CHAPTER TWELVE
CHEMICAL KINETICS
For Review
1.
The reaction rate is defined as the change in concentration of a reactant or product per unit
time. Consider the general reaction:
A → Products where rate =
 Δ[A]
Δt
If we graph [A] vs. t, it would usually look like the dark line in the following plot.
a
[A]
b
c
0
t1
t2
time
An instantaneous rate is the slope of a tangent line to the graph of [A] vs. t. We can determine
the instantaneous rate at any time during the reaction. On the plot, tangent lines at t ≈ 0 and t
= t1 are drawn. The slope of these tangent lines would be the instantaneous rates at t ≈ 0 and t
= t1. We call the instantaneous rate at t ≈ 0 the initial rate. The average rate is measured over
a period of time. For example, the slope of the line connecting points a and c is the average
rate of the reaction over the entire length of time 0 to t 2 (average rate = Δ[A]/Δt). An average
rate is determined over some time period while an instantaneous rate is determined at one
specific time. The rate which is largest is generally the initial rate. At t ≈ 0, the slope of the
tangent line is greatest, which means the rate is largest at t ≈ 0.
The initial rate is used by convention so that the rate of reaction only depends on the forward
reaction; at t  0, the reverse reaction is insignificant because no products are present yet.
408
CHAPTER 12
CHEMICAL KINETICS
409
2.
The differential rate law describes the dependence of the rate on the concentration of
reactants. The integrated rate law expresses reactant concentrations as a function of time. The
differential rate law is generally just called the rate law. The rate constant k is a constant that
allows one to equate the rate of a reaction to the concentration of reactants. The order is the
exponent that the reactant concentrations are raised to in the rate equation.
3.
The method of initial rates uses the results from several experiments where each experiment
is carried out at a different set of initial reactant concentrations and the initial rate is
determined. The results of the experiments are compared to see how the initial rate depends
on the initial concentrations. If possible, two experiments are compared where only one
reactant concentration changes. For these two experiments, any change in the initial rate must
be due to the change in that one reactant concentration. The results of the experiments are
compared until all of the orders are determined. After the orders are determined, then one can
go back to any (or all) of the experiments and set the initial rate equal to the rate law using
the concentrations in that experiment. The only unknown is k which is then solved for. The
units on k depend on the reactants and orders in the rate law. Because there are many
different rate laws, there are many different units for k.
Rate = k[A]n; For a first-order rate law, n = 1. If [A] is tripled, then the rate is tripled. When
[A] is quadrupled (increased by a factor of four), and the rate increases by a factor of 16, then
A must be second order (42 = 16). For a third order reaction, as [A] is doubled, the rate will
increase by a factor of 23 = 8. For a zero order reaction, the rate is independent of the
concentration of A. The only stipulation for zero order reactions is that the reactant or
reactants must be present; if they are, then the rate is a constant value (rate = k).
4.
Zero order:
 d[A]
 k,
dt
[ A ]t
[ A]
[ A ]t
t
[ A ]0
0
 d[A]    kdt
t
  kt , [A]t  [A]0   kt , [A]t = kt + [A]0
[ A ]0
0
First order:
 d[A]
 k[A],
dt
[ A ]t
ln[ A]
[ A ]t
t
d[A]
 [A]    kdt
[ A ]0
0
  kt , ln[ A]t  ln[ A]0   kt , ln[A]t = kt + ln[A]0
[ A ]0
Second order:
 d[A]
 k[A]2 ,
dt

1
[ A]
[ A ]t
[ A ]0
[ A ]t
t
d[A]
 [A]2    kdt
[ A ]0
0
  kt, 
1
1
1
1

  kt ,
 kt 
[A]t [A]0
[ A] t
[A]0
410
5.
CHAPTER 12
CHEMICAL KINETICS
The integrated rate laws can be put into the equation for a straight line, y = mx + b where x
and y are the x and y axes, m is the slope of the line and b is the y-intercept.
Zero order: [A] = kt + [A]
y = mx + b
A plot of [A] vs. time will be linear with a negative slope equal to k and a y-intercept equal
to [A].
First order: ln[A] = kt + ln[A]0
y = mx + b
A plot of ln[A] vs. time will be linear with a negative slope equal to k and a y-intercept
equal to ln[A]0.
Second order:
1
1
= kt +
[A ]
[ A ]0
y = mx + b
A plot of 1/[A] vs. time will be linear with a positive slope equal to k and a y-intercept equal
to 1/[A]0.
When two or more reactants are studied, only one of the reactants is allowed to change during
any one experiment. This is accomplished by having a large excess of the other reactant(s) as
compared to the reactant studied; so large that the concentration of the other reactant(s) stays
effectively constant during the experiment. The slope of the straight-line plot equals k (or –k)
times the other reactant concentrations raised to the correct orders. Once all the orders are
known for a reaction, then any (or all) of the slopes can be used to determine k.
6.
At t = t1/2, [A] = 1/2[A]0; Plugging these terms into the integrated rate laws yields the
following half-life expressions:
zero order
[ A ]0
t1/2 =
2k
first order
ln 2
t1/2 =
k
second order
1
t1/2 =
k[A]0
The first order half-life is independent of concentration, the zero order half-life is directly
related to the concentration, and the second order half-life is inversely related to concentration. For a first order reaction, if the first half-life equals 20. s, the second half-life will also
be 20. s because the half-life for a first order reaction is concentration independent. The
second half-life for a zero order reaction will be 1/2(20.) = 10. s. This is because the half-life
for a zero order reaction has a direct relationship with concentration (as the concentration
decreases by a factor of 2, the half-life decreases by a factor of 2). For a second order reaction
which has an inverse relationship between t1/2 and [A]0, the second half-life will be 40. s
(twice the first half-life value).
7.
a. An elementary step (reaction) is one for which the rate law can be written from the
molecularity, i.e., from coefficients in the balanced equation.
CHAPTER 12
CHEMICAL KINETICS
411
b. The molecularity is the number of species that must collide to produce the reaction
represented by an elementary step in a reaction mechanism.
c. The mechanism of a reaction is the series of proposed elementary reactions that may
occur to give the overall reaction. The sum of all the steps in the mechanism gives the
balanced chemical reaction.
d. An intermediate is a species that is neither a reactant nor a product but that is formed and
consumed in the reaction sequence.
e. The rate-determining step is the slowest elementary reaction in any given mechanism.
For a mechanism to be acceptable, the sum of the elementary steps must give the overall
balanced equation for the reaction and the mechanism must give a rate law that agrees
with the experimentally determined rate law. A mechanism can never be proven
absolutely. We can only say it is possibly correct if it follows the two requirements
described above.
Most reactions occur by a series of steps. If most reactions were unimolecular, then most
reactions would have a first order overall rate law, which is not the case.
8.
The premise of the collision model is that molecules must collide to react, but not all
collisions between reactant molecules result in product formation.
a. The larger the activation energy, the slower the rate.
b. The higher the temperature, the more molecular collisions with sufficient energy to
convert to products and the faster the rate.
c. The greater the frequency of collisions, the greater the opportunities for molecules to
react, and, hence, the greater the rate.
d. For a reaction to occur, it is the reactive portion of each molecule that must be
involved in a collision. Only some of all the possible collisions have the correct
orientation to convert reactants to products.
endothermic, E > 0
exothermic, E < 0
Ea
Ea
E
E
P

R

R
Reaction Progress
P
Reaction Progress
412
CHAPTER 12
CHEMICAL KINETICS
The activation energy for the reverse reaction will be the energy difference between
the products and the transition state at the top of the potential energy “hill.” For an
exothermic reaction, the activation energy for the reverse reaction (Ea,r) is larger than
the activation energy for the forward reaction (Ea), so the rate of the forward reaction
will be greater than the rate of the reverse reaction. For an endothermic reaction, Ea,r
< Ea so the rate of the forward reaction will be less than the rate of the reverse
reaction (with other factors being equal).
9.
 Ea
R
y= m
Arrhenius equation: k = Ae  E a / RT ; ln(k) =
1
  + lnA
T
x + b
The data needed is the value of the rate constant as a function of temperature. One would plot
lnk versus1/T to get a straight line. The slope of the line is equal to –Ea/R and the y-intercept
is equal to ln(A). The R value is 8.3145 J/Kmol. If one knows the rate constant at two
different temperatures, then the following equation allows determination of Ea:
k  E  1
1 

ln  2   a  
R  T1 T2 
 k1 
10.
A catalyst increases the rate of a reaction by providing reactants with an alternate pathway
(mechanism) to convert to products. This alternate pathway has a lower activation energy,
thus increasing the rate of the reaction.
A homogeneous catalyst is one that is in the same phase as the reacting molecules, and a
heterogeneous catalyst is in a different phase than the reactants. The heterogeneous catalyst is
usually a solid, although a catalyst in a liquid phase can act as a heterogeneous catalyst for
some gas phase reactions. Since the catalyzed reaction has a different mechanism than the
uncatalyzed reaction, the catalyzed reaction most likely will have a different rate law.
Questions
9.
In a unimolecular reaction, a single reactant molecule decomposes to products. In a
bimolecular reaction, two molecules collide to give products. The probability of the
simultaneous collision of three molecules with enough energy and proper orientation is very
small, making termolecular steps very unlikely.
10.
Some energy must be added to get the reaction started, that is, to overcome the activation
energy barrier. Chemically what happens is:
Energy + H2 → 2 H
The hydrogen atoms initiate a chain reaction that proceeds very rapidly. Collisions of H2 and
O2 molecules at room temperature do not have sufficient kinetic energy to form hydrogen
atoms and initiate the reaction.
11.
All of these choices would affect the rate of the reaction, but only b and c affect the rate by
affecting the value of the rate constant k. The value of the rate constant is dependent on
temperature. The value of the rate constant also depends on the activation energy. A catalyst
CHAPTER 12
CHEMICAL KINETICS
413
will change the value of k because the activation energy changes. Increasing the
concentration (partial pressure) of either H2 or NO does not affect the value of k, but it does
increase the rate of the reaction because both concentrations appear in the rate law.
12.
One experimental method to determine rate laws is the method of initial rates. Several
experiments are carried out using different initial concentrations of reactants, and the initial
rate is determined for each run. The results are then compared to see how the initial rate
depends on the initial concentrations. This allows the orders in the rate law to be determined.
The value of the rate constant is determined from the experiments once the orders are known.
The second experimental method utilizes the fact that the integrated rate laws can be put in
the form of a straight line equation. Concentration vs. time data is collected for a reactant as a
reaction is run. This data is then manipulated and plotted to see which manipulation gives a
straight line. From the straight line plot, we get the order of the reactant and the slope of the
line is mathematically related to k, the rate constant.
13.
The average rate decreases with time because the reverse reaction occurs more frequently as
the concentration of products increase. Initially, with no products present, the rate of the
forward reaction is at its fastest; but as time goes on, the rate gets slower and slower since
products are converting back into reactants. The instantaneous rate will also decrease with
time. The only rate that is constant is the initial rate. This is the instantaneous rate taken at t 
0. At this time, the amount of products is insignificant and the rate of the reaction only
depends on the rate of the forward reaction.
14.
The most common method to experimentally determine the differential rate law is the method
of initial rates. Once the differential rate law is determined experimentally, the integrated rate
law can be derived. However, sometimes it is more convenient and more accurate to collect
concentration versus time data for a reactant. When this is the case, then we do “proof” plots
to determine the integrated rate law. Once the integrated rate law is determined, the
differential rate law can be determined. Either experimental procedure allows determination
of both the integrated and the differential rate; which rate law is determined by experiment
and which is derived is usually decided by which data is easiest and most accurately
collected.
15.
 [ A] 
rate2
k[A]2x

  2 
x
rate1
k[A]1
 [A]1 
x
The rate doubles as the concentration quadruples:
2 = (4)x, x = 1/2
The order is 1/2 (the square root of the concentration of reactant).
For a reactant that has an order of 1 and the reactant concentration is doubled:
rate2
1
 (2) 1 
rate1
2
414
CHAPTER 12
CHEMICAL KINETICS
The rate will decrease by a factor of 1/2 when the reactant concentration is doubled for a 1
order reaction.
16.
A metal catalyzed reaction is dependent on the number of adsorption sites on the metal
surface. Once the metal surface is saturated with reactant, the rate of reaction becomes
independent of concentration.
17.
Two reasons are:
a. The collision must involve enough energy to produce the reaction, i.e., the collision
energy must be equal to or exceed the activation energy.
b. The relative orientation of the reactants when they collide must allow formation of any
new bonds necessary to produce products.
18.
The slope of the lnk vs. 1/T (K) plot is equal to Ea/R. Because Ea for the catalyzed reaction
will be smaller than Ea for the uncatalyzed reaction, the slope of the catalyzed plot will be
less negative.
Exercises
Reaction Rates
19.
The coefficients in the balanced reaction relate the rate of disappearance of reactants to the
rate of production of products. From the balanced reaction, the rate of production of P4 will
be 1/4 the rate of disappearance of PH3, and the rate of production of H2 will be 6/4 the rate
of disappearance of PH3. By convention, all rates are given as positive values.
Rate =
 Δ[PH3 ]
(0.0048 mol / 2.0 L)
= 2.4 × 10 3 mol/Ls

Δt
s
Δ[P4 ]
1 Δ[PH3 ]
= 2.4 × 10 3 /4 = 6.0 × 10 4 mol/Ls

Δt
4
Δt
Δ[H 2 ]
6 Δ[PH3 ]
= 6(2.4 × 10 3 )/4 = 3.6 × 10 3 mol/Ls

Δt
4 Δt
20.
Δ[ NH3 ]
Δ[H 2 ]
Δ[ N 2 ]
Δ[ N 2 ]
1 Δ[H 2 ] 1 Δ[ NH3 ]
3
and
 2
; So, 

Δt
Δt
Δt
Δt
3 Δt
2 Δt
or:
Δ[ NH3 ]
2 Δ[H 2 ]

Δt
3 Δt
Ammonia is produced at a rate equal to 2/3 of the rate of consumption of hydrogen.
CHAPTER 12
21.
CHEMICAL KINETICS
a. average rate =
415
 Δ[H 2O 2 ]  (0.500 M  1.000 M)
= 2.31 × 10 5 mol/Ls

Δt
(2.16  10 4 s  0)
From the coefficients in the balanced equation:
Δ[O 2 ]
1 Δ[H 2O 2 ]
= 1.16 × 10 5 mol/Ls

Δt
2
Δt
b.
 Δ[H 2O2 ]
 (0.250  0.500) M
= 1.16 × 10 5 mol/Ls

Δt
(4.32  10 4  2.16  10 4 ) s
Δ[O 2 ]
= 1/2 (1.16 × 10 5 ) = 5.80 × 10 6 mol/Ls
Δt
Notice that as time goes on in a reaction, the average rate decreases.
22.
0.0120/0.0080 = 1.5; Reactant B is used up 1.5 times faster than reactant A. This corresponds to a 3 to 2 mol ratio between B and A in the balanced equation. 0.0160/0.0080 = 2;
Product C is produced twice as fast as reactant A is used up. So the coefficient for C is twice
the coefficient for A. A possible balanced equation is: 2A + 3B → 4C
23.
a. The units for rate are always mol/Ls.
c. Rate = k[A],
mol
 mol 
 k

Ls
 L 
b. Rate = k; k must have units of mol/Ls.
d. Rate = k[A]2,
k must have units of s 1 .
mol
 mol 
 k

Ls
 L 
2
k must have units of L/mols.
e. L2/mol2s
1/ 2
24.
Rate = k[Cl]1/2[CHCl3],
mol
 mol 
 k

Ls
 L 
 mol 

 , k must have units of L1/2/mol1/2s.
L


Rate Laws from Experimental Data: Initial Rates Method
25.
a. In the first two experiments, [NO] is held constant and [Cl 2] is doubled. The rate also
doubled. Thus, the reaction is first order with respect to Cl 2. Or mathematically: Rate =
k[NO]x[Cl2]y
0.36
k (0.10) x (0.20) y (0.20) y


, 2.0 = 2.0y, y = 1
0.18
k (0.10) x (0.10) y (0.10) y
We can get the dependence on NO from the second and third experiments. Here, as the
NO concentration doubles (Cl2 concentration is constant), the rate increases by a factor of
four. Thus, the reaction is second order with respect to NO. Or mathematically:
416
CHAPTER 12
CHEMICAL KINETICS
1.45
k (0.20) x (0.20)
(0.20) x


, 4.0 = 2.0x, x = 2; So, Rate = k[NO]2[Cl2]
0.36
k (0.10) x (0.20)
(0.10) x
Try to examine experiments where only one concentration changes at a time. The more
variables that change, the harder it is to determine the orders. Also, these types of
problems can usually be solved by inspection. In general, we will solve using a
mathematical approach, but keep in mind you probably can solve for the orders by simple
inspection of the data.
b. The rate constant k can be determined from the experiments. From experiment 1:
2
0.18 mol
 0.10 mol   0.10 mol 
 k
 
 , k = 180 L2/mol2min
L min
L
L

 

From the other experiments:
k = 180 L2/mol2min (2nd exp.); k = 180 L2/mol2min (3rd exp.)
The average rate constant is kmean = 1.8 × 102 L2/mol2min.
26.
a. Rate = k[I]x[S2O82]y;
1.25  10 6
k (0.080 ) x (0.040 ) y

, 2.00 = 2.0x, x = 1
6.25  10 6
k (0.040 ) x (0.040 ) y
1.25  10 6
k (0.080 )(0.040 ) y

, 2.00 = 2.0y, y = 1; Rate = k[I][S2O82]
6.25  10 6
k (0.080 )(0.020 ) y
b. For the first experiment:
12.5  10 6 mol
 0.080 mol   0.040 mol 
3
 k

 , k = 3.9 × 10 L/mols
Ls
L
L



Each of the other experiments also gives k = 3.9 × 10 3 L/mols,
so kmean = 3.9 × 10 3 L/mols.
27.
a. Rate = k[NOCl]n; Using experiments two and three:
2.66  10 4
(2.0  1016 ) n

k
, 4.01 = 2.0n, n = 2; Rate = k[NOCl]2
6.64  10 3
(1.0  1016 ) n
2
b.
 3.0  1016 molecules 
5.98  10 4 molecules

 , k = 6.6 × 10 29 cm3/moleculess

k
3


cm3 s
cm


The other three experiments give (6.7, 6.6 and 6.6) × 10 29 cm3/moleculess,
respectively.
The mean value for k is 6.6 × 10 29 cm3/moleculess.
CHAPTER 12
c.
28.
CHEMICAL KINETICS
417
6.6  10 29 cm3
1L
6.022  10 23 molecules 4.0  10 8 L



molecules s
mol
mol s
1000 cm3
Rate = k[N2O5]x; The rate laws for the first two experiments are:
2.26 × 10 3 = k(0.190)x and 8.90 × 10 4 = k(0.0750)x
Dividing: 2.54 =
k=
29.
(0.190 ) x
= (2.53)x, x = 1; Rate = k[N2O5]
(0.0750 ) x
8.90  10 4 mol / L  s
Rate

= 1.19 × 10 2 s 1 ; kmean = 1.19 × 10 2 s 1
[N 2 O 5 ]
0.0750 mol / L
a. Rate = k[Hb]x[CO]y; Comparing the first two experiments, [CO] is unchanged, [Hb]
doubles, and the rate doubles. Therefore, x = 1 and the reaction is first order in Hb.
Comparing the second and third experiments, [Hb] is unchanged, [CO] triples. and the
rate triples. Therefore, y = 1 and the reaction is first order in CO.
b. Rate = k[Hb][CO]
c. From the first experiment:
0.619 µmol/Ls = k (2.21 µmol/L)(1.00 µmol/L), k = 0.280 L/µmols
The second and third experiments give similar k values, so kmean = 0.280 L/µmols.
d. Rate = k[Hb][CO] =
30.
0.280 L 3.36 μmol 2.40 μmol
= 2.26 µmol/Ls


μmol s
L
L
a. Rate = k[ClO2]x[OH-]y; From the first two experiments:
2.30 × 10 1 = k(0.100)x(0.100)y and 5.75 × 10 2 = k(0.0500)x(0.100)y
Dividing the two rate laws: 4.00 =
(0.100 ) x
= 2.00x, x = 2
(0.0500 ) x
Comparing the second and third experiments:
2.30 × 10 1 = k(0.100)(0.100)y and 1.15 × 10 1 = k(0.100)(0.0500)y
Dividing:
2.00 =
(0.100 ) y
= 2.00y, y = 1
(0.0500 ) y
The rate law is: Rate = k[ClO2]2[OH]
2.30 × 10 1 mol/Ls = k(0.100 mol/L)2(0.100 mol/L), k = 2.30 × 102 L2/mol2s = kmean
418
CHAPTER 12
CHEMICAL KINETICS
2
b. Rate = k[ClO2]2[OH] =
2.30  10 2 L2
0.0844 mol
 0.175 mol 
= 0.594 mol/Ls
 
 
2
L
L
mol s


Integrated Rate Laws
31.
The first assumption to make is that the reaction is first order because first-order reactions are
most common. For a first-order reaction, a graph of ln [H2O2] vs time will yield a straight
line. If this plot is not linear, then the reaction is not first order and we make another
assumption. The data and plot for the first-order assumption follows.
Time
(s)
[H2O2]
(mol/L)
ln H2O2]
0
120.
300.
600.
1200.
1800.
2400.
3000.
3600.
1.00
0.91
0.78
0.59
0.37
0.22
0.13
0.082
0.050
0.000
0.094
0.25
0.53
0.99
1.51
2.04
2.50
3.00
Note: We carried extra significant figures in some of the ln values in order to reduce roundoff error. For the plots, we will do this most of the time when the ln function is involved.
The plot of ln [H2O2] vs. time is linear. Thus, the reaction is first order. The rate law and integrated rate law are: Rate = k[H2O2] and ln [H2O2] = kt + ln [H2O2]o.
We determine the rate constant k by determining the slope of the ln [H2O2] vs time plot
(slope = k). Using two points on the curve gives:
slope = k =
Δy 0  (3.00)
= 8.3 × 10 4 s 1 , k = 8.3 × 10 4 s 1

Δx 0  3600 .
To determine [H2O2] at 4000. s, use the integrated rate law where at t = 0, [H2O2]o = 1.00 M.
 [H 2 O 2 ] 
 = kt
ln [H2O2] = kt + ln [H2O2]o or ln 
 [H 2 O 2 ]o 
 [H O ] 
ln  2 2  = 8.3 × 10 4 s 1 × 4000. s, ln [H2O2] = 3.3, [H2O2] = e 3.3 = 0.037 M
 1.00 
CHAPTER 12
32.
CHEMICAL KINETICS
419
a. Because the ln[A] vs time plot was linear, the reaction is first order in A. The slope of the
ln[A] vs. time plot equals k. Therefore, the rate law, the integrated rate law and the rate
constant value are:
Rate = k[A]; ln[A] = kt + ln[A]o; k = 2.97 × 10 2 min 1
b. The half-life expression for a first-order rate law is:
ln 2 0.6931
0.6931
t1/2 =
, t1/2 =
= 23.3 min

k
k
2.97  10 2 min 1
c. 2.50 × 10 3 M is 1/8 of the original amount of A present, so the reaction is 87.5%
complete. When a first-order reaction is 87.5% complete (or 12.5% remains), the reaction
has gone through 3 half-lives:
100%
→ 50.0% → 25%
t1/2
t1/2
→ 12.5%; t = 3 × t1/2 = 3 × 23.3 min = 69.9 min
t1/2
Or we can use the integrated rate law:
 2.50  10 3 M 
 [ A] 
2
1

   kt , ln 
ln 
 2.00  10 2 M  = (2.97 × 10 min ) t
[
A
]
o 



t=
33.
ln( 0.125)
= 70.0 min
 2.97  10 2 min 1
Assume the reaction is first order and see if the plot of ln [NO2] vs. time is linear. If this isn’t
linear, try the second-order plot of 1/[NO2] vs. time because second-order reactions are the
next most common after first-order reactions. The data and plots follow.
Time (s)
[NO2] (M)
ln [NO2]
1/[NO2] ( M 1 )
0.500
0.693
2.00
3
0.444
0.812
2.25
3.00 × 103
0.381
0.965
2.62
4.50 × 10
3
0.340
1.079
2.94
9.00 × 10
3
0.250
1.386
4.00
1.80 × 10
4
0.174
1.749
5.75
0
1.20 × 10
420
CHAPTER 12
CHEMICAL KINETICS
The plot of 1/[NO2] vs. time is linear. The reaction is second order in NO2. The rate law and
integrated rate law are: Rate = k[NO2]2 and
1
1
.
 kt 
[ NO2 ]
[ NO2 ]o
The slope of the plot 1/[NO2] vs. t gives the value of k. Using a couple of points on the plot:
Δy (5.75  2.00) M 1

slope = k =
= 2.08 × 10 4 L/mols
4
Δx (1.80  10  0) s
To determine [NO2] at 2.70 × 104 s, use the integrated rate law where 1/[NO2]o = 1/0.500 M
= 2.00 M 1 .
1
1
1
2.08  10 4 L
 kt 
,

× 2.70 × 104 s + 2.00 M 1
[ NO2 ]
[ NO2 ]o [ NO2 ]
mol s
1
= 7.62, [NO2] = 0.131 M
[ NO2 ]
34.
a. Because the 1/[A] vs. time plot was linear, the reaction is second order in A. The slope of
the 1/[A] vs. time plot equals the rate constant k. Therefore, the rate law, the integrated
rate law and the rate constant value are:
1
1
Rate = k[A]2;
; k = 3.60 × 10 2 L/mols
 kt 
[ A]
[A]o
1
b. The half-life expression for a second-order reaction is: t1/2 =
k [A]o
1
For this reaction: t1/2 =
= 9.92 × 103 s
2
3.60  10 L / mol  s  2.80  10 3 mol / L
Note: We could have used the integrated rate law to solve for t1/2 where
[A] = (2.80 × 10 3 /2) mol/L.
c. Since the half-life for a second-order reaction depends on concentration, we will use the
integrated rate law to solve.
CHAPTER 12
CHEMICAL KINETICS
421
1
3.60  10 2 L
1
1
1

t
,
 kt 
4
mol  s
2.80  10 3 M
[ A]
[A]o 7.00  10 M
1.43 × 103  357 = 3.60 × 10 2 t, t = 2.98 × 104 s
35.
a. Because the [C2H5OH] vs. time plot was linear, the reaction is zero order in C2H5OH. The
slope of the [C2H5OH] vs. time plot equals -k. Therefore, the rate law, the integrated rate
law and the rate constant value are: Rate = k[C2H5OH]0 = k; [C2H5OH] = kt +
[C2H5OH]o; k = 4.00 × 10 5 mol/Ls.
b. The half-life expression for a zero-order reaction is: t1/2 = [A]o/2k.
t1/2 =
[C 2 H 5 OH]o
1.25  10 2 mol / L

= 156 s
2k
2  4.00  10 5 mol / L  s
Note: we could have used the integrated rate law to solve for t1/2 where
[C2H5OH] = (1.25 × 10 2 /2) mol/L.
c. [C2H5OH] = kt + [C2H5OH]o , 0 mol/L = (4.00 × 10 5 mol/Ls) t + 1.25 × 10 2 mol/L
t=
36.
1.25  10 2 mol / L
= 313 s
4.00  10 5 mol / L  s
From the data, the pressure of C2H5OH decreases at a constant rate of 13 torr for every 100. s.
Since the rate of disappearance of C2H5OH is not dependent on concentration, the reaction is
zero order in C2H5OH.
k=
13 torr
1 atm
= 1.7 × 10 4 atm/s

100. s 760 torr
The rate law and integrated rate law are:
 1 atm 
 = kt + 0.329 atm
Rate = k = 1.7 × 10 4 atm/s; PC2 H5OH = kt + 250. torr 
 760 torr 
At 900. s: PC2 H5OH = -1.7 × 10 4 atm/s × 900. s + 0.329 atm = 0.176 atm = 0.18 atm = 130 torr
37.
The first assumption to make is that the reaction is first order. For a first-order reaction, a
graph of ln [C4H6] vs. t should yield a straight line. If this isn't linear, then try the secondorder plot of 1/[C4H6] vs. t. The data and the plots follow.
Time
195
604
2
1.5 × 10
1246
2
1.3 × 10
2180
2
1.1 × 10
6210 s
2
0.68 × 10 2 M
[C4H6]
1.6 × 10
ln [C4H6]
4.14
4.20
4.34
4.51
4.99
1/[C4H6]
62.5
66.7
76.9
90.9
147 M 1
Note: To reduce round-off error, we carried extra sig. figs. in the data points.
422
CHAPTER 12
CHEMICAL KINETICS
The ln plot is not linear, so the reaction is not first order. Since the second-order plot of
1/[C4H6] vs. t is linear, we can conclude that the reaction is second order in butadiene. The
rate law is:
Rate = k[C4H6]2
For a second order reaction, the integrated rate law is:
1
1
 kt 
[C4 H 6 ]
[C4 H 6 ]o
The slope of the straight line equals the value of the rate constant. Using the points on the
line at 1000. and 6000. s:
k = slope =
38.
144 L / mol  73 L / mol
= 1.4 × 10 2 L/mols
6000 . s  1000 . s
a. First, assume the reaction to be first order with respect to O. A graph of ln [O] vs. t
should be linear if the reaction is first order in O.
t(s)
0
10. × 10 3
20. × 10 3
30. × 10 3
[O] (atoms/cm3)
5.0 × 109
1.9 × 109
6.8 × 108
2.5 × 108
ln[O]
22.33
21.37
20.34
19.34
CHAPTER 12
CHEMICAL KINETICS
423
The graph is linear, so we can conclude that the reaction is first order with respect to O.
Note: by keeping the NO2 concentration so large we made this reaction into a pseudofirst-order reaction in O.
b. The overall rate law is: Rate = k[NO2][O]
Because NO2 was in excess, its concentration is constant. So for this experiment, the rate
law is: Rate = k[O] where k = k[NO2] . In a typical first-order plot, the slope equals
k. For this experiment, the slope equals k = k[NO2]. From the graph:
slope =
19.34  22.33
(30.  10
3
 0) s
= 1.0 × 102 s 1 , k = -slope = 1.0 × 102 s 1
To determine k, the actual rate constant:
k = k[NO2], 1.0 × 102 s 1 = k(1.0 × 1013 molecules/cm3)
k = 1.0 × 10 11 cm3/moleculess
39.
Because the 1/[A] vs. time plot is linear with a positive slope, the reaction is second order
with respect to A. The y-intercept in the plot will equal 1/[A]o. Extending the plot, the yintercept will be about 10, so 1/10 = 0.1 M = [A]o.
40.
The slope of the 1/[A] vs time plot in Exercise 12.39 with equal k.
slope = k =
a.
(60  20) L / mol
= 10 L/mols
(5  1) s
1
1
10 L
1
= 100, [A] = 0.01 M
 kt 

 9s 
[ A]
[A]o mol s
0.1 M
b. For a second-order reaction, the half-life does depend on concentration: t1/2 =
First half-life: t1/2 =
1
k [A]o
1
=1s
10 L 0.1 mol

mol s
L
Second half-life ([A]o is now 0.05 M): t1/2 = 1/(10 × 0.05) = 2 s
Third half-life ([A]o is now 0.025 M): t1/2 = 1/(10 × 0.025) = 4 s
41.
a. [A] =  kt + [A]o, [A] = (5.0 × 10 2 mol/Ls) t + 1.0 × 10 3 mol/L
b. The half-life expression for a zero-order reaction is: t1/2 =
t1/2 =
[ A ]o
2k
1.0  10 3 mol / L
= 1.0 × 10 2 s
2
2  5.0  10 mol / L  s
c. [A] = 5.0 × 10 2 mol/Ls × 5.0 × 10 3 s + 1.0 × 10 3 mol/L = 7.5 × 10 4 mol/L
424
CHAPTER 12
CHEMICAL KINETICS
Because 7.5 × 10 4 M A remains, 2.5 × 10 4 M A reacted, which means that 2.5 × 10 4 M
B has been produced.
42.
 [ A] 
ln 2 0.6931
   kt ; k 
= 4.85 × 10 2 d 1
ln 

t1/ 2 14.3 d
 [A]o 
If [A]o = 100.0, then after 95.0% completion, [A] = 5.0.
 5.0 
2
1
ln 
 = 4.85 × 10 d × t, t = 62 days
 100.0 
43.
a. When a reaction is 75.0% complete (25.0% of reactant remains), this represents two half
lives (100% → 50%→ 25%). The first-order half-life expression is: t1/2 = (ln 2)/k.
Because there is no concentration dependence for a first order half-life: 320. s = two halflives, t1/2 = 320./2 = 160. s. This is both the first half-life, the second half-life, etc.
ln 2
ln 2
ln 2
b. t1/2 =
= 4.33 × 10 3 s 1
, k 

k
t1/ 2
160. s
At 90.0% complete, 10.0% of the original amount of the reactant remains, so [A] =
0.100[A]0.
 [ A] 
0.100[A]0
ln 0.100
   kt , ln
= (4.33 × 10 3 s 1 )t, t =
= 532 s
ln 
[A]0
 4.33  10 3 s 1
 [A]o 
44.
For a first-order reaction, the integrated rate law is: ln([A]/[A]o) = kt. Solving for k:
 0.250 mol / L 
 = k × 120. s, k = 0.0116 s 1
ln 
 1.00 mol / L 
 0.350 mol / L 
 = - 0.0116 s 1 × t, t = 150. s
ln 
2
.
00
mol
/
L


45.
Comparing experiments 1 and 2, as the concentration of AB is doubled, the initial rate
increases by a factor of 4. The reaction is second order in AB.
Rate = k[AB]2, 3.20 × 10 3 mol / L  s = k1(0.200 M)2
k = 8.00 × 10 2 L / mol  s = kmean
For a second order reaction:
1
1
t1/2 =
= 12.5 s

2
k[AB]o 8.00  10 L / mol  s  1.00 mol / L
46.
a. The integrated rate law for a second-order reaction is: 1/[A] = kt + 1/[A]o, and the half-
CHAPTER 12
CHEMICAL KINETICS
425
life expression is: t1/2 = 1/k[A]o. We could use either to solve for t1/2. Using the
integrated rate law:
1
1
1.11 L / mol
= k × 2.00 s +
= 0.555 L/mols
, k
(0.900 / 2) mol / L
0.900 mol / L
2.00 s
b.
47.
1
1
8.9 L / mol
= 0.555 L/mols × t +
= 16 s
, t
0.100 mol / L
0.900 mol / L
0.555 L / mol  s
Successive half-lives double as concentration is decreased by one-half. This is consistent
with second-order reactions so assume the reaction is second order in A.
t1/2 =
1
1
1
= 1.0 L/molmin
, k

k[A]o
t1/ 2 [A]o 10.0 min (0.10 M )
1
1
1.0 L
1
× 80.0 min +
= 90. M 1 , [A] = 1.1 × 10 2 M
 kt 

[ A]
[A]o mol min
0.10 M
b. 30.0 min = 2 half-lives, so 25% of original A is remaining.
a.
[A] = 0.25(0.10 M) = 0.025 M
48.
Because [B]o>>[A]o, the B concentration is essentially constant during this experiment, so
rate = k[A] where k = k[B]2. For this experiment, the reaction is a pseudo-first-order
reaction in A.
a.
 3.8  10 3 M 
 [ A] 
 = k × 8.0 s, k = 0.12 s 1
 = kt, ln 
ln 
2


 [A]o 
 1.0  10 M 
For the reaction: k = k[B]2, k = 0.12 s 1 /(3.0 mol/L)2 = 1.3 × 10 2 L2/mol2s
b. t1/2 =
c.
ln 2
0.693
= 5.8 s

k'
0.12 s 1


[ A]
[A]
 = 0.12 s 1 × 13.0 s,
= e 0.12(13.0) = 0.21
ln 
2
2

1
.
0

10
M
1
.
0

10


[A] = 2.1 × 10 3 M
d. [A] reacted = 0.010 M  0.0021 M = 0.008 M
[C] reacted = 0.008 M ×
2 mol C
= 0.016 M ≈ 0.02 M
1 mol A
[C]remaining = 2.0 M - 0.02 M = 2.0 M; As expected, the concentration of C basically
426
CHAPTER 12
CHEMICAL KINETICS
remains constant during this experiment since [C]o>> [A]o.
Reaction Mechanisms
49.
50.
For elementary reactions, the rate law can be written using the coefficients in the balanced
equation to determine orders.
a. Rate = k[CH3NC]
b. Rate = k[O3][NO]
c. Rate = k[O3]
d. Rate = k[O3][O]
The observed rate law for this reaction is: Rate = k[NO]2[H2]. For a mechanism to be
plausible, the sum of all the steps must give the overall balanced equation (true for all of the
proposed mechanisms in this problem), and the rate law derived from the mechanism must
agree with the observed mechanism. In each mechanism (I - III), the first elementary step is
the rate-determining step (the slow step), so the derived rate law for each mechanism will be
the rate of the first step. The derived rate laws follow:
Mechanism I: Rate = k[H2]2[NO]2
Mechanism II: Rate = k[H2][NO]
Mechanism III: Rate = k[H2][NO]2
Only in Mechanism III does the derived rate law agree with the observed rate law. Thus,
only Mechanism III is a plausible mechanism for this reaction.
51.
A mechanism consists of a series of elementary reactions where the rate law for each step can
be determined using the coefficients in the balanced equations. For a plausible mechanism,
the rate law derived from a mechanism must agree with the rate law determined from
experiment. To derive the rate law from the mechanism, the rate of the reaction is assumed to
equal the rate of the slowest step in the mechanism.
Because step 1 is the rate-determining step, the rate law for this mechanism is: Rate =
[C4H9Br]. To get the overall reaction, we sum all the individual steps of the mechanism.
Summing all steps gives:
C4H9Br → C4H9+ + Br
C4H9+ + H2O → C4H9OH2+
C4H9OH2+ + H2O → C4H9OH + H3O+
____________________________________
C4H9Br + 2 H2O → C4H9OH + Br + H3O+
Intermediates in a mechanism are species that are neither reactants nor products, but that
are formed and consumed during the reaction sequence. The intermediates for this
mechanism are C4H9+ and C4H9OH2+.
52.
Because the rate of the slowest elementary step equals the rate of a reaction:
CHAPTER 12
CHEMICAL KINETICS
427
Rate = rate of step 1 = k[NO2]2
The sum of all steps in a plausible mechanism must give the overall balanced reaction.
Summing all steps gives:
NO2 + NO2 → NO3 + NO
NO3 + CO → NO2 + CO2
______________________
NO2 + CO → NO + CO2
Temperature Dependence of Rate Constants and the Collision Model
53.
In the following plot, R = reactants, P = products, Ea = activation energy and RC = reaction
coordinate which is the same as reaction progress. Note for this reaction that ΔE is positive
since the products are at a higher energy than the reactants.
E
Ea
P
E
R
RC
54.
When ΔE is positive, the products are at a higher energy relative to reactants and, when ΔE is
negative, the products are at a lower energy relative to reactants.
428
CHAPTER 12
CHEMICAL KINETICS
55.
125
kJ/mol
E
R
Ea, reverse
216
kJ/mol
P
RC
The activation energy for the reverse reaction is:
Ea,
reverse
= 216 kJ/mol + 125 kJ/mol = 341 kJ/mol
56.
When ΔE is negative, then Ea, reverse > Ea, forward (see energy profile in Exercise 12.55). When
ΔE is positive (the products have higher energy than the reactants as represented in the
energy profile for Exercise 12.53), then Ea, forward > Ea, reverse. Therefore, this reaction has a
positive ΔE value.
57.
The Arrhenius equation is: k = A exp (-Ea/RT) or in logarithmic form, ln k = -Ea/RT + ln A.
Hence, a graph of ln k vs. 1/T should yield a straight line with a slope equal to -Ea/R since the
logarithmic form of the Arrhenius equation is in the form of a straight line equation, y = mx +
b. Note: We carried extra significant figures in the following ln k values in order to reduce
round off error.
T (K)
1/T ( K 1 )
k ( s 1 )
ln k
CHAPTER 12
CHEMICAL KINETICS
2.96 × 10 3
3.14 × 10 3
3.36 × 10 3
338
318
298
Slope =
4.9 × 10 3
5.0 × 10 4
3.5 × 10 5
5.32
7.60
10.26
 10.76  (5.85)
= 1.2 × 104 K = Ea/R
(3.40  10 3  3.00  10 3 ) K 1
Ea = slope × R = 1.2 × 104 K ×
58.
429
8.3145 J
, Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol
K mol
From the Arrhenius equation in logarithmic form (ln k = -Ea/RT + ln A), a graph of ln k vs.
1/T should yield a straight line with a slope equal to Ea/R and a y-intercept equal to ln A.
a. slope = Ea/R, Ea = 1.10 × 104 K ×
8.3145 J
= 9.15 × 104 J/mol = 91.5 kJ/mol
K mol
b. The units for A are the same as the units for k( s 1 ).
y-intercept = ln A, A = e33.5 = 3.54 × 1014 s 1
c. ln k = Ea/RT + ln A or k = A exp(Ea/RT)
  9.15  10 4 J / mol 
 = 3.24 × 10 2 s 1
k = 3.54 × 1014 s 1 × exp 

8
.
3145
J
/
K

mol

298
K


59.
k = A exp(Ea/RT) or ln k =
 Ea
+ ln A (the Arrhenius equation)
RT
k  E
For two conditions: ln  2   a
 k1  R
1
1 
 
 (Assuming A is temperature independent.)
 T1 T2 
Let k1 = 3.52 × 10 7 L/mols, T1 = 555 K; k2 = ?, T2 = 645 K; Ea = 186 × 103 J/mol

k2
ln 
7
 3.52  10
 1.86  10 5 J / mol  1
1

 8.3145 J / mol  K  555 K  645 K



 = 5.6

430
CHAPTER 12
CHEMICAL KINETICS
k2
= e5.6 = 270, k2 = 270(3.52 × 10 7 ) = 9.5 × 10 5 L/mols
3.52  10 7
60.
k  E  1
1 
 (Assuming A factor is T independent.)
For two conditions: ln  2   a  
k
R
T
T
2 
 1
 1
2 1
 8.1  10 s 
 1
Ea
1 



ln 

 2 1 
 4.6  10 s  8.3145 J / mol  K  273 K 293 K 
0.57 =
61.
Ea
(2.5 × 10 4 ), Ea = 1.9 × 104 J/mol = 19 kJ/mol
83145
k  E
ln  2   a
 k1  R
ln(7.00) =
1
1  k2
 
;
= 7.00, T1 = 295 K, Ea = 54.0 × 103 J/mol
T
T
k
2 
1
 1
5.0  10 3 J / mol
8.3145 J / mol  K
 1
1 
1
1

,

 = 3.00 × 10 4
295
K
T
295
K
T
2 
2

1
= 3.09 × 10 3 , T2 = 324 K = 51°C
T2
62.
k  E
ln  2   a
 k1  R
1
1 
 
; Since the rate doubles, then k2 = 2 k1.
 T1 T2 
 1
Ea
1 

, Ea = 5.3 × 104 J/mol = 53 kJ/mol
ln (2.00) =

8.3145 J / mol  K  298 K 308 K 
63.
H3O+(aq) + OH(aq) → 2 H2O(l) should have the faster rate. H3O+ and OH will be electrostatically attracted to each other; Ce4+ and Hg22+ will repel each other (so Ea is much larger).
64.
Carbon cannot form the fifth bond necessary for the transition state because of the small
atomic size of carbon and because carbon doesn’t have low energy d orbitals available to
expand the octet.
Catalysts
65.
a. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side,
but it is not a reactant since it is regenerated in the second step.
b. NO2 is an intermediate. Intermediates also never appear in the overall balanced equation.
In a mechanism, intermediates always appear first on the product side while catalysts
always appear first on the reactant side.
c. k = A exp(Ea/RT);
k cat A exp [E a (cat) / RT]
 E (un )  E a (cat) 

 exp a

k un
A exp[E (un ) / RT]
RT


CHAPTER 12
CHEMICAL KINETICS
431


2100 J / mol
k cat
 = e0.85 = 2.3
 exp 
8
.
3145
J
/
mol

K

298
K
k un


The catalyzed reaction is 2.3 times faster than the uncatalyzed reaction at 25°C.
66.
The mechanism for the chlorine catalyzed destruction of ozone is:
O3 + Cl → O2 + ClO (slow)
ClO + O → O2 + Cl
(fast)
_________________________
O3 + O → 2 O2
Because the chlorine atom-catalyzed reaction has a lower activation energy, then the Cl
catalyzed rate is faster. Hence, Cl is a more effective catalyst. Using the activation energy, we
can estimate the efficiency with which Cl atoms destroy ozone as compared to NO molecules
(see Exercise 12.65c).
 (2100  11,900) J / mol  3.96
k Cl
  E a (Cl) E a ( NO) 
 = e = 52
At 25°C:
 exp 

  exp 
k NO
RT 
 RT
 (8.3145  298) J / mol 
At 25°C, the Cl catalyzed reaction is roughly 52 times faster than the NO catalyzed reaction,
assuming the frequency factor A is the same for each reaction.
67.
D
The reaction at the surface of the catalyst is assumed to follow the steps:
D
H
H
H
C
C
DCH 2
H
D
D
D
CH 2
D
D
CH 2DCH 2D(g)
metal surface
Thus, CH2D‒CH2D should be the product. If the mechanism is possible, then the reaction
must be:
C2H4 + D2 → CH2DCH2D
If we got this product, then we could conclude that this is a possible mechanism. If we got
some other product, e.g., CH3CHD2, then we would conclude that the mechanism is wrong.
Even though this mechanism correctly predicts the products of the reaction, we cannot say
conclusively that this is the correct mechanism; we might be able to conceive of other
mechanisms that would give the same products as our proposed one.
68.
a. W, because it has a lower activation energy than the Os catalyst.
b. kw = Aw exp[Ea(W)/RT]; kuncat = Auncat exp[Ea(uncat)/RT]; Assume Aw = Auncat
kw
  E a ( W) E a (uncat) 
 exp 


k uncat
RT
 RT

432
CHAPTER 12
CHEMICAL KINETICS
  163,000 J / mol  335,000 J / mol 
kw
 = 1.41 × 1030
 exp 
k uncat
8
.
3145
J
/
mol

K

298
K


The W-catalyzed reaction is approximately 1030 times faster than the uncatalyzed
reaction.
c. Because [H2] is in the denominator of the rate law, the presence of H2 decreases the rate
of the reaction. For the decomposition to occur, NH3 molecules must be adsorbed on the
surface of the catalyst. If H2 is also adsorbed on the catalyst surface, then there are fewer
sites for NH3 molecules to be adsorbed and the rate decreases.
69.
Assuming the catalyzed and uncatalyzed reactions have the same form and orders and because concentrations are assumed equal, the rates will be equal when the k values are equal.
k = A exp(Ea/RT), kcat = kun when Ea,cat/RTcat = Ea,un/RTun
4.20  10 4 J / mol
7.00  10 4 J / mol

, Tun = 488 K = 215C
8.3145 J / mol  K  293 K
8.3145 J / mol  K  Tun
70.
Rate =
 Δ[A]
 k[A] x
Δt
Assuming the catalyzed and uncatalyzed reaction have the same form and orders and because
1
concentrations are assumed equal, rate 
.
Δt
ratecat
Δt
ratecat
k
2400 yr
and
 un 
 cat
rateun
Δt cat
Δt cat
rateun
k un
A exp [E (cat) / RT]
exp[E a (cat)  E a (un )]
ratecat
k
a
=
 cat 
A exp [E (un ) / RT]
RT
rateun
k un
a
  5.90  10 4 J / mol  1.84  10 5 J / mol 
k cat
 = 7.62 × 1010
 exp

8
.
3145
J
/
mol

K

600
.
K
k un


Δt un
ratecat
k

 cat ,
Δt cat
rateun
k un
2400 yr
= 7.62 × 1010, tcat = 3.15 × 10 8 yr  1 sec
Δt cat
Additional Exercises
71.
Rate = k[NO]x[O2]y; comparing the first two experiments, [O2] is unchanged, [NO] is tripled,
and the rate increases by a factor of nine. Therefore, the reaction is second order in NO (32 =
9). The order of O2 is more difficult to determine. Comparing the second and third
experiments;
CHAPTER 12
CHEMICAL KINETICS
433
3.13  1017 k (2.50  1018 ) 2 (2.50  1018 ) y

, 1.74 = 0.694 (2.50)y, 2.51 = 2.50y, y = 1
1.80  1017 k (3.00  1018 ) 2 (1.00  1018 ) y
Rate = k[NO]2[O2]; From experiment 1:
2.00 × 1016 molecules/cm3s = k (1.00 × 1018 molecules/cm3)2 (1.00 × 1018 molecules/cm3)
k = 2.00 × 10 38 cm6/molecules2s = kmean
2
2.00  10 38 cm6  6.21  1018 molecules  7.36  1018 molecules
 
Rate =


molecules 2  s 
cm3
cm3

= 5.68 × 1018 molecules/cm3s
72.
The pressure of a gas is directly proportional to concentration. Therefore, we can use the
pressure data to solve the problem because Rate = Δ[SO2Cl2]/Δt   ΔPSO 2Cl 2 / Δt.
Assuming a first order equation, the data and plot follow.
Time (hour)
0.00
1.00
2.00
4.00
8.00
16.00
PSO 2Cl 2 (atm)
4.93
4.26
3.52
2.53
1.30
0.34
ln PSO 2Cl 2
1.595
1.449
1.258
0.928
0.262
1.08
434
CHAPTER 12
CHEMICAL KINETICS
Because the ln PSO 2Cl 2 vs. time plot is linear, the reaction is first order in SO2Cl2.
a. Slope of ln(P) vs. t plot is 0.168 hour 1 = k, k = 0.168 hour 1 = 4.67 × 10 5 s 1
Since concentration units don’t appear in first-order rate constants, this value of k
determined from the pressure data will be the same as if concentration data in molarity
units were used.
b. t1/2 =
c.
ln 2 0.6931
0.6931
= 4.13 hour


k
k
0.168 h 1
 PSO 2Cl 2
ln 
 Po

 = kt = 0.168 h 1 (20.0 h) = 3.36,


 PSO 2Cl 2

 P
o


 = e 3.36 = 3.47 × 10 2


Fraction left = 0.0347 = 3.47%
73.
From 338 K data, a plot of ln[N2O5] vs. t is linear and the slope = -4.86 × 10 3 (plot not
included). This tells us the reaction is first order in N2O5 with k = 4.86 × 10 3 at 338 K.
From 318 K data, the slope of ln[N2O5] vs t plot is equal to 4.98 × 10 4 , so k = 4.98 × 10 4 at
318 K. We now have two values of k at two temperatures, so we can solve for Ea.
k  E
ln  2   a
 k1  R
1
1
 
 T1 T2
 4.86  10 3 

 1
Ea
1 

, ln 



4 

 4.98  10  8.3145 J / mol  K  318 K 338 K 
Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol
74.
The Arrhenius equation is: k = A exp (-Ea/RT) or in logarithmic form, ln k = -Ea/RT + ln A.
Hence, a graph of ln k vs. 1/T should yield a straight line with a slope equal to -Ea/R since the
logarithmic form of the Arrhenius equation is in the form of a straight line equation, y = mx +
b. Note: We carried one extra significant figure in the following ln k values in order to
reduce round off error.
T (K)
195
230.
260.
298
1/T ( K 1 )
5.13 × 10 3
4.35 × 10 3
3.85 × 10 3
3.36 × 10 3
k (L/mols)
1.08 × 109
2.95 × 109
5.42 × 109
12.0 × 109
ln k
20.80
21.81
22.41
23.21
CHAPTER 12
CHEMICAL KINETICS
2.71 × 10 3
369
435
35.5 × 109
24.29
From "eyeballing" the line on the graph:
slope =
20.95  23.65

(5.00  10 3  3.00  10 3 ) K 1
Ea = 1.35 × 103 K ×
E a
 2.70
  1.35 103 K 
3
R
2.00 10
8.3145 J
= 1.12 × 104 J/mol = 11.2 kJ/mol
K mol
From a graphing calculator: slope = 1.43 × 103 K and Ea = 11.9 kJ/mol
75.
At high [S], the enzyme is completely saturated with substrate. Once the enzyme is
completely saturated, the rate of decomposition of ES can no longer increase, and the overall
rate remains constant.
76.
k = A exp (Ea/RT);
2.50 × 103 =
A cat exp(E a , cat / RT)
 E a , uncat 
E
k cat


 exp  a , cat

k uncat A uncat exp(E a , uncat / RT)
RT


  E a , cat  5.00  10 4 J / mol 
k cat

 exp 
 8.3145 J / mol  K  310 . K 
k uncat


ln (2.50 × 103 ) × 2.58 × 103 J/mol = Ea,cat + 5.00 × 104 J/mol
Ea, cat = 5.00 × 104 J/mol  2.02 × 104 J/mol = 2.98 × 104 J/mol = 29.8 kJ/mol
77.
a. Because [A]o < < [B]o or [C]o, the B and C concentrations remain constant at 1.00 M for
this experiment. So: rate = k[A]2[B][C] = k[A]2 where k = k[B][C]
For this pseudo-second-order reaction:
436
CHAPTER 12
CHEMICAL KINETICS
1
1
1
1
= kt +
= k(3.00 min) +
,
5
[A ]
[A] 0 3.26  10 M
1.00  10 4 M
k = 6890 L/molmin = 115 L/mols
k = k[B][C], k =
k'
115 L / mol  s
= 115 L3 / mol 3  s

[B][C]
(1.00 M ) (1.00 M )
b. For this pseudo-second-order reaction:
rate = k[A]2, t1/2 =
1
1
= 87.0 s

k'[A]0
115 L / mol  s (1.00  10 4 mol / L)
1
1
1
= kt +
= 115 L/mols × 600. s +
= 7.90 × 104 L/mol,
4
[A ]
[A] 0
1.00  10 mol / L
c.
[A] =
1
= 1.27 × 10–5 mol/L
7.90  10 4 L / mol
From the stoichiometry in the balanced reaction, 1 mol of B reacts with every 3 mol of A.
amount A reacted = 1.00 × 10–4 M – 1.27 × 10–5 M = 8.7 × 10–5 M
amount B reacted = 8.7 × 10–5 mol/L × 1 mol B / 3 mol A = 2.9 × 10–5 M
[B] = 1.00 M  2.9 × 10–5 M = 1.00 M
As we mentioned in part a, the concentration of B (and C) remains constant since the A
concentration is so small.
Challenge Problems
78.
 d[A]
= k[A]3,
dt
x
n
dx 
[ A ]t

[ A ]0
t
d[A]
   kdt
[A]3
0
xn 1
1
; So: 
n 1
2[A]2
[ A ]t
  kt , 
[ A ]0
1
1

  kt
2
2[A]t
2[A]o2
For the half-life equation, [A]t = 1/2[A]0:


1
1

2 [A]0 
2

3
2[A]02
2

1
  kt 1/ 2 ,
2[A]02
  kt 1/ 2 , t1/2 =
3
2[A]02 k

4
1

  kt 1/ 2
2
2[A]0
2[A]02
CHAPTER 12
CHEMICAL KINETICS
437
The first half-life is t1/2 = 40. s and corresponds to going from [A]0 to 1/2[A]0. The second
half-life corresponds to going from 1/2[A]0 to 1 / 4 [A]0 .
3
3
6
; Second half-life =
=
2
2
2[A]0 k
[A]02 k
1

2 [A]0  k
2

3
2[A]02 k
First half  life
= 3/12 = 1/4

6
Second half  life
[A]02 k
First half-life =
Because the first half-life is 40. s, the second half-life will be one-fourth of this or 10. s.
79.
Rate = k[I]x[OCl]y[OH]z; Comparing the first and second experiments:
18.7  10 3
k (0.0026 ) x (0.012 ) y (0.10) z

, 2.0 = 2.0x, x = 1
9.4  10 3
k (0.0013 ) x (0.012 ) y (0.10) z
Comparing the first and third experiments:
9.4  10 3
k (0.0013 )(0.012 ) y (0.10) z

, 2.0 = 2.0y, y = 1
4.7  10 3
k (0.0013 ) x (0.0060 ) y (0.10) z
Comparing the first and sixth experiments:
4.8  10 3
k (0.0013 )(0.012 )(0.20) z

, 1/2 = 2.0z, z = 1
9.4  10 3
k (0.0013 )(0.012 )(0.10) z
Rate =
k[ I  ][OCl ]
; The presence of OH decreases the rate of the reaction.
[OH  ]
For the first experiment:
9.4  10 3 mol
(0.0013 mol / L) (0.012 mol / L)
k
, k = 60.3 s 1 = 60. s 1
Ls
(0.10 mol / L)
For all experiments, kmean = 60. s 1 .
80.
For second order kinetics:
1
1
1

 kt and t1/ 2 
[A] [A]o
k[A]o
438
CHAPTER 12
a.
CHEMICAL KINETICS
1
1
1
1
= (0.250 L/mols)t +
= 0.250 × 180. s +
,
[A ]
[A]o
[ A]
1.00  10 2 M
1
= 145 M 1 , [A] = 6.90 × 10 3 M
[A ]
Amount of A that reacted = 0.0100
[A2] =
1
2
0.00690 = 0.0031 M
(3.1 × 10 3 M) = 1.6 × 10 3 M
b. After 3.00 minutes (180. s): [A] = 3.00 [B], 6.90 × 10 3 M = 3.00 [B]
[B] = 2.30 × 10 3 M
1
1
1
1
 k2t 
,
 k 2 (180. s) 
, k2 = 2.19 L/mols
3
[B]
[B]o 2.30  10 M
2.50  10 2 M
c. t1/2 =
81.
1
1
= 4.00 × 102 s

k[A]o 0.250 L / mol  s  1.00  10 2 mol / L
a. We check for first-order dependence by graphing ln [concentration] vs. time for each set
of data. The rate dependence on NO is determined from the first set of data because the
ozone concentration is relatively large compared to the NO concentration, so [O3] is
effectively constant.
Time (ms)
0
100.
500.
700.
1000.
[NO] (molecules/cm3)
6.0 × 108
5.0 × 108
2.4 × 108
1.7 × 108
9.9 × 107
ln [NO]
20.21
20.03
19.30
18.95
18.41
CHAPTER 12
CHEMICAL KINETICS
439
Because ln [NO] vs. t is linear, the reaction is first order with respect to NO.
We follow the same procedure for ozone using the second set of data. The data and
plot are:
Time (ms)
[O3] (molecules/cm3)
0
50.
100.
200.
300.
1.0 × 1010
8.4 × 109
7.0 × 109
4.9 × 109
3.4 × 109
ln [O3]
23.03
22.85
22.67
22.31
21.95
The plot of ln [O3] vs. t is linear. Hence, the reaction is first order with respect to ozone.
b. Rate = k[NO][O3] is the overall rate law.
c. For NO experiment, Rate = k[NO] and k = (slope from graph of ln [NO] vs. t).
k = -slope = 
18.41  20.21
= 1.8 s 1
3
(1000 .  0)  10 s
For ozone experiment, Rate = k[O3] and k = (slope from ln [O3] vs. t plot).
(21.95  23.03)
= 3.6 s 1
3
(300.  0)  10 s
d. From NO experiment, Rate = k[NO][O3] = k[NO] where k = k[O3].
k = slope = 
k = 1.8 s 1 = k(1.0 × 1014 molecules/cm3), k = 1.8 × 10 14 cm3/moleculess
We can check this from the ozone data. Rate = k[O3] = k[NO][O3] where k = k[NO].
k = 3.6 s 1 = k(2.0 × 1014 molecules/cm3), k = 1.8 × 10 14 cm3/moleculess
Both values of k agree.
440
82.
CHAPTER 12
CHEMICAL KINETICS
On the energy profile to the right, R =
reactants, P = products, Ea = activation
energy, ΔE = overall energy change
for the reaction and I = intermediate.
a-d.
See plot to the right.
e. This is a two-step reaction since an intermediate plateau appears between the reactant and
the products. This plateau represents the energy of the intermediate. The general reaction
mechanism for this reaction is:
R→I
I→P
R→P
In a mechanism, the rate of the slowest step determines the rate of the reaction. The
activation energy for the slowest step will be the largest energy barrier that the reaction
must over-come. Since the second hump in the diagram is at the highest energy, the
second step has the largest activation energy and will be the rate-determining step (the
slow step).
83.
a. If the interval between flashes is 16.3 sec, then the rate is:
1 flash/16.3 s = 6.13 × 10 2 s 1 = k
Interval
T
16.3 s
6.13 × 10 2 s 1
21.0°C (294.2 K)
13.0 s
7.69 × 10 2 s 1
27.8°C (301.0 K)
k
ln  2
 k1
b.
k
 Ea
 
 R
1
1 
 
 ; Solving: Ea = 2.5 × 104 J/mol = 25 kJ/mol
T
T
1
2



 2.5  10 4 J / mol
k

ln 
2 
 6.13  10  8.3145 J / mol  K

1
1


 294.2 K 303.2 K

 = 0.30

k = e0.30 × 6.13 × 10 2 = 8.3 × 10 2 s 1 ; Interval = 1/k = 12 seconds
c.
T
Interval
54-2(Intervals)
21.0 °C
27.8 °C
30.0 °C
16.3 s
13.0 s
12 s
21 °C
28 °C
30. °C
CHAPTER 12
CHEMICAL KINETICS
441
This rule of thumb gives excellent agreement to two significant figures.
84.
We need the value of k at 500. K.
k
ln  2
 k1
 Ea
 
 R
1
1
 
 T1 T2




 1.11  10 5 J / mol
k2

ln 
12

 2.3  10 L / mol  s  8.3145 J / mol  K
 1
1


 273 K 500. K

 = 22.2

k2
= e22.2, k2 = 1.0 × 10 2 L/mols
2.3  10 12
Because the decomposition reaction is an elementary reaction, then the rate law can be
written using the coefficients in the balanced equation. For this reaction: Rate = k[NO2]2. To
solve for the time, we must use the integrated rate law for second-order kinetics. The major
problem now is converting units so they match. Rearranging the ideal gas law gives n/V =
P/RT. Substituting P/RT for concentration units in the second-order integrated rate law
equation:
1
1
1
1
RT RT
RT  Po  P 


 kt 
,
 kt 
,

 kt , t 
[ NO2 ]
[ NO2 ]o P / RT
Po / RT
P
Po
k  P  Po 
t=
85.
(0.08206 L atm / K  mol )(500. K)  2.5 atm  1.5 atm 
 = 1.1 × 103 s
 
2
1
.
5
atm

2
.
5
atm
1.0  10 L / mol  s


a. [B] >> [A] so that [B] can be considered constant over the experiments. (This gives us a
pseudo-order rate law equation.)
b. Note in each data set that successive half-lives double (in expt. 1 the first half-life is 40.
sec, the second is 80. sec; in expt. 2 the first half-life is 20. sec, the second is 40. sec).
Thus, the reaction is second order in [A] since t1/2 for second order reactions is inversely
proportional to concentration. Between expt. 1 and expt. 2, we double [B] and the
reaction rate doubles, thus it is first order in [B]. The overall rate law equation is rate =
k[A]2[B].
1
1
Using t1/2 =
, we get k =
= 0.25 L/mols. But this is
k[A]0
(40. s) (10.0  10 2 mol / L)
actually k where rate = k [A]2 and k = k[B].
k=
86.
k'
0.25 L / mol  s
= 0.050 L2/ mol 2  s

[B]
5.0 mol / L
a. Rate = k[H2]x[NO]y; Looking at the data in experiment 2, notice that the concentration of
442
CHAPTER 12
CHEMICAL KINETICS
H2 is cut in half every 10. sec. Only first-order reactions have a half-life that is independent of concentration. The reaction is first order in H2. In the data for experiment 1,
notice that the half-life is 40. s. This indicates that in going from experiment 1 to
experiment 2 where the NO concentration doubled, the rate of reaction increased by a
factor of four. This tells us that the reaction is second order in NO.
Rate = k[H2][NO]2
b. This reaction is pseudo-first-order in [H2] as the concentration of NO is so large, it is
basically constant.
Rate = k[NO]2[H2] = k[H2] where k = k[NO]2
For a first-order reaction, the integrated rate law is:
 [H 2 ] 
 = kt
ln 
 [H 2 ]0 
Use any set of data you want to calculate k. For example, in experiment 1 from 0 to 20.
s the concentration of H2 decreased from 0.010 M to 0.0071 M.
 0.0071 
2 1
ln 
 = k(20. s), k = 1.7 × 10 s
 0.010 
k = k[NO]2, 1.7 × 10 2 s 1 = k(10.0 mol/L)2
k = 1.7 × 10 4 L2 / mol 2  s
We get similar values for k using other data from either experiment 1 or experiment 2.
c.
87.
 [H 2 ]
ln 
 0.010 M

 = kt = (1.7 × 10 2 L2 / mol 2  s ) × 30. s, [H2] = 6.0 × 10 3 M

Rate = k[A]x[B]y[C]z; During the course of experiment 1, [A] and [C] are essentially
constant, and Rate = k[B]y where k = k[A]0x [C]0z .
[B] (M)
time (s)
ln[B]
1/[B] (M -1)
1.0 × 10 3
2.7 × 10 4
1.6 × 10 4
1.1 × 10 4
8.5 × 10 5
6.9 × 10 5
5.8 × 10 5
0
1.0 × 105
2.0 × 105
3.0 × 105
4.0 × 105
5.0 × 105
6.0 × 105
6.91
8.22
8.74
9.12
9.37
9.58
9.76
1.0 × 103
3.7 × 103
6.3 × 103
9.1 × 103
12 × 103
14 × 103
17 × 103
CHAPTER 12
CHEMICAL KINETICS
443
A plot of 1/[B] vs. t is linear (plot not included). So the reaction is second order in B and the
integrated rate equation from the slope and y-intercept of the straight line in the plot is:
1/[B] = (2.7 × 10 2 L / mol  s ) t + 1.0 × 103 L/mol; k = 2.7 × 10 2 L / mol  s
For experiment 2, [B] and [C] are essentially constant and Rate = k[A]x where k =
k[B]0y [C]0z = k[B]02 [C]0z .
[A] (M)
1.0 × 10 2
8.9 × 10 3
7.1 × 10 3
5.5 × 10 3
3.8 × 10 3
2.9 × 10 3
2.0 × 10 3
time (s)
ln[A]
1/[A] ( M 1 )
0
1.0
3.0
5.0
8.0
10.0
13.0
4.61
4.72
4.95
5.20
5.57
5.84
6.21
1.0 × 102
110
140
180
260
340
5.0 × 102
A plot of ln[A] vs. t is linear. So the reaction is first order in A and the integrated rate law
from the slope and y-intercept of the straight line in the plot is:
ln[A] = (0.123 s 1 ) t  4.61; k = 0.123 s 1
Note: We will carry an extra significant figure in k.
Experiment 3: [A] and [B] are constant; Rate = k[C]z
The plot of [C] vs. t is linear. Thus, z = 0.
The overall rate law is: Rate = k[A][B]2
From Experiment 1 (to determine k):
k = 2.7 × 10 2 L / mol  s = k[A]0x [C]0z = k[A]o = k(2.0 M), k = 1.4 × 10 2 L2 / mol 2  s
From Experiment 2: k = 0.123 s 1 = k[B]02 , k =
88.
0.123 s 1
= 1.4 × 10 2 L2 / mol 2  s
(3.0 M ) 2
Thus, Rate = k[A][B]2 and k = 1.4 × 10-2 L2/mol2s.
a. Rate = (k1 + k2[H+])[I]m[H2O2]n
In all the experiments, the concentration of H2O2 is small compared to the concentrations
of I and H+. Therefore, the concentrations of I and H+ are effectively constant and the
rate law reduces to:
Rate = kobs[H2O2]n where kobs = (k1 + k2[H+])[I]m
Since all plots of ln[H2O2] vs. time are linear, the reaction is first order with respect to
H2O2 (n = 1). The slopes of the ln[H2O2] vs. time plots equal kobs which equals (k1 +
444
CHAPTER 12
CHEMICAL KINETICS
k2[H+])[I]m. To determine the order of I, compare the slopes of two experiments where
I changes and H+ is constant. Comparing the first two experiments:
 [k 1  k 2 (0.0400 M )][0.3000 M ) m
slope (exp . 2)
 0.360
,


slope (exp . 1)
 0.120
 [k 1  k 2 (0.0400 M )][0.1000 M ) m
m
 0.3000 
3.00 = 
 = (3.000)m, m = 1
0
.
1000


The reaction is also first order with respect to I.
b. The slope equation has two unknowns, k1 and k2. To solve for k1 and k2, we must have
two equations. We need to take one of the first set of three experiments and one of the
2nd set of three experiments to generate the two equations in k1 and k2.
Experiment 1: slope = (k1 + k2[H+])[I]
0.120 min 1 = [k1 + k2 (0.0400 M)] 0.1000 M or 1.20 = k1 + k2 (0.0400)
Experiment 4:
0.0760 min 1 =  [k1 + k2 (0.0200 M)] 0.0750 M or 1.01 = k1 + k2 (0.0200)
Subtracting 4 from 1:
1.20 = k1 + k2 (0.0400)
1.01 = k1  k2 (0.0200)
_____________________
0.19 =
k2 (0.0200), k2 = 9.5 L2/mol2min
1.20 = k1 + 9.5(0.0400), k1 = 0.82 L/ molmin
c. There are two pathways, one involving H+ with rate = k2[H+][I][H2O2] and another
pathway not involving H+ with rate = k1[I][H2O2]. The overall rate of reaction depends
on which of these two pathways dominates, and this depends on the H+ concentration.
Integrative Problems
89.
3600 s
ln 2
ln 2
= 3.15 × 104 s; k =
= 2.20 × 10 5 s 1

t1/ 2
h
3.15  10 4 s
The partial pressure of a gas is directly related to the concentration in mol/L. So instead of
using mol/L as the concentration units in the integrated first order rate law, we can use partial
pressures of SO2Cl2.
8.75 h ×
CHAPTER 12
CHEMICAL KINETICS
445
P
 P 
3600 s
 = (2.20 × 10 5 s 1 )(12.5 h ×
)
ln   = kt, ln 
h
 791 torr 
 P0 
1 atm
= 0.387 atm
PSO 2Cl 2 = 294 torr ×
760 torr
n=
PV
0.387 atm  1.25 L
= 9.94 × 10 3 mol SO2Cl2

0
.
08206
L
atm
RT
 593 K
mol K
9.94 × 10 3 mol ×
90.
k=
6.022  10 23 molecules
= 5.99 × 1021 molecules SO2Cl2
mol
ln 2
ln 2
= 1.04 × 10 3 s 1

t1/ 2
667 s
2.38 g InCl 
[In+]0 =
1 mol InCl 1 mol In 

150 .3 g
mol InCl
= 0.0317 mol/L
0.500 L
 [In  ] 
 [In  ]
ln      kt , ln 
 0.0317 M
 [In ]0 

3600 s
 = (1.04 × 10 3 s 1 ) × 1.25 h ×

h

[In+] = 2.94 × 10 4 mol/L
The balanced redox reaction is: 3 In+(aq) → 2 In(s) + In3+(aq)
mol In+ reacted = 0.500 L ×
1.57 × 10 2 mol In+ ×
91.
0.0317 mol
2.94  10 4 mol
 0.500 L 
= 1.57 × 10 2 mol In+
L
L
2 mol In
114.8 g In
= 1.20 g In


mol In
3 mol In
 1.7  10 2 s 1 
k  E  1
Ea
1 

ln  2   a    ; ln 
 4 1 
R  T1 T2 
 k1 
 7.2  10 s  8.3145 J / K  mol
 1
1 



 660 . K 720 . K 
Ea = 2.1 × 105 J/mol
For k at 325C (598 K):
5
 1.7  10 2 s 1 
  2.1  10 J / mol
ln 

 8.3145 J / K  mol
k


 1
1 

 , k = 1.3 × 10 5 s 1

 598 K 720. K 
For three half-lives, we go from 100% → 50% → 25% → 12.5%. After three half-lives,
12.5% of the original amount of C2H5I remains. Partial pressures are directly related to gas
concentrations in mol/L:
446
CHAPTER 12
CHEMICAL KINETICS
PC2 H5I = 894 torr × 0.125 = 112 torr after 3 half-lives
Marathon Problem
92.
a. Rate = k[CH3X]x[Y]y; For experiment 1, [Y] is constant so Rate = k[CH3X]x where
k = k(3.0 M)y.
A plot (not included) of ln[CH3X] vs t is linear (x = 1). The integrated rate law is:
ln[CH3X] = 0.93 t  3.99; k = 0.93 h 1
Fory Experiment 2, [Y] is again constant with Rate = k[CH3X]x where k = k(4.5
M) .
The ln plot is linear again with an integrated rate law:
ln[CH3X] = 0.93 t  5.40; k = 0.93 h 1
Dividing the rate constant values:
y
k'
0.93 k (3.0)
, 1.0 = (0.67)y, y = 0


k' '
0.93 k (4.5) y
The reaction is first order in CH3X and zero order in Y. The overall rate law is:
Rate = k[CH3X] where k = 0.93 h 1 at 25°C.
b. t1/2 = (ln 2)/k = 0.6931/(7.88 × 108 h 1 ) = 8.80 × 10 10 h
c.
ln
 7.88  108 
 1
Ea
k 2 Ea  1
1 
1 
 
  , ln 






k1
R  T1 T2 
8.3145 J / K  mol  298 K 358 K 
 0.93 
Ea = 3.0 × 105 J/mol = 3.0 × 102 kJ/mol
d. From part a, the reaction is first order in CH3X and zero order in Y. From part c, the
activation energy is close to the C-X bond energy. A plausible mechanism that explains
the results in parts a and c is:
CH3X → CH3 + X
(slow)
CH3 + Y → CH3Y
(fast)
Note: This is a possible mechanism since the derived rate law is the same as the
experimental rate law and the sum of the steps give the overall balanced equation.