Silberberg7e solution manual ch 16
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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 A
expressed as
, where a is the coefficient of the reactant or product A. The rate of the reactants is given a
a t
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.
O2
1 NO
1 N 2 O3
=
=
Rate =
t
t
4 t
2
b) Plan: Because NO is decreasing; its rate of concentration change is negative. Substitute the negative value into
the expression and solve for [O2]/t.
Solution:
O2
1
Rate = 1.60 x 104 mol/L • s =
= 4.00x10–5 mol/L·s
t
4
16.1B
Plan: Examine the equation that expresses the rate in terms of the change in concentration with time for each
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 A
term,
, describes a product (the term is positive) that has a coefficient of a in the balanced equation. In
a t
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 [O2]/t.
Solution:
1
Rate = (2.52x10–2 mol/L•s) = –
5
6
6
– (2.52x10–2 mol/L•s) =
∆[O2 ]
∆t
1 ∆[O2 ]
5
∆t
= –2.10x10–2 mol/L·s
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.
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.
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16-1
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:
rate 3 H 2 3
=
rate 1 H m
2 1
m
9.3 x 1023
1.9 x 1023
=
0.05503m
0.01131m
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:
rate 4 I 2 4
=
rate 2 I n
2 2
n
1.9 x 1022
1.1 x 1022
=
0.0056n4
0.0033n2
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
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
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16-2
[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 [H2 SeO3 ]m
3
=
m
[H
SeO
]
rate 1
2
3 1
m
–2
3.94x10–6 mol/L•s [1.0x10 mol/L]3
=
m
9.85x10–7 mol/L•s [2.5x10–3 mol/L]1
4 = (4)m
Therefore, m = 1
For the reaction order with respect to [I–], compare Experiments 1 and 2:
rate 2
rate 1
n
=
[I–]2
n
[I–]1
n
–2
7.88x10–6 mol/L•s [3.0x10 mol/L]2
=
n
9.85x10–7 mol/L•s [1.5x10–2 mol/L]
1
8 = (2)n
Therefore, n = 3
For the reaction order with respect to [H+], compare Experiments 2 and 4:
rate 4
rate 2
p
=
[H+]4
p
[H+]2
p
–2
3.15x 10–5 mol/L•s [3.0x10 mol/L]4
7.88x10–6 mol/L•s
=
p
[1.5x10–2 mol/L]2
4 = (2)p
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 105
0.25 x 105
4 =
=
x 2
32
x 2
9
x = 6 black spheres
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16-3
16.4B
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
A
t 0
1
1
= (2.4x10–21 L/mol•s)(t)
0.00900 mol/Lt 0.0100 mol/L0
t=
1
1
0.00900 mol/L 0.0100 mol/L
2.4 x1021 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)
[A]0
= kt
[A]t
1.28 M
ln
= (k)(10.0 min)
0.85M
0.4094
ln
k=
10.0 min
= 0.04094 = 0.041 min–1
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).
ln
1.00 M
= (0.041 min–1)(t)
0.75M
0.2877
t=
0.041 min–1
= 7.0171 = 7.0 min
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16-4
16.6A
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:
min
12 particles of X 2.5
min
6 particles of X 2.5
3 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:
min
12 particles of X 2.5
min
6 particles of X 2.5
3 particles of X 2.5
min
2.5 min
1.5 particles of X
0.75 particle of X
0.20 mol
Moles of X after 10.0 min = 0.75 particle
= 0.15 mol
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
16 A particles
8 A particles
4 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)
Number of half-lives at 72 min =
72 min
24 min/half-life
24 min
2 A particles
24 min
1 A particle
= 3 half-lives
According to the scheme above in part a), there are 2 A particles left after three half-lives.
Amount (mol) of A after 72 minutes = (2 particles A)
M=
16.7A
0.20 mol A
0.25 L
0.10 mol A
1 particle A
= 0.20 mol A
= 0.80 M
Plan: Rearrange the first-order half-life equation to solve for k.
Solution:
ln 2
ln 2
=
k=
= 0.0529119985 = 0.0529 h–1
t1/2
13.1 h
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16-5
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)
t1/2 =
ln 2
k
=
ln 2
9 x 10–2 day–1
= 7.7016 = 8 days
b) Number of half-lives at 40 days =
40 days
8 days/half-life
= 5 half-lives
After each half-life, ½ of the sample remains.
After five half-lives, ½ x ½ x ½ x ½ x ½ = (½)5 =
16.8A
1
32
of the sample remains.
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
T1 = 500. K
Ea = 1.00x102 kJ/mol
k2 = ? L/mol•s
T2 = 490. K
k2
1.00 x102 kJ/mol 1
1 103 J
ln
=
0.286 L/mol • s
8.314 J/mol • K 500.K 490.K 1 kJ
k2
ln
= –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:
ln
k2
E 1 1
= a
k1
R T1 T2
k1 = 7.02x10–3 L/mol•s
k2 = 2(7.02x10–3) L/mol•s
ln
2 7.02x10–3 L/mol•s
–3
7.02x10 L/mol•s
5.055110x10–5 K–1 =
1
T2
=
T1 = 500. K
T2 = unknown
1.14x105 J/mol
1
8.314 J/mol•K
500. K
1
500. K
–
–
Ea = 1.14x105 J/mol
1
T2
1
T2
= 1. 9494489x10–3 K–1
T2 = 513 K
16.9A
Plan: Begin by using Sample Problem 16.9 as a guide for labeling the diagram.
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16-6
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 Hrxn and Ea(rev).
Transition state:
O
O
H
H
Energy
Ea(rev)= +6 kJ
Ea(fwd)= +78 kJ
Hrxn= +72 kJ
Reaction progress
16.9B
Plan: Ea(fwd) is the sum of Hrxn and Ea(rev), so Hrxn can be calculated by subtracting Ea(fwd) – Ea(rev). Use Sample
Problem 16.9 as a guide for drawing the diagram.
Solution:
Hrxn = Ea(fwd) – Ea(rev)
Hrxn = 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:
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)
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16-7
b)
c)
(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)
(1) Unimolecular
(2) Bimolecular
(3) Bimolecular
(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)
(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
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16-8
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.
`
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
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16-9
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
16.4
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.
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
16.11
Concentration
Concentration
B(g)
A(g)
B(g)
Time
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.
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16-10
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 [AX 2 ]
=
= 0.00103 = 0.0010 mol/L•s
Rate =
2
2 t
20.0 s 0 s
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:
1 0.0088 mol/L 0.0249 mol/L
1 [AX 2 ]
a) Rate =
=
= 6.70833x10–4 = 6.71x10–4 mol/L•s
2
2 t
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 [A]
[B]
[C]
Rate =
=
=
2 t
t
t
Calculating the rate of change of [A]:
1 [A]
[C]
=
2 t
t
2 mol A/L•s
2 mol C/L•s
= –4 mol/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:
[D]
2 [E]
2 [F]
Rate =
=
=
t
3 t
5 t
Calculating the rate of change of [F]:
5/2 mol F/L•s
= 0.416667 = 0.42 mol/L•s
0.25 mol E/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
formed and [C] increases over time. The 1/2 factor is included for reactant B to account for the stoichiometric
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16-12
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:
[A]
1 [B]
[C]
Rate =
=
=
t
2 t
t
Calculating the rate of change of [A]:
1 mol A/L•s
0.5 mol B/L•s
= – 0.25 mol/L•s = – 0.2 mol/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 [D]
1 [E]
[F]
1 [G]
[H]
=
=
=
=
2 t
3 t
t
2 t
t
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
Rate =
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:
[NOBr]
0.0033 0.0100 mol/L
a) Rate =
–
= 6.7x10–4 mol/L•s
10.00 0.00 s
t
0.0055 0.0071 mol/L
= 8.0x10–4 mol/L•s
b) Rate = –
4.00 2.00 s
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16-13
c)
Initial Rate = – y/ x = – [(0.0040 – 0.0100) mol/L]/[4.00 – 0.00) s] = 1.5x10–3 mol/L•s
d) Rate at 7.00 s = – [(0.0030 – 0.0050) mol/L]/[11.00 – 4.00) s] = 2.857x10–4 = 2.9x10–4 mol/L•s
e) Average between t = 3 s and t = 5 s is:
Rate = – [(0.0050 – 0.0063) mol/L]/[5.00 – 3.00) s] = 6.5x10–4 mol/L•s
Rate at 4 s 6.7x10–4 mol/L•s, thus the rates are equal at about 4 seconds.
16.21
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 N2 and H2 since these substances are reacting
and [N2] and [H2] are decreasing over time. A positive sign is used for the rate in terms of the product NH3 since
it is being formed and [NH3] increases over time.
Solution:
[N 2 ]
1 [H 2 ] 1 [NH3 ]
Rate =
=
3 t
2 t
t
16.22
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 the reactant O2 since it is reacting and [O2] is
decreasing over time. A positive sign is used for the rate in terms of the product O3 since it is being formed and
[O3] increases over time. O3 increases 2/3 as fast as O2 decreases because two molecules of O3 are formed for
every three molecules of O2 that disappear.
Solution:
1 [O 3 ]
1 [O 2 ]
a) Rate =
=
2 t
3 t
b) Use the mole ratio in the balanced equation:
2.17x10 5 mol O 2 /L•s 2 mol O 3 /L•s
–5
= 1.45x10 mol/L•s
3
mol
O
/L•s
2
16.23
a) k is the rate constant, the proportionality constant in the rate law. k represents the fraction of successful
collisions which includes the fraction of collisions with sufficient energy and the fraction of collisions with
correct orientation. k is a constant that varies with temperature.
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.
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16-14
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
k=
=
=
1
2
[A]1[B]2
mol 3
mol mol
L L
L3
k=
mol L3
L•min mol 3
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.
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.
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16-15
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.
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.
16.33
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].
rateexp 2
rateexp 1
[A]exp 2
=
[A]exp 1
m
m
45.0 mol/L•min 0.300 mol/L
=
5.00 mol/L•min 0.100 mol/L
9.00 = (3.00)m
log (9.00) = m log (3.00)
m=2
Using experiments 3 and 4 also gives second order with respect to [A].
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].
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16-16
rateexp 3
rateexp 1
[B]exp 3
=
[B]exp 1
n
n
10.0 mol/L•min 0.200 mol/L
=
5.00 mol/L•min 0.100 mol/L
2.00 = (2.00)n
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
[A] [B]
[0.100 mol/L] 2 [0.100 mol/L]
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].
rateexp 2
rateexp 1
[A]exp 2
=
[A]exp 1
m
1.25x102 mol/L•min 0.1000 mol/L
=
6.25x103 mol/L•min 0.0500 mol/L
2.00 = (2.00)m
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].
m
rateexp 3
rateexp 2
[B]exp 3
=
[B]exp 2
n
5.00 x 102 mol/L•min 0.1000 mol/L
=
1.25 x 102 mol/L•min
0.0500 mol/L
n
4.00 = (2.00)
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].
n
rateexp 4
rateexp 1
[C]exp 4
=
[C]exp 1
p
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16-17
6.25x103 mol/L•min 0.0200 mol/L
=
6.25x103 mol/L•min 0.0100 mol/L
1.00 = (2.00)p
log (1.00) = p log (2.00)
p=0
The reaction is zero order with respect to C.
b) Rate = k[A]1[B]2[C]0
Rate = k[A][B]2
c) Using the data from experiment 1 to find k:
Rate = k[A][B]2
rate
6.25 x 103 mol/L min
k=
= 50.0 L2/mol2•s
2 =
2
[A][B]
[0.0500 mol/L][0.0500 mol/L]
p
16.36
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
L•time
x
k=
=
=
= 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•time
L
mol
mol
L2
L
L•time
x
k=
=
=
2
2
L•time mol
mol•time
mol
2
L
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
3
mol
mol
=k
L•time
L
mol
mol
L3
L2
L•time
x
k=
=
=
L•time mol3
mol 2 •time
mol3
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:
mol
mol
= k
L•time
L
5/2
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16-18
mol
mol
L5/2
L3/2
x
k = L•time
=
=
3/2
5/2
L•time mol
mol •time
mol5/2
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
L = mol
L•s
m
m
mol
L =1
b) Rate = k[A]m
m must be 0. The reaction is zero order.
m
mol
1 mol
=
L•yr
yr L
mol
m
mol
yr
L•yr
mol
L = 1 = L yr x 1
yr
mol
L
m
=
mol
L
m must be 1. The reaction is first order.
c) Rate = k[A]m
mol
mol1/2 mol
= 1/2
L•s
L •s L
mol
m
1/2
mol
L•s = mol x L •s
=
L
L•s
mol1/2
mol1/2
L1/2 •s
m
mol
L
m
=
mol1/2
m must be 1/2. The reaction is 1/2 order.
L1/2
d) Rate = k[A]m
mol
mol 5/2 mol
= 5/2
L• min
L • min L
mol
m
5/2
mol
L• min = mol x L •min
=
L
L•min
mol5/2
mol 5/2
5/2
L • min
m
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16-19
mol
L
16.38
m
=
mol7/2
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].
rateexp 1
rateexp 2
[CO]exp 1
=
[CO]exp 2
m
1.29x1029 mol/L•min 1.00 mol/L
=
1.33x1030 mol/L•min 0.100 mol/L
9.699 = (10.0)m
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].
m
rateexp 3
rateexp 2
[Cl2 ]exp 3
=
[Cl2 ]exp 2
n
1.30x10 29 mol/L•min 1.00 mol/L
=
1.33x1030 mol/L•min 0.100 mol/L
n
9.774 = (10.0)
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
n
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.
16.40
The half-life (t1/2) 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 onlyon 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
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16-20
[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
AB
t 0
1
1
AB AB
t
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
second-order 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
ABt AB0
1
ABt
= kt +
1
AB0
1
1
= (0.2 L/mol•s) (10.0 s) +
1.50
M
AB
t
1
1
= 2.66667
M
ABt
[AB]t = 0.375 = 0.4 M
16.43
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=
=
= 0.066014 = 0.0660 min–1
t1/2 =
t1/2
k
10.5 min
b) ln [A]t = ln [A]0 – kt
ln[A]t ln[A]0
=t
k
ln[0.25] ln[1]
=t
0.0660 min 1
t = 21.0045 = 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.
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16-21
16.44
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]t = 0.125 M k = 0.0012 yr–1
[A]0 = 1.00 M
1.00 M
–1
ln
= (0.0012 yr ) t
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.
16.45
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.)
ln[NH3]
x-axis (time, s)
0
1.000
2.000
[NH3]
4.000 M
3.986 M
3.974 M
y-axis (ln [NH3])
1.38629
1.38279
1.37977
1.387
1.386
1.385
1.384
1.383
1.382
1.381
1.38
1.379
0
0.5
1
1.5
2
Time, s
b) t1/2
16.46
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
=
= 212.62 = 2x102 s
=
k
3.260x103 s1
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.
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16-22
16.47
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.
16.48
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.
16.49
k = A e 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.
16.50
The Arrhenius equation, k = A e 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.
16.51
a) The value of k increases exponentially with temperature.
b) A plot of ln k vs. 1/T is a straight line whose slope is –Ea/R.
a)
b)
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16-23
The activation energy is determined from the slope of the line in the ln k vs. 1/T graph. The slope equals –Ea/R.
16.52
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.
16.53
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.
16.54
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
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.
16.55
Each A particle can collide with three B particles, so (4 x 3) = 12 unique collisions are possible.
16.56
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:
6.022x1023 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 = (6.08222 x 1023 particles of A)(1.279997 x 1024 particles of B)
= 7.76495x1047 = 7.76x1047 unique collisions
16.57
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
E
100. kJ/ mol
a =
= –40.362096
RT
8.314x103 kJ /mol•K 298 K
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16-24
Fraction = e Ea / RT = e–40.362096 = 2.9577689x10–18 = 2.96x10–18
16.58
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
E
100. kJ / mol
a =
= –37.238095
RT
8.314x103 kJ /mol•K 323 K
–37.238095
Fraction = e Ea / RT = e
= 6.725131x10–17
The fraction increased by (6.725131x10–17)/(2.9577689x10–18) = 22.737175 = 22.7
16.59
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
ln
33,600 J/mol 1
1
k2
=
3 1
8.314 J/mol•K 298 K 348 K
4.7x10 s
k2
= 1.948515 (unrounded)
Raise each side to ex
4.7x103 s1
k2
= 7.0182577
4.7x103 s1
k2 = (4.7x10–3 s–1)(7.0182577) = 0.0329858 = 0.033 s–1
ln
16.60
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 = ?
k
E 1 1
ln 2 = a
k1
R T1 T2
J 3.20x103 L/mol•s
8.314 mol•K ln
4.50x105 L/mol•s
=
1
1
468 K 531 K
Ea = 1.3984658x105 J/mol = 1.40x105 J/mol
k
R ln 2
k1
Ea =
1 1
T1 T2
16.61
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 Hrxn = 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)
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215This
kJ/mol
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16-25
