# NA Chemistry Chemical Kinetics Questions

Workshop #1 Chemical KineticsIntroduction: The study of the rate of chemical reactions is called kinetics. The
rate of chemical reactions is an important field of study and is relevant to the
lifetime of pollutants in the atmosphere and also the method of drug delivery to
name just two fields where kinetic studies are common.
For the simple chemical reaction:
C2H4 + HCl à C2H5Cl
+
The rate of the reaction is determined by observing the decrease in the
concentration of the reactants and/or the increase in the concentration of the
products. If the concentration of the reactants and products were measured as a
function of time the data may look like:
The rate of a reaction is defined as the change in concentration versus time. The
unit is Ms-1. For this reaction it can be expressed as:
36
−∆[𝐶% 𝐻’ ] −∆[𝐻𝐶𝑙] ∆[𝐶% 𝐻, 𝐶𝑙]
=
=
∆𝑡
∆𝑡
∆𝑡
Note that the rate is the negative of change in concentration of the reactants and
the positive of the change in concentration of the products.
Looking at the graph, it is apparent that the rate is not constant, but changes with
time. This is because reaction rates are typically functions of the concentration
of the reactants. There are several methods to determine a numeric value of
the rate. The first method is the method of initial rates, from the data table
[C2H5Cl] M
time (s) [C2H4] M
[HCl] M
0
0.1
0.2
0.3
0.4
0.5
0.6
0.1
0.098019867
0.096078944
0.094176453
0.092311635
0.090483742
0.088692044
𝑟𝑎𝑡𝑒 =
0.2
0.198019867
0.196078944
0.194176453
0.192311635
0.190483742
0.188692044
0
0.001980133
0.003921056
0.005823547
0.007688365
0.009516258
0.011307956
−∆ 𝐶% 𝐻’
0.09802 − 0.1000𝑀
=−
= 0.0198 𝑀𝑠 9:
∆𝑡
0.10𝑠 − 0.00𝑠
Two points are chosen at the start of the reaction, and the change in
concentration is divided by the change in time.
The initial rate may also be determined graphically by measuring the slope of a
tangent to the first few data points.
For reactions with different stoichiometry, the change in the concentration with
rate depends on the stoichiometry. For the reaction
Concentration (M)
3H2 + N2 à2NH3, a graph of concentration versus rate would appear as:
0.25
0.2
0.15
[H2] M
0.1
[N2] M
0.05
[NH3] M
0
0
1
2
3
4
5
6
Time (s)
Note that the concentration of H2 decreases three times faster than the
concentration of N2. Therefore the rate of this reaction may be expressed as:
37
𝑟𝑎𝑡𝑒 = −
∆[𝐻% ]
∆[𝑁% ] ∆[𝑁𝐻= ]
=−
=
3∆𝑡
∆𝑡
2∆𝑡
Problem 1: For each of the following reactions, write the expression for the rate
of the reaction in terms of the change in concentration of each reactant and
product.
Example
CH4(g) + 2O2(g) à CO2(g) + 2H2O(g)
𝑟𝑎𝑡𝑒 = −
∆[𝐶𝐻’ ]
∆ [𝑂% ] ∆[𝐶𝑂% ] ∆[𝐻% 𝑂]
=−
=
=
∆𝑡
2∆𝑡
∆𝑡
2∆𝑡
C3H8(g) + 5O2(g) à3CO2(g) + 4H2O(g)
Rate =
Fe(OH)3(s) + 3HBr(aq) à FeBr3(aq) + 3H2O(l)
Rate =
FeCl3(aq) + 3NaOH(aq) à Fe(OH)3(aq) + 3 NaCl(aq)
Rate =
2Fe(s) + 6HCl(aq) à 2FeCl3(aq) + 3H2(g)
Rate =
38
Once the rate of a reaction is determined the dependence of this rate on the
concentration of the various reactants may be determined. This must be
determined experimentally by measuring the rate at different initial
concentrations of the reactants. Knowing this about a reaction gives insight into
the fundamental steps of the reaction.
Example 1:
The reaction: C4H9Br + OH-àC4H9OH + Br The rate of the reaction may be expressed in terms of the concentration of the
reactants:
Rate = k[C4H9Br]a[OH-]b,
where a and b are exponents to be determined experimentally, and k is the rate
constant, which is specific for each reaction and must also be determined
experimentally. To determine these, the initial reaction rate must be determined
under a variety of initial conditions. The data are then usually presented in
tablular form:
Experiment
1
2
3
4
5
Initial
concentration of
C4H9Br (M)
0.10
0.20
0.30
0.10
0.10
Initial
concentration of
OH- (M)
0.10
0.10
0.10
0.20
0.30
Initial rate of
formation of
C4H9OH (M s-1)
0.0010
0.0020
0.0030
0.0010
0.0010
To find the exponent for C4H9Br, if we compare experiments 1 and 2, the
concentration of OH- does not change so any difference in the rate must be
caused by the different initial concentrations of C4H9Br. As the concentration
goes from 0.10 M to 0.20 M (it doubles), the rate goes from 0.0010Ms-1 to 0.0020
Ms-1 ( it also doubles). Therefore the rate depends linearly on [C4H9Br] and the
exponent a = 1. Mathematically the following operation may be done:
𝑅𝑎𝑡𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 2 0.0020 𝑀𝑠 9:
k[𝐶’ 𝐻F 𝐵𝑟]H [𝑂𝐻9 ]I
=
=
2
=
𝑅𝑎𝑡𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 1 0.0010 𝑀𝑠 9:
k[𝐶’ 𝐻F 𝐵𝑟]H [𝑂𝐻9 ]I
k[𝐶’ 𝐻F 𝐵𝑟]H [𝑂𝐻9 ]I (0.20 𝑀)H
=
=
= 2H
k[𝐶’ 𝐻F 𝐵𝑟]H [𝑂𝐻9 ]I (0.10 𝑀)H
The numeric value for the rate of experiment t 2 is divided by the rate of
experiment 1. The ratio is 2. These rates are equal to the expression,
39
k[C4H9Br]a[OH-]b with the values used in the experiments. Because k and the
concentration of OH- do not change, they are cancelled in the ratio, and the
value for the concentration of C4H9Br raised to a is left. This equation reduces to
2 = 2a. Therefore a = 1.
Comparing the rates for experiments 4 and 5, it is clear that the concentration of
OH- does not affect the rate, and its exponent, b must equal zero.
The rate law for this reaction is:
Rate = k[C4H9Br]1[OH-]0= k[C4H9Br]1
To determine the numeric value of k, the rate coefficient, substitute the values of
𝑟𝑎𝑡𝑒 = k[𝐶’ 𝐻F 𝐵𝑟]
𝑟𝑎𝑡𝑒
0.0010𝑀𝑠 9:
𝑘=
=
= 0.010 𝑠 9:
0.100 𝑀
k[𝐶’ 𝐻F 𝐵𝑟]
Example2:
The reaction: 2NO(g) + Br2(g) à2NOBr (g)
The following rate data were obtained for this reaction:
Experiment
1
2
3
4
5
Initial
concentration of
NO (M)
0.10
0.10
0.10
0.20
0.30
Initial
concentration of
Br2 (M)
0.10
0.20
0.30
0.10
0.10
Initial rate (M s-1)
12
24
36
48
108
The rate law has the form: rate = k[NO]a[Br2]b.
To find a:
𝑅𝑎𝑡𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 4 48 𝑀𝑠 9:
k[NO]H [𝐵𝑟% ]I k[NO]H [𝐵𝑟% ]I (0.20 𝑀)H
=
=
4
=
=
=
𝑅𝑎𝑡𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 1 12 𝑀𝑠 9:
k[NO]H [𝐵𝑟% ]I k[NO]H [𝐵𝑟% ]I (0.10 𝑀)H
= 2H
4 = 2a, therefore a =2.
To find b:
40
𝑅𝑎𝑡𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 2 24 𝑀𝑠 9:
k[NO]H [𝐵𝑟% ]I
k[NO]H [𝐵𝑟% ]I 0.20𝑀
=
=
2
=
=
=
𝑅𝑎𝑡𝑒 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡 1 12 𝑀𝑠 9:
k[NO]H [𝐵𝑟% ]I
k[NO]H [𝐵𝑟% ]I 0.10𝑀
= 2I
2 = 2b, therefore b =1.
The rate law is rate = k[NO]2[Br2].
To find k:
𝑟𝑎𝑡𝑒 = k[NO]% [𝐵𝑟% ]
𝑘=
𝑟𝑎𝑡𝑒
12 𝑀𝑠 9:
=
1.2 𝑥 10′ 𝑀9% 𝑠 9:
(0.10 𝑀)% (0.10𝑀)
[NO]% [𝐵𝑟% ]
The exponent is also called the order of the reaction. The reaction, C4H9Br +
OH-àC4H9OH + Br -, was found to 1st order with respect to C4H9Br, and zero
order with respect to OH –, and first order overall.
The reaction, 2NO(g) + Br2(g) à2NOBr (g), is 2nd order with respect to NO, 1st
order with respect to Br2 and 3rd order overall. The units of k are also dependent
on the order of the reaction.
Practice problems:
1. For the reaction, NO2(g) + O3(g) à NO3(g) + O2(g), the following data were
obtained:
Experiment
Initial
concentration of
NO2 (M)
Initial
concentration of
O3 (M)
Initial rate (M s-1)
1
2
3
5.0 x 10-5
5.0 x 10-5
2.5 x 10-5
1.0 x 10-5
2.0 x 10-5
2.0 x 10-5
0.022
0.044
0.022
Determine the rate law and k, the rate coefficient for this reaction.
41
Problem 2: For the reaction, 2NO(g) + Cl2 (g) à 2 NOCl, the following data
were obtained:
Experiment
1
2
3
4
Initial
concentration of
NO (M)
0.10
0.10
0.20
0.30
Initial
concentration of
Cl2 (M)
0.10
0.20
0.10
0.10
Initial rate (M s-1)
2.53 x 10-6
5.06 x 10-6
10.1 x 10-6
22.8 x 10-6
Determine the rate law and k, the rate coefficient.
Problem 3: For the reaction, 2NOCl à 2NO + Cl2, the following data were
obtained at 27 C.
Experiment
1
2
3
Initial
concentration of
NOCl (M)
0.30
0.60
0.90
Initial rate (M s-1)
3.60 x 10-9
1.44 x 10-8
3.24 x 10-8
Determine the rate law and k the rate coefficient.
42
Another method to determine the order of a reaction is to use the concentration
versus time data. Mathematically, for a simple reaction A à B
𝑟𝑎𝑡𝑒 =
−∆[𝐴]
= 𝑘[𝐴]H
∆𝑡
If we look at the instantaneous rate, then the rate is the derivative of the
concentration with respect to time.
𝑟𝑎𝑡𝑒 =
−∆[𝐴]
𝑑[𝐴]
=−
∆𝑡
𝑑𝑡
𝑟𝑎𝑡𝑒 =
−𝑑[𝐴]
= 𝑘[𝐴]H
𝑑𝑡
and the rate law becomes,
This rate law may be integrated for the different orders of the reaction.
For zero order reactions,
−𝑑 𝐴
= 𝑘[𝐴]H = 𝑘
𝑑𝑡
−𝑑 𝐴 = 𝑘𝑑𝑡
[T]U
R
−𝑑 𝐴 =
[T]V
[𝐴]R − 𝐴
𝑘𝑑𝑡
S
S
= −𝑘𝑡
A plot of concentration versus time gives a straight line and may look like this
43
The slope of the line = k.
Zero order reaction
0.0045
0.004
y = -7E-06x + 0.004
2
R = 0.9996
Concentration Bromine
0.0035
0.003
0.0025
0.002
0.0015
0.001
0.0005
0
0
50
100
150
200
250
300
350
400
450
500
Time
For a first order reaction,
−𝑑 𝐴
= 𝑘[𝐴]
𝑑𝑡
−𝑑 𝐴
= 𝑘𝑑𝑡
[𝐴]
[T]U
R
−𝑑 𝐴
= 𝑘𝑑𝑡
[𝐴]
[T]V
S
𝑙𝑛[𝐴]R − 𝑙𝑛 𝐴 S = −𝑘𝑡
For a first order reaction, a plot of ln [A] is linear and the slope is k. The plot
should resemble this:
44
First Order Reaction
-6.8
-7
y = -0.008x – 6.905
2
R = 0.9995
Natural log Concentration
-7.2
-7.4
-7.6
-7.8
-8
-8.2
0
20
40
60
80
100
120
140
Time
For a second order reaction,
−𝑑 𝐴
= 𝑘[𝐴]% ]
𝑑𝑡
−𝑑 𝐴
= 𝑘𝑑𝑡
[𝐴]%
[T]U
R
−𝑑 𝐴
=
𝑘𝑑𝑡
%
[T]V [𝐴] ]
S
1
1

= −𝑘𝑡
[𝐴]S [𝐴]R
This should resemble this:
Second Order Reaction
2500
y = 5.1391x + 498.79
2
R = 0.9999
1/Concentration
2000
1500
1000
500
0
0
50
100
150
200
250
Time
45
300
350
160
Example: For the reaction 2H2O2à 2 H2O + O2 the following data were obtained
Time (s)
0
120
300
600
1200
1800
2400
3000
3600
[H2O2] M
1.0
0.91
0.78
0.59
0.37
0.22
0.13
0.0830
0.050
First plot concentration versus time:
Concentration H2O2
[H2O2] M
1.2
1
0.8
0.6
0.4
0.2
0
0
1000
2000
3000
4000
Time (s)
This plot is not linear so the reaction is NOT zero order.
Now plot ln [H2O2] versus time.
46
Natural Log Concentration H2O2
[H2O2] M
0
y = -0.000834x – 0.005522
R² = 0.999747
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
0
500
1000
1500
2000
2500
3000
3500
4000
Time (s)
This reaction is first order with respect to H2O2, and the rate constant,
k=0.000834 s-1.
Problems: Using either graph paper or a spread sheet program, determine the
rate law and k.
1. For the reaction 2N2O5 à4NO2 + O2, Determine the order of the reaction
and k.
Time (s)
0
500
1000
1500
2000
2500
3000
[N2O5] M
5.0
3.52
2.48
1.75
1.23
0.87
0.61
47
2. For the reaction Aà B determine the order of the reaction and k.
time
concentration
0
20
40
60
80
100
120
140
160
180
200
0.2
0.153
0.124
0.104
0.09
0.079
0.07
0.063
0.058
0.053
0.049
Temperature dependence of reactions:
Lastly, increasing the temperature typically increases the reaction rate by
increasing k the rate coefficient. The equation is:𝑘 = 𝐴𝑒 9WX /Z[ . A is the
Arrhenius factor, and Ea is the activation energy, R is the ideal gas constant and
T is the temperature. The activation energy is the energy required to initiate the
reaction, and the Arrhenius factor gives some information on the type of
collisions required for the reaction to occur. To determine these values, k, the
rate constant is determined for several temperatures. The equation for k may be
written in log form:
ln k =ln A -Ea/RT
Typically, the natural log of k is plotted versus 1/T, the temperature units must be
Kelvin, the slope is -Ea/R and the y intercept is ln A.
Example, k, the rate constant, has been measured at several temperatures for
the reaction: H2 (g) + I2 (g) à 2HI (g), using this data determine Ea and A. The
initial data gives the temperature in C. A copy of the Excel spreadsheet data is
shown below.
Temperature (C)
283
302
355
393
430
k M-1s-1
1.20E-04
3.50E-04
6.80E-03
3.80E-02
1.70E-01
Temperature (K)
556
575
628
666
703
48
1/T K-1
0.001799
0.001739
0.001592
0.001502
0.001422
Natural log k
-9.028018815
-7.957577403
-4.990832667
-3.270169119
-1.771956842
The temperature is first convert to K, and then 1/T is calculated for each of these.
The rate constant is given in scientific notation, but for the spreadsheet, 1.20 x
10-4 is written, 1.20E-4. A plot is made of ln k versus 1/T, and a trendline is
0
-1
y = -19432x + 25.897
R² = 0.99979
-2
Natural log k
-3
-4
-5
-6
-7
-8
-9
-10
0.0013
0.0014
1/Temperature
0.0015
0.0016 K-1
0.0017
0.0018
0.0019
From the trendline, the slope is -19432 = Ea/R. The slope has units of Kelvin.
Therefore Ea= -19432K * 8.314 J/mol K =161558 J/mol = 162 kJ/mol. A = exp
(25.9) =1.8 x 1011M-1s-1. The Arrenhius factor has the same units as k, the rate
constant.
Problem: The rate constant has been measured as a function of Temperature for
the reaction AàB. Using either graph paper or a spreadsheet determine Ea and
A for this reaction.
Temperature C
205
210
215
220
225
230
k s-1
0.0191
0.0274
0.039
0.0551
0.0773
0.108
49

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