= 3.09×10-3 since although # is larger, the number of active rotation states of NO2 get squared!

[qEL]N2O4# / ([qEL]NO2)2
= [gEL]N2O4# / ( [gEL]NO2 )2 = 1/22 = ¼

Now the tedious one, [qVIB]N2O4# / ([qVIB]NO2)2
= Product (1-e-h nu NO2/kT)2 / Product (1-e-h nu #/kT) where h nu /kt = 1.4388 cm°K× nu /500°K = 0.00288 nu .

= [(0.978)(0.885)(0.990)]² / [(0.992)²(0.976)²(0.867)²(0.635)(0.513)] = 3.20

Putting it all together:

k2 = 1.04×1013 s-1 × 41.03 L/mol × 6.33×10-8 × 3.09×10-3 × ¼ × 3.20 × 2.53×10-12
k2 = 1.7×10-7 L mol-1 s-1

All of k2 except the exponential term is the "pre-exponential factor;" it's 6.7×104 L mol-1 s-1 for the equivalent of the Arrhenius A factor (from first principles, as it were).

(And if you think that was tedious, try coding it in HTML.)

  • The reverse of that reaction is termolecular. Its exothermic rate constant, k3 , can be obtained (pp. 876 & 279) given the k2 above and data in Table 2.12, p. C13. Remember that T=500° not 298°K. (An experimental value of k3 = 3.4×103 L2 mol-2 s-1 at 500°K.)

    • KC = [NO]2 [O2] / [NO2]2 = k2 / k3 by dynamic equilibrium. So k3 = k2 / KC.

    And KP = e- Delta G/RT = (pNO/pØ)2 (pO2/pØ) / (pNO2/pØ)2 = [p2NO pO2 / p2NO2] (1/pØ)

    [NO] = nNO (mol) / V (L) = pNO / RT
    or pNO = [NO] RT. Substitute that in KP:

    KP = ( [NO]2 [O2] / [NO2]2 ) (RT/pØ) =KC (RT/pØ)

    KC = (pØ/RT) KP = (pØ/RT) e- Delta G/RT

    where we can get Delta GØ from the thermo tables. We'll have to correct it from 298°K to 500°K using the CP values given there too. Of course, CP=f(T) also, but we can assume it's a weak function of temperature. After all, translations and rotations are already fully turned on, but vibrations (with the possible exception of NO2's 751 cm-1 mode) haven't yet seen the light of day. So CP ought to be pretty flat with T, or, better still, Delta CP ougth not to vary strongly with T.

    So Delta H = Delta HØ + Delta CP(T - 298°K)
    and
    Delta S = Delta SØ + Delta CP ln(T / 298°K).

    Then Delta G = Delta H - T Delta S all evaluated at 500°K.

    Delta CP = 2CP(NO) + CP(O2) - 2CP(NO2) = 2(29.844)+(29.355)-2(37.20) J/mol°K
    = 14.64 J/mol°K.

    Delta H = 2(90.25) - 2(33.18) + 14.64×10-3(500 - 298) kJ/mol = +117.1 kJ/mol
    showing the forward reaction to be very endothermic (no wonder k2 was miniscule), and hence the reverse reaction (k3) must be very exothermic.

    Delta S = 2(210.76) + (205.138) - 2(240.06) + 14.64 ln(500/298) J/mol°K = 154.1 J/mol°K
    showing that a pair of molecules shattering into three increases system entropy. Why are we not surprised?

    Thus Delta G = +117.1 - 500(0.1541) kJ/mol = 40.1 kJ/mol whereas Delta GØ = 70.5 kJ/mol showing that LeChatlier is right again. This endothermic reaction becomes more spontaneous (favors products more) as the temperature increases.

    So now we have all we need to estimate k3 as

    k3 = k2 / KC = k2 × (RT/pØ) e+ Delta G/RT
    = 1.7×10-7 L mol-1 s-1 (0.08206 L atm/mol°K)(500°K)/(1 atm) e+40,100/(8.314×500)
    = 0.11 L2 mol-2 s-1 some 30,000× smaller than the experimental value! Activated complex theory is not the most accurate on the planet.

  • By what factor does the NMR absorbance (pp. 548-9) of a mole of 14N nuclei (Table 18.1, p. C24) change (pp. 625-6) when the field doubles from 5 T to 10 T? (I.e., what's the ratio of absorbances?)

    • Comparing the rates of absorption to get the strength of the signal,

    Wnet(10 T)/Wnet(5 T) = (N-N')10 TB rho 10 T / [ (N-N')5 T B rho 5 T) where B is the Einstein coefficient (and cancels) and rho is the flux of stimulating radiation. In an NMR experiment, we control rho by setting the strength of the radio frequency coincident with the transition. And although that frequency doubles when the magnetic field, we can fix the flux independently of the frequency. So it is constant and cancels as well.

    We're then left with Wnet(10 T)/Wnet(5 T) = (N-N')10 T/(N-N')5 T, and since
    (N-N')/(N+N')= gamma hbar B/2kT where B is magnetic field, and N+N' = NAv,

    Wnet(10 T)/Wnet(5 T) = B(10 T)/B(5 T) = 10 T / 5 T = 2 or doubling the field, doubles the signal strength (by doubling the excess number of nuclei in the ground vs excited state).

  • Determine the symmetry (p. 519) of N2O4# given that the molecule is planar, both ONO angles are equal, and both NOO angles are equal. Is it polar or chiral (pp. 516-7)? 

    • O       O       N
 \     / \     / \
  \   /   ·   /   \
   \ /     \ /     \
    N       O       O
    where "·" marks the spot, the center of mass, the center of inversion, and the penetration point of the C2 symmetry axis (out of the screen). There is a mirror plane, but only one: the plane of the screen or sigma h.

    If we go through the algorithm,

    1. Linear? No
    2. Two or more Cn where n > 2? No
    3. Cn? Yes, C2
    4. 2 C2 perpendicular that C2? No
    5. sigma h? Yes; therefore C2h

    Since it has a symmetry axis, it can't have a dipole perpendicular to that axis. However, in addition, it has a sigma h; so it can't have a dipole along the C2 axis either. It's non-polar.

    Since S2=C2 sigma h, it must be achiral as well.

  • Liquid O2 is attracted to a magnet (p. 483) but liquid N2 is not. Compare and explain the magnetic properties, if any, of liquid NO.

    • O2(liquid) is magnetic due to those two unpaired electrons (by Hund's rule).

    N2(liquid) has all its electrons paired; so it is non-magnetic.

    NO(liquid), with an odd number of electrons (15), has a single unpaired electron which makes it magnetic but weaker so than O2(liquid).

    This exam was brought to you by the letters O and N and the number 6.022×1023
    Last modified 10 May 1997.