# Solved Examples in Chemical Kinetics1

```Example 1:
The rate of change of molar concentration of CH 3 radicals in the reaction 2πΆπ»3(π) →
πΆπ»3 πΆπ»3(π) was reported as d[CH3]/dt = −1.2 mol dm−3 s−1 under particular conditions.
What is (a) the rate of reaction and (b) the rate of formation of CH3CH3?
Solution: The reaction is given by
2πΆπ»3(π) → πΆπ»3 πΆπ»3(π)
From this, the rate of reaction can be written as
π1: πππ₯π = −
ππΆπ»3 (π)
2
= ππΆπ»3 πΆπ»3 (π)
That is, the rate of reaction, πππ₯π , is half the rate of disappearance of πΆπ»3(π) and is
equal to the rate of appearance of πΆπ»3 πΆπ»3(π). Note, the rate of disappearance of πΆπ»3 (π)
can also be written as
π2: − ππΆπ»3 (π) = −
π[πΆπ»3 ]
ππ‘
Thus, the rate of reaction can be calculated as
π3: πππ₯π = −
1 π [πΆπ»3 ]
2 ππ‘
1
π4: πππ₯π = − (−1.2 πππ ππ−3 π  −1 )
2
π5: πππ₯π = 0.6 πππ ππ−3 π  −1
The rate of formation of CH3CH3 can then be calculated as
ππΆπ»3 πΆπ»3 (π) = −
ππΆπ»3 (π)
2
1
ππΆπ»3 πΆπ»3 (π) = − (−1.2 πππ ππ−3 π  −1 )
2
ππΆπ»3 πΆπ»3 (π) = 0.6 πππ ππ−3 π  −1
Example 2:
The recombination of iodine atoms in the gas phase in the presence of argon was
investigated and the order of the reaction was determined by the method of initial
rates. The initial rates of reaction of 2πΌ(π) + π΄π(π) → πΌ2(π) + π΄ππ were as follows:
Experiment No.
1
2
3
4
5
6
7
8
9
10
11
12
[πΌ0 ], 10−5 πππ ππ−3
1.0
2.0
4.0
6.0
1.0
2.0
4.0
6.0
1.0
2.0
4.0
6.0
[π΄π], 10−3 πππ ππ−3
1.0
1.0
1.0
1.0
5.0
5.0
5.0
5.0
10.0
10.0
10.0
10.0
πππ‘π, πππ ππ−3 π  −1
8.70 x 10-4
3.48 x 10-3
1.39 x 10-2
3.13 x 10-2
4.35 x 10-3
1.74 x 10-2
6.96 x 10-2
1.57 x 10-1
8.69 x 10-3
3.47 x 10-2
1.38 x 10-1
3.13 x 10-1
Determine the orders of reaction with respect to the I and Ar atom concentrations and
the rate constant.
Solution 1: Using Ratio of Two Rates
The rate equation for this reaction can be written in the general form as
π1: πππ₯π = π [πΌ ]π [π΄π]π
The reaction order with respect to Iodine and Argon atom can be obtained by getting
the ratio of two rate equations.
π2:
πππ₯π1 π [πΌ ]1π [π΄π]1π
=
πππ₯π2 π [πΌ ]π2 [π΄π]π
2
Further simplification results to
π
πππ₯π 1
[πΌ ]1
[π΄π]1
π3:
=(
) (
)
[πΌ ]2
[π΄π]2
πππ₯π2
π
To simplify calculations in determining the reaction order of Iodine, n, choose
concentration pairs at which [Ar] is constant (e.g. experiment no. 1 and 2, 5 and 7,
etc.). The same is true for the determination of the reaction order of Argon, m, choose
concentration pairs at which [I] is constant. Doing so,
Determination of the reaction order of Iodine, n, using Experiment No. 1 and 2:
Substituting now the values in e3,
π
8.70 π₯ 10−4
1.0 π₯ 10−5
1.0 π₯ 10−3
π4:
=(
) (
)
3.48 π₯ 10−3
2.0 π₯ 10−5
1.0 π₯ 10−3
π5:
π
1
1 π
=( )
4
2
Solving now for n,
π6: π = 2
Checking the order: Use Experiment no. 9 and 11
π
π
8.69 π₯ 10−3
1.0 π₯ 10−5
10.0 π₯ 10−3
π7:
=
(
)
(
)
1.38 π₯ 10−1
4.0 π₯ 10−5
10.0 π₯ 10−3
869
1 π
π8:
=( )
13800
4
Solving now for n,
π9: π = 1.99 ≈ 2
Determination of the reaction order of Argon, m, using Experiment No. 1 and 5:
π
π
8.70 π₯ 10−4
1.0 π₯ 10−5
1.0 π₯ 10−3
π10:
=
(
)
(
)
4.35 π₯ 10−3
1.0 π₯ 10−5
5.0 π₯ 10−3
1
1 π
π11: = ( )
5
5
Solving now for m,
π12: π = 1
Checking the order: Use Experiment no. 4 and 12
π
π
3.13 π₯ 10−2
6.0 π₯ 10−5
1.0 π₯ 10−3
π10:
=(
) (
)
3.13 π₯ 10−1
6.0 π₯ 10−5
10.0 π₯ 10−3
1
1 π
π11:
=( )
10
10
π12: π = 1
Therefore, the reaction is second order with respect to Iodine and is first order with
respect to Argon. The rate equation can then be written as
π13: πππ₯π = π [πΌ ]2 [π΄π]
Since the data were obtained from experiment, you would expect that there some little
discrepancy between the rate constant, k, for each concentration pairs. Because of
this, the average rate constant should be used in the rate equation. The rate constant
for each concentration pair can be determined by manipulating e13 as follows.
π14: π =
πππ₯π
[πΌ ]2 [π΄π]
The unit of k is given by
π15: π =
πππ ππ−3 π  −1
(πππ ππ−3 )2 (πππ ππ−3 )
π16: π =
π  −1
πππ 2 ππ−6
The outline of the k-values is shown below.
Exp. No.
1
2
3
4
5
6
7
8
9
10
11
12
[I]
[Ar]
0.00001 0.001
0.00002 0.001
0.00004 0.001
0.00006 0.001
0.00001 0.005
0.00002 0.005
0.00004 0.005
0.00006 0.005
0.00001 0.01
0.00002 0.01
0.00004 0.01
0.00006 0.01
Average k
rate
8.70E-04
3.48E-03
1.39E-02
3.13E-02
4.35E-03
1.74E-02
6.96E-02
1.57E-01
8.69E-03
3.47E-02
1.38E-01
3.13E-01
k
8.70E+09
8.70E+09
8.69E+09
8.69E+09
8.70E+09
8.70E+09
8.70E+09
8.72E+09
8.69E+09
8.68E+09
8.63E+09
8.69E+09
8.69E+09
π  −1
Hence, the rate constant is 8.69 x 109 πππ2 ππ−6. The rate equation is then given by
π17: πππ₯π (πππ ππ−3 π  −1 ) = (8.69 x 109 )[πΌ ]2 [π΄π]
Solution 2: Using System of Linear Equations
The rate equation given in e1 can be converted to a linear form by taking the natural
logarithm (or even logarithm) of both sides. Doing so,
π18: πππ₯π = π[πΌ ]π [π΄π]π
π19: ln(πππ₯π ) = ln(π [πΌ ]π [π΄π]π)
Applying now the properties of logarithm,
π20: ln(πππ₯π ) = ln(π ) + π ln([πΌ ]) + π ln([π΄π])
Rewriting now,
π21: ln([πΌ ]) π + ln([π΄π]) π + ln(π ) = ln(πππ₯π )
In e21, the variables are n, m and ln(π ). The coefficients of the variables are ln([πΌ ]),
ln([π΄π]) and 1 for the variables n, m and ln(π ), respectively. The constant term is
ln(πππ₯π ). Since we have three unknowns, we also need three equations. From the data,
we can choose any three pairs provided that we use atleast two concentration values
for each specie (i.e. atleast two concentration values for Iodine and Argon). This is
because using the same specie concentration for the three equations would result to a
null matrix. For example, using experiment nos. 1, 2 and 3 would result to a math error.
Choosing now experiment nos. 1, 5 and 12, we can generate the following equations
Experiment no. 1: ln(0.00001) π + ln(0.001) π + ln(π ) = ln(0.00087)
Experiment no. 5: ln(0.00001) π + ln(0.005) π + ln(π ) = ln(0.00435)
Experiment no. 12: ln(0.00006) π + ln(0.01) π + ln(π ) = ln(0.313)
Solving now the three equations simultaneously, we obtain
π = 1.9996 ≈ 2, π = 1 and ln(π ) = 22.882
As for the value of the rate constant, the average k should be used as was discussed
and obtained in Solution 1.
Example 3:
The initial rate of a reaction depended on concentration of a substance J as follows:
Experiment No.
[π½]0 , ππππ ππ−3
πππ‘π, 10−7 πππ ππ−3 π  −1
1
5.0
3.6
2
8.2
9.6
3
17
41
4
30
130
Determine the order of the reaction with respect to J and calculate the rate constant.
Solution 1: Using Ratio of Two Rates
The rate law can be written as
π1: πππ₯π = π [π½]π
From this, the ratio of two rates can be written as
π
πππ₯π1
[π½]1
π2:
=(
)
[π½]2
πππ₯π2
From the data, we can pick any two data pairs, say, Experiment nos. 1 and 2.
Substituting this now in e2,
3.6 π₯ 10−7
5.0 π₯ 10−3
π3:
=
(
)
9.6 π₯ 10−7
8.2 π₯ 10−3
π
Solving now the n,
π4: π ≈ 1.98
Checking the reaction order using Experiment nos. 3 and 4,
π
41 π₯ 10−7
17 π₯ 10−3
π5:
=
(
)
130 π₯ 10−7
30 π₯ 10−3
π6: π ≈ 2.03
Thus, the order of reaction is second order with respect to J. The rate constant is
determined as the average rate constant. Its corresponding unit is obtained as follows
π7: π =
π8: π =
πππ₯π
[π½ ]2
πππ ππ−3 π  −1
(πππ ππ−3 )2
ππ3
π9: π =
πππ π
Note, the unit of concentration of J should be converted from mmol dm-3 to mol dm-3
to produce consistent units. The summary of the individual k values for each datum is
summarized below.
Exp. No.
1
2
3
4
[J]
5.00E-03
8.20E-03
1.70E-02
3.00E-02
Average k
rate
k
3.60E-07 0.0144
9.60E-07 0.0143
4.10E-06 0.0142
1.30E-05 0.0144
0.0143
Thus, the rate constant is 0.0143 dm3 mol-1 s-1. The rate law can then be written as
π10: πππ₯π (πππ ππ−3 π  −1 ) = 0.0143[π½]2
Solution 2: Solution 2: Using System of Linear Equations
The rate equation in e1 can be converted to the linear form by getting the natural
logarithm of both sides as shown below.
π11: ln(πππ₯π ) = ln(π [π½]π )
Applying now the properties of logarithm
π12: ln(πππ₯π ) = π ln([π½]] + ln(π )
Since there are two unknowns, we also need two equations. We can choose any two
pairs from the given data, say, experiment nos. 2 and 3. Substituting now the values in
e12,
Experiment no. 2: ln(9.6 π₯ 10−7 ) = ln(8.2 π₯ 10−3 ) π + ln(π )
Experiment no. 3: ln(4.1 π₯ 10−6 ) = ln(1.7 π₯ 10−2 ) π + ln(π )
Solving now the two equations simultaneously, we obtain
π ≈ 2 and ln(π ) ≈ −4.29
Again, the rate constant is evaluated from the average of the individual rate constant
as shown in Solution 1.
Example 4:
The variation in the partial pressure of azomethane with time was followed at 600 K,
with the results given below. Confirm that the decomposition
πΆπ»3 π2 πΆπ»3 (π) → πΆπ»3 πΆπ»3 (π) + π2(π)
is first-order in azomethane, and find the rate constant at 600 K.
Time, s
Partial
pressure, Pa
0
1000
2000
3000
4000
10.9
7.63
5.32
3.71
2.59
Solution:
Let [πΆπ»3 π2 πΆπ»3 ] = [π΄], for a first-order reaction, the rate equation can be written as
π1: πππ₯π = π [π΄]
where rrxn is the reaction rate, k is the rate constant and [A] is the concentration of
azomethane. Recall also that the reaction rate can be written as
π2: πππ₯π = −
π[πΆπ»3 π2 πΆπ»3 ] π [π΄]
=
ππ‘
ππ‘
Combining now e1 and e2,
π3: −
π [π΄]
= π [π΄]
ππ‘
Since the given data is in terms of pressure, we need to convert the concentration in
terms of pressure. Recall that for gases, its state parameters are related to each other
according to the modified ideal gas equation
π4: ππ = ππππ
where P is the gas pressure, V is the gas volume, Z is the compressibility constant, n is
the number of moles, R is the universal gas constant and T is the absolute temperature.
Note that the concentration of A can be written as
π5: [π΄] =
ππππ ππ π΄
π
=
π£πππ’ππ ππ π΄ π
Rearranging e4,
π
π6: π = π ( ) ππ
π
It follows then that
π7: π = π[π΄]ππ
Rearranging the terms,
π8: [π΄] =
π
πππ
Substituting e8 to e3,
π9: −
π
π (πππ)
ππ‘
= π(
π
)
πππ
Since Z, R and T are constants for this reaction, we can factor them out in the
differential term.
π10: −
1 ππ
π
=
π
πππ ππ‘ πππ
Simplifying further and applying variable separation,
π11:
ππ
= −πππ‘
π
Integrating now from t = 0 to time, t,
π
π‘
ππ
= −π ∫ ππ‘
π0 π
0
π12: ∫
π
π13: ln ( ) = −ππ‘
π0
where P0 is the pressure at t = 0 and P is the pressure at time, t. The plot of e13 is a
π
straight line if we plot ln ( ) vs. time, t. Thus, to say that the reaction is first-order
π0
with respect to azomethane, the plot (ln P/P0 vs. t ) of the experimental data should
be a straight line. From the data, note that P 0 = 10.9 Pa. The coordinates can then be
summarized as
Time, s ln(P/P0)
0
0
1000
-0.35667
2000
-0.71729
3000
-1.07773
4000
-1.4371
Plotting this now,
The data points fitted the linear trendline very well (R 2 = 1). Thus, we say that the
reaction is first-order with respect to azomethane. As for the value of the rate constant,
recall from e13 that it is the slope of the line. From the Trendline equation, it follows
that
π = π ππππ = 0.0004
Alternatively, we can calculate k from e13 for each data point, then getting the average
k.
π
ln ( )
π0
π14: π = −
π‘
Also, from e14, the unit of k is given by s-1.
Time, s ln(P/P0)
k
0
0
1000
-0.35667 0.000357
2000
-0.71729 0.000359
3000
-1.07773 0.000359
4000
-1.4371 0.000359
Average k
0.000358
Thus, the rate constant is 0.000358 s-1.
Example 5:
The rate of the second-order decomposition of acetaldehyde (ethanal, CH 3CHO) was
measured over the temperature range 700–1000 K, and the rate constants are reported
below. Determine the activation energy and the frequency factor.
T, K
k, dm3 mol-1 s-1
700
0.011
730
0.035
760
0.105
790
0.343
810
0.789
840
2.17
910
20.0
1000
145
Solution:
The effect of temperature on the reaction rate is on the rate constant. Arrhenius
equation provides a concrete relation between the two parameters. It is given by
πΈπ
π1: π = π΄π −ππ
where k is the rate constant, A is the pre-exponential factor (also called frequency
factor), Ea is the activation energy, R is the universal gas constant and T is the absolute
temperature. Activation energy is the energy needed to start a reaction.
To determine the values of the activation energy and the frequency factor, we can
convert e1 in the linear form by taking the natural logarithm of both sides.
πΈπ
π2: ln(π ) = ln (π΄π −ππ )
Applying now the properties of logarithm,
π3: ln(π ) = ln(π΄) −
πΈπ
ππ
Rewriting now in the form, π¦ = ππ₯ + π,
π4: ln(π ) = −
πΈπ 1
( ) + ln(π΄)
π π
It follows that
π=−
πΈπ
π
; π = ln(π΄)
From this,
π5: πΈπ = −ππ
π6: π΄ = π π
Thus, we can determine the activation energy and frequency factor by plotting ln(k) vs
1/T. Doing so,
T, K k, dm3 mol-1 s-1
700
0.011
730
0.035
760
0.105
790
0.343
810
0.789
840
2.17
910
20
1000
145
1/T, K-1
0.001429
0.00137
0.001316
0.001266
0.001235
0.00119
0.001099
0.001
ln(k)
-4.50986
-3.35241
-2.25379
-1.07002
-0.23699
0.774727
2.995732
4.976734
The slope of the line is 22651 Kelvin while its y-intercept is 27.707. The unit of the
slope is in Kelvin because −
πΈπ 1
π
(π) should be unitless since y (ln k) is also unitless. The
value of R in e5 depends on the desired unit of Ea. Since the data we’re given in SI
system, we can use R = 8.314 J/(mol K). Calculating now the activation energy,
π7: πΈπ = −(22651 πΎ ) (8.314
π8: πΈπ = 188,320
π½
)
πππ πΎ
π½
ππ½
= 188.32
πππ
πππ
From e1, it follows that the unit of A is also equal to that of k. Calculating now the
value of A,
π6: π΄ = π 27.707
π7: π΄ = 1.08 π₯ 1012 ππππ 3 πππ −1 π  −1
Therefore, the activation energy is 188.32 kJ/mol while the frequency factor is 1.08 x
1012 dmol3 mol-1 s-1.
Example 6:
Consider the trimolecular second-order reactions given by the reaction,