Reaction Energy
and Reaction
Kinetics
Reaction Rate
Objectives
1. Define the rate of reaction and describe how it can
be determined.
2. List and explain factors that affect the rate of
reaction.
3. Define a catalyst and describe how it can affect the
rate of reaction.
Reaction Rate
Reaction rate – change in
concentration of
reactants per unit time
as a reaction occurs.
Chemical kinetics – study
of reaction rates and
reaction mechanisms.
Rate-Influencing Factors
Nature of the Reactants
H2(g) + Cl2(g)  2 HCl(g)
3 H2(g) + N2(g)  2 NH3(g)
(fast)
(slow)
4 Na(s) + O2(g)  2 Na2O(s)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
(fast)
(slow)
Rate-Influencing Factors
Surface Area
• Heterogeneous reactions –
reactions with reactants in
two different phases
• Heterogeneous reactions
depend on contact between
the two phases.
• Increases in surface area
result in the increase in the
rate of heterogeneous
reactions.
Rate-Influencing Factors
Temperature
An increase in temperature causes:
• An increase in collision energy
• An increase in collision frequency
Both result in increased reaction rates
• In general, a 10oC increase in temperature causes the
reaction rate to double.
Rate-Influencing factors
Concentration
• Increasing the concentration
of the reactants may cause
an increase in the rate of
reaction for homogeneous
reactions.
Rate-Influencing Factors
Presence of a Catalyst
• Catalysts lower the
activation energy for a
reaction and, thus, increase
the reaction rate.
• Homogeneous vs.
Heterogeneous catalysts –
same or different phase as
reactants.
Rate Laws for Reactions
Objectives:
1. Explain the rate law for a chemical reaction.
2. Determine the rate law for a particular reaction given
appropriate kinetics data.
3. Discuss the relationship between the rate law and the
reaction mechanism.
Rate Law
Rate Law – an equation that relates reaction rate and
concentration of reactants
R = k[A]n[B]m…
R = reaction rate
k = rate constant that depends on temperature
n,m = must be determined experimentally
Rate Laws for Reactions
Example:
2 H2(g) + 2 NO(g)  N2(g) +2 H2O(g)
• Doubling the amount of H2 , while keeping the amount of
NO constant, causes the reaction rate to double. So…
R [H2]
• When the amount of NO is doubled, while keeping the
amount of H2 constant, causes the reaction rate to
increase fourfold. So…
R [NO]2
Overall rate law: R = k[H2][NO]2
Rate Laws for Reactions
For reactions that occur in a single step:
A + B  2C
R = k[A][B]
For the reverse reaction…. 2C  A + B
R = k[C]2
Rate Laws for Reactions
For multi-step reactions, the rate law depends on the
rate limiting step.
Example:
NO2(g) + CO(g)  NO(g) + CO2(g)
Step 1: NO2 + NO2  NO3 + NO
Step 2: NO3 + CO  NO2 + CO2
Rate Law:
R = k[NO2]2
(slow)
(fast)
Sample Problems
1. The rate law of a reaction is found to be R=k[X]3. By
what factor does the rate increase if the concentration of
X is tripled?
The rate will increase by a factor of 27.
2. The rate of reaction involving two reactants, X and Z, is
found to double when the concentration of X is doubled
and to quadruple when the concentration of Z is
doubled. Write the rate law for this reaction.
R = k[X][Z]2
3. The rate law for a single-step reaction that forms one
product, C, is R = k[A][B]2. Write the balanced reaction
of A and B to form C.
A + 2B  C
Sample Problem
• A particular reaction is found to have the following
rate law:
R = k[A][B]2
How is the rate affected by each of the following
changes?
a. The initial concentration of A is cut in half.
b. The initial concentration of B is tripled.
c. The concentration of A is doubled, but the
concentration of B is cut in half.
d. A catalyst is added.
Sample Problem
A chemical reaction is expressed by the balanced chemical
equation, A + 2B  C. Using the data below, answer the
following:
a. Determine the rate law for the reaction
b. Calculate the value of the specific rate constant.
c. If the initial concentrations of both A and B are 0.30 M, at
what initial rate is C formed?
Experiment #
Initial [A]
Initial [B]
Initial rate of
formation of C
1
0.20 M
0.20 M
2.0x10-4M/min
2
0.20 M
0.40 M
8.0x10-4M/min
3
O.40 M
0.40 M
1.6x10-3M/min