Louisiana Tech University, Chemistry 102 POGIL (Process Oriented Guided Inquiry
Learning) Exercise on Chapter 13(2). Chemical Kinetics 2
Why?
Chemical kinetics is important because it provides new models, concepts, and parameters to study the
rates of chemical reactions. Graphical and method of initial rates have been used to determine the rate law of a
chemical reaction. It is important to know different ways of expressing rate laws. The rate law in differential
gives the order of reactants, and in the integral form rate constant and concentration of initial and final reactants.
The half-life (t½) from is way to easily find how long it will take-to complete a reaction. Most of the chemical
reactions follow 1st order rate law; therefore we focus more on first order reactions. All nuclear reactions also
follow first order kinetics, i.e. decay of a radioactive material is directly proportional to the number or nuclei of
the element. Half-life (t½) from is used more frequently for radioactive decay in place of rate constant (k).
Learning Objectives
Understand rate laws
3. Determine reaction orders from a rate law, and use the integrated rate and initial rates methods to obtain orders
and rate constants (Section 13-3).
4. Calculate concentration from time, time to reach a certain concentration, half-life for a first-order reaction
(Section 13-3).
Success Criteria
 Identify rate law for a chemical reaction ( zero, first, or second order) from kinetic data using graphical
and initial rate methods.
 Calculations of reaction time, and final concentration of reactants given initial concentration and vice
versa.
Resources
Chemistry: The Molecular Science 4rd Edition, John W. Moore, Conrad L. Stanitski, Peter C. Jurs -ISBN-10:
1439049300 ISBN-13: 9781439049303 Prerequisites
New Concepts
Rate Law
The rate law is an expression that relates the rate of a chemical reaction to a constant (rate constant) and
concentration of reactants raised to a power. The power of a concentration is called the order with respect to the
particular reactant.
Every chemical reaction has a rate equation based on the rate law. The rate of a chemical reaction is directly
proportional to the concentration of reactants raised to a power, which could be an integer: 0, 1, 2, or 3.
E.g.
a A + b B -----> c C
rate [A]l[B]m
rate  k [A]l[B]m
k
= rate constant
[A]
= concentration of A
[B]
= concentration of B
l = order with respect to A ( nothing to do with a, the stoichiometric coefficient)
m = order with respect to B
First-Order Reactions
A first order reaction (order = 1) has a rate proportional to the concentration of one of the reactants.
rate = k[A] (or B instead of A),
Second-Order Reactions
A second-order reaction (order = 2) has a rate proportional to the concentration of the square of a single reactant
or the product of the concentration of two reactants:
rate = k[A]2 (or rate = k[A] [B])
Mixed-Order or Higher-Order Reactions
Mixed-order reactions have a fractional order for their rate:
e.g., rate = k[A]1/3
Orders of reactants have to be experimentally determined. There are two way to get it.
Graphical method:
Reaction: a A  b B
Order Differential Integrated Rate Law Graph (y) vs. (x) Slope
Half-life form t½
Rate Law
y =mx + b
0
rate = k
1
rate = k[A]
2
2
[A]t vs. t
-k
ln[A]t = -kt + ln[A]0
ln[A]t vs. t
-k
1/[A]t = kt + 1/[A]0
1/[A]t vs. t
k
[A]t = -kt + [A]0
rate=k[A]
t½ = [A]0/2k
t½ = 0.693/k
t½ = [A]0/2k
For simple reactions we can plot kinetic data. [A] vs. t to get a liner graph
For simple reactions we can plot kinetic data E. g.
2 N2O5 ----> 4 NO2 + O2
0.02
0
-2 0
0.015
0.01
[A]
0.005
50
100
-4
0
50
100
time
150
200
200
ln[A]
-6
0
150
-8
-10
time
[A] vs. t
ln [A] vs. t
Reaction rate law is: rate = k [N2O5] ; which is first order.
Method of Initial Rates:
Rate law of more complex reactions involving more than one reactant requires changing concentration of one
reactant while holding the other one to see the effect of order on the rate.
E.g. For the chemical reaction: A + B ----> C
Using the initial rates data given below deduce:
[A],mol/L
[B],mol/L
rate,mol/LS
-4
-5
4.6 x 10
3.1 x 10
2.08 x 10-3
-4
-5
4.6 x 10
6.2 x 10
4.16 x 10-3
-4
-5
9.2 x 10
6.2 x 10
1.664 x 10-2
The rate law : rate = k [A]x [B]y
a) the order of each reactant
Oder of A:
4 = (9.2 x 10-4/(4.6 x 10-4)x = 2x
4 = 2x
x = 2
Therefore, the order with respect to A is two or second order.
Order of B:
The rate law : rate = k [A]x [B]y
2 = (6.2 x 10-5/3.1 x 10-5) y = 2 y
2 = 2y
y = 1
Therefore, the order with respect to B is one or first order.
Rate law of the reaction: rate = k [A]2 [B]
Total order of the reaction 3 or third order.
b) the rate constant
The rate law for the reaction : A + B ----> C
rate = k [A]2 [B]1
plugging the values from one of the experiments:
rate
=
k
[A]2
[B]1
_________________________________________________________________________
2.08 x 10-3 =
2.08 x 10-3 =
k
k
(4.6 x 10-4)x
(4.6 x 10-4)2
(3.1 x 10-5)y
(3.1 x 10-5)1
_________________________________________________________________________
2.08 x 10-3
k =
----------------------------- = 3.17 x 108
(4.6 x 10-4)2 x (3.1 x 10-5)1
k = 3.17 x 108 L2/mol2 s
The rate constant for the reaction is 3.17 x 108 L2/mol2 s.
Calculation using half-life t1/2 form:
Half-life is the time required for a reactant to reach half its original concentration
( [A]t = ½ [A]0
First-order reaction is t½ = 0.693/k. The half-life of a first order reaction does not depend on concentration.
Integrated rate law: ln[A]t = -kt + ln[A]0; ln(½ [A]0) = -kt + ln[A]0; ln ([A]0/2)= ln [A]0 – ln 2 = -kt + ln[A]0; ln
2 = kt; 0.693 = k t½; t½ = 0.693/k
Zero-order rate law is given as t½ = [A]0/2k.
Second-order reaction is t½ = 1/(k[A]0).
GHW#2
Printed Name:_____________________
Group Name:__________
Chapter 13 (2) Key Questions (relatively simple to answer using the Focus Information)
1. The reaction A ---> B + C is known to follow the rate law: rate = k [A]1
What are the differential, integral and half-life (t½) form of this rate law?
2. Using graphical method, show that 2 N2O5 ---> 4 NO2 + O2, is a first order reaction.
[N2O5] /
Time / min
moldm-3
0
0.01756
20
0.00933
40
0.00531
60
0.00295
80
0.00167
100
0.00094
160
0.00014
3. For the reaction: A ---> D, Find the order of [A] for each case.
It was found in separate experiments that
a) The rate doubled when [A] doubled
b) The rate tripled when [A] tripled
c) The rate quadrupled when [A] doubled
d) The rate increased 8 times when [A] doubled
4. For the chemical reaction: A + B ----> C
Using the following initial data to deduce:
a) Order of each reactant
b) Rate constant
[A],mol/L [B],mol/L rate,mol/Ls
_____________________________
2.0
3.0
0.10
6.0
3.0
0.90
6.0
6.0
0.90
Exercises involving chemical and nuclear reactions
5. The rate constant for the first-order conversion of A to B is 2.22 hr-1. How much time will be required for the
concentration of A to reach 75% of its original value?
6. The half-life of a radioactive (follows first order rate law) isotope is 10 days. How many days would be
required for the isotope to degrade to one eighth of its original radioactivity?
7. The rate constant for the first order decomposition of SO2Cl2 (SO2Cl2-> SO2 +Cl2) at very high temperature is
1.37 × 10-3 min-1. If the initial concentration is 0.500 M, predict the concentration after five hours (300 min).