CHM 3312 Ch. 21 Solutions Spring 1997

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

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  1. Radical reactions don't respect mechanisms any better than any other kind of kinetics. For example, the oxidation of the free radical ·NO,

    2 ·NO + O2 arrow right 2 NO2k3

    displays 3rd order kinetics

    d[NO2]/dt = 2 k3 [·NO]2 [O2]

    as you might expect if the overall reaction were the elementary reaction. However, check out the following mechanism, which involves a rapid equilibrium, and confirm that it too is consistent with third order kinetics.

    2 ·NO = N2O2K
    N2O2 + O2 arrow right 2 NO2k2

Since the equilibrium is rapid, it is maintained at all times during the reaction.
Thus, K=[N2O2]/[·NO]2 or [N2O2] = K [·NO]2

d[NO2]/dt = 2 k2[N2O2][O2] = 2 k2 K [·NO]2 [O2], 3rd order QED.

  • When Rowland, Molina, and Cicerone won the Nobel Prize recently for analysis of the dangers to the Ozone Layer, another chemist was high on the list for consideration for exactly the same concerns. Professor H.S. "Hal" Johnston (UC Berkeley, retired) was warning the world of the delicacy of the Ozone Layer while those three were still in school. His article "Reduction of Stratospheric Ozone by Nitrogen Oxide Catalysis from Supersonic Transport Exhaust," Science 173, 517 (1971), discusses catalytic reactions and photochemically induced reactions in the stratosphere. After reading this paper, derive equation 2 in the article, which is the rate equation for [O3] + [O·]. Also discuss how the author obtains equation 7, the zero-order approximation for the steady state concentrations of O3, O·, NO2, and ·NO.

    • The first thing we must do is to correct the radical typo in reaction E. From the discussion of equation 1, it is clear that E should have been:

    O + O3 arrow right 2 O2 ke[O][O3]

    Page 520 of the article suggests some simplification of the symbols by collecting them together as single symbols. Actually, there is another purpose to this "simplification;" the letters A through K and M get used for quantities including species in high concentration while V through Y are species in low concentration.

    A = ja[O2] B = kb[M][O2] C = jc D = kd[M] E = ke F = kf
    G = kg H = jh I = ki[O2] J = kj[M] K = kk M = [M]
    V = [NO2] W = [NO] X = [O3] Y = [O] Z = [O2]


    The advantage is seen if we write the rate of appearance of O2 from equation E:

    d[O2]/dt = ke[O3][O] = EXY

    a parsimony of symbols there. Indeed, we should write out Dr. Johnston's mechanism (or that part that we'll be using) using these abbreviations:

    Elementary Reaction Rate as Symbol
    O2 + h nu (< 242 nm) arrow right O + O ja[O2] A (a)
    O + O2 + M arrow right O3 + M kb[O][O2][M] BY (b)
    O3 + h nu (190 to 350 nm, 450 to 650 nm) arrow right O + O2 jc[O3] CX (c)
    O + O3 arrow right 2 O2 ke[O][O3] EXY (e)
    NO + O3 arrow right NO2 + O2 kf[NO][O3] FWX (f)
    NO2 + O arrow right NO + O2 kg[NO2][O] GVY (g)
    NO2 + h nu (260 to 400 nm) arrow right NO + O jh[NO2] HV (h)
    NO + O + M arrow right NO2 + M kj[NO][O][M] JWY (j)
    HO + O3 arrow right HOO + O2 kx[HO][O3] none (x)
    O + HOO arrow right HO + O2 ky[O][HOO] none (y)
    O + HO arrow right H + O2 kz[O][HO] none (z)
    H + O2 + M arrow right HOO + M kaa[H][O2] none (aa)

    Now we can tackle d([O3]+[O])/dt which is just d[O3]/dt + d[O]/dt. Hal says we should ignore equation J as unimportant (compared to a-c, e-g, and x-aa); so we will skip J. Otherwise, we'll include all the elementary reactions which influence [O] or [O3]:

    d[O3]/dt = +BY-CX-EXY-FWX-kx[HO]X
    d[O]/dt = 2A-BY+CX-EXY-GVY+HV-ky[HOO]Y-kz[HO]Y
    d[both]/dt = 2A-2EXY-FWX-GVY+HV-kx[HO]X-ky[HOO]Y-kz[HO]Z

    But d[NO2] = FWX - GVY - HV = 0 by the steady-state approximation.

    That makes - FWX + HV = - GVY which simplifies that d[odd O]/dt by cancelling the terms in F and H and creating the term - 2GVY.

    Likewise, the sum of the steady-state approximations for [H] and [HOO] will simplify those final terms:

    d[H]/dt = kz[HO]Y - kaa[H]MZ = 0
    and
    d[HOO]/dt = kx[HO]X - ky[HOO]Y + kaa[H]MZ = 0

    which together mean that - kx[HO]X - kz[HO]Y = - ky[HOO]Y.

    And both simplifications substituted into the d[odd O]/dt expression make it:

    d([O3]+[O])/dt = 2A - 2EXY - 2GYV - 2ky[HOO]Y which is the article's equation 2 in its symbolically simplified form.

    Equation 7 is a little trickier. We're asked to find out what Hal threw away to get his "zeroth-order approximation" expressions. The hidden agenda is to get us to think about whether those expulsions were reasonable.

    The easiest one is W0 = alpha M - V0 where the 0 means zeroth order. The reason this is easy is that it isn't zeroth order at all; it's a tautology (always true) given the definition of alpha .

    alpha = [NOx]/[M] = ([NO]+[NO2])/[M] = (V+W)/M.

    So W0 = M(V0+W0)/M - V0 = W0 ... see, a tautology.

    OK.

    Dr. Johnston claims that the steady-state approximation for [NO2] gives rise to his expression for V0. Let's see...

    d[NO2]/dt = FW0X0 - GV0Y0 - HV0 + JW0Y0 = 0

    whereas V0 = alpha MFX0 / [FX0 + (G+J)Y0 + H] expands into

    FX0V0 + GV0Y0 + JV0Y0 + HV0 = alpha MFX0

    but that right-hand side expands into ([V0+W0]/M)MFX0 = FX0V0 + FX0W0.

    Substituting and rearranging gives us

    0 = FX0W0 - GV0Y0 - HV0 - JV0Y0

    but we can use our expression above to V0 to make this read:

    0 = FX0W0 - GV0Y0 - HV0 + JV0W0 - alpha MJY0

    That's the same as the [NO2] steady-state equation,

    0 = FX0W0 - GV0Y0 - HV0 + JV0W0

    in almost all terms. Apparently, in zeroth step, the mole fraction of NOx (namely, alpha ) is ignorable.

    Let us now follow Johnston's suggestion that the steady-state approximation for d([O3]-[O]+[NO2])/dt = 0 can be used to obtain Y0=X0C/B.

    Ignoring equation J again:

    d[O3]/dt = +BY0-CX0-EX0Y0 -FW0X0
    -d[O]/dt = -2A+BY0-CX0+EX0Y0 +GW0Y0-HV0
    d[NO2]/dt = +FW0X0-GV0Y0 -HV0
    Sum = 0 = -2A+2BY0-2CX0-2HV0

    But Johnston's approximation is really 2BY0 = 2CX0 (only the 2nd and 3rd terms kept) which means we're ignoring almost all photochemistry by setting the light flux terms (A and H) to zero in the zeroth approximation. But we're keeping the photochemical equation C since that range of near-UV light is the dominant UV in the stratosphere.

    Finally, the zeroth order [O3], viz., X0 = (AB/CE)½ which Johnston's says stems from 0 = d([O2]+[O]+[NO2])/dt.

    d[O2]/dt = -A-BY0+CX0+2EX0Y0 +FW0X0+GV0Y0
    d[O]/dt = 2A-BY0+CX0-EX0Y0 -GV0Y0+HV0
    d[NO2]/dt = +FW0X0-GV0Y0 -HV0
    Sum = 0 = A-2BY0+2CX0+EX0Y0 +2FW0X0-GV0Y0

    Only this time, we will keep the initial photolysis term, A, and the catalytic destruction term EX0Y0, but toss out all other terms which are products of small concentration species (the product of tiny numbers is smaller still) giving us:

    0 = A - 2BY0 + 2CX0 + EX0Y0

    and substitute the approximation for Y0 = CX0/B to get

    0 = A + ECX02/B which, ignoring the sign(!), rearranges to

    X0 = (AB/CE)½.

  • Consider the following chain reaction mechanism for the ortho-para hydrogen conversion:

    Initiation: M + H2(either) arrow right 2 H· + M ki
    Propagation: H· + H2(para) arrow right H2(ortho) + H· kp
    Termination: M + 2 H· arrow right H2(either) + M kt

    where you know from nuclear spin statistics that spin-paired para is only 1/3 as likely as spin-aligned ortho, e.g.,

    [H2(ortho)] = ¾ [H2(either)]
    [H2(para)] = ¼ [H2(either)]

    Determine the expression for the rate of formation of ortho H2.

    • d[H2(ortho)]/dt = - ki[M][H2(ortho)] + kp[H·][H2(para)] + ¾kt[M][H·]2

    but

    0 = d[H·]/dt = ki[M][H2] - kt[M][H·]2 or [H·]2 = ki[H2]/kt

    Hence [H·] = (ki / kt)½ [H2]½.

    Since [H2(ortho)] = ¾ [H2],

    d[H2(ortho)]/dt = - ki¾[H2][M] + ¼kp(ki / kt)½[H2]½[H2] + ¾kt(ki / kt)[H2][M]
    = ¼kp(ki / kt)½ [H2]3/2

  • In a photochemical experiment a quantum yield of 106 is found. What can we say immediately about its mechanism and why?

    • That each photon generates millions of products means that a radical chain mechanism is at work since there is no molecule which will fragment into 106 products on absorbing a photon, and the absorption process can only occur in a single molecule.

    Last modified 5 May 1997.