Unit two:
Chemical Kinetics
2.1.1 Introduction to Chemical Kinetics
 The rate at which a chemical reaction occurs is often very
important to us. A common task of chemists is to find ways
to change the rates of chemical reactions - we want our cars
and buildings to rust more slowly and our food to spoil more
slowly as well. But we need the reactions that produce
economically important substances, such as ammonia and
medical pharmaceuticals to occur quickly.
 In this unit we are concerned with how fast the reaction
occurs. We will be asking these key questions:
1. How do you measure the speed, or rate, of a reaction?
2. What factors influence how fast a reaction occurs?
2.1.2 Reaction Rates
 Reaction rate is a measure of how fast a reaction occurs, or
how something changes during a given time period.
 Lab
 Complete the cooling of water lab
 Consider the oxidation of glucose, C6H12O6 :
 C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
 How could we measure how fast this occurs?
 Here are some other possible things that could be measured
as they change:
 change in conductivity
 change in pH
 colour change
 change in pressure as gases are formed or used up
 A common measure of reaction rate is how the
concentration of a reaction participant changes over time
(reactants decrease, products increase.)
 We also must consider the second key aspect of determining
rate - time.
 Rate is a measure of how something changes over time. You
measure your rate of speed when driving by determining
how many kilometres you travel in an hour; you could
measure your rate of speed during a race by dividing the
distance of the course by the number of minutes it took to
run the rate.
Rate Equation
 the rate of a chemical reaction:
Rate
=
change in concentration
change in time
Chemistry Notation
 In chemistry, we typically represent concentration by using square
brackets around the chemical formula of the substance. For example
to indicate the concentration of SO2(g) in the following reaction we
would write it as [SO2].
 Also, the delta symbol, Δ is used to indicate a change. ΔT, for
example, means "the change in temperature."
Reaction rate over time
 When the reaction first starts, the reactants are plentiful so things
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move along pretty fast As the reaction continues, there will be less and less reactants
available so the reaction will slow down.
Eventually the reaction stops once all of the reactants are gone.
A graph illustrating reaction rate is often useful
graph
So, if we measured the rate at the start of the reaction, it would be
different than if we measured the rate near the end of the reaction.
Rate is not a constant - it changes during the course of the
reaction.
Average Rate
 For this reason, we will generally work with the average
rate of reaction, and measure rate over a longer period of
time.
 For example, if you were taking a long car trip, there might
be times when you are driving fast (on the highway) and also
times when you drive more slowly (through the cities). If you
calculated your average rate for the entire trip, you would
average out these fast and slow periods.
2.1.3 Calculating Reaction Rates
 Let's try an example of calculating a reaction rate. Consider the
following reaction:
 A→B
Time
(min)
[A]
(mol ·L-1)
[B]
(mol ·L-1)
0.0
1.000
0.000
3.0
0.400
0.600
6.0
0.250
0.750
 We could measure the rate of the reaction either by measuring
how the concentration of reactant A changes or how the
concentration of product B changes. Let's measure A's average rate
of change first:
 Rate=Δ [A] = (0.250 - 1.00) = – 0.125 mol · L-1· min -1
Δ time (6.00 - 0.00)
 If we calculate the average rate based on the production of product
B:
 Rate = Δ [B] = (0.750 - 0.00) = 0.125 mol · L-1· min -1
Δ time (6.00 - 0.00)
 Notice that we must compare rates measured during the same
time period.
 You will see that we got the same rate, except for the sign, when
we calculated the overall average rate for the disappearance of A as
for the formation of B. This is only true because our balanced
equation shows us that one mole of A produces one mole of B.
 Compare this rate to the rate of just the first three minutes of
the reaction:
 Rate = Δ [A] = (0.400 - 1.00) = - 0.200 mol · L-1· min -1
Δ time (3.000 - 0.000)
 We can see that the rate did slow down as the overall rate is
slower than the rate of the first three minutes of the reaction.
 We would not find the same rates if we did not have a 1:1
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relationship between reaction participants.
2 H2O2 → 2 H2O + O2
we find that only one mole of oxygen forms for every two
moles of hydrogen peroxide that decomposes.
Therefore, we could make the following relationships:
If 2 moles of H2O2 decompose then 1 mole of O2 is formed,
if the rate of decomposition of H2O2 is 4.00 mol · L-1· min 1 then the rate of formation of O is ½ × 4.00 or 2.00 mol ·
2
L-1· min -1
 1. In the following decomposition reaction,
2 N2O5 → 4 NO2 + O2
oxygen gas is produced at the average rate of 9.1 × 10-4 mol ·
L-1 · s-1. Over the same period, what is the average rate of the
following:
a)the production of nitrogen dioxide
b)the loss of nitrogen pentoxide
Solution #1:
 From the equation we see that for every 1 mole of oxygen
formed, four moles of nitrogen dioxide are produced. Thus,
the rate of production of nitrogen dioxide is four times that
of oxygen:
 A) rate NO2 production = 4 × (9.1 × 10-4 mol · L-1· s-1) =
3.6 × 10-3 mol · L-1· s-1 Nitrogen pentoxide is consumed at
twice the rate that oxygen is produced:
 B) rate loss of N2O5 = 2× (9.1 × 10-4 mol · L-1· s-1) = 1.8
× 10-3 mol · L-1· s-1
Example #2
 2. Consider the following reaction:
N2(g) + 3 H2(g) → 2 NH3(g)
If the rate of loss of hydrogen gas is 0.03 mol · L-1· s-1, what is
the rate of production of ammonia?
Solution #2:
 From the balanced equation we see that there are 2 moles
NH3 produced for every 3 moles H2 used. Thus:
 rate NH3 production =
2× (0.03 mol · L-1· s-1) = 0.02 mol · L-1· s-1
3
Example #3
 In a reaction between C4H9Cl and water, the concentration
of C4H9Cl is 0.220M at the beginning of the reaction. At
4.0s, the concentration of C4H9Cl is 0.100M. Calculate the
average reaction rate.
Solution #3
[C4H9Cl] at 0s= 0.220M
[C4H9Cl] at 4s= 0.100M
t1=0s
t2=4s
Rate= ∆ [C4H9Cl]/ ∆ t
= 0.100M-0.220M/ 4s- 0.0s
= -0.120M/ 4s
= 0.0300 mol/Ls
Assignment
 Assignment 2.1.3