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Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
434
15.1 In any sequence of steps, the slowest one will be rate–determining: (a) Pouring the coffeefrom the urn into the cup; (b) Entering the items on the cash register (if the market has agood laser scanner, paying and receiving change may be rate–determining); and (c)Preparing for the jump and passing through the door.
15.2 In any sequence of steps, the slowest one will be rate–determining: (a) Paying at the cashregister; (b) Going through the turnstile and having tickets collected; and (c) Paying thebill.
15.2 Every elementary reaction must depict actual molecular processes:(a) I2 → I + I(b) H2 + I2 → H2I2
(c) H2 + I2 → H + HI2
15.2 Every elementary reaction must depict actual molecular processes:(a) Cl2 → Cl + Cl(b) NO + Cl2 → NOCl + Cl(c) 2 NO + Cl2 → 2 NOCl
15.5 A molecular picture of an elementary reaction shows the reactants, the products, and (ifnecessary) the intermediate collision complex:
15.6 A molecular picture of an elementary reaction shows the reactants, the products, and (ifnecessary) the intermediate collision complex:
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.7 A satisfactory mechanism must consist entirely of reasonable elementary steps that sumto give the correct overall stoichiometry of the reaction:(a) I2 → I + I
I + H2 → HI + HH + I → HI
(b) H2 + I2 → H2I2
H2I2 → HI + HI(c) H2 + I2 → H + HI2
H + HI2 → HI + HI
15.8 A satisfactory mechanism must consist entirely of reasonable elementary steps that sumto give the correct overall stoichiometry of the reaction:(a) Cl2 → Cl + Cl
NO + Cl → NOClNO + Cl → NOCl
(b) NO + Cl2 → NOCl + ClNO + Cl → NOCl
(c) 2 NO + Cl2 → 2 NOCl
15.9 The rate of a reaction can be expressed using the general expression:
Rate = −1aΔ[A]Δt
(a) Rate = −Δ[Cl2 ]Δt
(b) −Δ[Cl2 ]Δt
= 12Δ[NOCl]
Δt(c) Use the result of (b) to calculate that NOCl appears at a rate of 94 M s1
15.10 The rate of a reaction can be expressed using the general expression:
Rate = −1aΔ[A]Δt
Reaction: 2 O3 → 3 O2
(a) Rate = 13Δ[O2 ]Δt
(b) 13Δ[O2 ]Δt
= −12Δ[O3]Δt
(c) Use the result of (b) to calculate that ozone disappears at a rate of 1.8 x 106 M s1
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15.11 In the reaction of NO and Cl2, two NO molecules react for every Cl2 that reacts,producing two NOCl molecules:
15.12 In the reaction of NO and O3, one NO molecule reacts for every O3 that reacts, producingone O2 and one NO2 molecule:
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.13 (a) To calculate the average rate of production, determine how many moles form duringthe time interval and divide by the time:
n =
PV
RT=
(0.15 atm)(5.0 L)(0.08206 L atm
mol K )(550 + 273.15)K= 1.11 x 102 mol
Δ[CO2 ]Δt
=1.11 x 102 mol
5.0 min= 2.2 x 103 mol/min
(b) The reaction is CaCO3 → CaO + CO2, so the amount of CaCO3 decomposing is thesame as the amount of CO2 produced, 1.1 x 102 mol.
15.14 (a) To calculate the average rate of reaction, determine how many moles are consumedduring the time interval and divide by the time:Δn = (0.25 mol/L) – (0.50 mol/L) = – 0.25 mol/L
Rate = −13Δ[H2 ]Δt
= −13
−0.25 M30 s
⎛⎝⎜
⎞⎠⎟= 2.77 x 103 M/s
Round to two significant figures: Rate = 2.8 x 103 M/s(b) The rate for any particular reagent is the coefficient for that reagent times the rate of
reaction: Rate(NH3) = 2(2.8 x 103 M/s) = 5.6 x 103 M/s(c) The concentration of any particular reagent is its initial concentration minus the
change during the time interval:Change = (Coeff)(Rate)(time) = –1(2.8 x 103 M/s)(30 s) = –8.4 x 102 MConcentration = (1.25 M) – (0.084 M) = 1.17 M
15.15 (a) Before the reaction begins:
(b) After 20 minutes, the amount reacted is (20 min)(0.25 molecules/min) = 5 molecules:
15.16 (a) Before the reaction begins:
(b) After 2.0 s, the amount reacted is (2.0 s)(1.5 molecules/min) = 3 molecules:
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.17 (a) The number of decays is directly proportional to the starting amount of the isotope infirst–order kinetics, therefore there would be 6.0 x 106 decays in the 5.00 nmol case.
(b) The fraction decaying is the same in both cases:
Fraction decaying = Number of decays
Number of nuclei present
F = 1.2 x 106 nuclei
(1.00 nmol)(109 mol/nm)(6.022 x 1023 nuclei/mol= 2.0 x 109
(c) In first–order kinetics, the number reacting is proportional to the amount present, butthe fraction reacting is independent of concentrations.
15.18 Rates of elementary "unimolecular" reactions depend on how many reacting items arepresent. When an amount changes by some factor, the rate changes by the same factor.(a) After 50 kernels have popped, 100 remain, so the number popping is 6(100/150) = 4.
(b) The fraction of kernels popping per second is 6
(150)(5 s)= 8 x 103/s initially, and it
is 4
(100)(5 s)= 8 x 103/s after 50 have popped. The fraction remains constant.
(c) After 50 kernels have popped, there are fewer unpopped kernels remaining in thepopcorn popper, so the overall rate of popping is lower. The fraction popping,however, is the same.
15.19 (a) The rate law for an elementary step contains the product of the reactantconcentrations:Rate = k[C][AB]
(b) The units of a rate constant have time in the denominator along with concentrationsone power less than the number of reactants: units = (conc.)1 (time)1
(c) The steps of the mechanism must sum to give the observed stoichiometry for thereaction. Intermediate A can react with AB, then B and C can react:C + AB → BC + AA + AB → B + A2
B + C → BCNet: 2 C + 2 AB → 2 BC + A2
15.20 (a) The rate law for an elementary step contains the product of the reactantconcentrations:Rate = k[AB][AB] = k[AB]2
(b) The units of a rate constant have time in the denominator along with concentrationsone power less than the number of reactants: units = (conc.)1 (time)1
(c) The steps of the mechanism must give the observed stoichiometry for the reaction. Abimolecular reaction between the intermediate (B) and C is the simplest possibility:2 AB →A2 + 2 BB + C → BC. This reaction occurs twice for each time the first step occurs.Net: 2 C + 2 AB → 2 BC + A2
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.21 (a) The rate law for an elementary step contains the product of the reactantconcentrations:Rate = k[C][C][AB] = k[C]2[AB]
(b) The units of a rate constant have time in the denominator along with concentrationsone power less than the number of reactants: units = (conc.)2 (time)1
(c) The steps of the mechanism must sum to give the observed stoichiometry for thereaction. A bimolecular reaction between the intermediate (AC) and AB is thesimplest possibility:2 C + AB → BC + ACAC + AB → BC + A2
Net: 2 C + 2 AB → 2 BC + A2
15.22 (a) The rate law for an elementary step contains the product of the reactantconcentrations:Rate = k[AB]
(b) The units of a rate constant have time in the denominator along with concentrationsone power less than the number of reactants: units = (time)1
(c) The steps of the mechanism must sum to give the observed stoichiometry for thereaction. Intermediate A can react with AB, then B and C can react:AB → A + BA + AB → B + A2
B + C → BC. This reaction occurs twice for each time the first step occurs.Net: 2 C + 2 AB → 2 BC + A2
15.23 According to the stated rate law, the rate of reaction is proportional to each concentration.A contains 6 molecules of each type, while B contains 10 NO and 2 O3:A: Rate = k(6)(6) = 36 kB: Rate = k(10)(2) = 20 kB will react slower by a factor of 20/36 = 0.56.
15.24 According to the stated rate law, the rate of reaction is proportional to each concentration.A contains 6 molecules of each type, while B contains 3 NO2 and 12 O2:A: Rate = k(6)(6) = 36 kB: Rate = k(3)(12) = 36 kB will react at the same rate as A.
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.25 (a) One way to treat experimental data is by plotting: For firstorder behavior, ln [A]0[A]
⎛⎝⎜
⎞⎠⎟
vs. t is a straight line, and for secondorder behavior, 1[A]
−1[A]0
vs. t is a straight
line. Here, the plot of ln [A] vs. t gives a straight line, so this reaction is first order.
(b) For a first–order reaction, k = slope of the graph:
k = slope = ΔyΔx
=0.491− 0
800 s − 0 s = 6.14 x 104 s1
(c) To find concentration at any particular time, use ln[A] = ln[A]o– kt
ln[A] = ln[2.50] – (6.14 x 104 s1)(1600 s) = – 0.0661[A] = e0.0661 = 0.936 atm
(d) To find the time at which concentration reaches a particular value, use
kt = ln [A]0[A]
⎛⎝⎜
⎞⎠⎟
t = ln [A]0
[A]( )k
=ln 2.50 atm
0.500 atm⎛⎝⎜
⎞⎠⎟
6.14 x 10−4s−1 = 2.62 x 103 s
15.26 (a) One way to treat experimental data is by plotting: For firstorder behavior, ln [A] vs. tis a straight line, and for secondorder behavior, 1/[A] vs. t is a straight line. Here,however, there is a quicker way, because the data show half–lives (the time requiredfor the amount to fall by a factor of two). It takes 8.0 days for mass to fall from 12.0µg to 6.0 µg and 8.0 days for the mass to fall from 6.0 to 3.0 µg. Thus, the half–lifeis independent of amount, a characteristic of first–order reactions.
(b) The half–life of a first–order reaction provides a convenient way to calculate a rate
constant. Use Equation 15–4: t1/2 =
ln2k
k =
ln2t1/2
;
k = 0.693
8.0 days = 8.7 x 102 days1
(c) To find the amount remaining at any particular time, use ln[A] = ln[A]o– kt
ln[A] = ln[12.0] – (8.7 x 102 days1)(32 days) = – 0.299[A] = e0.299 = 0.74 µg (a simple half–life analysis can also be used to give 0.75 µg)
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(d) To find the time at which the amount reaches a particular value, use
kt = ln [A]0[A]
⎛⎝⎜
⎞⎠⎟
t = ln [A]0
[A]( )k
=ln 12.0 µg
1.2 µg⎛⎝
⎞⎠
8.7 x 102days−1 = 27 days
15.27 This is stated to be a second–order reaction, so Rate = k[NOBr]2 and Equation 15–5
applies: 1[A]
−1[A]0
= kt
The problem states that k = 25 M1 min1.
(a) kt =1
(0.010 M)−
1(0.025 M)
= 60 M1
t = 60 M1
25 M1 min1 = 2.4 min
(b) 1[A]
= 1
(0.025 M) + (25 M1 min1)(125 min) = (40 + 3125) M1 = 3165 M1
[A] = 3.2 x 104 M
15.28 This is stated to be a second–order reaction, so Rate = k[C4H6]2 and Equation 15–5
applies: 1[A]
−1[A]0
= kt
The problem states that k = 0.93 M1 min1.
(a) kt =1
0.100 M−
10.240 M
= (10.0 – 4.17) M1 = 5.83 M1
t = 5.83 M1
0.93 M1 min1 = 6.3 min
(b) 1[A]
= 1
0.240 M + (0.93 M1 min1)(25 min) = 27.42 M1
[A] = 3.6 x 102 M
15.29 This is stated to be a first–order reaction, so Rate = k[C5H11Br] and Equation 15–3applies:
kt = ln [A]0[A]
⎛⎝⎜
⎞⎠⎟
(a) kt = ln0.125
1.25 x 10−3⎛⎝⎜
⎞⎠⎟
= 4.61
t = 4.61
0.385 hr1 = 12.0 hr
(b) ln[A] = ln[0.125] – 3.5 hr(0.385 hr1) = –3.43
[A] = e3.43 = 0.0324 M or 3.24 x 102 M
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.30 This is stated to be a first–order reaction, so Rate = k[Hg*] and Equation 15–3 applies:
kt = ln [A]0[A]
⎛⎝⎜
⎞⎠⎟
; The problem states that k = 1.65 x 106 s1.
(a) kt = ln4.5 x 10−6
4.5 x 10−7⎛
⎝⎜
⎞
⎠⎟ = 2.303
t = 2.303
1.65 x 106 s1 = 1.4 x 106 s
ln[A] = ln(4.5 x 106) – (1.65 x 106 s1)(2.5 x 106 s) = –16.436[A] = e16.436 = 7.3 x 108 M
15.31 The isolation method requires that one concentration be substantially smaller than all theothers, so the experimental reaction order follows the order with respect to that onereactant. We are told that it is possible to track the concentration of N2O5, so thisreactant has to be the one with the low concentration (otherwise the concentration wouldnot change enough to give good data).Do two experiments; in each [H2O]0 > 100 [N2O5]0 but with two different values for
[H2O]0. Plot ln N2O5[ ]0N2O5[ ]
⎛
⎝⎜
⎞
⎠⎟ and
1N2O5[ ] –
1N2O5[ ]0
vs. t to determine order with
respect to N2O5, and use the ratio of slope values and H2O concentrations to determineorder with respect to H2O.
15.32 The isolation method requires that one concentration be substantially smaller than all theothers, so the experimental reaction order follows the order with respect to that onereactant. We are told that it is possible to track the concentration of NO2, so this reactanthas to be the one with the low concentration (otherwise the concentration would notchange enough to give good data).Do two experiments; in each [CO]0 > 100 [NO2]0 but with two different values for
[CO]0. Plot ln NO2[ ]0NO2[ ]
⎛
⎝⎜
⎞
⎠⎟ and
1NO2[ ] –
1NO2[ ]0
vs. t to determine order with respect to
NO2, and use the ratio of slope values and CO concentrations to determine order withrespect to CO.
15.33 From the data provided, we recognize this as an initial rate problem. The essential featureof the initial rate method is that we can take ratios of initial rates under differentconditions. First, apply this technique to Experiments 1 and 2, which have the sameinitial concentration of S2O82:
Initial rate1 = 4.4 x 102 M/min = k (0.125 M)x(0.150 M)y
Initial rate2 = 1.3 x 101 M/min = k (0.375 M)x(0.150 M)y
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When we take the ratio of the second initial rate to the first, the rate constant and theinitial concentration term for S2O82 cancel:
Initial rate2Initial rate1
=1.3 x 101 M/min4.4 x 102 M/min
=k 0.375 M( )x 0.150 M( )y
k 0.125 M( )x 0.150 M( )y=
0.375 M( )x
0.125 M( )xSimplifying, we find:
3.0 = (3.0)x, from which x = 1.Now repeat this analysis for the third experiment and the first experiment, for which theinitial concentrations of I are the same:
Initial rate3Initial rate1
=1.5 x 102 M/min4.4 x 102 M/min
=k 0.125 M( )x 0.050 M( )y
k 0.125 M( )x 0.150 M( )y=
0.050 M( )y
0.150 M( )y0.34 = (0.33)y, from which y = 1
The reaction is first order in each reactant, so the rate law is Rate = k[S2O82][ I]Use any of the experiments to evaluate the rate constant k:
4.4 x 102 M/min = k (0.125 M)(0.150 M)
k = 4.4 x 102 M/min
0.125 M( )(0.150 M)= 2.3 M1 min1
15.34 From the data provided, we recognize this as an initial rate problem. The essential featureof the initial rate method is that we can take ratios of initial rates under differentconditions. First, apply this technique to Experiments 1 and 2, which have the sameinitial concentration of OH:
Initial rate1 = 2.9 x 102 M/s = k (0.050 M)x(0.050 M)y
Initial rate2 = 6.5 x 102 M/s = k (0.075 M)x(0.050 M)y
When we take the ratio of the second initial rate to the first, the rate constant and theinitial concentration term for OH cancel:
Initial rate2Initial rate1
=6.5 x 102 M/s2.9 x 102 M/s
=k 0.075 M( )x 0.050 M( )y
k 0.050 M( )x 0.050 M( )y=
0.075 M( )x
0.050 M( )xEvaluating the ratios gives 2.24 = (1.5)x, from which x = 2.Now repeat this analysis for the third experiment and the second experiment, for whichthe initial concentrations of ClO2 are the same:
Initial rate3Initial rate2
=9.8 x 102 M/min6.5 x 102 M/min
=k 0.075 M( )x 0.075 M( )y
k 0.075 M( )x 0.050 M( )y=
0.075 M( )y
0.050 M( )y1.5 = (1.5)y, from which y = 1.
The rate law is Rate = k[ClO2]2[ OH]Use any of the experiments to evaluate the rate constant k:
2.9 x 102 M/s = k (0.050 M)2(0.050 M) k = 2.9 x 102 M/s
0.050 M( )3= 2.3 x 102 M2 s1
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.35 The rate law should relate the rate of reaction to the concentration of the reactants.Reactive intermediates should not be shown in the rate law:
C + AB BC + A, followed byA + AB → B + A2
B + C → BCNet: 2 C + 2 AB → 2 BC + A2
The rate law is determined by the rate–determining step: Rate = k2[A][AB]. This is notsatisfactory, however, because it contains the concentration of an intermediate (A). Setthe rates equal for the forward and reverse first step:
k1[C] [AB] = k1[BC][A]
Solve this equality for [A]: [A] = k1k−1
⎛⎝⎜
⎞⎠⎟[C][AB][BC]
Substitute into the rate expression: Rate = k2k1k−1
⎛⎝⎜
⎞⎠⎟[C][AB]2
[BC]
15.36 The rate law should relate the rate of reaction to the concentration of the reactants.Reactive intermediates should not be shown in the rate law.(a) Cl2 Cl + Cl, followed by
NO + Cl → NOClThe rate law is determined by the rate–determining step: Rate = k2[NO][Cl]. This isnot satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step:k1[Cl2] = k1[Cl]2
Solve this equality for [Cl]: [Cl] = k1k−1
⎛⎝⎜
⎞⎠⎟
1/2[Cl2]1/2
Substitute into the rate expression: Rate = k2k1k−1
⎛⎝⎜
⎞⎠⎟
1/2[NO][Cl2]1/2
(b) NO + Cl2 NOCl + Cl, followed by NO + Cl → NOClThe rate law is determined by the rate–determining step: Rate = k2[NO][Cl]. This isnot satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step:
k1[NO][Cl2] = k1[NOCl][Cl]
Solve this equality for [Cl]: [Cl] = k1k−1
⎛⎝⎜
⎞⎠⎟
[NO] [Cl2 ][NOCl]
Substitute into the rate expression: Rate = k2k1k−1
⎛⎝⎜
⎞⎠⎟[NO]2[Cl2 ][NOCl]
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.37 (a) The steps of a mechanism must sum to give the observed overall stoichiometry of thereaction. For ozone decomposition, this is 2 O3 → 3 O2. The two steps proposed bythe student consume 1 O3, produce 1 O2, and generate an O atom, which must beconsumed. Thus, the third step is:
O3 + O → 2 O2(b) The rate law is determined by the rate–determining step: Rate = k2[O5]. This is not
satisfactory, however, because it contains the concentration of an intermediate. Setthe rates equal for the forward and reverse first step:
k1[O3][O2] = k1[O5]
Solve this equality for [O5]: [O5] = k1k−1
[O3][O2]
Substitute into the rate expression:
Rate = k2k1k−1
[O3][O2]
(c) Atmospheric chemists would consider this mechanism to be molecularly unreasonablebecause fragmentation of O5 in the second step (the breaking of two bondssimultaneously) is highly unlikely.
15.38 (a) The steps of a mechanism must sum to give the observed overall stoichiometry of thereaction. For phosgene formation, this is CO + Cl2 → COCl2. The first step of themechanism proceeds rapidly in both directions, leading to no net change. In thesecond and third steps, Cl is consumed and then produced, for no net change; andCOCl is produced and consumed, for no net change:
Cl + CO → COClCOCl + Cl2 → COCl2 + Cl
Net: CO + Cl2 → COCl2.(b) The rate law is determined by the rate–determining step: Rate = k2[CO][Cl]. This is
not satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step: k1[Cl2] = k1[Cl]2
Solve this equality for [Cl]: [Cl] = k1k−1
⎛⎝⎜
⎞⎠⎟
1/2[Cl2]1/2
Substitute into the rate expression: Rate = k2k1k−1
⎛⎝⎜
⎞⎠⎟
1/2[CO][Cl2]1/2
(c) Reactive intermediates are those species that are produced and consumed in themechanism but do not appear in the overall stoichiometry of the reaction: Cl and COCl.
Olmsted & Williams Fourth Edition Instructor’s Solutions Manual Chapter 15
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15.39 (a) The rate law is determined by the rate–determining step: Rate = k2[N2O2][O2]. Thisis not satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step:
k1[NO]2 = k1[N2O2]
Solve this equality for [N2O2]: [N2O2] = k1k−1
⎛⎝⎜
⎞⎠⎟
[NO]2
Substitute into the rate expression:
Rate = k2 k1k−1
⎛⎝⎜
⎞⎠⎟
[NO]2[O2]
(b) This rate expression has an overall order of (2 + 1) = 3, so the mechanism isconsistent with third–order behavior.
(c) The intermediate species is N2O2. Two NO molecules could bind in several ways:
In the second step, O2 collides with the intermediate and reacts to form two NO2molecules. The ONNO arrangement is the only intermediate for which the new set ofbonds can easily occur:
15.40 (a) Steps 2 and 3 of the mechanism consume one NO3 each, so step 1 occurs twice toproduce two molecules of NO3 , and the intermediates, NO3 and NO, cancel:
2 N2O5 2 NO2 + 2 NO3
NO2 + NO3 → NO + NO2 + O2
NO + NO3 → 2 NO2
Net: 2 N2O5 → 4 NO2 + O2
(b) The rate law is determined by the rate–determining step: Rate = k2[NO2][NO3]. Thisis not satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step:
k1[N2O5] = k1[NO2][NO3]
Solve this equality for [NO3]: [NO3] = k1k−1
⎛⎝⎜
⎞⎠⎟[N2O5][NO2]
Substitute into the rate expression: Rate = k2 k1k−1
⎛⎝⎜
⎞⎠⎟
[N2O5]
(c) The mechanism with step 2 as the rate–determining step predicts first–order behaviorwith respect to N2O5. If the first step is rate–determining, the rate law isRate = k1[N2O5]. Thus, although it is consistent with the mechanism, the informationdoes not prove that step 2 is rate–determining.
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15.41 The rate constant for a reaction depends on temperature according to the Arrheniusequation (Equation 15–6): k = AeEa/RT. When Ea = 0, the exponent is e0 = 1 and k isindependent of temperature. A zero activation energy exists when a reaction can occurwithout first breaking any chemical bonds. The most common example is thecombination of two free radicals, such as H3C• + •CH3 → H3C–CH3.
15.42 When Ea = 0, there is no energy "hump" to overcome before products form, so theactivation energy for the reverse reaction is the negative of the energy change for theforward reaction:
15.43 An exothermic reaction has products lower in energy than reactants. In the activatedcomplex A will be bonded to both B and C:
Ereaction
15.44 An endothermic reaction is "uphill" from reactants to products. In the activated complex,A will be bonded to both B and C:
Ereaction
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15.45 An exothermic reaction is "downhill" from reactants to products, and the activationenergy plot should show this:
15.46 The activation energy for the forward reaction can be found from the exothermicity of thereaction and Ea for the reverse reaction: Ea, forward = 70 – 57 = 13 kJ/mol
15.47 Assume that the ratio of the rate constants is proportional to the ratio of the number offlashes, then use the rearranged version of Equation 158 in your textbook:
Ea = R ln k2
k1
⎛⎝⎜
⎞⎠⎟
1T1
−1T2
⎛⎝⎜
⎞⎠⎟
−1= (8.314 J mol1K1)ln 2.7
3.3⎛⎝⎜
⎞⎠⎟
1(29 + 273)K
−1
(23+ 273)K⎛⎝⎜
⎞⎠⎟−1
Ea =−1.67 J mol−1 K−1
−6.7 x 10−5 K−1⎛
⎝⎜
⎞
⎠⎟
10−3 kJ 1 J
⎛
⎝⎜
⎞
⎠⎟ = 25 kJ/mol
15.48 Activation energies are calculated from the Arrhenius equation using Equation 15–8:
Ea = R ln k2
k1
⎛⎝⎜
⎞⎠⎟1T1
−1T2
⎛⎝⎜
⎞⎠⎟
−1
k1 = 1.9 x 102/min k2 = 39.6/minT1 = 28 + 273 = 301 K T2 = 5 + 273 = 278 K
1T1
−1T2
⎛⎝⎜
⎞⎠⎟
−1 = (3.322 x 103 – 3.597 x 103)1 = –3636 K
k2k1
=39.6 min−1
1.9 x 102min−1 = 0.208 ln
k2k1
⎛⎝⎜
⎞⎠⎟
= –1.568
Ea = 8.314 J mol1 K1 10−3 kJ1J
⎛
⎝⎜
⎞
⎠⎟
−1.5681
⎛⎝⎜
⎞⎠⎟
−3636 K1
⎛⎝⎜
⎞⎠⎟
= 47 kJ/mol
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15.49 Hydrogen gas adsorbs on the catalyst’s surface as H atoms, which then react with COmolecules much more easily (and quickly) than direct reaction of CO molecules with H2molecules. Bonds that must be broken for this reaction to occur (depicted by the squigglylines) are two H–H single bonds and both the π bonds in the CO molecule. Bonds thatare formed (dashed lines) are three C–H bonds and one O–H bond.
15.50 The C–O bond in methanol and one of the π bonds in CO need to be broken. The bondsto be formed are a C–C bond and a C–O bond. A catalyst would make it easier for thisreaction to occur by helping break the C–O bonds.
15.51 (a) The effect of a catalyst on a reaction is to reduce the activation energy withoutchanging the energy of the reactants or the products. Thus, the energies of reactantsand products are the same for both curves. Only the “bump” in the activation energydiagram changes, being lower in the presence of the catalyst than in its absence:
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(b) A catalyst is present at the beginning and end of the reaction but does not appear inthe net reaction. Here, Pd metal acts as a catalyst. An intermediate is anything that isproduced in one step of the mechanism and then used up in another step. Theintermediates for this reaction are the H atoms that form when H2 gas absorbs ontothe Pd metal surface. The valley on the activation energy diagram represents thisintermediate stage.
(c)
15.52 (a) The overall reaction, addition of O atoms to O3, is exothermic. The first step is slow,indicating a high activation energy barrier; the second is fast, indicating a low barrier:
(b) NO is present at the beginning and end, so it is a catalyst; NO2 is produced and thenconsumed, so it is an intermediate.
(c) Because the Cl–catalyzed reaction has a very low activation energy barrier, thisreaction can proceed effectively even at the low temperature of the upperstratosphere, so Cl is a more serious threat than NO.
15.53 (a) The second step must use up the intermediate, O:O + NO → NO2 .
(b)
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15.54 (a) Before determining the steps in a mechanism, it is necessary to know the balancedoverall reaction: O2 + 2 CO → 2 CO2The first step consumes one molecule each of O2 and CO and produces one moleculeof product and an O atom. The easiest way to complete the mechanism is bycombining the O atom with another CO molecule: CO + O → CO2.
(b) A molecular picture of a reaction shows reactants, reaction intermediate, andproducts:
15.55 (a) A rate expression relates rate to changes: Rate = Δ[C6H6]
Δt= −
13
Δ[C2H2]Δt
⎛⎝⎜
⎞⎠⎟
(b) Rate laws must always be determined experimentally. Thus, there is insufficientinformation to write the rate law. Experiments would have to be carried outmeasuring the rate as a function of [C2H2] and the data analyzed using techniquesdescribed in your textbook.
15.56 Rates of reaction of various reagents are related to one another through the stoichiometriccoefficients of the net chemical equation:
Rate (NH3) = −44
⎛⎝⎜
⎞⎠⎟ Rate (NO) = – 1.5 x 103 M s1
Rate (O2) = −54
⎛⎝⎜
⎞⎠⎟ Rate (NO) =
−54
⎛⎝⎜
⎞⎠⎟ (1.5 x 103 M s1) = – 1.9 x 103 M s1
Rate (H2O) = 64
⎛⎝⎜
⎞⎠⎟ Rate (NO) =
64
⎛⎝⎜
⎞⎠⎟ (1.5 x 103 M s1) = 2.3 x 103 M s1
15.57 The essential units of information needed to construct an activation energy diagram arethe energy change and activation energy. Here, the reaction is a formation reaction and
the change in moles of gas is zero during the reaction, so ΔE ≅ ΔHfo :
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15.58 The essential units of information needed to construct an activation energy diagram arethe energy change and activation energy. Two curves must be shown, one for theuncatalyzed and one for the catalyzed reaction:
15.59 When the concentration of a reactant increases by a factor of three (triples), the rate ofreaction changes by 3n, where n is the order with respect to that concentration.(a) nine–fold increase; (b) no change; (c) rate increases by 5.2 times.
15.60 When the concentration of a reactant increases by a factor of two (doubles), the rate ofreaction changes by 2n, where n is the order with respect to that concentration.(a) rate doubles; (b) rate increases by 1.4 times; (c) rate is cut in half.
15.61 To determine the order of a reaction from a set of experimental data, prepare plots of
ln [A]0[A]
⎛⎝⎜
⎞⎠⎟
vs. t and 1[A]
−1[A]0
vs. t:
The first–order plot is linear, while the second–order plot is not. This reaction is firstorder. Determine the rate constant from the slope:
k = Slope = ΔyΔx
=0.766 − 0
300 s − 0 s = 2.55 x 103 s1
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15.62 To determine the order of a reaction from a set of experimental data, prepare plots of
ln [A]0[A]
⎛⎝⎜
⎞⎠⎟
vs. t and 1[A]
−1[A]0
vs. t:
The first–order plot is linear, while the second–order plot is not. This reaction is firstorder. Determine the rate constant from the slope:
k = Slope = ΔyΔx
=0.491− 0
800 s − 0 s = 6.14 x 104 s1
15.63 The mechanism has the rate law, Rate = k[X2], first order in X2 and independent of Y.a contains 5 X2 and 8 Y, while b contains 10 X2 and 8 Y. The rate for b will be twice thatfor a, because the concentration of X2 is twice as great.
15.64 The mechanism has the rate law, Rate = k[X2], first order in X2 and independent of Y.A contains 5 X2 and 8 Y, while B contains 5 X2 and 16 Y. The rate for B will be the sameas that for A, because the concentration of X2 is unchanged.
15.65 The initial concentration of N2O5 is much smaller than the initial concentration of H2O inboth experiments, so this is an example of the isolation technique. Assuming that the ratelaw has the form, Rate = k[N2O5]x[H2O]y, we have
Experimental rate = kobs[N2O5]x, where kobs = k[H2O]0y
We need to plot the data: if x = 1, a log plot will be linear, while if x = 2, a 1/[N2O5] plotwill be linear. Here is the first–order plot for Experiment A:
This plot is linear, with slope = kobs = 0.0038 s1.
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We do not need to plot the data for Experiment B, in which the initial concentration ofH2O is twice as large as in Experiment A. Instead, we note that the reaction is goingtwice as fast: In Experiment B, it takes only 60 seconds for the concentration to fall to thevalue that is reached in 120 seconds in Experiment A. Doubling [H2O] leads to adoubling of the rate, so the reaction is first order in H2O as well as first order in N2O5.The rate law is:
Rate = k[N2O5][H2O]
To evaluate k, we can use either set of conditions. Here we use Experiment A:kobs = k[H2O]0 = 0.0038 s1 and [H2O]0= 0.025 M
k = kobsH2O[ ]0
= 0.0038 s1
0.025 M = 0.15 M1 s1
15.65 The initial concentration of NO2 is much smaller than the initial concentration of CO inboth experiments, so this is an example of the isolation technique. Assuming that the ratelaw has the form, Rate = k[NO2]x[CO]y, we haveExperimental rate = kobs[NO2]x, where kobs = k[CO]0
y.We need to plot the data: if x = 1, a log plot will be linear, while if x = 2, a 1/c plot willbe linear. Here is the second–order plot for Experiment A:
This plot is linear, with slope = kobs = 0.49 M1 s1.We do not need to plot the data for Experiment B, in which the initial concentration ofCO is 1.5 times as large as in Experiment A. Instead, we note that the reaction is going atthe same rate in both experiments. Changing [CO] has no effect on the rate, so thereaction is zero order in CO and second order in NO2. The rate law is:
Rate = k[NO2]2
Because there is no dependence on [CO], kobs = k = 0.49 M1 s1.
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15.67 Reaction times for first–order reactions can be calculated using Equation 15–3, suitablyrearranged:
t =ln [A]o
[A]⎛⎝⎜
⎞⎠⎟
k
For 10.0% decomposition, [A] = 0.900[A]o and ln[A]o[A]
⎛⎝⎜
⎞⎠⎟
= ln1.0000.900
⎛⎝⎜
⎞⎠⎟ = 0.105
t = 0.105
5.5 x 104s−1 = 1.9 x 102 s
For 50.0% decomposition, [A] = 0.500[A]o and ln[A]o[A]
⎛⎝⎜
⎞⎠⎟
= ln1.0000.500
⎛⎝⎜
⎞⎠⎟ = 0.693
t = 0.693
5.5 x 104s−1 = 1.3 x 103 s
For 99.9% decomposition, [A] = 0.001[A]o and ln[A]o[A]
⎛⎝⎜
⎞⎠⎟
= ln1.0000.001
⎛⎝⎜
⎞⎠⎟ = 6.91
t = 6.91
5.5 x 104s−1 = 1.3 x 104 s
15.68 The reaction is first order in each reactant, but the initial concentrations allow us to treatH2S as isolated, so the kinetics are described by Rate = kobs[H2S], with kobs = k[Cl2]o:
kobs = (3.5 x 102 M1 s1)(0.035 M) = 1.2 x 103 s1
(a) ln[A] = ln[A]o – kt = ln(5.0 x 105) – (1.2 x 103 s1)(225 s) = –10.173
[A] = e10.173 = 3.8 x 105 M
(b) t =ln [A]o
[A]⎛⎝⎜
⎞⎠⎟
k=
ln 5.0 x 105M1.0 x 105 M
⎛
⎝⎜
⎞
⎠⎟
1.2 x 103s−1 = 1.3 x 103 s
15.69 (a) False (overall reaction would give fourth–order kinetics); (b) False (rate constantsmust be measured for at least two different temperatures to calculate Ea); (c) True (ratesof reaction increase with increasing temperature); (d) True (a unimolecular step hasfirst–order kinetics).
15.70 (a) True (N2 is a reactant with coefficient of 1); (b) False (reaction order must always bedetermined by experiments); (c) True (3 molecules of H2 react for every 1 molecule ofN2); (d) False (3 H2 disappear for every N2); (e) False (reactants always have negativeconcentration changes as time increases); (f) True (simultaneous four–molecule collisionsdo not occur); (g) True (bonds must break for this reaction to proceed).
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15.71 Reaction orders are given by the exponents on the concentrations that appear in the ratelaw. Overall order is the sum of those exponents:(a) first order in N2O5 and first order overall; (b) second order in NO, first order in H2,and third order overall; and (c) first order in enzyme and first order overall.
15.72 Reaction orders are given by the exponents on the concentrations that appear in the ratelaw. Overall order is the sum of those exponents:(a) zero order overall (no concentration dependence); (b) first order in sucrose, H2O, andH3O+, and third order overall; and (c) first order in CHCl3, one–half order in Cl2, andthree–halves order overall.
15.73 The speed of a chemical reaction refers to how fast it proceeds. The spontaneity of achemical reaction refers to whether or not the reaction can go in the direction writtenwithout outside intervention. A spontaneous reaction may nevertheless have a very slowspeed.
15.74 In any reacting system, concentrations of reactants and products are changing with time,even though these changes may be slow for some species. A changing concentrationrequires a non–zero net rate for each process involved in the mechanism of the reaction.Thus, for a reacting system, forward and reverse rates can never be exactly equal. (Asdiscussed in Chapter 16 of your textbook, the system is at equilibrium when forward andreverse rates are exactly equal.)
15.75 Ultraviolet light causes chlorofluorocarbons to fragment, producing Cl atoms thatcatalyze the destruction of ozone:
CF2Cl2 hυ⎯ →⎯ CF2Cl + Cl
O3 hυ⎯ →⎯ O2 + O
Cl + O3 → ClO + O2
ClO + O → Cl + O2Because the fourth reaction regenerates a Cl atom, the second through fourth reactionsoccur many hundreds of times for every CF2Cl2 molecule that fragments.
15.76 In the unpolluted stratosphere, ozone is generated by the action of ultraviolet light on O2and is destroyed by the action of ultraviolet light on O3:
O2 hυ⎯ →⎯ O + O
O + O2 → O3
O3 hυ⎯ →⎯ O2 + O
O3 + O → 2 O2
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15.77 (a) Obtain the overall stoichiometry by adding the three steps, ignoring the reverse reaction ofStep one, which leads to no net change: 2 NO + 2 H2 → N2 + 2 H2O
(b) The rate law is determined by the rate–determining step: Rate = k2[N2O2][H2]. Thisis not satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step: k1[NO]2 = k1[N2O2]
Solve this equality for [N2O2]: [N2O2] = k1k−1
[NO]2
Substitute into the rate expression: Rate = k2k1k−1
[NO]2[H2]
15.78 (a) Obtain the overall stoichiometry by adding the three steps, ignoring the reverse reaction ofStep one, which leads to no net change: 2 B +D → 2 F
(b) The rate law is determined by the rate–determining step: Rate = k2[C][D]. This is notsatisfactory, however, because it contains the concentration of an intermediate. Setthe rates equal for the forward and reverse first step: k1[A][B] = k1[C]
Solve this equality for [C]: [C] = k1k−1
[A][B]
Substitute into the rate expression: Rate = k2k1k−1
[A][B][D]
(c) A is a catalyst (consumed in an early step and regenerated in a later step), while C and Eare intermediates (produced in early steps and consumed in later steps).
15.79 (a) At the molecular level, a catalyst binds to one or more of the reactants in a way thatweakens chemical bonds and makes it easier for bonds to rearrange to form theproducts.
(b) When temperature increases, the average energies of the molecules increase, with theresult that enough energy is present for a larger fraction of the molecules to havesufficient energy to overcome the activation energy barrier.
(c) When concentration increases, the molecular density increases. There are moremolecules to react, leading to a higher rate of molecular collisions. Both factorscontribute to a greater rate of reaction.
15.80 Elementary reactions must be processes that can actually occur at the molecular level.The Haber reaction involves one molecule of N2 and three molecules of H2, which wouldhave to collide in a simultaneous four–molecule process. In addition, four bonds mustbreak and six new bonds form. It is not reasonable for all this to occur in a single step.
15.81 Flask 1 contains 5 molecules, and Flask 2 contains 10 molecules, so the concentration isdoubled. The problem states that Flask 2 reacts four times as fast. This is (2)2, so therate law is Rate = k[A]2. At the molecular level, the rate–determining step could be areaction between two A molecules. For this process, the rate increases because the higherconcentration results in a higher rate of collisions.
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15.82 Each flask contains the same number of radioactive phosphorus atoms (six). Thequestion asks about the half–lives (how long it takes for the initial amount, six atoms, tofall to half that amount, three atoms). For first–order decay, the half–life is independentof concentration, so the times will be identical for the two flasks.
15.83 (a) The rate law is that for an elementary bimolecular reaction: Rate = k[H2][X2].(b) When a first step is rate–determining, it determines the rate law: Rate = k[X2].(c) The rate law is determined by the rate–determining step: Rate = k2[X][H2]. This is
not satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step: k1[X2] = k1[X]2
Solve this equality for [X]: [X] = k1k−1
⎛⎝⎜
⎞⎠⎟
1/2[X2]1/2
Substitute into the rate expression: Rate = k2k1k−1
⎛⎝⎜
⎞⎠⎟
1/2 [H2][X2]1/2
15.84 This reaction is 1/2 order in one reactant and first order in the other reactant. There ismore than one way to increase the concentrations so that the rate doubles. The simplestis to double the concentration of the reactant for which there is first–order behavior. Theillustration shows three diatomic molecules and six atoms, so the new view should showthree diatomic molecules and twelve atoms:
15.85 (a) The net reaction can be obtained by summing the three steps, because when thereverse step occurs there is no net change: Cl2 + CHCl3 → HCl + CCl4;
(b) Intermediates are produced in early steps and consumed in later steps: Cl and CCl3;(c) The rate law is determined by the rate–determining step: Rate = k2[CHCl3][Cl]. This
is not satisfactory, however, because it contains the concentration of an intermediate.Set the rates equal for the forward and reverse first step: k1[Cl2] = k1[Cl]2
Solve this equality for [Cl]: [Cl] = k1k−1
⎛⎝⎜
⎞⎠⎟
1/2[Cl2]1/2
Substitute into the rate expression: Rate = k2k1k−1
⎛⎝⎜
⎞⎠⎟
1/2 [CHCl3][Cl2]1/2
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15.86 (a) The half–life is not constant, so test for 2ndorder behavior by plotting 1/c –1/co vs. t:t hr 0 1.0 2.0 3.0 5.0 7.0 9.0c M 0.500 0.375 0.300 0.250 0.188 0.150 0.1251/c –1/co M1 0.00 0.67 1.33 2.00 3.32 4.67 6.00
The second–order plot is a straight line, so Rate = k[NH4NCO]2
(b) k = (Slope) = 6.00 M−1 − 0.00 M−1
9.0 hr − 0.0 hr = 0.67 M1 hr1
(c) At 50 oC, it takes 2.0 hr for concentration to fall from 0.500 to 0.300 M, while at25 oC, it takes 6.0 hr for the same change to occur. The reaction is three times as fastat 50 oC as at 25 oC, so k2/k1 = 3.0; Use Equation 15–8 to calculate Ea:
Ea = R ln k2k1
⎛⎝⎜
⎞⎠⎟1T1
−1T2
⎛⎝⎜
⎞⎠⎟
−1
T2 = 50 + 273 = 323 K and T1 = 25 + 273 = 298 K
Ea = 8.314 J1 mol K
⎛⎝⎜
⎞⎠⎟
10−3 kJ1 J
⎛
⎝⎜
⎞
⎠⎟
1298 K
– 1323 K
⎛⎝⎜
⎞⎠⎟
(ln 3.0) = 35 kJ/mol
15.87 Activation energies are calculated from the Arrhenius equation using Equation 15–8:
Ea = R ln k2k1
⎛⎝⎜
⎞⎠⎟1T1
−1T2
⎛⎝⎜
⎞⎠⎟
−1
Development time is inversely proportional to rate constant, so k2k1
=t1t2
.
(a) t1t2
= 2 T1 = 20 + 273 = 293 K T2 = 293 + 10 = 303 K
1T1
−1T2
⎛⎝⎜
⎞⎠⎟
−1 = (3.413 x 103 K1– 3.300 x 103 K1) 1 = 8880. K
Ea = 8.314 J1 mol K
⎛⎝⎜
⎞⎠⎟
10−3 kJ1 J
⎛
⎝⎜
⎞
⎠⎟ (ln2)(8880.K) = 51 kJ/mol
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(b) To determine the time it takes at 25 oC, use ln t1t2
⎛⎝⎜
⎞⎠⎟
= EaR
1T1
−1T2
⎛⎝⎜
⎞⎠⎟
:
Ea = 51 kJ/mol t1 = 10 minT1 = 20 + 273 = 293 K T2 = 25 + 273 = 298 K
ln 10 mint2
⎛⎝⎜
⎞⎠⎟
= 1
293 K– 1
298 K⎛⎝⎜
⎞⎠⎟
1 mol K8.314 x 10−3 kJ
⎛⎝⎜
⎞⎠⎟
51 kJ1 mol
⎛⎝⎜
⎞⎠⎟
= 0.351
10 mint2
= 1.42 from which t2 = 10min1.42
= 7.0 min
15.88 Figure 15–20 shows that while the activation energy for the uncatalyzed forwardreaction, O3 + O, is about 19 kJ/mol, the activation energy for the uncatalyzed reversereaction, O2 + O2, is about 410 kJ/mol. Adding the catalyst reduces both values by about15 kJ/mol, so the forward reaction has an activation energy of only about 4 kJ/mol, butthe reverse reaction still has an immense activation energy, about 395 kJ/mol. The rate ofthe forward reaction is increased immensely by the catalyst, while the reverse reactionhas essentially zero rate whether catalyzed or uncatalyzed.
15.89 (a) Prepare first–order and second–order plots and look for linear behavior:t s 0 1000 2000 3000 4000c M 0.250 0.118 0.0770 0.0572 0.0455ln ([A]o/[A]) 0.000 0.751 1.18 1.47 1.701/[A] – 1/[A]o M1 0.00 4.47 8.99 13.5 18.0
The second–order plot is linear, so Rate = k[CH3CHO]2
(b) Determine the rate constant from the slope of the second–order plot:
k = Slope = 18.0 M−1 − 0.00 M−1
4000 s – 0 s = 4.5 x 103 M1 s1
(c) Use Equation 15–5, suitably rearranged: kt = 1[A]
–
1[A]o
[A]o = 0.250 M and [A] = 0.250 M100% − 75%100%
⎛⎝⎜
⎞⎠⎟ = 0.0625 M
t = 16.0 M−1 − 4.00 M−1
4.5 x 103 M−1 s−1 = 2.7 x 103 s
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15.90 Here, there are two concentration variables, [A] and [B]. In the first experiment,[B] >> [A], so A is isolated and the data can be analyzed to determine the order withrespect to A. The data could be plotted, but half–lives can be recognized. It takes 30 sfor the concentration to fall from 0.050 M to 0.025 M, 30 s to fall from 0.040 to 0.020 M,and 30 s to fall from 0.032 to 0.016 M. Thus, the half–life is a constant value, 30 s, andthe reaction is first order in A. In the second experiment, the rate of disappearance of Adoes not change even though the initial concentration of B is increased. Consequently,the rate law does not depend on [B].(a) Rate = k[A](b) The half–life of a first–order reaction provides a convenient way to calculate a rate
constant. Use Equation 15–4: t1/2 = ln2k
, rearranged to give k =ln2t1/2
:
k = 0.69330 s
= 2.3 x 102 s1
15.91 The hint for this problem suggests using the Arrhenius equation, k = AeEa/RT. Evaluate
EaRT
for the catalyzed and uncatalyzed situations, using T = 21 + 273 = 294 K:
EaRT
⎛⎝⎜
⎞⎠⎟ uncat
=125 kJ mol1
(8.314 x 103 kJ mol1 K1)(294 K) = 51.1
EaRT
⎛⎝⎜
⎞⎠⎟ cat
=46 kJ mol1
(8.314 x 103 kJ mol1 K1)(294 K) = 18.8
kuncat = Ae–51.1 and kcat = Ae–18.8
Divide one of these by the other to eliminate A and find the ratio of rate constants:
kcatkuncat
= e(51.1 – 18.8) = e32.3 = 1.1 x 1014
15.92 Here, there are two concentration variables, [O2] and [NO]. In the first experiment,[NO] >> [O2], so O2 is isolated and the data can be analyzed to determine the order withrespect to O2. The data could be plotted, but half–lives can be recognized. It takes 3.0 sfor the concentration to fall from 4.1 to 2.05 x 104 M, 3.0 s to fall from 2.05 to1.02 x 104 M, and 3.0 s to fall from 0.51 to 0.25 x 104 M. Thus, the half–life is aconstant value, 3.0 s, and the reaction is first order in O2. In the second experiment, [O2]>> [NO], so NO is isolated and the data can be analyzed to determine the order withrespect to NO. The half–life is not constant, so try a second–order plot:
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This plot is a straight line, so the reaction is second order with respect to NO.The rate law is Rate = k[O2][NO]2
The half–life of a first–order reaction provides a convenient way to calculate a rateconstant. Apply Equation 15–4 to the first experiment:
t1/2 =
ln2k
, which can be rearranged to give k =
ln 2t1/2
k = 0.6933.0 s
= 2.3 x 101 s1
This is not the true rate constant, because it describes the conditions under which O2 was
isolated: k = ktrue[NO]o2 and ktrue =
k
[NO]o2 =
0.23 s−1
(9.63 x 10−3 M)2
ktrue = 2.5 x 103 M2 s1
15.93 Neither intermediates nor catalysts appear in the overall stoichiometry of the reaction, soany species that appears in the mechanism but not in the overall stoichiometry is either anintermediate or a catalyst. Catalysts are consumed in early steps and regenerated in latersteps, while intermediates are produced in early steps and consumed in later steps.
15.94 To determine which set of concentrations leads to the fastest rate, determine the productof concentrations for each set:
(a) 0.5 mol2.0 L
⎛⎝⎜
⎞⎠⎟
0.5 mol2.0 L
⎛⎝⎜
⎞⎠⎟ = 0.06 mol2/L2
(b) 0.5 mol1.0 L
⎛⎝⎜
⎞⎠⎟
0.5 mol1.0 L
⎛⎝⎜
⎞⎠⎟ = 0.25 mol2/L2
(c) 0.1 mol1.0 L
⎛⎝⎜
⎞⎠⎟
2.0 mol1.0 L
⎛⎝⎜
⎞⎠⎟ = 0.20 mol2/L2
The set of conditions described by (b) has the largest product and the fastest rate.
15.95 The only true statements are (d) and (f). Statements (a) and (b) are false becauseΔEreaction = C – A. Here is an energy diagram showing the quantities involved:
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15.96 Plots of ln c vs. t and 1/c vs. t are not linear, showing that the reaction is neither firstorder or second order. Further examination of the data shows that the concentrationchange per unit time is independent of the concentration:
0 – 10 s,
Δc
Δt =
0.12 M10 s
= 0.012 M/s 10 – 20 s,
Δc
Δt =
0.12 M10 s
= 0.012 M/s
20 – 30 s,
Δc
Δt =
0.12 M10 s
= 0.012 M/s
A rate that is totally independent of concentration indicates zero–order behavior:Rate = k
15.97 When an enzyme binds a reactant, bondsbetween enzyme and reactant result in areduction in the strength of the bonds thatneed to be broken in order for thecatalyzed reaction to occur. Thisreduction in strength in turn reduces theamount of energy that must be suppliedfor the reaction to occur, accounting forthe decrease in the activation energy forthe enzyme–catalyzed reaction.
15.98 (a) The first step in the proposed mechanism generates an intermediate which must beconsumed in the second step. To give the correct overall stoichiometry, the secondstep also must consume a Cl ion: [Co(NH3)5]3+ + Cl → [Co(NH3)Cl]2+
(b) The rate law is determined by the rate–determining step: Rate = k2[I][Cl],where I = [Co(NH3)5]3+. This is not satisfactory, however, because it contains theconcentration of an intermediate. Set the rates equal for the forward and reverse firststep:
k1[[Co(NH3)5H2O]3+]= k1[I][H2O]
Solve this equality for [I]: [I] = k1k−1
[[Co(NH)5H2O]3+]
[H2O]
Substitute into the rate expression: Rate = k2k1k−1
[[Co(NH)5H2O]3+][Cl ]
[H2O]
(c) The observed rate is consistent with the predicted rate law. Although H2O appears in therate expression, its concentration in aqueous solution is so great that [H2O] = constant.Also, in 1 M HCl solution, [Cl] = 1 M, so the complex is isolated when it is present at1 mM concentration. The observed rate constant and rate expression are:
kobs = k2k1k−1
[Cl ][H2O]
Rate = kobs [[Co(NH)5H2O]3+]
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15.99 Your explanation should include the main features of the induced–fit model: a cavity intowhich the molecule fits, that adjusts its shape to the target molecule (square), then distortsto catalyze decomposition of the square:
15.100 Rates of elementary two–body reactions depend on concentration of each type of body.When a concentration is increased by some factor, the rate increases by the same factor:
(a) There are half as many red balls, so half as many pairs will form: 2.(b) There are 1.5 times as many white balls, so 1.5 times as many pairs will form: 6.(c) There are half as many red but twice as many white balls: Pairs = (1/2)(2)(4) = 4.(d) There are twice as many of each type of balls: Pairs = (2)(2)(4) = 16.(e) When the concentration of either type of ball changes, the frequency of collisions with
other balls changes in the same proportion, so the rate of formation of pairs, whichdepends on collision rate, also changes proportionally.
15.101 (a) The catalyst, Br, is consumed in the first step, so it must be regenerated in a laterstep. Thus, A = Br. For the stoichiometry to be correct, B = O2;
(b) For the overall reaction:ΔE ≅ ΔH = [2 mol(–285.83 kJ/mol) + 1 mol(0 kJ/mol)]
– 2 mol(–187.8 kJ/mol) = – 196.1 kJFor the first step (Δng = 0):
ΔE = ΔH = [1 mol(–285.83 kJ/mol) + 1 mol(–94.1 kJ/mol)]– [1 mol(–187.8 kJ/mol) + 1 mol(–121.4 kJ/mol) = – 70.73 kJ
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15.102 (a) The information provided suggests that transfer of an O atom between NO2 and O3,giving NO3, is a reasonable first step. Then NO3 could react with NO2, giving thecorrect overall stoichiometry. If the first reaction is rate–determining, the observedrate law results:
NO2 + O3 → NO3 + O2 (slow, rate determining)NO3 + NO2 → N2O5 (rapid)
(b) A molecular picture of a reaction shows reactants, reactive intermediate, andproducts:
(c) The activation energy diagram should indicate the exothermicity and activationenergy of the overall process:
15.103 (a) The data are in the form of initial rate information, so use the ratios of initial rates todetermine the order with respect to each reagent:
In Experiments 1 and 2, only [I] changes: 0.0060 M0.0025 M
⎛⎝⎜
⎞⎠⎟x=
9.1 x 102 M/s3.8 x 102 M/s
⎛
⎝⎜
⎞
⎠⎟
Simplifying, (2.4)x = 2.4 x = 1The reaction is first order in I.
In Experiments 1 and 3, only [OCl] changes: 0.0037 M0.0025 M
⎛⎝⎜
⎞⎠⎟y=
5.6 x 102 M/s3.8 x 102 M/s
⎛
⎝⎜
⎞
⎠⎟
Simplifying, (1.5)y = 1.5 y = 1The reaction is first order in OCl.
In Experiments 3 and 4, only [OH] changes: 0.10 M0.20 M
⎛⎝⎜
⎞⎠⎟z=
5.6 x 102 M/s2.8 x 102 M/s
⎛
⎝⎜
⎞
⎠⎟
Simplifying, (0.50)z = 2.0; the rate doubles when the concentration is cut in half.This means that the rate is inversely dependent on [OH]: z = –1
The rate law is:
Rate = k[I ][OCl ][OH]
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(a) Use any of the initial rate data to evaluate k:
k = Initial rate( ) OH⎡
⎣⎤⎦
OCl⎡⎣
⎤⎦ I⎡⎣
⎤⎦
= 3.8 x 102 M/s( ) 0.10 M( )
0.025 M( ) 0.0025 M( ) = 61 s1
(b) The inverse dependence indicates a complicated mechanism. The problem states thatone reactant is in rapid equilibrium with water. Because HI is a strong acid, I is not agood choice:
OCl + H2O k1⎯ →⎯k−1
← ⎯⎯⎯ HOCl + OH (rapid equilibrium)
Now let HOCl react with I in a slow, rate–determining step:
I + HOCl k2⎯ →⎯ HOI + Cl (slow)
HOI can react with hydroxide to give the correct overall stoichiometry:OH + HOI →H2O + OI (fast)
Rate = k2[HOCl][I]Set the rates of the forward and reverse first reaction equal to each other and solve for[HOCl]:
k1[OCl][H2O] = k1[OH][HOCl] so [HOCl] = k1k−1
OCl⎡⎣
⎤⎦ H2O[ ]OH⎡⎣
⎤⎦
Substitute this into the rate expression:
Rate = k2k1k−1
OCl⎡⎣
⎤⎦ I
⎡⎣
⎤⎦ H2O[ ]
OH⎡⎣
⎤⎦
This rate law has the correct first–order dependence on the two reactants and aninverse dependence on [OH], in agreement with the experimental rate law.
(d) The predicted rate law is first order in water concentration. In order to test for thisdependence, we could do kinetic experiments in mixed solvents. For example, wecould use a 1:1 molar mixture of ethanol and water. In this solvent, the rate should behalf as fast, if the proposed mechanism is the correct one.