Chapter 14
Homework Answers
14.1
Student responses will vary.
(a)
combustion of gasoline
(b)
cooking an egg in boiling water
(c)
curing of cement
14.2
A collision between only two molecules is much more probable than the simultaneous collision of three
molecules. We therefore conclude that a reaction involving a two–body collision is faster (i.e. will occur
more frequently) than one requiring a three–body collision.
14.3
The instantaneous rate of reaction is the rate of the reaction at a particular moment. The average rate of
reaction is the average rate for the reaction over the time of the whole reaction. This includes the very
rapid rates when the concentration of reactants is high and the very slow rates when the concentration of
reactants is low.
14.4
A tangent line to the curve of concentration as a function of time at time zero is drawn. The slope of this
line is determined which is the initial instantaneous rate of reaction.
14.5
A homogeneous reaction is one in which all reactants and products are in the same phase. An example
would be:
2H2(g) + O2(g) 2H2O(g)
A heterogeneous reaction is one in which all reactants and products are not in the same phase. An example
would be:
25O2(g) + 2C8H18(l) 16CO2(g) + 18H2O(g)
14.6
Chemical reactions that are carried out in solution take place smoothly because the reactants can mingle
effectively at the molecular level.
14.7
Heterogeneous reactions are most affected by the extent of surface contact between the reactant phases.
14.8
In a heterogeneous reaction, the smaller the particle size, the faster the rate. This is because decreasing
the particle size increases the surface area of the material, thereby increasing contact with another
reactant phase.
14.9
This illustrates the effect of concentration.
14.10
Reaction rate generally increases with increasing temperature.
14.11
In cool weather, the rates of metabolic reactions of cold–blooded insects decrease, because of the effect
of temperature on rate.
14.12
The low temperature causes the rate of metabolism to be very low.
14.13
Reaction rate has the units mol L s , or molar per second (M s ). Reaction rates are reported as
positive values regardless of whether the species increases or decreases during the recaction.
14.14
The units are, in each case, whatever is required to give the units of rate (mol L s ) to the overall rate
law:
–1
–1 –1
2
–2 –1
(a)
s
(b)
L mol s
(c)
L mol s
–1 –1
–1
–1 –1
14.15
A zero–order reaction has no dependence on concentration while the first–order reaction is linearly
dependent on concentration and the second-order reaction is inversely dependent on concentration.
14.16
No, the coefficients of a balanced chemical reaction do not predict with any certainty the exponents in a
rate law. These exponents must be determined through experiment.
14.17
This is the first case in Table 14.4 of the text, that is a zero-order reaction.
14.18
This is the fourth case Table 14.4 of the text, that is a first-order reaction.
14.19
This is the seventh case in Table 14.4, and the rate increases by a factor of 2 = 4.
14.20
This is the eleventh case in Table 14.4, and the order of the reaction with respect to the reactant is 3.
14.21
Since the substrate concentration does not influence rate, its concentration does not appear in the rate
law, and the order of the reaction with respect to substrate is zero.
14.22
The half–life of a first–order reaction is unaffected by the initial concentration.
14.23
The half–life of a second–order reaction is inversely proportional to the initial concentration, as expressed
in equation 14.11.
14.24
The half-life of a zeroth-order reaction is directly proportional to initial concentration.
14.25
First order
[ A]0
ln
= kt
[ A]t
2
Second order
1
1
−
= kt
[ A]t [ A]0
t1/2 =
[ A]t
1/2
=
1
[ A]0
2
ln
[ A]0
1
[ A]0
2
= kt1/2
t1/ 2 =
ln 2 0.693
=
k
k
1
1
−
= kt1/ 2
1
[ A]0 [ A]0
2
1
1
2
2
1
−
= kt1/2 =
−
=
1
2
[ A]0 2 [ A]0 [ A]0
[ A]0 [ A]0
2
2
1
[ A]0 = −kt1/2 + [ A]0
2
kt1/2 = [ A]0 −
1
k [ A]0
Zeroth order
[ A]t = −kt + [ A]0
1
[ A]0
2
t1/2 =
[ A]0
14.26
For the zero–order reaction, substitute half the value of [A]0 for the value of [A]t in the equation
[A]0 – [A]t = kt
[A]0 – 1/2[A]0 = kt1/2
1/2[A]0 = kt1/2
[A]0
t 12 =
2k
14.27
To answer this question refer to the integrated equations for each order.
Graph (b) represents a kinetics plot for a first order reaction.
Graph (c) represents a kinetics plot for a second order reaction.
Graph (a) represents a kinetics plot for a zeroth order reaction.
2k
14.28
According to collision theory, the rate is proportional to the number of collisions per second among the
reactants.
14.29
The effectiveness of collisions is influenced by the orientation of the reactants and by the activation
energy.
14.30
This happens because a larger fraction of the reactant molecules possess the minimum energy necessary
to surpass Ea.
Potential Energy
14.31
E a (forward)
Ea (reverse)
Products
∆Hreaction
Reactants
Reaction Coordinate
14.32
Potential Energy
Transition State
Reactants
∆H
Products
Reaction Coordinate
14.33
Breaking a strong bond requires a large input of energy, hence a large Ea.
14.34
An elementary process is an actual collision event that occurs during the reaction. It is one of the key
events that moves the reaction along in the stepwise process that leads to the overall reaction that is
observed. It is thus one step in a potentially multi–step mechanism.
14.35
The rate–determining step in a mechanism is the slowest step.
14.36
The rate law for a reaction is based on the rate–determining step.
14.37
Adding all of the steps gives:
2NO + 2H2 N2 + 2H2O
The intermediates are N2O2 and N2O.
14.38
For such a mechanism, the rate law should be:
rate = k[NO2][CO]
14.39
Since this is not the same as the observed rate law, this is not a reasonable mechanism to propose.
Add all of the steps together:
CO + O2 + NO CO2 + NO2
14.40
On adding together all of the steps in each separate mechanism, we get, in each case:
–
–
–
OCl + I Cl + OI
14.41
–
The predicted rate law is based on the rate–determining step:
2
rate = k[NO2]
14.42
A catalyst changes the mechanism of a reaction, and provides a reaction path having a smaller activation
energy.
14.43
A homogeneous catalyst is one that is present in the same phase as the reactants. It is used in one step of
a cycle, but regenerated in a subsequent step, so that in a net sense, it is not consumed.
14.44
Adsorption is a clinging to a surface. Absorption involves a penetration below the surface, as in the action
of a sponge. Heterogeneous catalysis involves adsorption.
14.45
The catalytic converter promotes oxidation of unburned hydrocarbons, as well as the decomposition of
nitrogen oxide pollutants. Lead poisons the catalyst and renders it ineffective.
14.46
Since they are in a 1-to-1 mol ratio, the rate of formation of SO2 is equal and opposite to the rate of
consumption of SO2Cl2. This is equal to the slope of the curve at any point on the graph (see below). At
–4
–5
200 min, we obtain a value of about 1 × 10 M/min. At 600 minutes, this has decreased to about 7 × 10
M/min.
14.48
This is determined by the coefficients of the balanced chemical equation. For every mole of N2 that
reacts, 3 mol of H2 will react. Thus the rate of disappearance of hydrogen is three times the rate of
disappearance of nitrogen. Similarly, the rate of disappearance of N2 is half the rate of appearance of
NH3, or NH3 appears twice as fast as N2 disappears.
14.49
From the coefficients in the balanced equation we see that, for every mole of B that reacts, 2 mol of A are
consumed, and three mol of C are produced. This means that A will be consumed twice as fast as B, and C
will be produced three times faster than B is consumed.
–1 –1
rate of disappearance of A = 2(–0.30) = –0.60 mol L s
–1 –1
rate of appearance of C = 3(0.30) = 0.90 mol L s
14.51
We rewrite the balanced chemical equation to make the problem easier to answer: N2O5 2NO2 +
1/2O2. Thus, the rates of formation of NO2 and O2 will be, respectively, twice and one half the rate of
disappearance of N2O5.
–6
–6
–1 –1
rate of formation of NO2 = 2(2.5 × 10 ) = 5.0 × 10 mol L s
–6
–6
–1 –1
rate of formation of O2 = 1/2(2.5 × 10 ) = 1.3 × 10 mol L s
14.55
rate = (1.3 × 10 L mol s )(1.0 × 10 mol/L)(1.0 × 10 mol/L)
–3
–1 –1
rate = 1.3 × 10 mol L s
14.62
The reaction is first–order in OCl , because an increase in concentration by a factor of 4.75, while holding
–
the concentration of I constant (compare the first and second experiments of the table), has caused an
1
–
increase in rate by a factor of 4.75 = 4.75. The order of reaction with respect to I is also 1, as is
demonstrated by a comparison of the first and third experiments.
11
–1 –1
–7
–7
–
–
–
rate = k[OCl ][I ]
Using the last data set:
5
–1 –1
–3
–3
1.05 × 10 mol L s = k[1.6 × 10 mol/L][9.6 × 10 mol/L]
9
–1 –1
k = 6.8 × 10 L mol s
14.65
–
In the first, fourth, and fifth experiments, the concentration of OH has been made to increase while the
[(CH3)3CBr] is left unchanged. In each of these experiments, there is no change in rate. This means that
–
–
the rate is independent of [OH ], and the order of reaction with respect to OH is zero. The concentration
of (CH3)3CBr increases by a factor of 2.42 from the first to the second experiment, and by a factor of 3.84
–
from the first to the third experiment, as the OH concentration is held constant. There is a corresponding
1
2.4–fold (i.e. 2.4 ) increase in rate from the first to the second experiment, and there is a 3.8–fold (i.e.
1
3.8 ) increase in rate from the first to the third experiment. In both cases, we conclude that the order
with respect to (CH3)3CBr is one.
rate = k[(CH3)3CBr]
Using the third set of data gives:
–3
–1 –1
–1
3.8 × 10 mol L s = k[7.3 × 10 mol/L]
–3 –1
k = 5.2 × 10 s
14.67
Since it is the plot of 1/conc. that gives a straight line, the order of the reaction with respect to CH3CHO is
two. The rate constant is given by the slope directly:
–1 –1
k = 0.0771 M s
14.73
ln
ln
[ A]0
[ A]t
= kt
1.1 x 10−6 g L−1
0.30 x 10−6 g L−1
= k x 180 s
k = 7.2 x 10−3 s −1
14.75
1
1
−
= kt
A
A
[ ]t [ ]0
14.79
Since 1/2 of the Sr–90 decays every half–life, it will take 5 half–lives, or 5 × 28 yrs = 140 yrs, for the Sr–90
to decay to 1/32 of its present amount.
14.83
We can use the half-life to determine the value of k.
3
4.88 x 10 s = 0.693/k
[ A]0
ln
[ A]t
= kt
-4 -1
k = 1.42 x 10 s
0.012 M
= 1.42 x 10 −4 x t
0.0020 M
t = 1.3 x 10 4 s
ln
14.86
Use equation 14.8 and 14.9.
r
ln 0 = kt
rt
 1.2 × 10−12 
ln 
 = 1.21 × 10−4 y−1 9.0 × 103 y


r
t


−
12
 1.2 × 10


 = exp  1.21 × 10−4 y −1 9.0 × 103 y 




rt


 1.2 × 10−12 

 = rt = 4.0 × 10−13


2.97


(
)(
(
14.88
)
)(
)
The graph is prepared exactly as in example 14.12 of the text. The slope is found using linear regression,
3
3
to be: –9.5 × 10 K. Thus –9.5 × 10 K = –Ea/R
3
–1
–1
4
Ea = –(–9.5 × 10 K)(8.314 J K mol ) = 7.9 × 10 J/mol = 79 kJ/mol
Using the equation, we proceed as follows:
−E a  1
k
1
ln 2 =
−


k1
R  T2
T1 
 1.94 × 10−3 L mol−1 s−1 
−E a
1 
 1
ln 
−
 =
−4
−1 −1
−1 −1  673 K
593 K 

 2.88 × 10 L mol s  8.314 J mol K
2.00 × 10−4 K −1
× Ea
8.314 J mol−1 K −1
4
Ea = 7.93 × 10 J/mol = 79.3 kJ/mol
1.907 =
14.90
Using the equation we have:
−E a  1
k2
1
=
−


k1
R  T2
T1 
 1.0 × 10−3 L mol−1 s−1 
−E a
1 
 1
ln 
−
 =
−5
−1 −1
−1 −1  403 K
373 K 

 9.3 × 10 L mol s  8.314 J mol K
ln
2.37 =
2.00 × 10 −4 K −1
8.314 J mol−1 K −1
× Ea
4
Ea = 9.89 × 10 J/mol = 99 kJ/mol
 −E 
Equation states k = A exp  a 
 RT 
A=
=
k
 −E 
exp  a 
 RT 
9.3 × 10 −5 L mol−1 s −1
 −9.89 × 10 4 J

mol 
exp 
 8.314 J mol K ( 373 K ) 


(
)
= 6.6 × 109 L mol−1 s −1


−9.89 × 104 J / mol
k = 6.6 × 109 L mol−1 s −1 exp 

−
1
−
1
 8.314 J mol K x 473 K 


-2
-1 -1
k = 7.9 x 10 L mol s
14.96
An intermediate is produced in one step and used up in another. The intermediate in this reaction is N2O2.
The overall reaction is the sum of the two steps
2NO(g) + O2(g)
2NO2(g)
The rate law is determined from the slow step.
Rate = k[N2O2][O2]
An intermediate cannot be part of the rate law expression. We can use the first step, which is an
equilibrium step to solve for the concentration of N2O2.
Rate =
k f [NO]2 = kr [N 2O2 ]
Solve for [N2O2] to get
[N2O2] =
kf
kr
[NO]2
Rate = k[NO]2 [O 2 ] where all of the constants have been combined into a new constant
14.98
From the tangent lines we can obtain the instantaneous rate at 1000 s and 9000 s.
For the decomposition of cyclobutane:
-7
-1
-7
-1
At 1000 s the slope is − 9.6 x 10 M s which is the rate of decomposition.
At 9000 s the slope is − 3.8 x 10 M s which is the rate of decomposition
For the formation of ethylene:
-6
-1
-7
-1
At 1000 s the slope is 1.9 x 10 M s which is the rate of formation of.
At 9000 s the slope is 7.5 x 10 M s
14.103 (a) The reaction rate is first order so as the concentration of reactants triples the rate will triple.
(b) The reaction rate is second order so as the concentration of reactants triples the rate will increase by a
factor of nine.
(c) The reaction rate is zeroth order so the rate is independent of concentration. There is no change in the
rate.
(d) The reaction rate is second order so as the concentration of reactants triple the rate increases by a
factor
of nine
(e) The reaction rate is third order so as the concentration of reactants triple the rate increases by a factor
of twenty seven.
2
14.105 (a)
(b)
(c)
(d)
(e)
rate = k1[A]
1
rate = k–1[A2]
1
1
rate = k2[A2] [E]
2A + E B + C
The rates for the forward and reverse directions of step one are set equal to each other in order
2
to arrive at an expression for the intermediate [A2] in terms of the reactant [A]: k1[A] = k–1[A2]
k
2
[A2] = 1 [A]
k −1
This is substituted into the rate law for question (c) above, giving a rate expression that is written
k
2
1
using only observable reactants: rate = k2 1 [A] [E]
k −1
14.111 To solve this problem, plot the data provided as 1/T vs 1/t where T is the absolute temperature and 1/t is
proportional to the rate constant.
t (min)
10
9
8
7
6
T (K)
291
293
294
295
297
1/T
0.003436
0.003412
0.003401
0.003389
0.003367
ln(1/t)
–2.302585
–2.197224
–2.079441
–1.945910
–1.791759
13.6
288
0.003472
–2.608
The slope of the graph is equal to –Ea/R, therefore:
–7,704 = –Ea/R
7,704R = Ea
7,704 K(8.314 J/mol·K) = Ea
Ea = 64,050 J/mol
Ea = 64 kJ/mol
From the straight–line equation, we can determine the time needed to develop the film at 15 °C is 14 min.
14.117
We can determine the temperature that the reaction mixture would achieve if all of the heat is supplied
at once by using the enthalpy of reaction and the heat capacity of the system.
q = Ccal∆T
-1
-1
6
o -1
40 kJ mol x 1000 J kJ x 10,000 mol = 7.5 x 10 J C ∆T
o
∆T = 53.3 C
o
Thus, the final temperature of the mixture would be 78 C or 351 K
ln
ln
ln
−E a  1
k2
1
=
−


k1
R  T2
T1 
1000 J
1 
 1
kJ
 351 K − 298 K 
−1


mol
−50 kJ mol−1 x
k2
1x10−5 s −1
k2
1x10−5 s −1
=
8.314 J K −1
= 3.047
-4 -1
k2 = 2 x 10 s
o
If the temperature of the reaction mixture was 182 C and all of the heat released at once the final
temperature would be:
o
o
o
o
182 C + 53 C = 235 C so the reaction would indeed reach 199 C.