CHAPTER 16 KINETICS: RATES AND
MECHANISMS OF CHEMICAL REACTIONS
FOLLOW–UP PROBLEMS
16.1A
Plan: Balance the equation. The rate in terms of the change in concentration with time for each substance is
1 Δ[ A ]
expressed as
, where a is the coefficient of the reactant or product A. The rate of the reactants is given a
a Δt
negative sign.
Solution:
a) The balanced equation is 4NO(g) + O2(g) → 2N2O3.
Choose O2 as the reference because its coefficient is 1. Four molecules of NO (nitrogen monoxide) are consumed
for every one O2 molecule, so the rate of O2 disappearance is 1/4 the rate of NO decrease. By similar reasoning,
the rate of O2 disappearance is 1/2 the rate of N2O3 (dinitrogen trioxide) increase.
Δ [O 2 ]
1 Δ[ NO ]
1 Δ [N 2 O3 ]
Rate = −
=−
= +
Δt
4 Δt
2 Δt
b) Plan: Because NO is decreasing; its rate of concentration change is negative. Substitute the negative value into
the expression and solve for Δ[O2]/Δt.
Solution:
Δ [O 2 ]
1
−1.60 x 10−4 mol/L • s = −
= 4.00x10–5 mol/L·s
Δt
4
Plan: Examine the equation that expresses the rate in terms of the change in concentration with time for each
Rate = −
16.1B
(
)
substance. The number in the denominator of each fraction is the coefficient for the corresponding substance in
the balanced equation. The terms that are negative in the rate equation represent reactants in the balanced
equation, while the terms that are positive represent products in the balanced equation. For example, the following
1 Δ[ A ]
term,
, describes a product (the term is positive) that has a coefficient of a in the balanced equation. In
a Δt
part b), use the rate equation to compare the rate of appearance of H2O with the rate of disappearance of O2.
Solution:
a) The balanced equation is 4NH3 + 5O2 → 4NO + 6H2O.
b) Plan: Because H2O is increasing; its rate of concentration change is positive. Substitute its rate into the
expression and solve for Δ[O2]/Δt.
Solution:
1
1 Δ[O2 ]
Rate = (2.52x10 –2 mol / L • s) = –
6
5 Δt
Δ
[O
]
5
2
= –2.10x10–2 mol/L·s
– (2.52x10 –2 mol / L • s) =
Δt
6
The negative value indicates that [O2] is decreasing as the reaction progresses. The rate of reaction is always
expressed as a positive number, so [O2] is decreasing at a rate of 2.10x10–2 mol/L·s.
16.2A
Plan: The reaction orders of the reactants are the exponents in the rate law. Add the individual reaction orders
to obtain the overall reaction order. Use the rate law to determine how the changes listed in the problem will affect
the rate.
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16-1
Solution:
a) The exponent of [I–] is 1, so the reaction is first order with respect to I–. Similarly, the reaction is first order
with respect to BrO3–, and second order with respect to H+. The overall reaction order is (1 + 1 + 2) = 4, or
fourth order overall.
b) Rate = k[I–][BrO3–][H+]2. If [BrO3–] and [I–] are tripled and [H+] is doubled, rate = k[3 x I–][3 x BrO3–][2 x H+]2,
then rate increases to 3 x 3 x 22 or 36 times its original value. The rate increases by a factor of 36.
16.2B
Plan: The reaction orders of the reactants are the exponents in the rate law. Add the individual reaction orders
to obtain the overall reaction order. Use the rate law to determine how the changes listed in the problem will affect
the rate.
Solution:
a) The exponent of [ClO2] is 2, so the reaction is second order with respect to ClO2. Similarly, the reaction is
first order with respect to OH–. The overall reaction order is (1 + 2) = 3, or third order overall.
b) Rate = k[ClO2]2[OH–]. If [ClO2] is halved and [OH–] is doubled, rate = k[1/2 x ClO2]2[2 x OH–], then rate
increases to (1/2)2 x 2 or 1/2 its original value. The rate decreases by a factor of 1/2.
16.3A
Plan: Assume that the rate law takes the general form rate = k[H2]m[I2]n. To find how the rate varies with respect to
[H2], find two experiments in which [H2] changes but [I2] remains constant. Take the ratio of rate laws for those
two experiments to find m. To find how the rate varies with respect to [I2], find two experiments in which [I2]
changes but [H2] remains constant. Take the ratio of rate laws for those two experiments to find n. Add m and n to
obtain the overall reaction order. Use the rate law to solve for the value of k.
Solution:
For the reaction order with respect to [H2], compare Experiments 1 and 3:
m
rate 3
[H ]
= 2 3m
rate 1
[ H 2 ]1
9.3 x 10−23
1.9 x 10−23
m
=
[0.0550 ]3
m
[ 0.0113]1
4.8947 = (4.8672566)m
Therefore, m = 1
If the reaction order was more complex, an alternate method of solving for m is:
log (4.8947) = m log (4.8672566); m = log (4.8947)/log (4.8672566) = 1
For the reaction order with respect to [I2] , compare Experiments 2 and 4:
n
rate 4
[I ]
= 2 4n
rate 2
[ I 2 ]2
1.9 x 10−22
1.1 x 10−22
n
=
[0.0056 ]4
n
[ 0.0033]2
1.72727 = (1.69697)n
Therefore, n = 1
The rate law is rate = k[H2][I2] and is second order overall.
Calculation of k:
k = Rate/([H2][I2])
k1 = (1.9x10–23 mol/L∙s)/[(0.0113 mol/L)(0.0011 mol/L)] = 1.5 x 10–18 L/mol∙s
k2 = (1.1x10–22 mol/L∙s)/[(0.0220 mol/L)(0.0033 mol/L)] = 1.5 x 10–18 L/mol∙s
k3 = (9.3x10–23 mol/L∙s)/[(0.0550 mol/L)(0.0011 mol/L)] = 1.5 x 10–18 L/mol∙s
k4 = (1.9x10–22 mol/L∙s)/[(0.0220 mol/L)(0.0056 mol/L)] = 1.5 x 10–18 L/mol∙s
Average k = 1.5x10–18 L/mol∙s
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16-2
16.3B
Plan: Assume that the rate law takes the general form rate = k[H2SeO3]m[I–]n[H+]p. To find how the rate varies with
respect to [H2SeO3], find two experiments in which [H2SeO3] changes but [I–] and [H+] remain constant. Take the
ratio of rate laws for those two experiments to find m. To find how the rate varies with respect to [I–], find two
experiments in which [I–] changes but [H2SeO3] and [H+] remain constant. Take the ratio of rate laws for those two
experiments to find n. To find how the rate varies with respect to [H+], find two experiments in which [H+]
changes but [H2SeO3] and [I–] remain constant. Take the ratio of rate laws for those two experiments to find p.
Add m, n, and p to obtain the overall reaction order. Use the rate law to solve for the value of k.
Solution:
For the reaction order with respect to [H2SeO3], compare Experiments 1 and 3:
rate 3
[H SeO3 ]3m
= 2
[H2 SeO3 ]1m
rate 1
[1.0x10 –2 mol/L]3m
3.94x10 –6 mol/L • s
=
9.85x10 –7 mol/L • s [2.5x10 –3 mol/L]1m
4 = (4)m
Therefore, m = 1
For the reaction order with respect to [I–], compare Experiments 1 and 2:
[I – ]n
rate 2
= – 2n
rate 1
[I ]1
[3.0x10 –2 mol/L]2n
7.88x10 –6 mol/L • s
=
9.85x10 –7 mol/L • s [1.5x10 –2 mol/L]1n
8 = (2)n
Therefore, n = 3
For the reaction order with respect to [H+], compare Experiments 2 and 4:
[H+]4p
rate 4
=
rate 2
[H+]2p
[3.0x10 –2 mol/L]4p
3.15x 10 –5 mol/L • s
=
7.88x10 –6 mol/L • s
[1.5x10 –2 mol/L]2p
p
4 = (2)
Therefore, p = 2
The rate law is rate = k[H2SeO3][I–]3[H+]2 and is sixth order overall.
b) Calculation of k:
k = Rate/([H2SeO3][I–]3[H+]2)
k1 = (9.85x10–7 mol/L∙s)/[(2.5x10–3 mol/L)(1.5x10–2 mol/L)3(1.5x10–2 mol/L)2] = 5.2x105 L5/mol5∙s
16.4A
Plan: The reaction is second order in X and zero order in Y. For part a), compare the two amounts of reactant X.
For part b), compare the two rate values.
Solution:
a) Since the rate law is rate = k[X]2, the reaction is zero order in Y. In Experiment 2, the amount of reactant X
has not changed from Experiment 1 and the amount of Y has doubled. The rate is not affected by the doubling
of Y since Y is zero order. Since the amount of X is the same, the rate has not changed. The initial rate of
Experiment 2 is also 0.25x10–5 mol/L·s.
b) The rate of Experiment 3 is four times the rate in Experiment 1. Since the reaction is second order in X, the
concentration of X must have doubled to cause a four-fold increase in rate. There should be 6 black spheres and 3
green spheres in Experiment 3.
1.0 x 10−5
[ x ]2
= 2
−5
0.25 x 10
[3]
4 =
[ x ]2
9
x = 6 black spheres
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16-3
16.4B
Plan: Examine how a change in the concentration of the different reactants affects the rate in order to determine
the rate law. Then use the rate law to determine the number of particles in the scene for Experiment 4.
Solution:
a) Comparing Experiments 1 and 2, we can see that the number of blue A spheres does not change while the
number of yellow B spheres changes from 4 to 2. Although the number of yellow spheres changes, the rate does
not change. Therefore, B has no effect on the rate, which suggests that the reaction is zero order with respect to B.
Now that we know that the yellow B spheres do not affect the rate, we can look at the influence of the blue A
spheres. Comparing Experiments 2 and 3, we can see that the number of blue spheres changes from 4 to 2
(experiment 3 concentration of A is half of that of experiment 2) while the rate changes from 1.6x10–3 mol/L∙s to
8.0x10–4 mol/L∙s (the rate of experiment 3 is half of the rate of experiment 2). The fact that the concentration of
blue spheres changes in the same way the rate changes suggests that the reaction is first order with respect to the
blue A spheres. Therefore, the rate law is: rate = k[A].
b) The rate of Experiment 4 is twice the rate in Experiment 1. Since the reaction is first order in A, the
concentration of A must have doubled to cause a two-fold increase in rate. There should be 2 x 4 particles =
8 particles of A in the scene for Experiment 4.
16.5A
Plan: The rate expression indicates that the reaction order is two (exponent of [HI] = 2), so use the integrated
second-order law. Substitute the given concentrations and the rate constant into the expression and solve for time.
Solution:
1
1
−
= kt
[ A ]t
[ A ]0
1
1
= (2.4x10–21 L/mol∙s)(t)
−
⎡ 0.00900 mol/L ⎤
⎡ 0.0100 mol/L ⎤
⎢⎣
⎥⎦ t
⎢⎣
⎥⎦ 0
1
1
−
0.00900 mol/L 0.0100 mol/L
t=
2.4 x10−21 L /mol • s
t = 4.6296x1021 s = 4.6x1021 s
16.5B
Plan: The problem states that the decomposition of hydrogen peroxide is a first-order, reaction, so use the firstorder integrated rate law. Substitute the given concentrations and the time into the expression and solve for the
rate constant.
Solution:
[A]0
= kt
a)
ln
[A]t
ln
1.28 M
= (k)(10.0 min)
0.85 M
k=
0.4094
= 0.04094 = 0.041 min–1
10.0 min
b) If you start with an initial concentration of hydrogen peroxide of 1.0 M ([A]0 = 1.0 M) and 25% of the sample
decomposes, 75% of the sample remains ([A]t = 0.75 M).
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16-4
ln
1.00 M
= (0.04094 min–1)(t)
0.75M
t=
16.6A
0.2877
= 7.02736 = 7.0 min
0.04094 min –1
Plan: The initial scene contains 12 particles of Substance X. In the second scene, which occurs after 2.5 minutes
have elapsed, half of the particles of Substance X (6 particles) remain. Therefore, 2.5 min is the half-life. The
half-life is used to find the number of particles present at 5.0 min and 10.0 min. To find the molarity of X, moles
of X is divided by the given volume.
Solution:
a) Since 2.5 minutes is the half-life, 5.0 minutes represents two half-lives:
2.5 min
2.5 min
⎯→ 6 particles of X ⎯⎯
⎯→ 3 particles of X
12 particles of X ⎯⎯
After 5.0 minutes, 3 particles of X remain; 9 particles of X have reacted to produce 9 particles of
Y. Draw a scene in which there are 3 black X particles and 9 red Y particles.
b) 10.0 minutes represents 4 half-lives:
2.5 min
2.5 min
2.5 min
⎯→ 6 particles of X ⎯⎯
⎯→ 3 particles of X ⎯⎯
⎯→
12 particles of X ⎯⎯
2.5 min
1.5 particles of X ⎯⎯ ⎯→ 0.75 particle of X
⎛ 0.20 mol ⎟⎞
⎟ = 0.15 mol
Moles of X after 10.0 min = (0.75 particle)⎜⎜
⎜⎝1 particle ⎟⎟⎠
mol X
0.15 mol
=
= 0.30 M
Molarity =
volume
0.50 L
16.6B
Plan: The initial scene contains 16 particles of Substance A. In the second scene, which occurs after 24 minutes have
elapsed, half of the particles of Substance A (8 particles) remain. Therefore, 24 min is the half-life. The half-life is used
to find the amount of time that has passed when only one particle of Substance A remains and to find the number of
particles of A present at 72 minutes. To find the molarity of A, moles of A is divided by the given volume.
Solution:
a) The half-life is 24 minutes. We can use that information to determine the amount of time that has passed when
one particle of Substance A remains:
24 min
24 min
24 min
24 min
→ 8 A particles ⎯⎯⎯⎯
→ → 4 A particles ⎯⎯⎯⎯
→ 2 A particles ⎯⎯⎯⎯
→ 1 A particle
16 A particles ⎯⎯⎯⎯
After 4 half-lives, only one particle of Substance A remains.
4 half-lives = 4 x (24 min/half-life) = 96 min
b)
72 min
Number of half-lives at 72 min =
= 3 half-lives
24 min/half-life
According to the scheme above in part a), there are 2 A particles left after three half-lives.
⎛ 0.10 mol A ⎟⎞
⎟⎟ = 0.20 mol A
Amount (mol) of A after 72 minutes = (2 particles A) ⎜⎜
⎝⎜1 particle A ⎟⎠
M=
0.20 mol A
= 0.80 M
0.25 L
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16-5
16.7A
Plan: Rearrange the first-order half-life equation to solve for k.
Solution:
ln 2
ln 2
=
k=
= 0.0529119985 = 0.0529 h–1
13.1
h
t1/2
16.7B
Plan: Rearrange the first-order half-life equation to solve for half-life. Determine the number of half-lives that
pass in the 40 day period described in the problem and use this number to determine the amount of pesticide
remaining.
Solution:
a)
ln2
ln2
t½ =
=
= 7.7016 = 8 days
k
9 x 10 –2 day –1
40 days
= 5 half-lives
b) Number of half-lives at 40 days =
8 days/half-life
After each half-life, ½ of the sample remains.
1
of the sample remains.
After five half-lives, ½ x ½ x ½ x ½ x ½ = (½)5 =
32
16.8A
Plan: The activation energy, rate constant at T1, and a second temperature, T2, are given. Substitute these values
into the Arrhenius equation and solve for k2, the rate constant at T2.
Solution:
k
E ⎛1 1⎞
ln 2 = a ⎜⎜⎜ − ⎟⎟⎟
k1
R ⎝ T1 T2 ⎠⎟
k1 = 0.286 L/mol • s
k2 = ? L/mol • s
T1 = 500. K
Ea = 1.00x102 kJ/mol
T2 = 490. K
3
1.00 x10 2 kJ /mol ⎛⎜ 1
k2
⎟⎟⎜⎜10 J ⎞⎟⎟
1 ⎞⎛
⎜⎜
=
ln
−
⎟⎟⎜
⎟
0.286 L /mol • s
8.314 J /mol • K ⎜⎜⎝ 500.K 490. K ⎟⎠⎝⎜⎜ 1 kJ ⎟⎟⎠
ln
k2
= –0.49093
0.286 L /mol • s
k2
= 0.612057
0.286 L/mol • s
k2 = (0.612057)(0.286 L/mol ∙ s) = 0.175048 = 0.175 L/mol∙s
16.8B
Plan: The activation energy and rate constant at T1 are given. We are asked to find the temperature at which the
rate will be twice as fast (i.e., the temperature at which k2 = 2 x k1). Substitute the given values into the Arrhenius
equation and solve for T2.
Solution:
k
E ⎛1 1⎞
ln 2 = a ⎜⎜ − ⎟⎟⎟
k1
R ⎜⎝ T1 T2 ⎟⎠
k1 = 7.02x10–3 L/mol ∙ s
T1 = 500. K
k2 = 2(7.02x10–3) L/mol ∙ s
T2 = unknown
2 (7.02x10 L/mol • s)
–3
ln
(7.02x10
–3
L/mol • s)
=
Ea = 1.14x105 J/mol
1.14x105 J/mol ⎛⎜ 1
1⎞
– ⎟⎟⎟
⎜
8.314 J/mol • K ⎜⎝ 500. K T2 ⎟⎠
⎛ 1
1⎞
5.055110x10–5 K–1 = ⎜⎜
– ⎟⎟
⎜⎝ 500. K T2 ⎟⎟⎠
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16-6
1
= 1. 9494489x10–3 K–1
T2
T2 = 513 K
16.9A
Plan: Begin by using Sample Problem 16.9 as a guide for labeling the diagram.
Solution:
The reaction energy diagram indicates that O(g) + H2O(g) → 2OH(g) is an endothermic process, because the
energy of the product is higher than the energy of the reactants. The highest point on the curve indicates the
transition state. In the transition state, an oxygen atom forms a bond with one of the hydrogen atoms on the H2O
molecule (hashed line) and the O-H bond (dashed line) in H2O weakens. Ea(fwd) is the sum of ΔHrxn and Ea(rev).
Transition state:
O
Energy
O
H
H
Ea(rev)= +6 kJ
Ea(fwd)= +78 kJ
ΔHrxn= +72 kJ
Reaction progress
16.9B
Plan: Ea(fwd) is the sum of ΔHrxn and Ea(rev), so ΔHrxn can be calculated by subtracting Ea(fwd) – Ea(rev). Use Sample
Problem 16.9 as a guide for drawing the diagram.
Solution:
ΔHrxn = Ea(fwd) – Ea(rev)
ΔHrxn = 7 kJ – 72 kJ = –65 kJ
The negative sign of the enthalpy indicates that the reaction is exothermic. This means that the energy of the
products is lower than the energy of the reactants. The highest point on the curve indicates the transition state. In
the transition state, a bond is forming between the chlorine and the hydrogen (hashed line) while the bond
between the hydrogen and the bromine weakens (another hashed line):
Transition state:
Cl
H
Br
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16-7
16.10A Plan: The overall reaction can be obtained by adding the three steps together. The molecularity of each step is the
total number of reactant particles; the molecularities are used as the orders in the rate law for each step.
Solution:
a)
(1) H2O2(aq) → 2OH(aq)
(2) H2O2(aq) + OH(aq) → H2O(l) + HO2(aq)
(3) HO2(aq) + OH(aq) → H2O(l) + O2(g)
Total: 2H2O2(aq) + 2OH(aq) + HO2(aq) → 2OH(aq) + 2H2O(l) + HO2(aq) + O2(g)
(overall) 2H2O2(aq) → 2H2O(l) + O2(g)
b)
(1) Unimolecular
(2) Bimolecular
(3) Bimolecular
c)
(1) Rate1 = k1 [H2O2]
(2) Rate2 = k2 [H2O2][OH]
(3) Rate3 = k3 [HO2][OH]
16.10B Plan: The overall reaction can be obtained by adding the three steps together. The molecularity of each step is the
total number of reactant particles; the molecularities are used as the orders in the rate law for each step.
Solution:
a)
(1) 2NO(g) ⇆ N2O2(g)
(2) N2O2(g) + H2(g) → N2O(g) + H2O(g)
(3) N2O(g) +H2(g) → N2(g) + H2O(g)
Total: 2NO(g) + N2O2(g) + 2 H2(g) + N2O(g) → N2O2(g) + N2O(g) + 2H2O(g) + N2(g)
(overall) 2NO(g) + 2H2(g) → 2H2O(g) + N2(g)
b)
(1) Bimolecular
(2) Bimolecular
(3) Bimolecular
c)
(1) Rate1 = k1 [NO]2
(2) Rate2 = k2 [N2O2][H2]
(3) Rate3 = k3 [N2O][H2]
16.11A Plan: The overall reaction can be obtained by adding the three steps together. An intermediate is a substance that
is formed in one step and consumed in a subsequent step. The overall rate law for the mechanism is determined
from the slowest step (the rate-determining step) and can be compared to the experimental rate law.
Solution:
a)
(1) H2O2(aq) → 2OH(aq)
(2) H2O2(aq) + OH(aq) → H2O(l) + HO2(aq)
(3) HO2(aq) + OH(aq) → H2O(l) + O2(g)
Total: 2H2O2(aq) + 2OH(aq) + HO2(aq) → 2OH(aq) + 2H2O(l) + HO2(aq) + O2(g)
(overall) 2H2O2(aq) → 2H2O(l) + O2(g)
2OH(aq) and HO2(aq) are intermediates in the given mechanism. 2OH(aq) are produced in the first step and
consumed in the second and third steps; HO2(aq) is produced in the second step and consumed in the third step.
Notice that the intermediates were not included in the overall reaction.
b) The observed rate law is: rate = k[H2O2]. In order for the mechanism to be consistent with the rate law, the
first step must be the slow step. The rate law for step one is the same as the observed rate law.
16.11B Plan: The overall reaction can be obtained by adding the three steps together. An intermediate is a substance that
is formed in one step and consumed in a subsequent step. The overall rate law for the mechanism is determined
from the slowest step (the rate-determining step) and can be compared to the experimental rate law.
Solution:
a)
(1) 2NO(g) ⇆ N2O2(g)
(2) N2O2(g) + H2(g) → N2O(g) + H2O(g)
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16-8
(3) N2O(g) +H2(g) → N2(g) + H2O(g)
Total: 2NO(g) + N2O2(g) + 2 H2(g) + N2O(g) → N2O2(g) + N2O(g) + 2H2O(g) + N2(g)
(overall) 2NO(g) + 2 H2(g) → 2H2O(g) + N2(g)
N2O2(g) and N2O(g) are intermediates in the given mechanism. N2O2(g) is produced in the first step and consumed
in the second step; N2O(g) is produced in the second step and consumed in the third step. Notice that the
intermediates were not included in the overall reaction.
b) The observed rate law is: rate = k[NO]2[H2], and the second step is the slow, or rate-determining, step. The rate
law for step two is: rate = k2[N2O2][H2]. This rate law is NOT the same as the observed rate law. However, since
N2O2 is an intermediate, it must be replaced by using the first step. For an equilibrium, rateforward rxn = ratereverse rxn. For
step 1 then, k1[NO]2 = k–1[N2O2]. Rearranging to solve for [N2O2] gives [N2O2] = (k1/k–1)[NO]2. Substituting this
value for [N2O2] into the rate law for the second step gives the overall rate law as rate = (k2k1/k–1)[NO]2[H2] or
rate = k[NO]2[H2], which is consistent with the observed rate law.
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B16.1
Plan: Add the two equations, canceling substances that appear on both sides of the arrow. The
rate law for each step follows the format of rate = k[reactants]. An initial reactant that appears as a
product in a subsequent step is a catalyst; a product that appears as a reactant in a subsequent step
is an intermediate (produced in one step and consumed in a subsequent step).
Solution:
a) Add the two equations:
(1)
X(g) + O3(g) → XO(g) + O2(g)
(2)
XO(g) + O(g) → X(g) + O2(g)
Overall O3(g) + O(g) → 2O2(g)
Rate laws:
Step 1 Rate1 = k1[X][O3]
Step 2 Rate2 = k2[XO][O]
b) X acts as a catalyst; it is a reactant in step 1 and a product in step 2. XO acts as an intermediate; it was
produced in step 1 and consumed in step 2.
B16.2
Plan: Replace X in the mechanism in B16.1 with NO, the catalyst. To find the rate of ozone
depletion at the given concentrations, use step 1) since it is the rate-determining (slow) step.
Solution:
a) (1) NO(g) + O3(g) → NO2(g) + O2(g)
(2) NO2(g) + O(g) → NO(g) + O2(g)
Overall O3(g) + O(g) → 2O2(g)
b) Rate1 = k1[NO][O3]
= (6x10–15 cm3/molecule ∙ s)[1.0x109 molecule/cm3][5x1012 molecule/cm3]
= 3x107 molecule/s
B16.3
Plan: The p factor is the orientation probability factor and is related to the structural complexity of
the colliding particles. The more complex the particles, the smaller the probability that collisions
will occur with the correct orientation and the smaller the p factor.
Solution:
a) Step 1 with Cl will have the higher value for the p factor. Since Cl is a single atom, no matter
how it collides with the ozone molecule, the two particles should react, if the collision has enough
energy. NO is a molecule. If the O3 molecule collides with the N atom in the NO molecule,
reaction can occur as a bond can form between N and O; if the O3 molecule collides with the O atom in the NO
molecule, reaction will not occur as the bond between N and O cannot form. The probability of a successful
collision is smaller with NO.
b) The transition state would have weak bonds between the chlorine atom and an oxygen atom in ozone, and
between that oxygen atom and a second oxygen atom in ozone.
`
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16-9
O
Cl
O
O
END–OF–CHAPTER PROBLEMS
16.1
Changes in concentrations of reactants (or products) as functions of time are measured to determine the reaction rate.
16.2
Rate is proportional to concentration. An increase in pressure will increase the number of gas molecules per
unit volume. In other words, the gas concentration increases due to increased pressure, so the reaction rate
increases. Increased pressure also causes more collisions between gas molecules.
16.3
The addition of more water will dilute the concentrations of all solutes dissolved in the reaction vessel. If any of
these solutes are reactants, the rate of the reaction will decrease.
16.4
An increase in solid surface area would allow more gaseous components to react per unit time and thus would
increase the reaction rate.
16.5
An increase in temperature affects the rate of a reaction by increasing the number of collisions, but more
importantly, the energy of collisions increases. As the energy of collisions increases, more collisions result in
reaction (i.e., reactants → products), so the rate of reaction increases.
16.6
The second experiment proceeds at the higher rate. I2 in the gaseous state would experience more collisions with
gaseous H2.
16.7
The reaction rate is the change in the concentration of reactants or products per unit time. Reaction rates change
with time because reactant concentrations decrease, while product concentrations increase with time.
16.8
a) For most reactions, the rate of the reaction changes as a reaction progresses. The instantaneous rate is the rate
at one point, or instant, during the reaction. The average rate is the average of the instantaneous rates over a period
of time. On a graph of reactant concentration vs. time of reaction, the instantaneous rate is the slope of the
tangent to the curve at any one point. The average rate is the slope of the line connecting two points on the curve.
The closer together the two points (shorter the time interval), the more closely the average rate agrees with the
instantaneous rate.
b) The initial rate is the instantaneous rate at the point on the graph where time = 0, that is when reactants are mixed.
16.9
The calculation of the overall rate is the difference between the forward and reverse rates. This complication may
be avoided by measuring the initial rate, where product concentrations are negligible, so the reverse rate is
negligible. Additionally, the calculations are simplified as the reactant concentrations can easily be determined
from the volumes and concentrations of the solutions mixed.
16.10
At time t = 0, no product has formed, so the B(g) curve must start at the origin. Reactant concentration (A(g))
decreases with time; product concentration (B(g)) increases with time. Many correct graphs can be drawn. Two
examples are shown below. The graph on the left shows a reaction that proceeds nearly to completion, i.e.,
[products] >> [reactants] at the end of the reaction. The graph on the right shows a reaction that does not proceed
to completion, i.e., [reactants] > [products] at reaction end.
A(g)
Time
Concentration
Concentration
B(g)
A(g)
B(g)
Time
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16-10
16.11
a) Calculate the slope of the line connecting (0, [C]o) and (tf, [C]f) (final time and concentration). The negative of
this slope is the average rate.
b) Calculate the negative of the slope of the line tangent to the curve at t = x.
c) Calculate the negative of the slope of the line tangent to the curve at t = 0.
d) If you plotted [D] vs. time, you would not need to take the negative of the slopes in a)-c) since [D]
would increase over time.
16.12
Plan: The average rate is the total change in concentration divided by the total change in time.
Solution:
a) The average rate from t = 0 to t = 20.0 s is proportional to the slope of the line connecting these two points:
1 (0.0088 mol/L − 0.0500 mol/L )
1 Δ[AX 2 ]
=−
= 0.00103 = 0.0010 mol/L • s
Rate = −
2 Δt
(20.0 s − 0 s)
2
The negative of the slope is used because rate is defined as the change in product concentration with time. If a
reactant is used, the rate is the negative of the change in reactant concentration. The 1/2 factor is included to
account for the stoichiometric coefficient of 2 for AX2 in the reaction.
b)
[AX2] vs time
0.06
0.05
[AX2]
0.04
0.03
0.02
0.01
0
0
5
10
15
20
25
time, s
The slope of the tangent to the curve (dashed line) at t = 0 is approximately –0.004 mol/L∙s. This initial rate is
greater than the average rate as calculated in part a). The initial rate is greater than the average rate because
rate decreases as reactant concentration decreases.
16.13
Plan: The average rate is the total change in concentration divided by the total change in time.
Solution:
a) Rate = −
1 (0.0088 mol/L − 0.0249 mol/L)
1 Δ[AX 2 ]
=−
= 6.70833x10–4 = 6.71x10–4 mol/L • s
2
2 Δt
(20.0 s − 8.0 s)
b) The rate at exactly 5.0 s will be higher than the rate in part a).
The slope of the tangent to the curve at t = 5.0 s (the rate at 5.0 s) is approximately –2.8x10–3 mol/L • s.
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16-11
0.06
0.05
[AX2]
0.04
0.03
0.02
0.01
0
0
5
10
15
20
25
time, s
16.14
Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance. A negative sign is used for the rate in terms of reactant A since A is reacting and [A] is decreasing
over time. Positive signs are used for the rate in terms of products B and C since B and C are being formed and
[B] and [C] increase over time. Reactant A decreases twice as fast as product C increases because two molecules
of A disappear for every molecule of C that appears.
Solution:
Expressing the rate in terms of each component:
1 Δ[A]
Δ[B]
Δ[C]
Rate = −
=
=
2 Δt
Δt
Δt
Calculating the rate of change of [A]:
1 Δ[A]
Δ[C]
−
=
2 Δt
Δt
⎛ 2 mol A/L • s ⎞⎟
= –4 mol/L ∙ s
(2 mol C/L • s)⎜⎜
⎝ 1 mol C/L • s ⎠⎟⎟
The negative value indicates that [A] is decreasing as the reaction progresses. The rate of reaction is always
expressed as a positive number, so [A] is decreasing at a rate of 4 mol/L • s.
16.15
Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance. A negative sign is used for the rate in terms of reactant D since D is reacting and [D] is decreasing
over time. Positive signs are used for the rate in terms of products E and F since E and F are being formed and [E]
and [F] increase over time. For every 3/2 mole of product E that is formed, 5/2 mole of F is produced.
Solution:
Expressing the rate in terms of each component:
Δ[D]
2 Δ[E]
2 Δ[F]
=
=
Rate = −
Δt
3 Δt
5 Δt
Calculating the rate of change of [F]:
⎛ 5/2 mol F/L • s ⎞⎟
(0.25 mol E/L • s)⎜⎜
⎟ = 0.416667 = 0.42 mol/L • s
⎝ 3/2 mol E/L • s ⎠⎟
16.16
Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance. A negative sign is used for the rate in terms of reactants A and B since A and B are reacting and [A]
and [B] are decreasing over time. A positive sign is used for the rate in terms of product C since C is being
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16-12
formed and [C] increases over time. The 1/2 factor is included for reactant B to account for the stoichiometric
coefficient of 2 for B in the reaction. Reactant A decreases half as fast as reactant B decreases because one
molecule of A disappears for every two molecules of B that disappear.
Solution:
Expressing the rate in terms of each component:
Δ[A]
1 Δ[B]
Δ[C]
=−
=
Rate = −
Δt
2 Δt
Δt
Calculating the rate of change of [A]:
⎛1 mol A/L • s ⎞⎟
= – 0.25 mol/L • s = – 0.2 mol/L • s
(0.5 mol B/L • s)⎜⎜
⎝ 2 mol B/L • s ⎟⎟⎠
The negative value indicates that [A] is decreasing as the reaction progresses. The rate of reaction is always
expressed as a positive number, so [A] is decreasing at a rate of 0.2 mol/L • s.
16.17
Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance. A negative sign is used for the rate in terms of reactants D, E, and F since these substances are
reacting and [D], [E], and [F] are decreasing over time. Positive signs are used for the rate in terms of products G
and H since these substances are being formed and [G] and [H] increase over time. Product H increases half as
fast as reactant D decreases because one molecule of H is formed for every two molecules of D that disappear.
Solution:
Expressing the rate in terms of each component:
1 Δ[D]
1 Δ[E]
Δ[F]
1 Δ[G]
Δ[H]
Rate = −
=−
=−
=
=
2 Δt
3 Δt
Δt
2 Δt
Δt
Calculating the rate of change of [H]:
⎛ 1 mol H/L • s ⎞⎟
(0.1 mol D/L • s)⎜⎜
= 0.05 mol/L • s
⎝ 2 mol D/L • s ⎟⎠⎟
16.18
Plan: A term with a negative sign is a reactant; a term with a positive sign is a product. The inverse of the fraction
becomes the coefficient of the molecule.
Solution:
N2O5 is the reactant; NO2 and O2 are products.
2N2O5(g) → 4NO2(g) + O2(g)
16.19
Plan: A term with a negative sign is a reactant; a term with a positive sign is a product. The inverse of the fraction
becomes the coefficient of the molecule.
Solution:
CH4 and O2 are the reactants; H2O and CO2 are products.
CH4 + 2O2→ 2H2O + CO2
16.20
Plan: The average rate is the total change in concentration divided by the total change in time. The initial rate is
the slope of the tangent to the curve at t = 0.0 s and the rate at 7.00 s is the slope of the tangent to the curve at
t = 7.00 s
Solution:
Δ[NOBr] 0.0033 − 0.0100 mol/L
–
= 6.7x10–4 mol/L • s
Δt
10.00 − 0.00 s
0.0055 − 0.0071 mol/L
= 8.0x10–4 mol/L • s
b) Rate = –
4.00 − 2.00 s
a) Rate = −
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16-13
c))
R = – Δ y/Δ
Δ x = – [(0.00440 – 0.0100) mol/L]/[4.00
m
– 0.00)
0
s] = 1.5xx10–3 mol/L∙s
Initial Rate
–4
d)) Rate at 7.00 s = – [(0.0030 – 0.0050) mol//L]/[11.00 – 4..00) s] = 2.8577x10 = 2.9x100–4 mol/L∙s
e)) Average betw
ween t = 3 s and t = 5 s is:
Rate = – [(0.0050 – 0.0063)
0
mol/L]]/[5.00 – 3.00) s] = 6.5x10–4 mol/L∙s
m
–4
Rate att 4 s ≈ 6.7x10 mol/L∙s, thuss the rates are equal
e
at about 4 seconds.
16.21
Pllan: Use Equattion 16.2 to desscribe the rate of this reactionn in terms of reeactant disappeearance and prooduct
apppearance. A negative
n
sign iss used for the raate in terms off reactants N2 and H2 since theese substances are reacting
annd [N2] and [H2] are decreasinng over time. A positive sign is used for thee rate in terms of
o the product NH3 since it
is being formed and [NH3] incrreases over tim
me.
Soolution:
Δ
Δ[N 2 ]
1 Δ[H 2 ] 1 Δ[NH
3]
Rate = −
=−
Δt
3 Δt
2 Δt
16.22
Pllan: Use Equattion 16.2 to desscribe the rate of this reactionn in terms of reeactant disappeearance and prooduct
apppearance. A negative
n
sign iss used for the raate in terms off the reactant O2 since it is reaacting and [O2] is
deecreasing over time. A positivve sign is usedd for the rate inn terms of the product
p
O3 since it is being formed and
[O
O3] increases ov
ver time. O3 inncreases 2/3 as fast as O2 decrreases because two moleculess of O3 are form
med for
evvery three moleecules of O2 thhat disappear.
Soolution:
O3 ]
Δ 2]
1 Δ[O
1 Δ[O
a)) Rate = −
=
Δ
3 Δt
2 Δt
b)) Use the molee ratio in the baalanced equatioon:
−5
⎛ 2.17x10
m O 3 /L • s ⎞⎟
moll O 2 /L • s ⎞⎟⎜⎛ 2 mol
⎜⎜ 2
⎟ = 1.446667x10
0–5 = 1.45x10–55 mol/L∙s
⎟⎟⎜
⎜⎝
⎟⎜ 3 mol
m O 2 /L • s ⎟⎟⎠
⎠⎝
16.23
a)) k is the rate co
onstant, the prooportionality constant
c
in the rate
r law. k reprresents the fracction of successsful
coollisions which
h includes the fraction
f
of collisions with suffficient energy and the fractioon of collisionss with
coorrect orientation. k is a consttant that variess with temperatture.
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ducation. This is proprietary mateerial solely for au
uthorized instructtor use. Not authoorized for sale or distribution in
any manner.. This document may
m not be copied
d, scanned, dupliccated, forwarded
d, distributed, or posted
p
on a websiite, in whole or paart.
16-14
b) m represents the order of the reaction with respect to [A] and n represents the order of the reaction with respect
to [B]. The order is the exponent in the relationship between rate and reactant concentration and defines how
reactant concentration influences rate.
The order of a reactant does not necessarily equal its stoichiometric coefficient in the balanced equation. If a
reaction is an elementary reaction, meaning the reaction occurs in only one step, then the orders and
stoichiometric coefficients are equal. However, if a reaction occurs in a series of elementary reactions, called a
mechanism, then the rate law is based on the slowest elementary reaction in the mechanism. The orders of the
reactants will equal the stoichiometric coefficients of the reactants in the slowest elementary reaction but may not
equal the stoichiometric coefficients in the overall reaction.
c) For the rate law rate = k[A] [B]2 substitute in the units:
Rate (mol/L·min) = k[A]1[B]2
mol/L • min
rate
mol/L • min
=
1
2 =
1
2
mol3
[A] [B]
⎡ mol ⎤ ⎡ mol ⎤
⎢
⎥ ⎢
⎥
⎢⎣ L ⎥⎦ ⎢⎣ L ⎥⎦
L3
mol ⎛⎜ L3 ⎞⎟
k=
⎟
⎜
L • min ⎜⎝ mol3 ⎟⎟⎠
k=
k = L2/mol2∙ min
16.24
a) Plot either [A2] or [B2] vs. time and determine the negative of the slope of the line tangent to the curve at t = 0.
b) A series of experiments at constant temperature but with different initial concentrations are run to determine
different initial rates. By comparing results in which only the initial concentration of a single reactant is changed,
the order of the reaction with respect to that reactant can be determined.
c) When the order of each reactant is known, any one experimental set of data (reactant concentration and reaction
rate) can be used to determine the reaction rate constant at that temperature.
16.25
a) The rate doubles. If rate = k[A]1 and [A] is doubled, then the rate law becomes rate = k[2 x A]1. The rate
increases by 21 or 2.
b) The rate decreases by a factor of four. If rate = k[B]2 and [B] is halved, then the rate law becomes
rate = k[1/2 x B]2. The rate decreases to (1/2)2 or 1/4 of its original value.
c) The rate increases by a factor of nine. If rate = k[C]2 and [C] is tripled, then the rate law becomes
rate = k[3 x C]2. The rate increases to 32 or 9 times its original value.
16.26
Plan: The order for each reactant is the exponent on the reactant concentration in the rate law. The individual
orders are added to find the overall reaction order.
Solution:
The orders with respect to [BrO3–] and to [Br–] are both 1 since both have an exponent of 1. The order with respect
to [H+] is 2 (its exponent in the rate law is 2). The overall reaction order is 1 + 1 + 2 = 4.
first order with respect to BrO3–, first order with respect to Br– , second order with respect to H+,
fourth order overall
16.27
Plan: The order for each reactant is the exponent on the reactant concentration in the rate law. The individual
orders are added to find the overall reaction order.
Solution:
The rate law may be rewritten as rate = k[O3]2[O2] –1. The order with respect to [O3] is 2 since it has an
exponent of 2. The order with respect to [O2] is –1 since it has an exponent of –1. The overall reaction order
is 2 + (–1) = 1.
second order with respect to O3, (–1) order with respect to O2, first order overall
16.28
a) The rate is first order with respect to [BrO3–]. If [BrO3–] is doubled, rate = k[2 x BrO3–], then rate increases to 21
or 2 times its original value. The rate doubles.
b) The rate is first order with respect to [Br–]. If [Br–] is halved, rate = k[1/2 x Br–], then rate decreases by a factor
of (1/2)1 or 1/2 times its original value. The rate is halved.
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16-15
c) The rate is second order with respect to [H+]. If [H+] is quadrupled, rate = k[4 x H+]2, then rate increases to 42 or
16 times its original value.
16.29
a) The rate is second order with respect to [O3]. If [O3] is doubled, rate = k[2 x O3]2, then rate increases to 22 or 4
times its original value. The rate increases by a factor of 4.
b) [O2] has an order of –1. If [O2] is doubled, rate = k[2 x O2] –1, then rate decreases to 2–1 or 1/2 times its original
value. The rate decreases by a factor of 2.
c) [O2] has an order of –1. If [O2] is halved, rate = k[1/2 x O2] –1, then rate decreases by a factor of (1/2)–1 or 2 times
its original value. The rate increases by a factor of 2.
16.30
Plan: The order for each reactant is the exponent on the reactant concentration in the rate law. The individual
orders are added to find the overall reaction order.
Solution:
The order with respect to [NO2] is 2, and the order with respect to [Cl2] is 1. The overall order is: 2 + 1 = 3 for
the overall order.
16.31
Plan: The order for each reactant is the exponent on the reactant concentration in the rate law. The individual
orders are added to find the overall reaction order.
Solution:
The rate law may be rewritten as rate = k[HNO2]4[NO] –2.
The order with respect to [HNO2] is 4, and the order with respect to [NO] is –2. The overall order is:
4 + (– 2) = 2 for the overall order.
16.32
a) The rate is second order with respect to [NO2]. If [NO2] is tripled, rate = k[3 x NO2]2, then rate increases to 32 or
9 times its original value. The rate increases by a factor of 9.
b) The rate is second order with respect to [NO2] and first order with respect to [Cl2]. If [NO2] and [Cl2] are
doubled, rate = k[2 x NO2]2[2 x Cl2]1, then the rate increases by a factor of 22 x 21 = 8.
c) The rate is first order with respect to [Cl2]. If Cl2 is halved, rate = k[1/2 x Cl2]1, then rate decreases to 1/2 times
its original value. The rate is halved.
16.33
a) The rate is fourth order with respect to [HNO2]. If [HNO2] is doubled, rate = k[2 x HNO2]4, then rate increases
to 24 or 16 times its original value. The rate increases by a factor of 16.
b) [NO] has an order of –2. If [NO] is doubled, rate = k[2 x NO]–2, then rate increases to 2–2 or 1/(2)2 = 1/4 times
its original value. The rate decreases by a factor of 4.
c) The rate is fourth order with respect to [HNO2]. If [HNO2] is halved, rate = k[1/2 x HNO2]4, then rate decreases
to (1/2)4 or 1/16 times its original value. The rate decreases by a factor of 16.
16.34
Plan: The rate law is rate = [A]m[B]n where m and n are the orders of the reactants. To find the order of each
reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once
the rate law is known, any experiment can be used to find the rate constant k.
Solution:
a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B]
is constant. Use experiments 1 and 2 (or 3 and 4 would work) to find the order with respect to [A].
Set up a ratio of the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and
solve for m, the order with respect to [A].
m
⎛[A]exp 2 ⎞⎟
rate exp 2
⎜
⎟
⎜
=⎜
⎟
⎜⎝ [A] ⎠⎟⎟
rate
exp 1
exp 1
m
45.0 mol/L • min
⎛ 0.300 mol/L ⎞⎟
= ⎜⎜
⎝ 0.100 mol/L ⎟⎠⎟
5.00 mol/L • min
m
9.00 = (3.00)
log (9.00) = m log (3.00)
m=2
Using experiments 3 and 4 also gives second order with respect to [A].
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16-16
To find the order for reactant B, first identify the reaction experiments in which [B] changes but [A]
is constant. Use experiments 1 and 3 (or 2 and 4 would work) to find the order with respect to [B].
Set up a ratio of the rate laws for experiments 1 and 3 and fill in the values given for rates and concentrations and
solve for n, the order with respect to [B].
n
⎛[B]exp 3 ⎞⎟
rate exp 3
⎜
⎟
⎜
=⎜
⎟
⎜⎝ [B] ⎟⎟⎠
rate
exp 1
exp 1
n
10.0 mol/L • min
⎛ 0.200 mol/L ⎞⎟
= ⎜⎜
⎝ 0.100 mol/L ⎟⎠⎟
5.00 mol/L • min
n
2.00 = (2.00)
log (2.00) = n log (2.00)
n=1
The reaction is first order with respect to [B].
b) The rate law, without a value for k, is rate = k[A]2[B].
c) Using experiment 1 to calculate k (the data from any of the experiments can be used):
Rate = k[A]2[B]
rate
5.00 mol/L • min
k=
=
= 5.00x103 L2/mol2 •min
2
[0.100 mol/L] 2 [0.100 mol/L]
[A] [B]
16.35
Plan: The rate law is rate = k [A]m[B]n[C]p where m, n, and p are the orders of the reactants. To find the order of
each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes.
Once the rate law is known, any experiment can be used to find the rate constant k.
Solution:
a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B] and [C]
are constant. Use experiments 1 and 2 to find the order with respect to [A]. Set up a ratio of the rate laws for
experiments 1 and 2 and fill in the values given for rates and concentrations and solve for m, the order with
respect to [A].
m
⎛[A]exp 2 ⎞⎟
rate exp 2
⎜
⎟⎟
= ⎜⎜
⎜⎝ [A]exp 1 ⎠⎟⎟
rate exp 1
m
1.25x10−2 mol/L • min
⎛ 0.1000 mol/L ⎞⎟
= ⎜⎜
⎝ 0.0500 mol/L ⎟⎟⎠
6.25x10−3 mol/L • min
m
2.00 = (2.00)
log (2.00) = m log (2.00)
m=1
The order is first order with respect to A.
To find the order for reactant B, first identify the reaction experiments in which [B] changes but [A] and [C]
are constant. Use experiments 2 and 3 to find the order with respect to [B]. Set up a ratio of the rate laws for
experiments 2 and 3 and fill in the values given for rates and concentrations and solve for n, the order with respect
to [B].
n
⎛ [B]exp 3 ⎞⎟
rate exp 3
⎜
⎟
= ⎜⎜
⎟
rate
⎝⎜[B] ⎟⎠⎟
exp 2
exp 2
n
5.00 x 10−2 mol/L • min
⎛ 0.1000 mol/L ⎞⎟
= ⎜⎜
−2
⎝ 0.0500 mol/L ⎠⎟⎟
1.25 x 10 mol/L • min
4.00 = (2.00)n
log (4.00) = n log (2.00)
n=2
The reaction is second order with respect to B.
To find the order for reactant C, first identify the reaction experiments in which [C] changes but [A] and [B]
are constant. Use experiments 1 and 4 to find the order with respect to [C]. Set up a ratio of the rate laws for
experiments 1 and 4 and fill in the values given for rates and concentrations and solve for p, the order with respect
to [C].
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16-17
rate exp 4
rate exp 1
⎛[C]exp 4 ⎟⎞
⎟⎟
= ⎜⎜⎜
⎜⎝ [C]exp 1 ⎟⎟⎠
p
6.25x10−3 mol/L • min
⎛ 0.0200 mol/L ⎞⎟
= ⎜⎜
−3
⎝ 0.0100 mol/L ⎟⎠⎟
6.25x10 mol/L • min
1.00 = (2.00)p
p
log (1.00) = p log (2.00)
p=0
The reaction is zero order with respect to C.
b) Rate = k[A] [B] [C]
Rate = k[A][B]2
1
2
0
c) Using the data from experiment 1 to find k:
Rate = k[A][B]2
k=
16.36
6.25 x 10−3 mol/L • min
rate
= 50.0 L2/mol2 • s
=
2
2
[0.0500 mol/L][0.0500 mol/L]
[A][B]
Plan: Write the appropriate rate law and enter the units for rate and concentrations to find the units of k. The units
of k are dependent on the reaction orders and the unit of time.
Solution:
a) A first-order rate law follows the general expression, rate = k[A]. The reaction rate is expressed as a change in
concentration per unit time with units of mol/L∙time. Since [A] has units of mol/L, k has units of time–1:
Rate = k[A]
mol
mol
=k
L • time
L
mol
mol
L
1
x
k = L • time =
=
= time–1
mol
L • time mol
time
L
b) A second-order rate law follows the general expression, rate = k[A]2. The reaction rate is expressed as a change
in concentration per unit time with units of mol/L·time. Since [A] has units of mol2/L2, k has units of L/mol∙time:
Rate = k[A]2
2
mol
⎛ mol ⎞⎟
= k ⎜⎜
⎟
⎟
⎝ L ⎠
L • time
mol
L
mol
L2
k = L • time
=
=
x
2
2
mol
mol • time
L • time mol
L2
c) A third-order rate law follows the general expression, rate = k[A]3. The reaction rate is expressed as a change in
concentration per unit time with units of mol/L∙time. Since [A] has units of mol3/L3, k has units of L2/mol2∙time:
Rate = k[A]3
mol
⎛ mol ⎞⎟
= k ⎜⎜
⎝ L ⎟⎟⎠
L • time
mol
mol
L3
L2
k = L • time
=
=
x
3
3
2
mol
L • time mol
mol • time
L3
d) A 5/2-order rate law follows the general expression, rate = k[A]5/2. The reaction rate is expressed as a change
in concentration per unit time with units of mol/L·time. Since [A] has units of mol5/2/L5/2, k has units of
L3/2/mol3/2· time:
3
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16-18
5/2
mol
⎛ mol ⎞⎟
= k ⎜⎜
⎝ L ⎟⎟⎠
L • time
mol
mol
L5/2
L3/2
k = L • time
=
=
x
5/2
mol
L • time mol 5/2
mol 3/2 • time
5/2
L
16.37
Plan: Write the appropriate rate law and enter the units for rate and the rate constant to find the units of
concentration. The units of concentration will give the reaction order.
Solution:
a) Rate = k[A]m
mol
mol ⎡ mol ⎤
=
⎢
⎥
L•s
L • s ⎢⎣ L ⎥⎦
mol
m
⎡ mol ⎤
⎢
⎥ = L•s
mol
⎣⎢ L ⎦⎥
L•s
m
⎡ mol ⎤
⎢
⎥ =1
⎢⎣ L ⎥⎦
b) Rate = k[A]m
m
m must be 0. The reaction is zero order.
m
mol
1 ⎡ mol ⎤
=
⎢
⎥
L • yr
yr ⎢⎣ L ⎥⎦
mol
m
⎡ mol ⎤
mol
yr
L • yr
x
=
⎢
⎥ =
1
L • yr
1
⎣⎢ L ⎦⎥
yr
⎡ mol ⎤
⎢
⎥
⎣⎢ L ⎦⎥
m
=
mol
L
m must be 1. The reaction is first order.
c) Rate = k[A]m
m
mol
mol1/ 2 ⎡ mol ⎤
= 1/ 2 ⎢
⎥
L•s
L • s ⎣⎢ L ⎦⎥
mol
m
mol
L1/ 2 • s
⎡ mol ⎤
•s
L
=
x
⎢
⎥ =
1/ 2
⎢⎣ L ⎥⎦
mol
L•s
mol1/2
L1/ 2 • s
m
⎡ mol ⎤
mol1/2
m must be 1/2. The reaction is 1/2 order.
⎢
⎥ =
⎢⎣ L ⎥⎦
L1/2
d) Rate = k[A]m
mol
mol−5 / 2 ⎡ mol ⎤
= −5 / 2
⎢
⎥
L • min
L
• min ⎣⎢ L ⎦⎥
m
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16-19
16.38
⎡ mol ⎤
⎢
⎥
⎣⎢ L ⎦⎥
m
⎡ mol ⎤
⎢
⎥
⎢⎣ L ⎥⎦
m
mol
mol
L−5 / 2 • min
• min
L
=
x
=
−5 / 2
mol
L • min
mol−5/2
−5 / 2
• min
L
mol 7/2
m must be 7/2. The reaction is 7/2 order.
=
L7/2
Plan: The rate law is rate = k [CO]m[Cl2]n where m and n are the orders of the reactants. To find the order of each
reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once
the rate law is known, the data in each experiment can be used to find the rate constant k.
Solution:
a) To find the order for CO, first identify the reaction experiments in which [CO] changes but [Cl2] is constant.
Use experiments 1 and 2 to find the order with respect to [CO]. Set up a ratio of the rate laws for experiments 1
and 2 and fill in the values given for rates and concentrations and solve for m, the order with respect to [CO].
m
⎛ [CO]exp 1 ⎞⎟
rate exp 1
⎜
⎟⎟
= ⎜⎜
⎜⎝[CO]exp 2 ⎟⎟⎠
rateexp 2
n
1.29x10−29 mol/L • min
⎛ 1.00 mol/L ⎞⎟
= ⎜⎜
⎝ 0.100 mol/L ⎟⎟⎠
1.33x10−30 mol/L • min
m
9.699 = (10.0)
log (9.699) = m log (10.0)
m = 0.9867 = 1
The reaction is first order with respect to [CO].
To find the order for Cl2, first identify the reaction experiments in which [Cl2] changes but [CO] is constant. Use
experiments 2 and 3 to find the order with respect to [Cl2]. Set up a ratio of the rate laws for experiments 2 and 3
and fill in the values given for rates and concentrations and solve for n, the order with respect to [Cl2].
n
⎛ [Cl2 ]exp 3 ⎞⎟
rate exp 3
⎜
= ⎜⎜
⎟⎟
⎜⎝[Cl2 ]exp 2 ⎟⎟⎠
rate exp 2
n
1.30x10−29 mol/L • min
⎛ 1.00 mol/L ⎞⎟
= ⎜⎜
−30
⎝ 0.100 mol/L ⎟⎟⎠
1.33x10 mol/L • min
9.774 = (10.0)n
log (9.774) = n log (10.0)
n = 0.9901 = 1
The reaction is first order with respect to [Cl2].
Rate = k[CO][Cl2]
b) k = rate/[CO][Cl2]
Exp 1: k1 = (1.29x10–29 mol/L∙s)/[1.00 mol/L][0.100 mol/L] = 1.29x10–28 L/mol∙s
Exp 2: k2 = (1.33x10–30 mol/L∙s)/[0.100 mol/L][0.100 mol/L] = 1.33x10–28 L/mol∙s
Exp 3: k3 = (1.30x10–29 mol/L∙s)/[0.100 mol/L][1.00 mol/L] = 1.30x10–28 L/mol∙s
Exp 4: k4 = (1.32x10–31 mol/L∙s)/[0.100 mol/L][0.0100 mol/L] = 1.32x10–28 L/mol∙s
kavg = (1.29x10–28 + 1.33x10–28 + 1.30x10–28 + 1.32x10–28) L/mol∙s/4 = 1.31x10–28 L/mol∙s
16.39
The integrated rate law can be used to plot a graph. If the plot of [reactant] vs. time is linear, the order is zero. If
the plot of ln[reactant] vs. time is linear, the order is first. If the plot of inverse concentration (1/[reactant]) vs.
time is linear, the order is second.
a) The reaction is first order since ln[reactant] vs. time is linear.
b) The reaction is second order since 1/[reactant] vs. time is linear.
c) The reaction is zero order since [reactant] vs. time is linear.
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16-20
16.40
The half-life (t½) of a reaction is the time required to reach half the initial reactant concentration. For a first-order
process, no molecular collisions are necessary, and the rate depends only on the fraction of the molecules having
sufficient energy to initiate the reaction.
16.41
Plan: The rate expression indicates that this reaction is second order overall (the order of [AB] is 2), so use the
second-order integrated rate law to find time. We know k (0.2 L/mol∙s), [AB]0 (1.50 M), and
[AB]t (1/3[AB]0 = 1/3(1.50 M) = 0.500 M), so we can solve for t.
Solution:
1
1
−
= kt
[ AB]t [ AB]0
⎛ 1
1 ⎞⎟
⎟
−
⎜⎜⎜
⎜⎝[ AB]t [ AB]0 ⎟⎟⎠
t=
k
⎛
1
1 ⎞⎟⎟
⎜⎜
−
⎟
⎜⎜
⎜⎝ 0.500 M 1.50 M ⎟⎟⎠
t=
0.2 L/mol • s
t = 6.6667 = 7 s
16.42
Plan: The rate expression indicates that this reaction is second order overall (the order of [AB] is 2), so use the secondorder integrated rate law. We know k (0.2 L/mol∙s), [AB]0 (1.50 M), and t (10.0 s), so we can solve for [AB]t.
Solution:
1
1
−
= kt
[ AB]t [ AB]0
1
1
= kt +
[ AB]t
[ AB]0
1
1
= (0.2 L/mol ∙ s) (10.0 s) +
[ AB]t
1.50 M
1
1
= 2.66667
[ AB]t
M
[AB]t = 0.375 = 0.4 M
16.43
16.44
Plan: The rate constant is known; use the first-order half-life equation to calculate the half-life, t1/2.
Count the reactant particles in the two scenes to determine the fraction of particles remaining after the
elapsed unknown time. Determine the number of half-lives that have passed and multiply by the half-life
Solution:
ln 2
ln 2
t1/2 =
=
= 11 min
0.063 min−1
k
There are 20 A particles at t = 0 and 5 A particles at t = ?. Thus, the fraction of A remaining is ¼ of the
initial amount which is two half-lives (½ x ½). Two half-lives is 2 x 11 min = 22 min.
value.
Plan: The half-life is the time required for one-half of the reactant to be consumed. Count the reactant molecules
in the three scenes to determine the number of molecules remaining after each time period. This information is
used to determine the half-life. Once t1/2 is known, use the first-order half-life equation to calculate the rate
constant, k.
Solution:
a) There are twelve molecules of reactant at t = 0, eight reactant molecules after 20 min, and three reactant
molecules after 60 min. After 60 min, one-fourth (three of twelve molecules) of the initial amount of
cyclopropane remains unreacted. Therefore, 60 min represents two half-lives. The half-life is 30 min.
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16-21
b) t1/2 =
16.45
ln 2
k
k=
ln 2
ln 2
= 0.023104906 = 0.023 min–1
=
t1/2
30 min
Plan: This is a first-order reaction so use the first-order integrated rate law. In part a), we know t (10.5 min). Let
[A]0 = 1 M and then [A]t = 50% of 1 M = 0.5 M. Solve for k. In part b), use the value of k to find the time
necessary for 75.0% of the compound to react. If 75.0% of the compound has reacted, 100–75 = 25% remains at
time t. Let [A]0 = 1 M and then [A]t = 25% of 1 M = 0.25 M.
Solution:
a) ln [A]t = ln [A]0 – kt
ln [0.5] = ln [1] – k(10.5 min)
–0.693147 = 0 – k(10.5 min)
0.693147 = k(10.5 min)
k = 0.0660 min–1
Alternatively, 50.0% decomposition means that one half-life has passed. Thus, the first-order half-life equation may
be used:
ln 2
ln 2
ln 2
k=
t1/2 =
= 0.066014 = 0.0660 min–1
=
10.5 min
k
t1/2
b) ln [A]t = ln [A]0 – kt
ln[A]t − ln[A]0
=t
−k
ln[0.25] − ln[1]
=t
−0.066014 min−1
t = 21.0000 = 21.0 min
If you recognize that 75.0% decomposition means that two half-lives have passed, then
t = 2 (10.5 min) = 21.0 min.
16.46
Plan: This is a first-order reaction so use the first-order integrated rate law (the units of k, yr–1, indicates first
order). In part a), the first-order half-life equation may be used to solve for half-life since k is known. In part b),
use the value of k to find the time necessary for the reactant concentration to drop to 12.5% of the initial
concentration. Let [A]0 = 1.00 M and then [A]t = 12.5% of 1 M = 0.125 M.
Solution:
ln 2
ln 2
a) t1/2 =
=
= 577.62 = 5.8x102 yr
k
0.0012 yr−1
b) ln [A]t = ln [A]0 – kt
[A]0
ln
= kt
[A]t
[A]0 = 1.00 M
[A]t = 0.125 M k = 0.0012 yr–1
⎛ 1.00 M ⎞⎟
= (0.0012 yr–1) t
ln ⎜⎜
⎝ 0.125 M ⎟⎟⎠
t = 1732.86795 = 1.7x103 yr
If the student recognizes that 12.5% remaining corresponds to three half-lives; then simply multiply the answer in
part a) by three.
16.47
Plan: In a first-order reaction, ln [NH3] vs. time is a straight line with slope equal to k. The half-life can be
determined using the first-order half-life equation.
Solution:
a) A new data table is constructed: (Note that additional significant figures are retained in the calculations.)
x-axis (time, s)
[NH3]
y-axis (ln [NH3])
0
4.000 M
1.38629
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16-22
b)
1.000
3.986 M
1.38279
2.000
3.974 M
1.37977
k = slope = rise/run = (y2 – y1)/(x2 – x1)
k = (1.37977 – 1.38629)/(2.000 – 0) = (0.00652)/(2) = 3.260x10–3 s–1 = 3x10–3 s–1
(Note that the starting time is not exact, and hence, limits the significant figures.)
ln 2
ln 2
t1/2 =
=
= 212.62 = 2x102 s
3.260x10−3 s−1
k
16.48
The central idea of collision theory is that reactants must collide with each other in order to react. If reactants must
collide to react, the rate depends on the product of the reactant concentrations.
16.49
No, collision frequency is not the only factor affecting reaction rate. The collision frequency is a count of the total
number of collisions between reactant molecules. Only a small number of these collisions lead to a reaction. Other
factors that influence the fraction of collisions that lead to reaction are the energy and orientation of the collision. A
collision must occur with a minimum energy (activation energy) to be successful. In a collision, the orientation, that
is, which ends of the reactant molecules collide, must bring the reacting atoms in the molecules together in order for
the collision to lead to a reaction.
16.50
At any particular temperature, molecules have a distribution of kinetic energies, as will their collisions have a
range of energies. As temperature increases, the fraction of these collisions which exceed the threshold energy,
increases; thus, the reaction rate increases.
16.51
k = Ae− Ea / RT
The Arrhenius equation indicates a negative exponential relationship between temperatures and the rate constant,
k. In other words, the rate constant increases exponentially with temperature.
16.52
The Arrhenius equation, k = Ae− Ea / RT , can be used directly to solve for activation energy at a specified
temperature if the rate constant, k, and the frequency factor, A, are known. However, the frequency factor is
usually not known. To find Ea without knowing A, rearrange the Arrhenius equation to put it in the form of a
linear plot: ln k = ln A – Ea/RT where the y value is ln k and the x value is 1/T. Measure the rate constant at a series
of temperatures and plot ln k vs. 1/T. The slope equals –Ea/R.
16.53
a) The value of k increases exponentially with temperature.
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16-23
b) A plot of ln k vs. 1/T is a straight line whose slope is –Ea/R.
The activation energy is determined from the slope of the line in the ln k vs. 1/T graph. The slope equals –Ea/R.
16.54
a) As temperature increases, the fraction of collisions which exceed the activation energy increases; thus, the
reaction rate increases.
b) A decrease in activation energy lowers the energy threshold with which collisions must take place to be
effective. At a given temperature, more collisions occur with the lower energy so rate increases.
16.55
No. For 4x10–5 moles of EF to form, every collision must result in a reaction and no EF molecule can decompose
back to AB and CD. Neither condition is likely. All collisions will not result in product as some collisions will
occur with an energy that is lower than the activation energy. In principle, all reactions are reversible, so some EF
molecules decompose. Even if all AB and CD molecules did combine, the reverse decomposition rate would
result in an amount of EF that is less than 4x10–5 moles.
16.56
Collision frequency is proportional to the velocity of the reactant molecules. At the same temperature, both
reaction mixtures have the same average kinetic energy, but not the same velocity. Kinetic energy equals 1/2 mv2,
where m is mass and v velocity. The methylamine (N(CH3)3) molecule has a greater mass than the ammonia
molecule, so methylamine molecules will collide less often than ammonia molecules, because of their slower
velocities. Collision energy thus is less for the N(CH3)3(g) + HCl(g) reaction than for the NH3(g) + HCl(g)
reaction. Therefore, the rate of the reaction between ammonia and hydrogen chloride is greater than the
rate of the reaction between methylamine and hydrogen chloride.
The fraction of successful collisions also differs between the two reactions. In both reactions the hydrogen from
HCl is bonding to the nitrogen in NH3 or N(CH3)3. The difference between the reactions is in how easily the H can
collide with the N, the correct orientation for a successful reaction. The groups (H) bonded to nitrogen in
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16-24
ammonia are less bulky than the groups bonded to nitrogen in trimethylamine (CH3). So, collisions with correct
orientation between HCl and NH3 occur more frequently than between HCl and N(CH3)3 and the reaction
NH3(g) + HCl(g) → NH4Cl(s) occurs at a higher rate than N(CH3)3(g) + HCl(g) → (CH3)3NHCl(s). Therefore, the
rate of the reaction between ammonia and hydrogen chloride is greater than the rate of the reaction between
methylamine and hydrogen chloride.
16.57
Each A particle can collide with three B particles, so (4 x 3) = 12 unique collisions are possible.
16.58
Plan: Use Avogadro’s number to convert moles of particles to number of particles. The number of unique
collisions is the product of the number of A particles and the number of B particles.
Solution:
⎛ 6.022x10 23 A particles ⎞⎟
23
Number of particles of A = (1.01 mol A)⎜⎜
⎟⎟ = 6.08222x10 particles of A
⎜⎝
⎟⎠
1 mol A
⎛ 6.022x1023 B particles ⎞⎟
24
Number of particles of B = (2.12 mol B)⎜⎜
⎟⎟ = 1.279997x10 particles of B
⎜⎝
⎟
1 mol B
⎠
Number of collisions = (6.08222x1023 particles of A)(1.279997x1024 particles of B)
= 7.76495x1047 = 7.76x1047 unique collisions
16.59
Plan: The fraction of collisions with a specified energy is equal to the e−Ea / RT term in the Arrhenius equation.
Solution:
f = e−Ea / RT
T = 25°C + 273 = 298 K
Ea = 100. kJ/mol
R = 8.314 J/mol∙K = 8.314x10–3 kJ/mol∙K
−
100. kJ/ mol
Ea
= –40.362096
=−
RT
8.314x10−3 kJ/mol • K 298 K
(
Fraction = e
16.60
−Ea / RT
)(
–40.362096
=e
)
= 2.9577689x10–18 = 2.96x10–18
Plan: The fraction of collisions with a specified energy is equal to the e−Ea /RT term in the Arrhenius equation.
Solution:
f = e−Ea / RT
T = 50.°C + 273 = 323 K
Ea = 100. kJ/mol
R = 8.314 J/mol∙K = 8.314x10–3 kJ/mol∙K
−
100. kJ/ mol
Ea
= –37.238095
=−
RT
8.314x10−3 kJ/mol • K 323 K
(
)(
Fraction = e−Ea / RT = e
–37.238095
)
= 6.725131x10–17
The fraction increased by (6.725131x10–17)/(2.9577689x10–18) = 22.737175 = 22.7
16.61
Plan: You are given one rate constant k1 at one temperature T1 and the activation energy Ea. Substitute these values
into the Arrhenius equation and solve for k2 at the second temperature.
Solution:
k1 = 4.7x10–3 s–1
T1 = 25°C + 273 = 298 K
k2 = ?
T2 = 75°C + 273 = 348 K
Ea = 33.6 kJ/mol = 33,600 J/mol
k
E ⎛1 1⎞
ln 2 = a ⎜⎜ − ⎟⎟⎟
k1
R ⎜⎝ T1 T2 ⎟⎠
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16-25
ln
ln
33,600 J /mol ⎛⎜ 1
k2
1 ⎟⎟⎞
⎜⎜
=
−
⎟
4.7x10−3 s−1
8.314 J /mol • K ⎜⎜⎝ 298 K 348 K ⎟⎟⎠
k2
−3
−1
4.7x10 s
k2
Raise each side to ex
= 1.948515 (unrounded)
= 7.0182577
−3
4.7x10 s−1
k2 = (4.7x10–3 s–1)(7.0182577) = 0.0329858 = 0.033 s–1
16.62
Plan: You are given the rate constants, k1 and k2, at two temperatures, T1 and T2. Substitute these values into the
Arrhenius equation and solve for Ea.
Solution:
k1 = 4.50x10–5 L/mol∙s
T1 = 195°C + 273 = 468 K
k2 = 3.20x10–3 L/mol∙s
T2 = 258°C + 273 = 531 K
Ea = ?
k2
E ⎛1 1⎞
ln
= a ⎜⎜ − ⎟⎟⎟
k1
R ⎜⎝ T1 T2 ⎠⎟
Ea =
⎛ k ⎞
R ⎜⎜⎜ln 2 ⎟⎟⎟
⎝ k1 ⎟⎠
⎛ 1 1 ⎟⎞
⎜⎜ − ⎟
⎜⎝ T1 T2 ⎟⎟⎠
=
−3
J ⎞⎟⎜⎛⎜ 3.20x10 L/mol • s ⎞⎟⎟
⎛
⎜⎜8.314
ln
⎟
⎟⎜
⎝
mol • K ⎠⎜⎜⎝ 4.50x10−5 L /mol • s ⎠⎟⎟
⎛ 1
1 ⎞⎟⎟
⎜⎜
−
⎟
⎜⎜
⎜⎝ 468 K 531 K ⎠⎟⎟
Ea = 1.3984658x105 J/mol = 1.40x105 J/mol
16.63
Plan: The reaction is exothermic (ΔH is negative), so the energy of the products must be lower than
that of the reactants. Use the relationship ΔHrxn = Ea(fwd) – Ea(rev) to solve for Ea(rev). To draw the transition
state, note that the bond between B and C will be breaking while a bond between C and D will be forming.
Solution:
a)
Ea (fwd)
215 kJ/mol
Energy
Ea(rev)
ABC + D
ΔHrxn
AB + CD
–55 kJ/mol
Reaction coordinate
b) ΔHrxn = Ea(fwd) – Ea(rev)
Ea(rev) = Ea(fwd) – ΔHrxn = 215 kJ/mol – (–55 kJ/mol) = 2.70x102 kJ/mol
c)
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16-26
bond forming
B
A
C
D
bond weakening
16.64
Plan: The forward activation energy Ea(fwd) is larger than the reverse activation energy Ea(rev) which indicates that
the energy of the products must be higher than that of the reactants. Use the relationship
ΔHrxn = Ea(fwd) – Ea(rev) to solve for ΔHrxn. To draw the transition state, note that the bonds in the A2 and B2 molecules
will be breaking while bonds between A and B will be forming.
Solution:
a)
Ea(rev)
Energy
Ea(fwd)
2AB
A2 + B2
ΔHrxn
Reaction coordinate
b) ΔHrxn = Ea(fwd) – Ea(rev) = 125 kJ/mol – 85 kJ/mol = 40 kJ/mol
c)
A.....B
|
|
A.....B
16.65
Plan: You are given the rate constants, k1 and k2, at two temperatures, T1 and T2. Substitute these values into the
Arrhenius equation and solve for Ea.
Solution:
k1 = 0.76/s
T1 = 727°C + 273 = 1000. K
k2 = 0.87/s
T2 = 757°C + 273 = 1030. K
Ea = ?
k
E ⎛1 1⎞
ln 2 = a ⎜⎜ − ⎟⎟⎟
k1
R ⎜⎝ T1 T2 ⎠⎟
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16-27
J ⎞⎟⎜⎛⎜ 0.87 / s ⎞⎟⎟
⎛
⎜⎜8.314
⎟
⎟⎜ln
⎝
mol • K ⎠⎜⎝⎜ 0.76 / s ⎟⎠⎟
⎛ 1
1 ⎞⎟⎟
⎜⎜
−
⎟
⎜⎜
⎜⎝1000. K 1030. K ⎠⎟⎟
Ea = 3.8585x104 J/mol = 3.9x104 J/mol
⎛ k ⎞
R ⎜⎜ln 2 ⎟⎟⎟
⎜⎝ k1 ⎠⎟
Ea =
=
⎛ 1 1 ⎞⎟
⎜⎜ − ⎟
⎜⎝ T1 T2 ⎠⎟⎟
16.66
Plan: The reaction is endothermic (ΔH is positive), so the energy of the products must be higher than
that of the reactants. Use the relationship ΔHrxn = Ea(fwd) – Ea(rev) to solve for Ea(rev). To draw the transition
state, note that the bond in Cl2 will be breaking while the bond between N and Cl will be forming.
Solution:
NOCl + Cl
Ea(rev)
Energy
a)
ΔHrxn = +83 kJ
Ea(fwd) = +86 kJ
NO + Cl2
Reaction coordinate
b) ΔHrxn = Ea(fwd) – Ea(rev)
Ea(rev) = Ea(fwd) – ΔHrxn = 86 kJ – 83 kJ = 3 kJ.
c) To draw the transition state, look at structures of reactants and products:
Cl
Cl
+
N
O
N
Cl
+
Cl
O
The collision must occur between one of the chlorine atoms and the nitrogen. The transition state would have
weak bonds between the nitrogen and chlorine and between the two chlorine atoms.
N
Cl
Cl
O
16.67
The rate of an overall reaction depends on the slowest step. Each individual step’s reaction rate can vary widely,
so the rate of the slowest step, and hence the overall reaction, will be slower than the average of the individual
rates because the average contains faster rates as well as the rate-determining step.
16.68
An elementary step is a single molecular event, such as the collision of two molecules. Since an elementary step
occurs in one step, its rate must be proportional to the product of the reactant concentrations. Thus, the exponents
in the rate of an elementary step are identical to the coefficients in the equation for the step. Since an overall
reaction is generally a series of elementary steps, it is not necessarily proportional to the product of the overall
reactant concentrations.
16.69
Yes, it is often possible to devise more than one mechanism since the rate law for the slowest step determines the
rate law for the overall reaction. The preferred mechanism will be the one that seems most probable, where
molecules behave in their expected fashion.
16.70
Reaction intermediates have some stability, however limited, but transition states are inherently unstable.
Additionally, unlike transition states, intermediates are molecules with normal bonds.
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16-28
16.71
A bimolecular step (a collision between two particles) is more reasonable physically than a termolecular step
(a collision involving three particles) because the likelihood that two reactant molecules will collide with the
proper energy and orientation is much greater than the likelihood that three reactant molecules will collide
simultaneously with the proper energy and orientation.
16.72
No, the overall rate law must contain reactants only (no intermediates) and is determined by the slow step. If the
first step in a reaction mechanism is slow, the rate law for that step is the overall rate law.
16.73
If the slow step is not the first one, the faster preceding step produces intermediates that accumulate before being
consumed in the slow step. Substitution of the intermediates into the rate law for the slow step will produce the
overall rate law.
16.74
Plan: The overall reaction can be obtained by adding the three steps together. An intermediate is a substance
that is formed in one step and consumed in a subsequent step. The molecularity of each step is the total number of
reactant particles; the molecularities are used as the orders in the rate law for each step. The overall rate law for
the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the actual
rate law.
Solution:
a)
(1) A(g) + B(g) → X(g) fast
(2) X(g) + C(g) → Y(g) slow
(3) Y(g) → D(g) fast
Total: A(g) + B(g) + X(g) + C(g) + Y(g) → X(g) + Y(g) + D(g)
Overall:
A(g) + B(g) + C(g) → D(g)
b) Both X and Y are intermediates in the given mechanism. Intermediate X is produced in the first step and
consumed in the second step; intermediate Y is produced in the second step and consumed in the third step.
Notice that neither X nor Y were included in the overall reaction.
c)
A(g) + B(g) →X(g)
Molecularity
bimolecular
Rate law
rate1 = k1[A][B]
X(g) + C(g) → Y(g)
bimolecular
rate2 = k2[X][C]
Y(g) → D(g)
unimolecular
rate3 = k3[Y]
Step
d) Yes, the mechanism is consistent with the actual rate law. The slow step in the mechanism is the second step
with rate law: rate = k2[X][C]. Since X is an intermediate, it must be replaced by using the first step. For an
equilibrium, rateforward rxn = ratereverse rxn. For step 1 then, k1[A][B] = k–1[X]. Rearranging to solve for [X] gives
[X] = (k1/k–1)[A][B]. Substituting this value for [X] into the rate law for the second step gives the overall rate law
as rate = (k2k1/k–1)[A][B][C] which is identical to the actual rate law with k = k2k1/k–1.
e) Yes, The one step mechanism A(g) + B(g) + C(g) → D(g) would have a rate law of rate = k[A][B][C], which
is the actual rate law.
16.75
Plan: The overall reaction can be obtained by adding the three steps together. An intermediate is a substance
that is formed in one step and consumed in a subsequent step. The molecularity of each step is the total number of
reactant particles; the molecularities are used as the orders in the rate law for each step. The overall rate law for
the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the actual
rate law.
Solution:
fast
a)
(1) ClO–(aq) + H2O(l) → HClO(aq) + OH–(aq)
slow
(2) I–(aq) + HClO(aq) → HIO(aq) + Cl–(aq)
fast
(3) OH–(aq) + HIO(aq) → H2O(l) + IO–(aq)
–
Total: ClO (aq) + H2O(l) + I–(aq) + HClO(aq) + OH–(aq) + HIO(aq) → HClO(aq) + OH–(aq) + HIO(aq)
+ Cl–(aq) + H2O(l) + IO–(aq)
–
–
–
–
(overall)
ClO (aq) + I (aq) → Cl (aq) + IO (aq)
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16-29
b) HClO(aq), OH –(aq), and HIO(aq) are intermediates in the given mechanism. HClO(aq) is produced in the
first step and consumed in the second step; OH –(aq) is produced in the first step and consumed in the third step;
HIO(aq) is produced in the second step and consumed in the third step. Notice that the intermediates were not
included in the overall reaction.
c)
(1) Bimolecular; Rate1 = k1 [ClO –] [H2O]
(2) Bimolecular; Rate2 = k2 [I –][HClO]
(3) Bimolecular; Rate3 = k3 [OH –][HIO]
d) The slow step in the mechanism is the second step with rate law: rate = k2[I–][HClO]. Since HClO is an
intermediate, it must be replaced by using the first step. For an equilibrium, rateforward rxn = ratereverse rxn. For step 1
then, leaving out the water, k1[ClO –] = k–1[HClO][ OH–]. Rearranging to solve for [HClO] gives
[HClO] = (k1/k–1)[ClO–]/[OH–]. Substituting this value for [HClO] into the rate law for the second step gives the
overall rate law as rate = (k2k1/k–1)[I–][ClO–]/[OH–] or rate = k[I–][ClO–]/[OH–].
This is not the observed rate law. The mechanism is not consistent with the actual rate law.
16.76
Plan: Use the rate-determining step to find the rate law for the mechanism. The concentration of the
intermediate in the rate law must be expressed in terms of a true reactant which is then substituted into the rate
law for the concentration of the intermediate.
Solution:
Nitrosyl bromide is NOBr(g). The reactions sum to the equation 2NO(g) + Br2(g) → 2NOBr(g), so criterion
1 (elementary steps must add to overall equation) is satisfied. Both elementary steps are bimolecular and
chemically reasonable, so criterion 2 (steps are physically reasonable) is met. The reaction rate is determined by
the slow step; however, rate expressions do not include reaction intermediates (NOBr2). The slow step in the
mechanism is the second step with rate law: rate = k2[NOBr2][NO]. Since NOBr2 is an intermediate, it must be
replaced by using the first step. For an equilibrium like step 1, rateforward rxn = ratereverse rxn.
Solve for [NOBr2] in step 1:
Rate1 (forward) = rate1 (reverse)
k1[NO][Br2] = k–1[NOBr2]
[NOBr2] = (k1/k–1)[NO][Br2]
Rate of the slow step: Rate2 = k2[NOBr2][NO]
Substitute the expression for [NOBr2] into this equation, the slow step:
Rate2 = k2(k1/k–1)[NO][Br2][NO]
Combine the separate constants into one constant: k = k2(k1/k–1)
Rate2 = k[NO]2[Br2]
The derived rate law equals the known rate law, so criterion 3 is satisfied. The proposed mechanism is valid.
16.77
Plan: Use the rate-determining step to find the rate law for each mechanism. If there is an intermediate in the
rate law, the concentration of the intermediate must be expressed in terms of a true reactant which is then
substituted into the rate law for the concentration of the intermediate.
Solution:
I.
2NO(g) + O2(g) → 2NO2(g)
Rate = k[NO]2[O2]
II.
(1) 2NO(g) → N2O2(g)
(2) N2O2(g) + O2(g) → 2NO2(g)
slow
Rate = k2[N2O2][O2]
N2O2 is an intermediate
Rate1 (forward) = rate1 (reverse)
k1[NO]2 = k–1[N2O2]
[N2O2] = (k1/k–1)[NO]2
Substitute this for [N2O2] in the rate law:
Rate = k[N2O2][O2]
Rate = k2(k1/k–1)[NO]2[O2]
Rate = k[NO]2[O2]
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16-30
(1) 2NO
O(g) → N2(g) + O2(g)
(2) N2(g
g) + 2O2(g) → NO2(g)
slow
Rate = k2[N2][O2]2
i
N2 is an intermediate
Rate1 (fforward) = ratee1 (reverse)
k1[NO]2 = k–1[N2][O2]
[N2] = (k1/k–1)[NO]2/[O
O2] = (k1/k–1)[N
NO]2[O2]–1
Substittute this for [N2] in the rate laaw:
Rate = k2[N2][O2]2
Rate = k2(k1/k–1)[NO]2 [O2]–1[O2]2
Rate = k[NO]2[O2]
a)) All the mechaanisms are connsistent with thee rate law.
b)) The most reasonable mechaanism is II, sincce none of its elementary
e
stepps are more complicated thann being
biimolecular. Meechanism I andd III have stepss that are termoolecular.
IIII.
16.78
Pllan: Review th
he definitions of
o homogeneouus vs. heterogeeneous catalysts. To draw the reaction energgy diagrams,
reecall that additiion of a catalysst to a reaction results in a low
wer activation energy.
Soolution:
a)) A heterogeneeous catalyst sppeeds up a reacction that occurrs in a differentt phase. Gold is
i a heterogeneeous
caatalyst since itt is a solid and is catalyzing a reaction in thee gas phase.
b)) The activatio
on energy for thhe uncatalyzedd reaction mustt be greater thaan the activatioon for the catalyyzed
reeaction.
16.79
No, a catalyst ch
N
hanges the mecchanism of a reeaction to one with
w lower actiivation energy. Lower activation energy
m
means
a faster reeaction. An inccrease in tempeerature does noot influence thee activation energy, but instead increases
thhe fraction of collisions with sufficient energy to react.
16.80
a)) No, by definition, a catalyst is a substannce that increases reaction raate without beeing consumedd. The spark
prrovides energy
y that is absorbed by the H2 and O2 moleccules to achievve the thresholld energy needded for
reeaction.
b)) Yes, the powdered metal accts like a heteroogeneous catalyyst, providing a surface uponn which the reaaction
beetween O2 and H2 becomes more
m
favorable because the acctivation energyy is lowered.
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ducation. This is proprietary mateerial solely for au
uthorized instructtor use. Not authoorized for sale or distribution in
any manner.. This document may
m not be copied
d, scanned, dupliccated, forwarded
d, distributed, or posted
p
on a websiite, in whole or paart.
16-31
16.81
Catalysts decrease the amount of energy required for a reaction. To carry out the reaction less energy must be
generated. The generation of less energy means that fewer by-products of energy production will be released into
the environment.
16.82
a) Enzymes stabilize a reaction’s transition state to a remarkable degree which lowers the activation energy and thus
greatly increase the reaction rate.
b) Enzymes are extremely specific and have the ability to change shape to adopt a perfect fit with the substrate.
16.83
Plan: For parts a) and b), find the order of the given reactant; the order of the other reactant is found by realizing
that the orders m + n = 2 since the reactions are second order overall. For parts c)-e), use the rate laws to
determine the rate changes that result from the given changes in reactant concentrations.
Solution:
a) The rate law is rate = [NO2]m[CO]n. When [NO2] increases by a factor of 2, the rate increases by a factor
of 4:
m
⎛[NO 2 ]2 ⎞⎟
rate 2
⎟⎟
= ⎜⎜
⎜⎝ [NO 2 ]1 ⎟⎠
rate1
4 = [2]m
log (4.00) = m log (2.00)
m=2
The reaction is second order with respect to NO2. Since the overall order is 2, m + n = 2:
m+n=2
n=2–m
n=2–2=0
The reaction is zero order with respect to CO.
Rate = k[NO2]2[CO]0 or Rate = k[NO2]2
b) The rate law is rate = [NO]m[O3]n. When [NO] increases by a factor of 2, the rate increases by a factor
of 2:
m
⎛ [NO]2 ⎞⎟
rate 2
⎟⎟
= ⎜⎜
⎜⎝ [NO]1 ⎟⎠
rate1
2 = [2]m
log (2.00) = m log (2.00)
m=1
The reaction is first order with respect to NO. Since the overall order is 2, m + n = 2:
m+n=2
n=2–m
n=2–1=1
The reaction is first order with respect to O3.
Rate = k[NO][O3]
c) The concentrations of the reactants are one-half of the original when the reaction is 50% complete.
2
k [ NO 2 ]
Rate initial
=4
=
Reaction 1:
2
Rate 50%
k ⎡⎢ 0.5 (NO 2 )⎤⎥
⎣
⎦
Reaction 2:
k [ NO ][O3 ]
Rate initial
=4
=
Rate 50%
k ⎡⎢0.5(NO)⎤⎥ ⎡⎢0.5(O3 )⎤⎥
⎣
⎦⎣
⎦
d) [NO2]initial = 2[CO]initial; when the reaction is 50% complete, [CO] = 1/2[CO]initial and [NO2] = 0.75[NO2]initial.
2
k [ NO 2 ]
Rate initial
=
= 1.7778 = 1.8
2
Rate 50%
k ⎡⎢ 0.75 (NO 2 )⎤⎥
⎣
⎦
e) [NO]initial = 2[O3]initial, when the reaction is 50% complete, [O3] = 1/2[O3]initial and [NO] = 0.75[NO]initial.
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16-32
k [ NO ][O3 ]
Rate initial
=
= 2.6667 = 2.7
Rate 50%
k ⎡⎢0.75(NO)⎤⎥ ⎡⎢ 0.5(O3 )⎤⎥
⎣
⎦⎣
⎦
16.84
a) There are two elementary steps since there are two different values of activation energy.
b) The second step is the rate-limiting step since it has the greater activation energy and would therefore be the
slower of the two steps.
c) The reaction is exothermic since a loss of energy results in the products having lower energy than the
reactants.
16.85
Plan: An intermediate is a substance that is formed in one step and consumed in a subsequent step. The
coefficients of the reactants in each elementary step are used as the orders in the rate law for the step. The overall
rate law for the mechanism is determined from the slowest step (the rate-determining step) and can be compared
to the actual rate law. Reactants that appear in the mechanism after the slow step do not determine the rate and
therefore do not appear in the rate law.
Solution:
a) Water does not appear as a reactant in the rate-determining step.
Note that as a solvent in the reaction, the concentration of the water is assumed not to change even though some
water is used up as a reactant. This assumption is valid as long as the solute concentrations are low (∼1 M or less).
So, even if water did appear as a reactant in the rate-determining step, it would not appear in the rate law. See rate
law for step 2 below.
b)
Rate law for step (1):
rate1 = k1[(CH3)3CBr]
Rate law for step (2):
rate2 = k2[(CH3)3C+]
Rate law for step (3):
rate3 = k3[(CH3)3COH2+]
+
c) The intermediates are (CH3)3C and (CH3)3COH2+. (CH3)3C+ is formed in step 1 and consumed in step 2;
(CH3)3COH2+ is formed in step 2 and consumed in step 3.
d) The rate-determining step is the slow step, (1). The rate law for this step is rate = k1[(CH3)3CBr] since the
coefficient of the reactant in this slow step is 1. The rate law for this step agrees with the actual rate law
with k = k1.
16.86
Plan: Radioactive decay is first order so the first-order integrated rate law is used. First use the first-order half-life
equation to solve for k since the half-life is given. Since 25% of the 14C remains at time t, let [14C]0 = 100% and
then [14C]t = 15%.
Solution:
ln 2
ln 2
ln 2
= 1.20968x10–4 yr–1 (unrounded)
t1/2 =
k=
=
k
5730 yr
t1/2
Use the first-order integrated rate law ln [14C]t = ln [14C]0 – kt or
⎡ 14 C ⎤
ln ⎣ 14 ⎦ t = – kt
⎡ C⎤
⎣ ⎦0
[15.5%]t
ln
= –(1.20968x10–4 yr–1) t
[100%]0
t = 1.541176x104 = 1.54x104 yr
16.87
Plan: The activation energy can be calculated using the Arrhenius equation. Although the rate constants, k1 and k2,
are not expressly stated, the relative times give an idea of the rate. The reaction rate is proportional to the rate
constant. At T1 = 20°C, the rate of reaction is 1 apple/4 days while at T2 = 0°C, the rate is 1 apple/16 days.
Therefore, rate1 = 1 apple/4 days and rate2 = 1 apple/16 days are substituted for k1 and k2, respectively.
Solution:
k1 = 1/4
T1 = 20°C + 273 = 293 K
k2 = 1/16
T2 = 0°C + 273 = 273 K
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16-33
Ea = ?
k
E ⎛1 1⎞
ln 2 = a ⎜⎜⎜ − ⎟⎟⎟
k1
R ⎝ T1 T2 ⎠⎟
J ⎞⎟⎜⎛⎜ 1/16 ⎟⎟⎞
⎛
⎛ k ⎞
⎜⎜8.314
⎟
⎟⎜ln
R ⎜⎜ln 2 ⎟⎟⎟
⎝
mol • K ⎠⎜⎜⎝ 1/4 ⎟⎟⎠
⎜⎝ k1 ⎠⎟
Ea =
=
⎛ 1
⎛ 1 1 ⎞⎟
1 ⎞⎟⎟
⎜
⎜⎜ − ⎟
−
⎟
⎜⎜⎜
⎜⎝ T1 T2 ⎠⎟⎟
⎜⎝ 293 K 273 K ⎠⎟⎟
Ea = 4.6096266x104 J/mol = 4.61x104 J/mol The significant figures are based on the Kelvin temperatures.
16.88
Plan: Use the first-order integrated rate law. First use the first-order half-life equation to solve for k since the halflife is given. Since 95% of the benzoyl peroxide (BP) remains at time t, let [BP]0 = 100% and then [BP]t = 95%.
Solution:
ln 2
ln 2
k=
= 7.07293x10–5 d–1 (unrounded)
t1/2 =
k
t1/2
Use the first-order integrated rate law with BP = benzoyl peroxide:
[ BP ]t
ln
= – kt
[ BP ]0
ln
[ 95%]t
= – (7.07293x10–5 d–1) t
[100%]0
t = 725.2057 = 7.3x102 d
Plan: The reaction is exothermic (ΔH is negative), so the energy of the products must be lower than
that of the reactants in the reaction energy diagram. Since there are two steps in the proposed mechanism,
the diagram must show two transition states. The first step in the mechanism is the slower step so its Ea value
is larger than the Ea value of the second step in the mechanism. The overall rate law for the alternative mechanism
is determined from the slowest step (the rate-determining step) and can be compared to the actual
rate law.
Solution:
a)
Energy
16.89
CO + NO2
ΔH = –226kJ
CO2 + NO
Reaction coordinate
b) The rate of the mechanism is based on the slowest step, 2NO2(g) → N2(g) + 2O2(g). The rate law for this step is
rate = k1[NO2]2 which is consistent with the actual rate law. The alternative mechanism includes an elementary
reaction (step 2) that is a termolecular reaction. Thus, the original mechanism given in the text is more reasonable
physically since it involves only bimolecular reactions.
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16-34
16.90
Plan: The rate law is rate = [A]x[B]y[C]z where x, y, and z are the orders of the reactants. To find the order of each
reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once
the rate law is known, any experiment can be used to find the rate constant k. For part d), use Equation 16.2 to
describe the rate of this reaction in terms of reactant disappearance and product appearance. A negative sign is
used for the rate in terms of reactants A, B, and C since these substances are reacting and [A], [B], and [C] are
decreasing over time. Positive signs are used for the rate in terms of products D and E since these substances are
being formed and [D] and [E] increase over time.
Solution:
a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B] and [C]
are constant. Use experiments 1 and 2 to find the order with respect to [A]. Set up a ratio of the rate laws for
experiments 1 and 2 and fill in the values given for rates and concentrations and solve for x, the order with respect
to [A].
x
z
k [ A ]2 [ B]2y [ C ]2
Rate 2
=
x
z
y
Rate1
k [ A ]1 [ B]1 [ C ]1
[B]1 = [B]2
[C]1 = [C]2
x
k [ A ]2
Rate 2
=
x
Rate1
k [ A ]1
x
k ⎡⎢ 0.096 mol/L ⎤⎥
⎣
⎦2
=
x
⎡
6.0x10−6 mol/L • s
k ⎢ 0.024 mol/L ⎤⎥
⎣
⎦1
16 = 4x
x=2
Second order with respect to A
In finding the order with respect to B, there are no experiments in which only [B] changes. Experiments 2 and 4
can be used as [C] is constant while [A] and [B] change. Since we already know the order with respect to A, we
can determine the order with respect to B in these two experiments. Set up a ratio of the rate laws for experiments
2 and 4 and fill in the values given for rates and concentrations and solve for y, the order with respect to [B].
2
z
k [ A ]2 [ B]2y [ C ]2
Rate 2
=
z
2
y
Rate 4
k [ A ]4 [ B]4 [ C ]4
[C]2 = [C]4
2
k [ A ]2 [ B]2y
Rate 2
=
2
k [ A ]4 [ B]4y
Rate 4
9.6x10−5 mol/L • s
y
k ⎡⎢ 0.096 mol/L⎤⎥ ⎡⎢ 0.085 mol/L ⎤⎥
⎣
⎦2 ⎣
⎦2
=
y
2
1.5x10−6 mol/L • s
⎡
⎤
⎡
k ⎢0.012 mol/L ⎥ ⎢ 0.170 mol/L ⎤⎥
⎣
⎦4 ⎣
⎦4
64 = (8)2 (0.5)y
y=0
Zero order with respect to B
In finding the order with respect to C, there are no experiments in which only [C] changes. Experiments 1 and 3
can be used as [A] is constant while [B] and [C] change. Since we already know the order with respect to B, we
can determine the order with respect to C in these two experiments. Set up a ratio of the rate laws for experiments
1 and 3 and fill in the values given for rates and concentrations and solve for z, the order with respect to [C].
2
z
k [ A ]1 [ B]10 [ C ]1
Rate1
=
z
2
k [ A ]3 [ B]30 [ C ]3
Rate3
2
9.6x10−5 mol/L • s
[A]1 = [A]3
z
k [ B]1 [ C ]1
Rate1
=
z
k [ B]30 [ C ]3
Rate3
0
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16-35
z
k ⎡⎢ 0.085 mol/L ⎤⎥ ⎡⎢ 0.032 mol/L ⎤⎥
⎦1 ⎣
⎦1
= ⎣
z
0
1.5x10−5 mol/L • s
⎡
⎤
⎡
k ⎢ 0.034 mol/L ⎥ ⎢ 0.080 mol/L ⎤⎥
⎣
⎦3 ⎣
⎦3
Since y = 0, the B term may be ignored, it is only shown here for completeness.
0.4 = 0.4z
z=1
First order with respect to C
b) You can use any trial to calculate k, with the rate law: Rate = k[A]2[B]0[C]1 = k[A]2[C]
Using experiment 1:
rate = k[A]2[C]
6.0x10−6 mol/L • s
rate
k=
=
= 0.32552 = 0.33 L2/mol2s
[A]2 [C]
[0.024 mol/L] 2 [0.032 mol/L]
This value will need to be divided by the coefficient of the substance to which the initial rate refers.
If the initial rate refers to the disappearance of A or B then the constant (k') is:
k' = k/2 = 0.32552/2 = 0.16276 = 0.16 L2/mol2s
If the initial rate refers to the disappearance of C or the formation of D then the constant (k') is:
k' = k = 0.33 L2/mol2s
If the initial rate refers to the formation of E then the constant (k') is:
k' = k/3 = 0.32552/3 = 0.108507 = 0.11 L2/mol2s
2
c) Rate = = k[A] [C] (substitute the appropriate k value from part b))
Δ[ C ]
Δ[ D ]
1 Δ [ B]
1 Δ[ E ]
1 Δ[ A ]
d) Rate = −
=−
=−
=
=
2 Δt
3 Δt
2 Δt
Δt
Δt
6.0x10−6 mol/L • s
0
16.91
Plan: Use the given rate law, rate = k[H+][sucrose], and enter the given values. The glucose and fructose are not in
the rate law, so they may be ignored.
Solution:
a) The rate is first order with respect to [sucrose]. The [sucrose] is changed from 1.0 M to 2.5 M, or is increased
by a factor of 2.5/1.0 or 2.5. Then the rate = k[H+][2.5 x sucrose]; the rate increases by a factor of 2.5.
b) The [sucrose] is changed from 1.0 M to 0.5 M, or is decreased by a factor of 0.5/1.0 or 0.5. Then the
rate = k[H+][0.5 x sucrose]; the rate decreases by a factor of ½ or half the original rate.
c) The rate is first order with respect to [H+]. The [H+] is changed from 0.01 M to 0.0001 M, or is decreased by a
factor of 0.0001/0.01 or 0.01. Then the rate = k[0.01 x H+][sucrose]; the rate decreases by a factor of 0.01.
Thus, the reaction will decrease to 1/100 the original.
d) The [sucrose] decreases from 1.0 M to 0.1 M, or by a factor of (0.1 M/1.0 M) = 0.1. [H+] increases from 0.01 M
to 0.1 M, or by a factor of (0.1 M/0.01 M) = 10. Then the rate will increase by k[10 x H+][0.1 x sucrose]= 1.0
times as fast. Thus, there will be no change.
16.92
Plan: The overall order is equal to the sum of the individual orders. Since the reaction is eleventh order overall, the sum
of the exponents equals eleven. Add up the known orders and subtract that sum from eleven to find the unknown order.
Solution:
Sum of known orders = 1 + 4 + 2 + 2 = 9
Overall order – sum of known orders = 11 – 9 = 2.
The reaction is second order with respect to NAD.
16.93
Plan: The reaction is A + B → products. Assume the reaction is first order with respect to A and first order with
respect to B. Find the concentration of each reactant in Mixture I and use those values and the initial rate to
calculate k, the rate constant, for the reaction. Knowing k, the initial rate in Mixture II can be calculated using the
rate law and the reactant concentrations.
Solution:
Rate = k[A][B]
Mixture I:
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16-36
⎛ 0.010 mol A ⎞⎟
⎟
⎝⎜ 1 sphere ⎠⎟⎟
(6 spheres A)⎜⎜⎜
Concentration of A =
Concentration of B =
0.50 L
⎛ 0.010 mol B ⎞⎟
⎟
(5 spheres B)⎜⎜⎜
⎝ 1 sphere ⎟⎠⎟
= 0.12 mol/L
= 0.10 mol/L
0.50 L
Use the rate law to find the value of k, the rate constant:
Rate = k[A][B]
8.3x10−4 mol/L • min
rate
k=
=
= 0.069167 L/mol·min
[0.12 mol/L][0.10 mol/L]
[A][B]
Use this value of k to find the initial rate in Mixture II:
⎛ 0.010 mol A ⎞⎟
⎟
(7 spheres A)⎜⎜
⎜⎝ 1 sphere ⎟⎠⎟
Concentration of A =
= 0.14 mol/L
0.50 L
⎛ 0.010 mol B ⎞⎟
⎟
(8 spheres B)⎜⎜
⎜⎝ 1 sphere ⎟⎟⎠
Concentration of B =
= 0.16 mol/L
0.50 L
Rate = k[A][B]
Rate = 0.069167 L/mol·min[0.14 mol/L][0.16 mol/L]
Rate = 1.5493x10–3 = 1.5x10–3 mol/L∙min
16.94
Plan: The overall rate law for the mechanism is determined from the slowest step (the rate-determining step). An
overall rate law can only include reactants and products; intermediates cannot be included in the rate law. Express
[intermediate] in terms of [reactant]. Compare the resulting rate law from the mechanism to the actual rate law.
Solution:
Initially, the slow step in the mechanism gives:
Rate = k2[CHCl3][Cl]
However, Cl is an intermediate, and should not be in the final rate law.
For an equilibrium, rateforward rxn = ratereverse rxn. For step 1, k1[Cl2] = k–1[Cl]2. Rearranging to solve for [Cl] gives
[Cl]2 = (k1/k–1)[Cl2]
[Cl] = (k1/k–1)1/2[Cl2]1/2
Substituting into the rate law from the slow step:
Rate = k2[CHCl3][Cl]
Rate = k2(k1/k–1)1/2[CHCl3][Cl2]1/2
Combining k’s:
Rate = k[CHCl3][Cl2]1/2
The rate law from the mechanism is consistent with the actual rate law.
16.95
Plan: First, find the rate constant, k, for the reaction by solving the first-order half-life equation for k. Then use the
first-order integrated rate law expression to find t, the time for decay.
Solution:
ln 2
ln 2
to k =
Rearrange t1/2 =
t1/2
k
ln 2
k=
= 5.7762x10–2 yr–1
12 yr
Use the first-order integrated rate law: ln
[ DDT ]t
[ DDT ]0
= – kt
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16-37
ln
[10. ppbm ]t
= – (5.7762x10–2 yr–1) t
[ 275 ppbm ]0
t = 57.3765798 = 57 yr
16.96
Plan: Solve the first-order half-life equation for k and then take the reciprocal of that value of k.
Solution:
ln 2
ln 2
to k =
Rearrange t1/2 =
k
t1/2
ln 2
k=
= 0.19804 min–1 (unrounded)
3.5 min
The problem states that the interval t = 1/k:
t = 1/(0.19804 min–1) = 5.04943 = 5.0 min
16.97
Plan: The rate law for the reaction is given. Use the given values of rate, [A]0, and [B]0 to find the value of
the rate constant, k, for the reaction. The value of k can be used to calculate the rate at the second set of reactant
concentrations.
Solution:
The rate law is: Rate = k[A]2[B]
Calculate k:
0.20 mol/L • s
rate
k=
=
= 0.20 L2/mol2s
2
2
[A] [B]
⎡1.0 mol/L ⎤ ⎡1.0 mol/L ⎤
⎣⎢
⎦⎥ ⎣⎢
⎦⎥
Using the k just calculated with the rate law:
Rate = k[A]2[B]
Rate = (0.20 L2/mol2s) [2.0mol/L]2[3.0mol/L] = 2.4 mol/L∙s
16.98
Plan: The rate constant can be determined from the slope of the integrated rate law plot. To find the correct order,
the data should be plotted as 1) [sucrose] vs. time – linear for zero order, 2) ln [sucrose] vs. time – linear for first
order, and 3) 1/[sucrose] vs. time – linear for second order. Once the order is established, use the appropriate
integrated rate law to find the time necessary for 75.0% of the sucrose to react. If 75.0% of the sucrose has reacted,
100–75 = 25% remains at time t. Let [sucrose]0 = 100% and then [sucrose]t = 25%.
Solution:
a) All three graphs are linear, so picking the correct order is difficult. One way to select the order is to compare
correlation coefficients (R2) — you may or may not have experience with this. The best correlation coefficient is
the one closest to a value of 1.00. Based on this selection criterion, the plot of ln [sucrose] vs. time for the firstorder reaction is the best.
Another method when linearity is not obvious from the graphs is to examine the reaction and decide which order
fits the reaction. For the reaction of one molecule of sucrose with one molecule of liquid water, the rate law would
most likely include sucrose with an order of one and would not include water.
The plot for a first-order reaction is described by the equation ln [A]t = –kt + ln [A]0. The slope of the plot of
ln [sucrose] vs. t equals –k. The equation for the straight line in the first-order plot is y = –0.21x – 0.6936.
So, k = – (–0.21 h–1) = 0.21 h–1.
Solve the first-order half-life equation to find t1/2:
ln 2
t1/2=
= 3.3007 = 3.3 h
0.21 hr−1
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16-38
Integrated rate law plots
4
y = 0.5897x + 1.93
R2 = 0.9935
(see legend)
3
2
y = −0.077x + 0.4896
1
R2 = 0.9875
0
0
-1
0.5
1
1.5
2
2.5
3
3.5
y = −0.21x − 0.6936
R2 = 0.9998
-2
time, h
Legend: ♦ y-axis is [sucrose]
■ y-axis is ln [sucrose]
▲ y-axis is 1/[sucrose]
b) If 75% of the sucrose has been reacted, 25% of the sucrose remains. Let [sucrose]0 = 100% and
[sucrose]t = 25% in the first-order integrated rate law equation:
[ sucrose ]t
= – kt
ln
[ sucrose ]0
[ 25%]t
ln
= – (0.21 h–1) t
[100%]0
t = 6.6014 = 6.6 h
c) The reaction might be second order overall with first order in sucrose and first order in water. If the
concentration of sucrose is relatively low, the concentration of water remains constant even with small changes in
the amount of water. This gives an apparent zero-order reaction with respect to water. Thus, the reaction
appears to be first order overall because the rate does not change with changes in the amount of water.
16.99
Plan: You are given the ratio of the rate constants, k1 and k2, at a particular temperature and you want to find the
difference in Ea between the uncatalyzed and catalyzed processes. Write the Arrhenius equation twice, once for
the uncatalyzed process and once for the catalyzed process and divide the two equations so that the constant A
factor divides out.
Solution:
k2
= 2.3x1014
T = 37°C + 273 = 310 K
k1
ΔEa = ?
k1 = Ae–Ea1/RT
and k2 = Ae–Ea2/RT
− Ea 2
( Ea1 − Ea2 )
k2
Ae RT
=
= e
−Ea1
RT
k1
Ae RT
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16-39
ln
k2
E − Ea2
= a1
k1
RT
or
⎛ k ⎞
RT ⎜⎜⎜ ln 2 ⎟⎟⎟ = Ea1 – Ea2
⎝ k1 ⎠⎟
2.3x1014
= Ea1 – Ea2
1
4
4
Ea1 – Ea2 = 8.5230x10 = 8.5x10 J/mol
(8.314 J/mol∙K) (310 K) ln
16.100 a) False, at any particular temperature, molecules have a range of kinetic energies.
b) False, at reduced pressure, the number of collisions per unit time is reduced, as is the reaction rate.
c) True
d) False, the increase in rate depends on the activation energy for the reaction. Also, biological catalysts
(enzymes) may decompose on heating, reducing their effectiveness.
e) False, they also must have the correct orientation.
f) False, the activation energy is unique to the mechanism of a particular reaction.
g) False, since most reaction rates depend to some extent on the reactant concentrations, as these decrease during
the course of the reaction, the reaction rate also decreases.
h) False, see part f).
i) False, a catalyst speeds up the reaction by lowering the activation energy.
j) False, the speed of a reaction (kinetics) is separate from the stability of the products (thermodynamics).
k) False, the frequency factor, A, is the product of the collision frequency which is affected by temperature and an
orientation probability factor.
l) True
m) False, the catalyst changes the activation energy, not ΔH of reaction.
n) True
o) True
p) False, bimolecular and unimolecular refer to the molecularity or the number of reactant particles involved in
the reaction step. There is no direct relationship to the speed of the reaction.
q) False, molecularity and molecular complexity are not related.
16.101 Plan: To find concentration of reactant at a later time, given the initial concentration and the rate constant k, use
an integrated rate law expression. Since the units on k are s–1, this is a first-order reaction. Use the first-order
integrated rate law. Since the time unit in k is seconds, time t must also be expressed in units of seconds. To find
the fraction of reactant that has decomposed, divide that amount of reactant that has decomposed
([N2O5]0 – [N2O5]t) by the initial concentration.
Solution:
⎛ 60 s ⎞⎟
a) Converting t in min to s: (5.00 min )⎜⎜
= 300. s
⎝1 min ⎟⎠⎟
[ N2 O5 ]t
ln
= – kt or
[ N 2 O5 ]0
ln [N2O5]t = ln [N2O5]0 – kt
ln [N2O5]t = ln [1.58 mol/L] – (2.8x10–3 s–1)(300. s)
ln [N2O5]t = – 0.382575
[N2O5]t = 0.68210 = 0.68 mol/L
[N 2 O5 ]0 − [N 2 O5 ]t
1.58 − 0.68210 mol/L
=
= 0.56829 = 0.57
b) Fraction decomposed =
1.58 mol/L
[N 2 O5 ]0
16.102 Plan: The overall rate law for the mechanism is determined from the slowest step (the rate-determining step). An
overall rate law can only include reactants and products; intermediates cannot be included in the rate law. Express
[intermediate] in terms of [reactant]. Compare the resulting rate law from the mechanism to the actual rate law.
Solution:
Rate law for slow step (Step 3):
Rate = k3[H2I][I]
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16-40
Both H2I and I are intermediates and cannot be in the final rate law. For an equilibrium, rateforward rxn = ratereverse rxn.
From first two steps:
From step 1:
k1[I2] = k–1[I]2; [I] = (k1/k–1)1/2[I2]1/2
From step 2:
k2[H2][I] = k–2[H2I]; [H2I] = k2/k–2[H2][I]
Substituting for [H2I] in rate = k3[H2I][I]
:
Rate = k3[k2/k–2[H2][I]][I]
Rate = k3k2/k–2[H2][I]2
Substituting for [I] in rate = k3k2/k–2[H2][I]2:
Rate = k3k2/k–2[H2][(k1/k–1)1/2[I2]1/2]2
Rate = k3k2/k–2(k1/k–1)[H2][I2]
Combining k values:
Rate = k[H2][I2] which is consistent with the known rate law.
16.103 a) Conductometric method. The HBr that forms is a strong acid in water, so it dissociates completely into ions. As
time passes, more ions form, so the conductivity of the reaction mixture increases.
b) Manometric method. The reaction involves a reduction in moles of gas, so the rate can be determined from the
change in pressure (at constant volume and temperature) over time.
16.104 Plan: To solve this problem, a clear picture of what is happening is useful. Initially only N2O5 is present at a
pressure of 125 kPa. Then a reaction takes place that consumes the gas N2O5 and produces the gases NO2 and O2.
The balanced equation gives the change in the number of moles of gas as N2O5 decomposes. Since the number of
moles of gas is proportional to the pressure, this change mirrors the change in pressure. The total pressure at the
end, 178 kPa, equals the sum of the partial pressures of the three gases.
Solution:
Balanced equation: N2O5(g) → 2NO2(g) + 1/2O2(g)
Therefore, for each mole of dinitrogen pentaoxide that is consumed, 2.5 moles of gas are produced.
N2O5(g) → 2NO2(g) + 1/2O2(g)
Initial P (kPa)
125
0
0
total Pinitial = 125 kPa
Final P (kPa)
125 – x
2x
1/2x
total Pfinal = 178 kPa
Solve for x:
PN2 O5 + PNO2 + PO2 = (125 – x) + 2x + 1/2x = 178
x = 35.3333 kPa (unrounded)
Partial pressure of NO2 equals 2x = 2(35.3333) = 70.667 = 71 kPa.
Check: Substitute values for all partial pressures to find total final pressure:
(125 – 35.3333) + (2 x 35.3333) + ((1/2) x 35.3333) = 178 kPa
The result agrees with the given total final pressure.
16.105 Plan: For part a), first use the first-order half-life equation to find the rate constant k for the reaction. Then use the
first-order integrated rate law to find concentration of reactant at a later time, given the initial concentration. Since
the time unit in the rate constant is minutes, t must be expressed in units of minutes. For part b), use the first-order
integrated rate law to solve for the time required for 2/3 of the pill to decompose, leaving 1/3 pill at time t. For
part c) use the Arrhenius equation to calculate Ea.
Solution:
ln 2
ln 2
a) Rearrange t1/2 =
to k =
k
t1/2
ln 2
k=
= 7.7016x10–3 min–1 (unrounded)
90 min
⎛ 60 min ⎞⎟
Converting t from h to min: (2.5 h )⎜⎜
⎟ = 150 min
⎝ 1 h ⎟⎠
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16-41
ln
[aspirin ]t
= – kt or
[aspirin ]0
ln [aspirin]t = ln [aspirin]0 – kt
ln [aspirin]t = ln [2 mg/100 mL] – (7.7016x10–3 min–1)(150 min)
ln [aspirin]t = – 5.06729
[aspirin]t = 6.29964x10–3 mg/mL or
6.29964x10−3 mg
x 100 mL = 0.62996 mg/100 mL = 0.6 mg/100 mL
mL
b) The antibiotic pill = PILL. The pill is taken at the fever temperature, so use the fever k.
[ PILL ]t
= – kt
ln
[ PILL ]0
⎡1/3 PILL ⎤
⎢
⎦⎥ t = – (3.9x10–5 s–1) t
ln ⎣
[1 PILL ]0
⎛ 1 h ⎞⎟
t = 28169.55 s or (28169.55 s)⎜⎜
⎟ = 7.8 h
⎝ 3600 s ⎠⎟
c)
Pills should be taken at about eight hour intervals.
k1 = 3.1x10–5 s–1
T1 = [98.6°F – 32](5/9) + 273.15 = 310.15 K
k2 = 3.9x10–5 s–1
T2 = [101.9°F – 32](5/9) + 273 = 311.98 K
Ea = ?
k
E ⎛1 1⎞
ln 2 = a ⎜⎜ − ⎟⎟⎟
k1
R ⎜⎝ T1 T2 ⎟⎠
−5 −1 ⎞
⎛
J ⎟⎞⎜⎜ 3.9x10 s ⎟⎟⎟
⎛
⎛ k2 ⎟⎞
⎜⎜8.314
ln
⎜
⎜
⎟
R ⎜⎜ln ⎟⎟
⎝
⎠⎜ 3.1x10−5 s−1 ⎟⎟⎟⎟
mol • K ⎟⎜
⎝⎜
⎠
⎝ k1 ⎟⎠
Ea =
=
⎛
⎛ 1 1 ⎟⎞
⎟⎞
1
1
⎜
⎜⎜ − ⎟
−
⎟⎟⎟
⎜⎜⎜
⎜⎝ T1 T2 ⎟⎟⎠
⎜⎝ 310.15 K 311.98 K ⎟⎠
5
Ea = 1.0092x10 J/mol = 1x105 J/mol
The subtraction of the 1/T terms leaves only one significant figure.
(
(
)
)
16.106 No. The uncertainty in the pressure, P, is 5%. The reaction rate is proportional to [P]4. The relative reaction
rate with 5% error would be [1.05]4 = 1.22 or 22% in error. The rate measurement has an uncertainty of 22% so a
10% change in rate is not significant.
16.107 a) The iodide ion approaches from the side opposite the relatively large chlorine.
H
I
_
C
Cl
H
H
b) The “backside attack” of the I– inverts the geometry at the carbon bearing the Cl, producing this product:
CH2CH3
I
C
CH3
H
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16-42
c) The planar intermediate can be attached from either side, producing a racemic mixture (that is, an equal mixture
of two optical isomers):
CH2CH3
H3CH2C
I
C
C
I
and
CH3
H3C
H
H
16.108 Plan: The activation energy can be calculated using the Arrhenius equation. Although the rate constants, k1 and k2,
are not expressly stated, the relative times give an idea of the rate. The reaction rate is proportional to the rate
constant. At T1 = 90.0°C, the rate of reaction is 1 egg/4.8 min while at T2 = 100.0°C, the rate is 1 egg/4.5 min.
Therefore, rate1 = 1 egg/4.8 min and rate2 = 1 egg/4.5 min are substituted for k1 and k2, respectively.
Solution:
k1 = 1 egg/4.8 min
T1 = 90.0°C + 273.2 = 363.2 K
k2 = 1 egg/4.5 min
T2 = 100.0°C + 273.2 = 373.2 K
Ea = ?
k
E ⎛1 1⎞
ln 2 = a ⎜⎜ − ⎟⎟⎟
k1
R ⎜⎝ T1 T2 ⎟⎠
The number of eggs (1) is exact, and has no bearing on the significant figures.
⎛
⎞
J ⎟⎞⎜⎜ 1 egg/4.5min ⎟⎟⎟
⎛
⎛ k2 ⎞⎟
⎜⎜8.314
ln
⎜
⎜
⎟
⎟⎠
R ⎜⎜ln ⎟⎟
⎝
mol • K ⎟⎜
⎜⎜ 1 egg/4.8min ⎟⎟⎟
⎝
⎠
⎝ k1 ⎠⎟
Ea =
=
⎛ 1
⎛ 1 1 ⎞⎟
1 ⎞⎟⎟
⎜⎜
⎜⎜ − ⎟
−
⎟
⎜⎜
⎜⎝ T1 T2 ⎟⎟⎠
⎜⎝ 363.2 K 273.2 K ⎟⎟⎠
(
(
)
)
Ea = 7.2730x103 J/mol = 7.3x103 J/mol
16.109 Plan: The overall reaction can be obtained by adding the three steps together. The overall rate law for
the mechanism is determined from the slowest step (the rate-determining step). An overall rate law can only
include reactants and products; intermediates cannot be included in the rate law. Express [intermediate] in terms
of [reactant].
Solution:
(1) 2H2SO4 → H3O+ + HSO4– + SO3
[fast]
(2) SO3 + C6H6 → H(C6H5+)SO3–
[slow]
(3) H(C6H5+)SO3– + HSO4– → C6H5SO3– + H2SO4
[fast]
[fast]
(4) C6H5SO3– + H3O+ → C6H5SO3H + H2O
a) Add the steps together and cancel:
C6H6 + H2SO4 → C6H5SO3H + H2O
b) Initially: Rate = k2[SO3][C6H6]
(from the slow step)
SO3 is an intermediate and cannot be included in the overall rate law. SO3 is produced in step 1 and
its concentration is dependent on k1 and [H2SO4]:
[SO3] = k1[H2SO4]2
Substituting for [SO3] in the rate law from the slow step:
Rate = k2[k1[H2SO4]2][C6H6]
Rate = k[H2SO4]2[C6H6]
16.110 Plan: Starting with the fact that rate of formation of O (rate of step 1) equals the rate of consumption of O (rate of
step 2), set up an equation to solve for [O] using the given values of k1, k2, [NO2], and [O2].
Solution:
Rate2 = k2[O][O2}
a)
Rate1 = k1[NO2]
Rate1 = rate2
k1[NO2] = k2[O][O2]
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16-43
(6.0x10
)
s−1 ⎡⎢ 4.0x10−9 M ⎤⎥
⎣
⎦
= 2.4x10–15 M
6
−2
⎡
⎤
k2 [ O 2 ]
1.0x10 L/mol • s ⎢1.0x10 M ⎥
⎣
⎦
b) Since the rate of the two steps is equal, either can be used to determine rate of formation of ozone.
Rate2 = k2[O][O2] = (1.0x106 L/mol∙s)(2.4x10–15 M)(1.0x10–2 M) = 2.4x10–11 mol/L∙s
[O] =
k1 [ NO 2 ]
=
(
−3
)
16.111 Plan: At time = 0.00 min, assume [A]0 = 1.00; use the given equation for % inactivation to calculate [A]t at
3.00 min. Now knowing [A]0 and [A]t, use the first-order integrated rate law to calculate the rate constant, k in
part a). For part b), use the equation for % inactivation to calculate [A]t and then use the k value from part a) to
calculate time, t.
Solution:
a) Calculating [A]t at 3.00 min (99.9% inactivation):
% inactivation = 100 x (1 – [A]t/[A]0)
At 3.00 min: 99.9% = 100 x (1 – [A]t/1.00]
99.9% = 100 – 100[A]t
[A]t = 0.001
[ A ]t
ln
= – kt
[ A ]0
[ A ]t
[ 0.001]t
ln
ln
[ A ]0
[1.00 ]0
k= −
=−
= 2.302585 = 2.3 min–1
t
3.00 min
b) Calculating [A]t at 95% inactivation:
% inactivation = 100 x (1 – [A]t/[A]0)
95% = 100 x (1 – [A]t/1.00)
95% = 100 – 100[A]t
[A]t = 0.05
[ A ]t
ln
= – kt
[ A ]0
[ A ]t
[0.05]t
ln
ln
[ A ]0
[1.00 ]0
t= −
=−
= 1.30103 = 1.3 min
k
2.302585 min −1
16.112 Plan: The overall rate law for the mechanism is determined from the slowest step (the rate-determining step). An
overall rate law can only include reactants and products; intermediates cannot be included in the rate law. Express
[intermediate] in terms of [reactant]. Compare the resulting rate law from the mechanism to the actual rate law.
Solution:
I. Since there is only one step in the mechanism, it must be the rate-determining step:
Rate = k1[N2O5]2
This rate law does not match the actual rate law so the proposed mechanism is not valid.
II. The first step is the rate-determining step. From step 1:
Rate = k1[N2O5]2
This rate law does not match the actual rate law so the proposed mechanism is not valid.
III. The second step is the rate-determining step. From step 2:
Rate = k2[NO2][N2O5]
NO2 is an intermediate and cannot be in the final rate law. For an equilibrium, rateforward rxn = ratereverse rxn.
From step 1:
k1[N2O5] = k–1[NO3][NO2]; [NO2] = (k1/k–1)[N2O5]/[NO3]
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16-44
Substituting for [NO2]:
Rate = k2(k1/k–1)[N2O5]2[NO3]–1 = k[N2O5]2[NO3]–1
This rate law does not match the actual rate law so the proposed mechanism is not valid.
IV. The second step is the rate-determining step. From step 2:
Rate = k2[N2O3][O]
N2O3 and O are intermediates and cannot be in the final rate law. For an equilibrium, rateforward rxn = ratereverse rxn.
From step 1:
k1[N2O5]2 = k–1[NO2] 2[N2O3][O]3;
[O] = (k1/k–1) 1/3[N2O5] 2/3/[NO2] 2/3[N2O3]1/3
Substituting for [O]:
Rate = k2(k1/k–1)1/3[N2O5]2/3[NO2]–2/3[N2O3]2/3
This rate law does not match the actual rate law so the proposed mechanism is not valid.
V. The first step is rate determining:
Rate = k1[N2O5]2
This rate law does not match the actual rate law so the proposed mechanism is not valid.
16.113 Plan: This problem involves the first-order integrated rate law (ln [A]t/[A]0 = –kt). The temperature must be part
of the calculation of the rate constant. The concentration of the ammonium ion is directly related to the ammonia
concentration. Use the given values of [NH3]0 and [NH3]t and the calculated values of k to find time, t.
Solution:
a) [NH3]0 = 3.0 mol/m3
[NH3]t = 0.35 mol/m3
T = 20°C
0.095(T – 15°C)
k1 = 0.47e
= 0.47e0.095(20 – 15°C) = 0.75576667 d–1
[ NH3 ]t
ln
= – kt
[ NH3 ]0
ln
t= −
t= −
[ NH3 ]t
[ NH3 ]0
k
⎡ 0.35 mol/m 3 ⎤
⎦t
ln ⎣
⎡3.0 mol/m 3 ⎤
⎣
⎦0
= 2.84272 = 2.8 d
0.75576667 d−1
b) Repeating the calculation at the different temperature:
[NH3]0 = 3.0 mol/m3
[NH3]t = 0.35 mol/m3
T = 10°C
0.095(T – 15°C)
k1 = 0.47e
= 0.47e0.095(10 – 15°C) = 0.292285976 d–1
[ NH3 ]t
ln
[ NH3 ]0
t= −
k
⎡ 0.35 mol/m 3 ⎤
⎦t
ln ⎣
⎡3.0 mol/m 3 ⎤
⎦0
t= − ⎣
0.292285976 d−1
t = 7.35045 = 7.4 d
c) For NH4+ the rate = k1[NH4+]
From the balanced chemical equation:
Δ[ NH 4 + ]
1 Δ[O2 ]
−
=−
2 Δt
Δt
Thus, for O2: Rate = 2 k1[NH4+]
Rate = (2)(0.75576667) ⎡⎢3.0 mol/m 3 ⎤⎥ = 4.5346 = 4.5 mol/m3
⎣
⎦
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16-45
16.114 Plan: The rate law is rate = [CS2]m where m is the order of the reactant. To find the order of the reactant, take the
ratio of the rate laws for two experiments. Once the rate law is known, any experiment can be used to find the rate
constant k.
Solution:
m
⎛ [CS2 ]exp 1 ⎞⎟
Rate exp 1
⎜
⎟
⎜
=⎜
a)
⎟
⎜⎝[CS ] ⎠⎟⎟
Rate
exp 4
2 exp 4
m
−7
2.7x10 mol/L • s ⎛ 0.100 mol/L ⎟⎞
= ⎜⎜
⎝ 0.044 mol/L ⎟⎟⎠
1.2x10−7 mol/L • s
2.25 = (2.27273)m
log (2.25) = m log (2.27273)
m=1
Rate = k [CS2]
b) First, calculate the individual k values; then average the values.
k = rate/[CS2]
k1 = (2.7x10–7 mol/L∙s)/(0.100 mol/L) = 2.7x10–6 s–1
k2 = (2.2x10–7 mol/L∙s)/(0.080 mol/L) = 2.75x10–6 = 2.8x10–6 s–1
k3 = (1.5x10–7 mol/L∙s)/(0.055 mol/L) = 2.7272x10–6 = 2.7x10–6 s–1
k4 = (1.2x10–7 mol/L∙s)/(0.044 mol/L) = 2.7272x10–6 = 2.7x10–6 s–1
kavg = [(2.7x10–6 s–1) + (2.75x10–6) + (2.7272x10–6) + (2.7272x10–6)]/4 = 2.7261 10–6 = 2.7x10–6 s–1
16.115 a) The reaction is C=C + H2(g) ↔ C − C. Since the hydrogenation and dehydrogenation reactions are reversible,
the direction of reaction is determined by the hydrogen pressure.
b) Dehydrogenation will require a higher temperature. Hydrogenation, adding hydrogen to the double bond in the
alkene, is exothermic. The hydrogenated product is of lower energy than the dehydrogenated reactant. The
reaction pathways are the same but in reverse order so the hydrogenated material has a larger activation energy
and thus a higher temperature is needed to obtain a useful reaction rate for dehydrogenation.
Energy
Ea (hydrogenation)
Ea(dehydrogenation)
Alkene + H2
ΔHrxn
Hydrogenated product
Reaction coordinate
c) In the hydrogenation process, when the double bond has been broken and one hydrogen atom has been added to
the bond, the molecule can rotate around the resulting single bond and then lose a hydrogen atom (since
hydrogenation and dehydrogenation are reversible) to restore the double bond and produce the trans fat.
16.116 Plan: Rate is the change in concentration divided by change in time. To find the average rate for each
trial in part a), the change in concentration of S2O32– is divided by the time required to produce the color. The rate
law is rate = k[I–]m[S2O82–]n where m and n are the orders of the reactants. To find the order of each reactant, take
the ratio of the rate laws for two experiments in which only the reactant in question changes. Once the rate law is
known, any experiment can be used to find the rate constant k. Since several solutions are mixed, final
concentrations of each solution must be found with dilution calculations using MiVi = MfVf in the form:
Mf = MiVi/Vf.
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16-46
Solution:
a) Mf S2O32– = [(10.0 mL)(0.0050 M)]/50.0 mL = 0.0010 M S2O32–
2−
Δ[ I 2 ]
1 Δ ⎡⎣S2 O3 ⎤⎦
−
= [1/2(0.0010 M)]/time = [0.00050 M]/time = rate
=−
Δt
2
Δt
Average rates:
Rate1 = (0.00050 M)/29.0 s = 1.724x10–5 = 1.7x10–5 M s−1
Rate2 = (0.00050 M)/14.5 s = 3.448x10–5 = 3.4x10–5 M s−1
Rate3 = (0.00050 M)/14.5 s = 3.448x10–5 = 3.4x10–5 M s−1
b) Mf KI = Mf I– = [(10.0 mL)(0.200 M)]/50.0 mL = 0.0400 M I– (Experiment 1)
Mf I– = [(20.0 mL)(0.200 M)]/50.0 mL = 0.0800 M I– (Experiments 2 and 3)
Mf Na2S2O8 = Mf S2O82– = [(20.0 mL)(0.100 M)]/50.0 mL = 0.0400 M S2O82– (Experiments 1 and 2)
Mf S2O82– = [(10.0 mL)(0.100 M)]/50.0 mL = 0.0200 M S2O82– (Experiment 3)
Generic rate law equation: Rate = k [I–]m[S2O82–]n
To find the order for I–, use experiments 1 and 2 in which [S2O82–] is constant while [I–] changes. Set up a ratio of
the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and solve for m, the
order with respect to [I–].
n
m
k2 [ I− ] ⎡⎣S2 O82− ⎤⎦
Rate 2
Molarity of S2O82–is constant.
=
n
m
Rate1
k1 [ I− ] ⎡⎣S2 O82− ⎤⎦
m
3.4x10−5 Ms−1
[ 0.0800 ]
=
m
1.7x10−5 Ms−1
[ 0.0400 ]
2.0 = (2.00)m
m=1
The reaction is first order with respect to I–.
To find the order for S2O82–, use experiments 2 and 3 in which [I–] is constant while [S2O82–] changes. Set up a ratio
of the rate laws for experiments 2 and 3 and fill in the values given for rates and concentrations and solve for n,
the order with respect to [S2O82–].
n
m
k 3 [ I− ] ⎡⎣S2 O82− ⎤⎦
Rate3
Molarity of I– is constant.
=
− m ⎡
2− ⎤ n
Rate 2
k 2 [ I ] ⎣S2 O8 ⎦
n
3.4x10−5 Ms−1
[ 0.0200 ]
=
n
3.4x10−5 Ms−1
[ 0.0400 ]
1.0 = (0.500)n
n=0
The reaction is zero order with respect to S2O82–.
c) Rate = k[I–]
k = rate/[I–]
Using experiment 2 (unrounded rate value)
k = (3.448x10–5 M s–1)/(0.0800 M) = 4.31x10–4 = 4.3x10–4 s–1
d) Rate = (4.3x10–4 s–1)[I–]
16.117 The ability of the bath to remove heat is proportional to ΔT, the difference between the bath temperature and
the temperature inside the flask. Therefore when the temperature is reached such that the rate of heat increase
exceeds the rate of heat loss, the reaction “runs away.”
The problem with the scale-up is that the heat transfer from the flask to the cooling bath is proportional to the
shared surface area of the reactant solution and the cooling bath, A, while the heat given off by the reaction is
proportional to the volume of the reactants. The volume increases as the cube of the radius of the flask increases
while the area increases as the square of the radius increases. Therefore the heat generation will exceed the
cooling capacity at a lower temperature in the larger flask and the reaction will run away.
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16-47
16.118 Plan: This is a first-order process so use the first-order integrated rate law. The increasing cell density changes the
integrated rate law from –kt to +kt. In part a), we know t(2 h), k, and [A]0 so [A]t can be found. Treat the
concentration of the cells as you would molarity. Since the rate constant is expressed in units of min–1, the time interval
of 2 h must be converted to a time in minutes. In part b), [A]0, [A]t, and k are known and time t is calculated.
Solution:
⎛ 60 min ⎞⎟
a) Converting time in h to min: (2 h )⎜⎜
⎟ = 120 min
⎝ 1 h ⎟⎠
ln [A]t = ln [A]0 + kt
ln [A]t = ln [1.0x103] + (0.035 min–1)(120 min)
ln [A]t = 11.107755
[A]t = 6.6686x104 = 7x104 cells/L
b) ln [A]t = ln [A]0 + kt
ln [A]t − ln [A]0
=t
k
ln [2x103 cells/L] − ln [1x103 cells/L]
=t
0.035 min−1
t =19.804 = 2.0x101 min
16.119 a) The shape is tetrahedral and the hybridization of C1 is sp3.
H
C
H
1
I
CH3
b) The shape is trigonal bipyramidal. Because an unhybridized p orbital is needed to overlap p orbitals on I and
Br, the hybridization around C1 is sp2.
H
1
Br
H
C
I
CH3
–
c) After Br is replaced with I– in the initial replacement reaction, the ethyl iodide is optically active. However, as
other I– ions react with the ethyl iodide by the same mechanism the molecules change from one isomer to the
other. Eventually, equal portions of each isomer exist and the ethyl iodide is optically inactive.
16.120 Plan: This is a first-order reaction so use the first-order integrated rate law. The first-order half-life equation
is used to find the rate constant, k. Then, knowing k, time, and [A]0, [A]t can be calculated. The fraction
remaining is [A]t/[A]0.
Solution:
ln 2
ln 2
Rearrange t1/2 =
to k =
k
t1/2
ln 2
k=
= 0.086212 d–1 (unrounded)
8.04 d
ln [A]t = ln [A]0 – kt
ln [A]t = ln [1.7x10–4] – (0.086212 d–1)(30. d)
ln [A]t = –11.26607212
[A]t = 1.27999x10–5 M (unrounded)
Fraction remaining = [A]t/[A]0 = (1.27999x10– 5 M)/(1.7x10–4 M) = 0.0752936 = 0.075
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16-48
16.121 Plan: For part a), use the Monod equation to calculate μ for values of S between 0.0 and 1.0 kg/m3 and then
graph. For parts b) and c), use the Monod equation to calculate μ at the given conditions. The value of μ is
the rate constant k in the first-order integrated rate law. The increasing population density changes the integrated
rate law from –kt to +kt. We know t(1 h), k, and [A]0 so [A]t can be found. Treat the density of the cells as you
would molarity.
Solution:
(
)(
) = 1.34x10 s
(
) (
)
(1.5x10 s )(0.50 kg/m ) = 1.42x10 s
μ S
μ=
=
K +S
(0.03 kg/m ) +(0.50 kg/m )
(1.5 x10 s )(0.75 kg/m ) = 1.44x10 s
μ S
μ=
=
K +S
(0.03 kg/m ) +(0.75 kg/m )
(1.5x10 s )(1.0 kg/m ) = 1.46x10 s
μ S
μ=
=
K +S
(0.03 kg/m ) +(1.0 kg/m )
(1.5x10 s )(0.30 kg/m ) = 1.3636x10
μ S
b) μ =
=
K +S
(0.03 kg/m ) +(0.30 kg/m )
μ=
1.5x10−4 s−1 0.25 kg/m 3
μmax S
=
Ks + S
0.03 kg/m 3 + 0.25 kg/m 3
−4
−1
– 4
–1
3
max
3
– 4
–1
– 4
–1
3
s
−4
−1
3
max
3
3
s
−4
−1
3
– 4
max
3
–1
3
s
−4
−1
3
–4
max
3
3
s–1
s
⎛ 3600 s ⎟⎞
= 3600 s
Converting time in h to seconds: (1 h )⎜⎜
⎝ 1 h ⎟⎟⎠
ln [A]t = ln [A]0 + kt
ln [A]t = ln [5.0x103] + (1.3636x10–4 s–1)(3600 s)
ln [A]t = 9.00808919
[A]t = 8.1689x103 = 8.2x103 cells/m3
c) μ =
μmax S
Ks + S
(1.5x10 s )(0.70 kg/m )
(0.03 kg/m ) +(0.70 kg/m )
−4
=
−1
3
3
3
= 1.438356x10–4 s–1
ln [A]t = ln [A]0 + kt
ln [A]t = ln [5.0x103] + (1.438356x10–4 s–1)(3600 s)
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16-49
ln [A]t = 9.03500135
[A]t = 8.39172x103 = 8.4x103 cells/m3
16.122 Plan: The rate law is rate = k[A]m[B]n where m and n are the orders of the reactants. The initial rate for each
reaction mixture is given. Calculate the concentration of A and B in each mixture. To find the order of each
reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes. Once
the rate law is known, Reaction Mixture I, II, or III can be used to find the rate constant k for the reactions without
the solid and Reaction Mixture IV can be used to find the rate constant k for the reaction with a solid present.
Solution:
a) Reaction Mixture I
⎛ 0.01 mol ⎟⎞⎛ 1 ⎞
⎟ = 0.10 mol/L A
⎟⎜
Concentration of A = (5 spheres)⎜⎜
⎜⎝ 1 sphere ⎟⎟⎠⎜⎝ 0.50 L ⎟⎟⎠
⎛ 0.01 mol ⎞⎟⎛ 1 ⎞
⎟ = 0.10 mol/L B
⎟⎜
Concentration of B = (5 spheres)⎜⎜
⎜⎝ 1 sphere ⎟⎟⎠⎜⎝ 0.50 L ⎟⎠⎟
Reaction Mixture II
⎛ 0.01 mol ⎞⎟⎛ 1 ⎞
⎟ = 0.16 mol/L A
⎟⎜
Concentration of A = (8 spheres)⎜⎜
⎜⎝ 1 sphere ⎠⎟⎟⎜⎝ 0.50 L ⎟⎟⎠
⎛ 0.01 mol ⎞⎟⎛ 1 ⎞
⎟ = 0.10 mol/L B
⎟⎜
Concentration of B = (5 spheres)⎜⎜
⎜⎝ 1 sphere ⎟⎟⎠⎜⎝ 0.50 L ⎟⎟⎠
Reaction Mixture III
⎛ 0.01 mol ⎞⎟⎛ 1 ⎞
⎟ = 0.16 mol/L A
⎟⎜
Concentration of A = (8 spheres)⎜⎜
⎜⎝ 1 sphere ⎟⎠⎟⎜⎝ 0.50 L ⎟⎟⎠
⎛ 0.01 mol ⎞⎟⎛ 1 ⎞
⎟ = 0.14 mol/L B
⎟⎜
Concentration of B = (7 spheres)⎜⎜
⎜⎝ 1 sphere ⎟⎟⎠⎜⎝ 0.50 L ⎟⎠⎟
To find the order for A, use Mixtures I and II in which [B] is constant while [A] changes. Set up a ratio of the rate
laws for Mixtures I and II and fill in the values given for rates and concentrations and solve for m, the order with
respect to [A].
m
n
Rate II
k [ A ] [ B]
The concentration of B is constant
= 2 m n
Rate I
k1 [ A ] [ B]
5.6x10−4 mol/L • s
3.5x10−4 mol/L • s
m
=
[ 0.16 ]
m
[ 0.10 ]
1.6 = (1.6)m
m=1
The reaction is first order with respect to A.
To find the order for B, use Mixtures II and III in which [A] is constant while [B] changes. Set up a ratio of the
rate laws for Mixtures II and III and fill in the values given for rates and concentrations and solve for n, the order
with respect to [B].
m
Rate III
k [ A ] [ B]n
= 2 m n
Rate II
k1 [ A ] [ B]
5.6x10−4 mol/L • s
5.6x10−4 mol/L • s
The concentration of A is constant.
n
=
[ 0.14 ]
n
[ 0.10 ]
1 = (1.4)n
n=0
The reaction is zero order with respect to B.
Rate law: Rate = k[A][B]0 = k[A]
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16-50
b) The overall reaction order is 1 + 0 = 1.
c) Use Reaction Mixture I:
Rate = k[A]
3.5x10–4 mol/L·s = k[0.10]
k = 3.5x10–3 s–1
d) The catalyst was used in Reaction Mixture IV.
⎛ 0.01 mol ⎞⎟⎛ 1 ⎞
⎟ = 0.10 mol/L A
⎟⎜
Concentration of A = (5 spheres)⎜⎜
⎜⎝ 1 sphere ⎟⎟⎠⎜⎝ 0.50 L ⎟⎟⎠
⎛ 0.01 mol ⎞⎟⎛ 1 ⎞
⎟ = 0.16 mol/L B
⎟⎜
Concentration of B = (8 spheres)⎜⎜
⎜⎝ 1 sphere ⎟⎠⎟⎜⎝ 0.50 L ⎟⎠⎟
Rate = k[A]
4.9x10–4 = k[0.10]
k = 4.9x10–3 s–1
Yes, the gray pellets had a catalytic effect. The rate of reaction and the rate constant are greater with the pellets
than without.
ΔI
ΔI
= k I or –
= k Δ,
Δ,
I
where k = fraction of light removed per unit length, <, and I = light intensity.
At length , = 0, I0 = intensity of light entering the solution.
At some later length <, I< = intensity of light leaving the solution.
I, ΔI
,
–∫
= k ∫ Δ,
I0
,= 0
I
I0
ln
= k ( , − 0)
I,
ln I0 – ln I< = k x <
ln I< – ln I0 = –k x <
I
ln , = − k x , = – fraction of light removed per unit length x distance (length) traveled
I0
ΔS
ΔS
= k S or –
= k Δt
b) Rate of savings decline = –
Δt
S
where k = fraction of savings lost per unit of time, t and S = savings.
At time t = 0, S0 = initial value of savings.
At some later time t, St = value of savings remaining.
t
St ΔS
–∫
= k ∫ Δt
t =0
S0 S
S
ln 0 = k (t − 0)
St
16.123 a) Rate of light intensity decrease = –
ln S0 – ln St = k x t
ln St – ln S0 = –k x t
S
ln t = − k x t = – fraction of savings lost per unit time x savings time interval
S0
16.124 Plan: The figure shows the H2 molecule adsorbing to the metal surface with H2 bond breakage. The
individual H atoms form bonds to the metal catalyst atoms. Ethylene also adsorbs to the metal catalyst and then
the two C–H bonds form, one at a time. The resulting C2H6 leaves the metal surface. The overall reaction can be
obtained by adding the steps of the mechanism together.
Solution:
H2(g) → H2(ads)
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16-51
H2(ads) + 2M → 2M–H
C2H4(g) → C2H4(ads)
C2H4(ads) + M–H → C2H5(ads) + M
C2H5(ads) + M–H → C2H6(g) + M
C2H4(g) + H2(g) → C2H6(g)
16.125 Plan: Rate is proportional to the rate constant, so if the rate constant increases by a certain factor, the rate
increases by the same factor. Thus, to calculate the change in rate the Arrhenius equation can be used and
substitute ratecat/rateuncat = kcat/kuncat.
Solution:
k = Ae–Ea/RT
ΔEa = 5 kJmol = 5000 J/mol
T = 37°C + 273 = 310 K
− Ea 2
( Ea1 − Ea2 )
Ae RT
k2
=
= e
−Ea1
RT
k1
Ae RT
k
E − Ea2
5000 J/mol
ln 2 = a1
=
= 1.93998
J ⎞⎟
⎛
k1
RT
⎜⎜8.314
⎟(310 K )
⎝
mol • K ⎠
k2
= e1.93998 = 6.9586 = 7
The rate of the enzyme-catalyzed reaction occurs at a rate 7 times faster
k1
than the rate of the uncatalyzed reaction at 37°C.
16.126 Plan: This is a first-order reaction so use the first-order integrated rate law. In part a), use the first-order half-life
equation and the given value of k to find the half-life. In parts b) and c), solve the first-order integrated rate law to
find the time necessary for 40.% and 90.% of the acetone to decompose. If 40.% of the acetone has decomposed,
100 – 40. = 60.% remains at time t. If 90.% of the acetone has decomposed, 100 – 90. = 10.% remains at time t.
Solution:
ln 2
ln 2
a) t1/2 =
=
= 79.672 = 8.0x101 s
k
8.7x10−3 s−1
b) ln [acetone]t = ln [acetone]0 – kt
ln [acetone]t − ln [acetone]0
=t
40.% of acetone has decomposed; [acetone]t = 100 – 40. = 60.%
−k
ln [60.] − ln [100]
= t = 58.71558894 = 59 s
−8.7x10−3 s−1
c) ln [acetone]t = ln [acetone]0 – kt
ln [acetone]t − ln [acetone]0
=t
90.% of acetone has decomposed; [acetone]t = 100 – 90. = 10.%
−k
ln [10.] − ln [100]
= t = 264.66 = 2.6x102 s
−8.7x10−3 s−1
16.127 Plan: The reaction begins with B as the reactant; the final product is A. Isomer C is formed during the reaction
but is not a final product.
Solution:
a) B → C
C→A
B → A
b) C is an intermediate.
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16-52