Integrated Rate Laws
There are two types of unimolecular process (processes with only a single type of reactant molecule):
aA ö bB + cC (decomposition)
aA ö Aa (dimerization)
The generic rate law for a unimolecular process would be rate = k@ADx . So, we assume an order (‘x’)
and integrate the rate law. The end result of each assumed order of reaction (zero, first, second and
third) are given below. Note that the ‘a’ in each integrated rate law represents the number of stoichiometric moles
0th Order : @ADt - @AD0 = -a k t
@AD
1st Order : ln@ADt - ln@AD0 = -a k t or lnJ t N = -a k t
@AD0
2
nd
Order :
3rd Order :
1
@ADt
1
@AD2t
-
1
@AD0
1
@AD20
=ak t
=2ak t
Each of the integrated rate laws contains two variables: @ADt and t. By inspection, we can linearize
each of the integrated rate laws with the purpose of knowing how data of that order would need to be
plotted to appear linear. For example, the 1st Order rate law would linearize as such
ln@ADt = -a k t + ln@AD0
y
= mx + b
So, plotting experimental data as ln@ADt vs. time should yield a straight line with slope = -k and b =
ln@AD0 , if the order of the unimolecular process is first order. However, should the reaction be any other
order, the resulting plot would certainly be nonlinear. Notice how the axis that are plotted for each order
are different in Figure I. Now, compare the axis of each graph to the integrated rate law. The necessary axis to plot are present in the integrated rate law, and, since the integrated rate laws are given to
you, there is no need to memorize each graph.
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Zero Order
First Order
0.0
0.7
0.6
-0.5
0.5
@AD 0.4
ln@AD
-1.0
0.3
0.2
-1.5
0.1
0.0
0
10
20
30
40
50
60
0
10
20
timeHsL
30
40
50
60
timeHsL
Second Order
Third Order
30
5
25
4
1
1
3
@AD2
@AD
2
20
15
10
1
5
0
0
0
10
20
30
timeHsL
40
50
60
0
10
20
30
40
50
60
timeHsL
Figure I: Presented are four plots of the data represented in Table I, below. The four plots represent
the unimolecular reaction order 0 - 3rd . The dependent axis for each plot is chosen so as to represent a
linear relationship of data if the unimolecular reaction matches the order. The data in Table I is seen to
be second order, as the 0, 1st , and 3rd data points are all nonlinear, while the 2nd order data poins are
clearly linear. Best fit straight lines are show for each plot to clarify the nonlinearity of the actual data
points.
Table I: Sample concentration vs. time data for the decomposition of A ö Products.
Analysis of the data shows that the process is second order and, thus, has a rate law of rate = k @AD2 .
Time (s)
0.
5.
10.
15.
20.
25.
30.
35.
40.
45.
50.
55.
60.
[A]
0.762
0.598
0.493
0.416
0.361
0.317
0.289
0.262
0.234
0.216
0.198
0.183
0.178
Once the true rate law is known, the integrated rate law now allows (a) the prediction of the time it takes
for a reaction to use a certain amount of reactant (or make product), or (b) the prediction of the amount
of reactant still remaining (or used) in a certain amount of time.
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Once the true rate law is known, the integrated rate law now allows (a) the prediction of the time it takes
for a reaction to use a certain amount of reactant (or make product), or (b) the prediction of the amount
of reactant still remaining (or used) in a certain amount of time.
† Exercise 1: The reverse Haber proces is 2 NH3 HgL ö3 H2 HgL + N2 HgL. Describe how you would
analyze the data from a single experiment to verify that this process is zero order.
The experiment would do be to measure the concentration of one of the species as time
progresses. Since the zero order integrated rate law is @ADt = - 2 kt + @AD0 , I would plot [A] vs.
time and see if the result was linear data. If linear, then the process is zero order. If nonlinear (a
curve), then the process is some other order.
† Exercise 2: In figure I, the slope of the best fit line is 0.0843 M -1 s-1 . If the same reaction were run
with an intial concentration of A of 0.250M, how long would it take for the @ADt to reach 0.095M?
1
1
=k
@ADt
@AD0
1
N = 77.4 s
0.25 M
The reaction from figure I is 2 nd order, so
t=
1
k
J
1
@ADt
-
1
1
N=
@AD0
0.0843 M -1 s-1
J
1
0.095 M
-
t. Thus,
† Exercise 3: How long would it take for 75% of X to decay from an initial concentration of 1.000M, if
the reaction is zero order with respect to X and k = 0.147M/s?
[XDt = 0.25 (1.000M) = 0.250M after decay. Rearranging zero order integrated rate law gives,
t=
-1
k
H@XDt - @XDo L =
-1
0.147 Mês
H0.250 M - 1.000 ML = 5.10 s
† Exercise 4: A unimolecular dimerization of substance M has a rate constant of k = 12.08 s-1 . If 80.0
moles of this substance are in a 12.00L container, how many moles of M will remain after the reaction
proceeds for 30 seconds?
@MD0 = 80.0 mol ê 12.00 L = 6.67 M. The rate constant’s units are s -1 , so the reaction is first order.
Thus, ln@MDt - ln@MD0 = -kt and ln@MDt = -k t + ln@MD0 .
ln@MDt = -I12.08 s -1 M H30.00 sL + lnH6.667 ML = -360.5.
Subscript@@MD, tD = ‰ -360.5 = 2.73 ä 10 -157 M. This represents essentially 0M and we expect no
moles of M remain after 30 seconds.
Half Lives
One method of simplifying the integrated rate laws for analysis is to look for a specific time that it takes
to reach some defined concentration. The half life is the time it takes for a reaction to reach half the
1
2
initial concentration. That is, t = t1ê2 when @ADt = @AD0 . Substituting the previous condition into each
integrated rate law yields the following, with only the zero order example being worked for you. The half
life equation for each order of reaction can be memorized or generated from the given integrated rate
laws.
Zero Order Half Life:
@ADt - @AD0 = -a k t
:at t = t1ê2 , @ADt =
1
2
@AD0 ... substitute both in >
4|
1
2
@AD0 - @AD0 = -a k t1ê2
1
- @AD0 = -a k t1ê2
2
@AD0
t1ê2 =
2ak
Using the same method, the other half life equations can be determined and are listed in Table II,
below. Notice that the 1st order reaction does not depend on how much stuff we initially have. That
means that 20g of substance A and 2000lbs of substance A will reach their half life in the same time!
The other orders all depend on the concentration initially in the system.
Table II: Half life equations for unimolecular
reactions of orders 0 through 3.
Order Half Life Equation
0
t1ê2 =
1
t1ê2 =
2
t1ê2 =
3
t1ê2 =
@AD0
2ak
ln H2L
ak
1
a k @AD0
3
2 a k @AD02
† Exercise 5: When 1.0M of compound G decomposes (G ö products) at 400K, the rate constant is
0.0405s-1 . (a) What is the order of reaction with respect to G and how do you know? (b) What is
the half life for the decomposition at 400K. (c) How would your answer to (b) change if you started
with 2.5M of G?
a) the units of the rate constant tell me it’s 1st order. The units of the rate constant are always
M 1-order s -1 , where the ‘order’ is the overall order of the reaction.
b) t1ê2 =
lnH2L
k
=
0.6931
0.0405 s-1
= 17.11 s
c) For first order reaction, the half life is independent of the initial concentration (nowhere in the
equation is the concentration seen), so there would be no difference.
† Exercise 6: Like most natural processes, radioactive decay is a first order process. Beryllium-11 has
a half life of 13.81 seconds. (a) What is the rate constant for the radioactive decay of 11 Be? (b) How
long does it take for X-grams of 11 Be to decay into a lump with only X/4 grams?
a) k =
lnH2L
t1ê2
=
0.6931
13.81 s
= 0.05019 s -1
b) Since it’s a first order process, lnJ
masst
N=
mass0
-k t. We don’t need to use concentration now
because the conversion of the mass to concentration is identical to both the numerator and
denominator of the expression, so the conversions would simply cancel out. Thus,
masst
-1
-1
X ê4
t =
lnJ
N=
lnJ N = 27.62 s
-1
k
mass0
0.05019 s
X
b) Since it’s a first order process, lnJ
masst
N=
mass0
-k t. We don’t need to use concentration now
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because the conversion of the mass to concentration is identical to both the numerator and
denominator of the expression, so the conversions would simply cancel out. Thus,
masst
-1
-1
X ê4
t =
lnJ
N=
lnJ N = 27.62 s
-1
k
mass0
X
0.05019 s
† Exercise 7: Substance Z dimerizes into Z2 in a third order process. When a 2.1M sample is allowed
to dimerize, it takes 3.87hrs to reach 0.50M in Z. (a) How many seconds did it take to reach 0.50M?
(b) What is the rate constant for the dimerization of Z? (c) What is the half life of the dimerization of
Z? (d) How long will it take a 2.0M sample to dimerize to 0.50M in Z?
a) #s = 3.87 hrs ä
60 min
1 hr
ä
60 s
1 min
= 13 932 s = 1.39 ä 10 4 s
b) Using the given information and the 3 rd order integrated rate law,
1
@ADt2
-
1
@AD02
= 2 k t, we can find
the rate
constant.
k=
1
2t
K
1
@ADt2
c) t1ê2 =
-
1
@AD02
3
2 k @AD02
=
O=
1
2 µ 13 932 s
J
1
0.50 M 2
-
1
2.1 M 2
3
2 I1.354ä 10
-4
M -2 s-1 M H2.1 ML2
N = 1.354 ä 10 -4 M -2 s -1 = 1.4 ä 10 -4 M -2 s -1
= 837.3 s = 8.4 ä 10 2 s
d) Almost the same problem, but not exactly. We have k, so the rest is just using the proper
equation:
t=
1
2k
K
1
@ADt2
1
@ADt2
-
1
@AD02
1
@AD02
O=
= 2 k t.
1
2 I1.354ä 10
-4
M -2 s -1 M
J
1
0.50 M 2
-
1
2.0 M 2
N = 13 849 s = 1.4 ä 10 4 s. Note, just a little
longer than the initial reaction.
Temperature Dependence
To increase the number of collisions that have the proper orientation and enough energy to break any
existing bond (the conditions for a reaction to occur), we determined that two experimental parameters
could be adjusted: the molar concentration and the temperature. The temperature component, until
now, has been ignored by imagining the reactions were run at the same temperature, thus allowing the
use of the same rate constant.
While the concentration dependence found for the rate law tends to remain the same as temperature
changes, the rate constant does not. Thus, the rate constant will change depending upon the temperature at which we do our experiments. Svante Arrhenius determined that the temperature dependence
can be modeled with
Ea
k = A ‰- RT
which is known as the Arrhenius rate law. Here, Ea is the activation energy (or Energy of Activation), A
is the frequency factor (or pre-exponential constant), k is the rate constant, and R is the universal gas
law constant J8.314
J
N.
Kÿmole
Figure II displays The activation energy is an important quantity to know for
a reaction, and it is noteworthy that this barrier to a successful reaction is not dependent on temperature. That is, the forces that need to be broken require a certain energy to break them, and adjusting
the temperature does nothing to alter how much energy is need to break whatever bonds/interactions.
Instead, a temperature increase causes the average translational kinetic energy of the system to
increase. Thus, molecules, on average, move quicker and collide more often and with more energy.
The higher temperature conditions improve the chances of a successful collision and, in turn, increase
the rate of reaction.
law constant J8.314
6|
J
N.
Kÿmole
Figure II displays The activation energy is an important quantity to know for
a reaction, and it is noteworthy that this barrier to a successful reaction is not dependent on temperature. That is, the forces that need to be broken require a certain energy to break them, and adjusting
the temperature does nothing to alter how much energy is need to break whatever bonds/interactions.
Instead, a temperature increase causes the average translational kinetic energy of the system to
increase. Thus, molecules, on average, move quicker and collide more often and with more energy.
The higher temperature conditions improve the chances of a successful collision and, in turn, increase
the rate of reaction.
Figure II: Potential energy curves for an endothermic (left) and exothermic (right) reaction. The left
diagram identifies the energy aspects in terms of their chemical names or symbols, such as activation
energy and enthalpy change (DHrxn ). The right diagram displays the same properties but in terms of
where the energy changes occur and what causes the energy change.
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Experimental determination of the activation energy is usually done graphically. By taking the natural
log of both sides of the Arrhenius equation, a linear form of the equation is generated. A plot of ln(k) vs.
1/T should yield a straight line, as in figure III.
lnHkL =
-Ea 1
+ ln@AD
R T
=
m x + b
y
0
-5
lnHkL
-10
-15
0.0018
0.0020
0.0022
0.0024
0.0026
0.0028
0.0030
0.0032
1êTHKL
Figure III: Plot of ln(k) vs. 1/T displaying the linear nature of the data in this form.
The slope of an Arrhenius plot is m = -Ea ê R, allowing the computation of the
activation energy graphically.
To ensure accuracy, it is always best to utilize the graphical method when obtaining an activation
energy. However, for expensive reactions where 5 or more trials would be costly, a good approximation
can be found with only two trials. Thus, in the absence of sufficient data to make a plot similar to Figure
III, a ratio of rate constants taken at two different temperatures will give an approximate (and less
accurate) activation energy.
kT1
kT2
=
A ‰-Ea êR T1
A‰
-Ea êR T2
=‰
-Ea êR J
1
T1
-
1
T2
N
, which can be written as lnJ
kT1
kT2
N=
-Ea
R
J
1
T1
-
1
N.
T2
† Exercise 8: A good rule of thumb is that increasing the temperature of a reaction by 10°C will double
the rate of reaction. Thus, the rate constant, k, will double for most reactions if the temperature
change is 10°C. That said, what is a good estimate to the size of the activation energy for reactions
that run near and around room temperature (say, 20°C and 30°C)?
Using the ratio of rate constants of the Arrhenius rate law, we have
2
1
lnJ N =
- Ea
8.314 JêK ÿmol
J
1
303 K
-
1
N,
293 K
which yields Ea = 51 151 J ê mol º 50 kJ ê mol.