Chapter 15 Chemical Kinetics Notes
Chemical Kinetics
The study of how fast chemical reactions occur.
How fast species are appearing and disappearing.
It has nothing to do with the extent to which a reaction will occur – thermodynamics.
There is a difference between the probability that molecules will react when they meet each other and once
they do react how stable is the product.
 First, probability of reaction has to do with overcoming an energy barrier or activation energy.
 Second, has to do with the differences between reactants and products.
Five factors affecting rate of reactions:
In order to truly understand them you must - Understand chemical reactions at the molecular level.
1. Surface area of solid or liquid reactants or catalyst
 The greater degree the reactants are exposed to each other, the faster the reaction will proceed.
2. Temperature of reactions
 At higher temperatures, bonds within the reacting molecules are vibrating more and are closer to the
breaking point than at lower temperatures. Reactants are less stable
 At higher temperatures, the average velocity of molecules goes up and therefore when potentially
reactive molecules collide there is a greater probability they will react. Getting over the activation
barrier has a lot to do with how fast molecules slam into one another.
3. Pressure of gaseous reactants or catalyst
4. Presence of a catalyst (+ or -) also called an inhibitor
 Catalysts affect the rate of the reaction by placing one or more of the reactants in an environment in
which they are more susceptible to chemical reaction. +Catalysts lower the activation energy.
5. Concentration of reactants
 This is usually the most important factor, a lot of kinetics is about determining the Rate Law for a
specific reaction. The Rate Law is how the reaction rate depends upon concentration of reactants.
Reaction Rates
Use stoichiometric ratios that relate rates just like you would relate reactants used or products formed.
Rate = [product] = - [reactant]
t
t
For Example 1):
2A + 3B  C + 4D
- [A] = - [B] =  [C] =  [D] : notice 1 over the stochiometric value
2t
3t
t
4t
For reaction rate = r
: related to A
rA = 2 rB = -2rC = -1rD
3
2
For Example 2):
Rate = [product] = - [reactant]
t
t
AB
Average Rate = moles of B = - moles of A
t
t
Change in mole of B = (moles B final) - (moles B initial)
Let’s use some numbers:
If we begin with 1.00mole A
At t=0 there is 1.00mole A and no B present
At t=10min, if there is 0.74mole A then there is 0.26mole B
At t=20min, there is 0.54mole A and 0.46mole B
So t=30min, how much A and B? Do you notice there is a relationship?
Average RateB = 0.26molB - 0molB = 0.26molB
10min - 0min
min
Average RateA = 0.74molA – 1.00molA = -0.26molA note the sign of negative appearance, disappearance
10min - 0min
min
rB = -rA








A plot of moles verses time shows how reactant A disappears with a mathematical relationship to the
appearance of B.
Since volume is constant and rate in terms of moles are directly proportional molarity is also
proportional and more useful. Making the units mol/Ls or M/s.
Plots then are in terms of changes in concentration over changes in time [A]/t.
The rate at any time is called the instantaneous rate.
It is the slope of the straight line tangent to the curve we plot for the reaction.
We call this the Instantaneous Rate. That’s why we graph!!!! To find rates at any time t.
We use math for a few reactions to define the change with time for the rate law or proportionality
constant we call k or the rate constant. k is a description of how the reaction proceeds.
Once we have determined the rate law and the rate constant FOR THAT reaction we can use them to
calculate reaction rate under any set of initial concentrations
Consider the reaction:
NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l)
We measure initial reaction rates. The initial rate is the instantaneous rate at time t=0
We find at various initial concentrations of each reactant.
As [NH4+] doubles, with [NO2-] constant the rate is found to double.
As [NO2-] doubles, with [NH4+] constant the rate is found to double.
We conclude that the rate of NH4+ is proportional to NO2-.
The overall concentration dependence of reaction rate is given in a rate law or rate expression:
R = k [NO2-][NH4+]
The proportionality constant k is called the rate constant.
Rate Laws:
The Rate Law or rate expression r = k[A]n describes the reaction under any concentrations at that
temperature and pressure.
For example: aA + bB  cC + dD
 Zero Order: r = k
 First Order in A: r = k [A]
 First Order in B: r = k [B]
 Second Order in A: r = k [A]2
 Second Order in B: r = k [B]2
 First Order in A, first Order in B, for a second order overall: r = k [A][B]
For: aAB
Zero Order
First Order
Second Order
Rate Law
r=k[A]o
r=k[A]1
r=k[A]2
d[A] = k[A]m
y=mx +b
[A]=-kt +[A]o
ln[A]=-kt+ln[A]o
1 = kt + 1
[A]
[A]o
Plot for a linear
graph
[A] vs t
ln[A] vs t
1 vs t
[A]
slope of linear plot
-k
-k
k
half-life
t1/2when
t1/2= [A]o
2k
t1/2= 0.693
k
t1/2=
1
k[A]o
Using Initial Rates to determine Rate Laws
To determine the rate law, we observe the effect of changing initial concentrations in several experiments at
the same chemical reaction at the same conditions of temperature and pressure.
 If a reaction is zero order in a reactant, then changing the initial concentration of that reactant
will have no effect on the rate (as long as some reactant is present.)
 If a reaction is first order in a reactant, then doubling the initial concentration will cause the
rate to double.
 If a reaction is second order in a reactant, then doubling the initial concentration will cause a
22 increase in the rate of the reaction. Similarly, tripling the initial concentration will result in
a 32 increase in rate.
 A reaction is nth order if doubling the concentration causes a 2n increase in rate.
Note that the rate, NOT the rate constant, depends on concentration.
Let’s use some numbers once again. For a reaction:
A + B  Products
Exp1
Exp2
Exp3
Exp4
[A]
0.1
0.2
0.3
0.1
[B]
0.1
0.1
0.1
0.2
Rate (mol/Ls) or (M/s)
0.00046
0.00091
0.0013
0.0018
r = k[A]m[B]n
Select sets of data that allow you to account for changes in rate because of changes in concentration.
r4 = [A]2
r1 [A]1
m
[B]2
[B]1
n
If you examine experiments 1 & 4 where [A] is the same, then any change in the rate must be caused by the
change in [B].
So, since the [B] from experiment 1 to 4 doubled the change experienced in rate was not simply a
doubling it was a doubling squared or what we did to the [ ] change raised to 2
r4 = 0.0018 = 3.9 ≈ 4 = 0.2 n = [2]n
r1 0.00046
0.1
so n= 2, meaning that wrt B it is a second order reaction
If you examine experiments 1 & 2 where [B] is the same, then any change in the rate must be caused by the
change in [A].
So, since the [A] from experiment 1 to 2 doubled the change experienced in rate was doubling then
the rate is raised to 1
r2 = 0.00091 = 1.9 ≈ 2 = 0.2 m = [2]m
r1 0.00046
0.1
so m= 1, meaning that wrt A it is a first order reaction
We can use another set of data to determine k, let’s use experiment 3,
r = k[A]m[B]n
0.0013 = k (0.3)1 (0.1)2
So for this reaction is k = 0.43, now we can calculate the rate for any initial concentrations without
performing the experiment.
Collision Theory, Activation Energy and Activated Complex
This is just about trying to explain how the actual reaction occurs. What happens when tow reactant
molecules collide? Sometimes they react, sometimes they don’t. It is explained in terms of activation energy
that is required to trigger the reaction and an intermediate molecule, called an activated complex that is the
unstable intermediate species occurring when the molecules collide.
Activated Complex
E
n
e
r
g
y
Ea
Inhibitor raises Ea
Reactants
Activation Energy
Hrxn
Catalyst lowers Ea
Products
E
n
e
r
g
y
Products
Ea
Hrxn
Reactants
Reaction Coordinate
Reaction Coordinate
Exothermic Reaction (H< 0)
Endothermic Reaction (H> 0)
Activation Energy
With the idea of activation energy, temperature really comes into play. The higher the temperature, the more
energy that’s in the system and the closer the molecules are to getting over the activation energy hump.
The important equation here is the Arrhenius Equation that relates the rate constant k to temperature.
Arrhenius Equation
k = Ae (-Ea/RT)
k = rate constant
A = total collisions ≈ frequency factor
e = Euler’s number (opposite of ln)
Ea = activation energy
R = Universal gas constant
T = temperature in Kelvin
Linearized Arrhenius Equation
Slope = -Ea
R
lnk
lnk = lnA – (Ea/RT)
1/T
Using the linear equation we can find a number of things
Plot lnk vs 1/T the slope will give you the activation energy
If we know Ea and k at one temperature then you can find k at a different temperature
ln k2 - ln k1 = ln k2 = - Ea
k1
R
1 - 1
T2 T1
Reaction Mechanisms
Reactions follow a path of multiple steps, you need to be able to derive a rate expression form a reaction
mechanism.
3 Basic Rules
1. The Slow-Step is always rate determining
2. For any step in the reaction mechanism you can just use its stoichiometric coefficient as its rate
order
3. The final rate expression must include only the species that are in the overall reaction, get rid of
intermediate steps
For Example: This is the information you will be given.
2NO(g) + Cl2(g)  2 NOCl(g)
The rate law is found to be second order in NO(g) and first order in Cl2(g).
Derive the expression from the proposed reaction mechanism:
1.
NO(g) + Cl2(g)  NOCl2(g)
Fast
2.
NOCl2(g) + NO(g)  2NOCl(g)
Slow
Slow step is rate determining: r = k [NOCl2][NO]
NOCl2 is a reaction intermediate so use a combination of the reactions
r = k[NO]2[Cl2]