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Unformatted text preview: CHAPTER Stratospheric Chemistry:
The Ozone Layer Problem 11 From the text, E = 119627/2» where K is in nm and the units of the constant are nm k] mol~l (a) E = 119627 nm k] moH/280 nm = 427 k} moll
From Figure 1—2, 280 nm lies in the UV region, at the junction between UV‘B and UV—C (b) E = 119627 nm k] mol—1/4OO nm = 299 k] mol—l
Junction ofUV(A) and visible regions (c) E = 119627 nm k] mol—1/750 nm = 160 k] mol“l Junction of visible and infrared regions ((1) E = 119627 nm k] moi—V4000 nm = 29.9 1(Jmol'l
Beginning of thermal IR region Problem 12 Rearrange the formula from Problem 1'1 to obtain 2.: K = 119627/E = 119627 nm k] mol1/105 nm = 1140 nm, i.e.,IR light. Problem 13 Obtain E as AH° of the reaction
NO2 ——) NO + 0 AH“ =AH;(N0) + AH;(0) — Ammoz)
=9o.2 + 249.2 — 33.2
= 306.2 k] moi—I K = 119627/E = 119627 nm k] mol'1/306.2 k] moltl = 390.7 nm
1 2 Chapter 1 For the complete dissociation, we need AH° for NoZ ——> N+ZO AH" = AH;(N) + 2AH;(0) _ AHﬂNOZ)
= 472.7 + 2 x 249.2 _ 33.2
= 937.9 7» = ll9627fE = ll9627l9379 = l275 nm Problem 14 Consider an altitude slightly lower than that at which the absolute concentration peaks. Here the
number of ozone molecules is less but the number ofair molecules is greater, so the relative concentra tion falls more than does the absolute. In contrast, just above the altitude for the peak, the ozone level has fallen but that of the air has also fallen, so the relative concentration falls less than does the ab solute. From the trends, we conclude that the peak for the relative concentration will lie at a higher
altitude than for the absolute. Problem 15 The reaction for which we. need the AH is
03 —> 0* + 02* The reactions for which we have AH values are 1) 03 —> o + oZ
2) o —> 0*
3) 02 —> 02* If we add the three reactions together, we obtain the needed reaction, so by Hess’ Law,
AH = AH, + AHZ + AH}
= 105 + 180 + 95 k] mol‘l = 390 k] mol—l The wavelength corresponding to this energy can be obtained by rearranging the formula (see text) E = 119627/7»
to solve for 7»: 7» = 119627/E 119627 nm kJmol1/39O k] mol“ 307 nm Slralospherit [hemislry: The Ozone layer 3 Problem 16 oj+No —> OZ+NOZ
NoZ —>“““““‘ NO + o
o + 02 —> 03 Net: Nil. This sequence does not destroy ozone. Problem 1—7 Since X = OH here, we obtain by substituting it for X in the two steps of Mechanism I,
OH + O3 ——) HOO + O2
HOO + O ——) OH + O2 Adding these two reactions, and cancelling the common OH and HOO terms, we obtain 03 + _ o —> 202 Problem 1—8
0* + H20 ——> 20H Box 11 Problem 1 Labelling the reactions given in the text as 1, 2, and 3, then the expression for the rate of change of the concentration of O‘ with time is
d [0*1/dt= kl i021 — kZ [0*] [M1 — k, [0*] [H201
But d [0*] /dt = 0 at steady state. Thus, setting the righthand side of this equation to zero, and collecting terms involving 0* on the left, we obtain
[0*] (k2 [M] + k3 [HZOl ) = ‘9 [Oz]
and thus [0*] = k1 [Oz] / (k2 [M] + k3 [HZO]) Box 1—1 Problem 2 Since Cl is formed in steps 1 and 3 and destroyed in step 2, we obtain d [Cl] / dt = 2 k1 [C12] — k2 [Cl] [OJ] + 2 k3 [ClO]2 = 0 at steady state 4 Chapter I where the coefﬁcients of Z are needed since two Cl's are formed. and the square .t (1" ix required since two of them are reactants in the step.
Similarly, d [ClO] /dt = l<2 [Cl] [03] — 2 k1 [C10]; — k4 [CEO] [530$  C lfwe add the two equations together. the middle two terms wrll cancel out. Lea» 1n: the simple equation
2 k1 [Clz] — k4 [ClO] [NOZ] = O
which we can use to solve for [C10]:
[C10] = 2 k1 [C12] / k4 [NOZ] Since the equation for d [Cl] /dt contains only one term for [ClO], we can substitute this expres' sion into it to obtain an equation in [Cl]:
2 k1 [Clz] — l<2 [Cl] [03] + 2 k3 X 4 klz [C12]2 / k42 [NOZ]Z = O The rate of destruction ofO3 is simply the middle term, l<2 [Cl] [03], of this equation, so by re—
arrangement we obtain for its net rate ofchange ( = — k2 [Cl] [03]). Rate of change in [03] = — 2 k1 [C12] — 8 k3 1‘12 [C12]2 /1<42 [NOZ]2 Alternatively, the equation for [C1] can be obtained by solving the equation for the middle term:
1<2 [C1] [03] = 2 k1 [C12] + 8 1<3l<1Z [C12]Z/l<4Z [NOZ]Z
[C1] = 2 k1 [C12] /l<2 [03] + 8 k3 [<12 [C12]2 / k2 k42 [O3] [N02]2 Box 12, Problem 1 Bonding Energy H20 + 0 (ground state)
H20 2 OH Extent of Reaction The reaction involving ground’state O is slow since E,a > 69 k], but that involving 0* is fast since its
E3 is ~ 0. Stratospheric Chemistry: The Ozone layer 5 Additional Problems 1. Let us denote the vibrational~excited oxygen molecules as 02*. Thus the ﬁrst two steps in the proposed mechanism are: 03 “is“ 02* + o 02* + 02 —) 03 + 0
Thus the net result of these two steps is: x 24}
net 02 :9" 20 The most likely fate of the oxygen atoms is to collide with other 02 molecules and form ozone: o+o2 —> 03 To obtain the overall reaction, we need to include this last reaction twice to convert all the oxy—
gen atoms. If we add twice this reaction to the proceeding one, we obtain the overall reaction: A 2)
net 302 14—"9‘ 203 (Other, but less likely fates, for the oxygen atoms are to react with each other to produce 02, or with O} molecules to destroy them.) 2. Although the ﬁrst step is consistent with the general one for Mechanism 1, the second step is not since HOO here reacts with 03, not with atomic oxygen. The sum of the two steps is
OH+OJ+HOO+OJ—)
HOO + 02 + OH + 202
After cancelling common terms from left and right sides, the overall reaction is
203 —> 302
3. The reactions as described are as follows: CIONOZ Ph—S (:1 + No, photon No3 —> NO + 02 Both Cl and NO are known to destroy ozone, so presumably their reactions with O3 follow these two:
Cl+O3 —> ClO+O2 NO+O3 —> NOZ+O2 Chapter I To complete the cycle, we reform the ClONO2 by the known combination reaction ofClO
with N02: C10 + Mo2 ~—> CIONOZ Summing all ﬁve reactions, we obtain: photons 203 ——> 302 Since the ﬁrst step destroys ozone, subsequent steps must reform it because the overall reaction is a null cycle. The only ozone creation reaction we know is the combination of atomic oxygen withOZ:
O + 02 ——> 03 Thus the mechanism must involve other steps that create atomic oxygen from XO. Since mecha' nism II is involved, two XO units must collide; one such reaction that produces atomic oxygen willbe:
XOOX ——> 2X + 20 Thus the overall sequence would be 2fX+O3 ———> XO+OZ]
ZXO ———> XOOX
XOOX ——> 2X+ZO
2[O+O2 ——> 03] net null The process of concern transforms diatomic chlorine gas, C12, into atomic chlorine and can be written as:
ClZ (g) ._;. 2 Cl (g)
Thus, AH = 2AH{(C1, g) — AH,(C1,, g) We are given AH{(C1, g) = +121.7 k] mole—1, and since Cl2 (g) is stated to be the stablest form
of the element, by convention in thermochemistry, AH{ (Clz, g) = 0. Thus, AH = 2 X 121.7 — O
= 243.4 k] mol’l To convert to wavelength, we use the formula: E = 1196270» Stratospheric Chemistry: The Ozone layer 7 so A = 119627/E = 119627 nm k] mol‘l/Z43.4 kJ mol—1
= 491 nm This wavelength lies in the visible region. (a) If the N02 concentration is increased, the rate of its reaction with HOO will increase,
thereby decreasing the H00 concentration in the lower stratosphere. Since HOO is re
quired to complete the catalyst ozone destruction by the OH/HOO cycle, it will be slowed and the rate of ozone destruction by them would decrease. (b) In the midvstratosphere and high'stratosphere, NO2 would return to NO by reaction with atomic oxygen, and the NO returns to NO2 by destruction of ozone: NOZ+O ——~> NO+O2
NO+O3 ——~> NOZ+O2 net 03 + O —> 2 02
Thus any increase in the N02 concentration would increase the rate of ozone destruction at these levels. (c) Supersonic airplanes that emit substantial amounts of nitrogen oxides should ﬂy in the lower
atmosphere only, because they would destroy ozone at higher levels in the stratosphere. Recall that the rate for a reaction step equals the rate constant k times the concentration of each reactant molecule. Thus for the reaction between 0* and CH4, the rate is Rate = k [0"] [CH4]
= 3 X 1010 cm3molecules1s'l X 100 molecules cm3 X 1011 molecules cm3 = 3 X 103 molecules cm—3 s—l
To deduce the destruction over one year, we deduce the number of seconds in 1 year: 60 seconds X 60 minutes X 24 hours X 365 days = 315 X 107 seconds/year 1 minute 1 hour 1 day 1 year Thus the rate of methane destruction per year is: 3 X 103molecules cm) X 3.15 X 107seconds
second year = 9.46 X 1010 molecules cm‘3 year ’1 8 Chapter 1 Finally, to convert to grams of methane, we convert molecules to grams: 1 mole CH4 = 6.02 X 1023 molecules CH4 = 16.0 grams CH4 Thus
16.0 g CH4 9.46 x lOlOmOlecules x ———23——
6.02 X 10 'molecules CH4 = 2.52 x ioIZgCH4 Thus, in every cubic centimeter, a total of 2.52 X 10‘12 grams of methane are destroyed per year. 8. The reactions of the two catalysts with atomic oxygen are: C1 + O, —> ClO + 02 ...(1)
and OH + 03 ——> HOO + 02 ...(2) The rate laws for these reactions will each be ﬁrst—order in ozone and in catalyst, because 1 molecule of each is involved.
Rate (1) = kl [Cl] [03]
Rate (2) = k2 [OH] [0}] The ratio of rates is then given by: Rate (1) k1 [Cl] Rate (2) = k2 [OH]
Given the information that [OH] = 100 [C1], thus
Rate (1) k1 [Cl] k1 Rama) k2100[C1] 100k2 — —250
Now £1— —————3 X 10 H e [T — 156+®0IT k2 2 X 10—12 e—940/T SubstirutingT = 273 — 50 = 223K, thus EL = 156690/223 = 330
k2
Thus Rate“) = L x 330 = 3.3 Rate (2) 100 Stratospheric Chemistry: The Ozone layer 9 The rate of ozone destruction in the ozone hole conditions described will be given by: Rate (1) = kl [Cl] [03]
= 3 x 10‘”e‘250fr x4 x 105 x 2 x 1012
= 3 x 10‘11e‘250/193 x4 x 105 x 2 x 1012 3 l = 7 X 106 molecules cm‘ sec— The Arrhenius equation is
k = Ae—E/RT The ratio of rate constants for the two reactions is therefore: kl Al e‘El/RT k2 A2 e—Ez/RT but, because we are given that A1 = A2, therefore k _1 _ e(El—EZVRT
k2 Now E1 = 301(Jmol‘l = 30,000] mol—l and E2 = 3 k] mol‘l = 3,000] mol—l
R = 8.314JK‘1mol'l,and T = —30 + 273 = 243 K, k 30000—3000 831 23
therefore —1 : elv , )/, 4x 4 k2
= 1.5 x 104’ Since Since we are told that the reactant concentrations are equal, their values cancel from the ratio of
the rate equations, and the ratio of rates is equal to that of the rate constants, namely 1.5 X 10—6. ...
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