# 1023-L10-070129

```Chemical Kinetics
Chapter 14
Reminders
• Assignment 2 due today (in class)
• Assignment 3 up now and will be due Mon.,
Feb. 05
• Assignment 4 (Ch. 15) will not be due before
Midterm 1, but Ch. 15 will be on the midterm
Energy diagram for a chemical reaction
SAMPLE EXERCISE Relating Energy Profiles to Activation Energies and Speeds of Reaction
Consider a series of reactions having the following energy profiles:
Assuming that all three reactions have nearly the same frequency factors, rank
the reactions from slowest to fastest.
Solution The lower the activation energy, the faster the reaction. The value of
H does not affect the rate. Hence the order is (2) < (3) < (1).
Reaction Mechanisms
The sequence of events that describes
the actual process by which reactants
become products is called the reaction
mechanism.
Reaction Mechanisms
• Remember that rate laws for reactions must be determined
experimentally; they cannot be predicted from the coefficients of
the balanced chemical equations.
• Reactions may occur all at once or through several discrete
steps.
• Each of these processes is known as an elementary reaction or
elementary process.
• Rate laws and relative speeds of these steps will dictate the
overall rate law.
• The next challenge in kinetics is to arrive at reaction
mechanisms that lead to rate laws that are consistent with those
observed experimentally.
Reaction Mechanisms
• The molecularity of a process tells how many molecules are
involved in the process.
• The rate law of any elementary reaction is based directly on its
molecularity.
• Remember that we cannot tell by merely looking at a balanced
chemical equation whether the reaction involves one or several
elementary steps.
SAMPLE EXERCISE Predicting the Rate Law for an Elementary Reaction
If the following reaction occurs in a single elementary reaction, predict the rate
law:
Solution
The reaction is bimolecular, involving one molecule of H2 with one molecule
of Br2. Thus, the rate law is first order in each reactant and second order
overall:
Rate = k[H2][Br2]
Comment: Experimental studies of this reaction show that the reaction
actually has a very different rate law:
Rate = k[H2][Br2]1/2
Because the experimental rate law differs from the one obtained by assuming a
single elementary reaction, we can conclude that the mechanism must involve
two or more elementary steps.
SAMPLE EXERCISE Determining the Rate Law for a Multistep Mechanism
The decomposition of nitrous oxide, N2O, is believed to occur by a two-step
mechanism:
(a) Write the equation for the overall reaction. (b) Write the rate law for the
overall reaction.
Solution
Solve: (a) Adding the two elementary reactions gives
Omitting the intermediate, O(g), which occurs on both sides of the equation,
gives the overall reaction:
(b) The rate law for the overall reaction is just the rate law for the slow, ratedetermining elementary reaction. Because that slow step is a unimolecular
elementary reaction, the rate law is first order:
Rate = k[N2O]
PRACTICE EXERCISE Predicting Relative Rates for a Two-step Mechanism
Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and
oxygen:
The reaction is believed to occur in two steps
The experimental rate law is rate = k[O3][NO2]. What can you say about the
relative rates of the two steps of the mechanism?
Answer: Because the rate law conforms to the molecularity of the first step,
that must be the rate-determining step. The second step must be much faster
than the first one.
Multistep Mechanisms
• In a multistep process, one of the steps will
be slower than all others.
• The overall reaction cannot occur faster than
this slowest, rate-determining step.
Slow Initial Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)
• The rate law for this reaction is found
experimentally to be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but the
rate of the reaction does not depend on its
concentration.
• This suggests the reaction occurs in two steps.
Slow Initial Step
• A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
• The NO3 intermediate is consumed in the second step.
• As CO is not involved in the slow, rate-determining step, it does
not appear in the rate law.
Fast Initial Step
2 NO (g) + Br2 (g)  2 NOBr (g)
• The rate law for this reaction is found to
be
Rate = k [NO]2 [Br2]
• Because termolecular processes are
rare, this rate law suggests a two-step
mechanism.
Fast Initial Step
• A proposed mechanism is
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
(fast)
(slow)
Step 1 includes the forward and reverse reactions.
Fast Initial Step
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
(fast)
(slow)
• The rate of the overall reaction depends
upon the rate of the slow step.
• The rate law for that step would be
Rate = k2 [NOBr2] [NO]
• But how can we find [NOBr2]?
Fast Initial Step
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
(fast)
(slow)
• NOBr2 can react two ways:
– With NO to form NOBr
– By decomposition to reform NO and Br2
• The reactants and products of the first
step are in equilibrium with each other.
• Therefore,
Ratef = Rater
Fast Initial Step
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
• Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
• Solving for [NOBr2] gives us
k1
[NO]
[Br
]
=
[NOBr
]
2
2
k−1
(fast)
(slow)
Fast Initial Step
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
(fast)
(slow)
Substituting this expression for [NOBr2]
in the rate law for the rate-determining
step gives
Rate =
k 2k 1
[NO] [Br2] [NO]
k−1
= k [NO]2 [Br2]
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
k = A • exp( -Ea/RT )
Ea
uncatalyzed
k
catalyzed
ratecatalyzed > rateuncatalyzed
Ea‘ < Ea
14.6
In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
•
Haber synthesis of ammonia
•
Ostwald process for the production of nitric acid
•
Catalytic converters
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
•
Acid catalysis
•
Base catalysis
14.6
Haber Process
N2 (g) + 3H2 (g)
Fe/Al2O3/K2O
catalyst
2NH3 (g)
14.6
Ostwald Process
4NH3 (g) + 5O2 (g)
Pt catalyst
2NO (g) + O2 (g)
2NO2 (g) + H2O (l)
4NO (g) + 6H2O (g)
2NO2 (g)
HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
Hot Pt wire
over NH3 solution
14.6
Catalytic Converters
CO + Unburned Hydrocarbons + O2
2NO + 2NO2
catalytic
converter
catalytic
converter
CO2 + H2O
2N2 + 3O2
14.6
Enzyme Catalysis
14.6
uncatalyzed
enzyme
catalyzed
[P]
rate =
t
rate = k [ES]
14.6
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