Chapter 13 Rates of Reaction

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Chapter 13 Rates of Reaction

• Dr. Peter Warburton
• peterw_at_mun.ca
• http//www.chem.mun.ca/zcourses/1011.php

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Reaction Rates

• Reaction rate is concentration change divided by
  time change
• Reaction rate DX / Dt

Changes! Final minus initial
DX Xfinal Xinitial Dt tfinal
tinitial
We most often use mol?L-1 as units of
concentration This means that rate often has
units mol?L-1?s-1
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Reaction Rate

• The reaction rate is defined either as the
  increase in the concentration of a product over
  time, or the decrease in the concentration of a
  reactant over time.

A B ? C D rate -DA / Dt -DB / Dt
DC / Dt DD / Dt Rate is always
positive, so we must put negative signs in front
of reactant concentration changes!
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2 N2O5 (g) ? 4 NO2 (g) O2 (g)
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2 N2O5 (g) ? 4 NO2 (g) O2 (g)
Between 300 and 400 seconds
Rate of decomposition of N2O5 - DN2O5 / Dt
- (0.0101 mol?L-1 0.0120 mol?L-1) / (400
s 300 s) 1.9 x 10-5 mol?(Ls)-1
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2 N2O5 (g) ? 4 NO2 (g) O2 (g)
Between 300 and 400 seconds
Rate of formation of NO2 DNO2 / Dt
(0.0197 mol?L-1 0.0160 mol?L-1) / (400 s
300 s) 3.7 x 10-5 mol?(Ls)-1
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2 N2O5 (g) ? 4 NO2 (g) O2 (g)
Between 300 and 400 seconds
Rate of formation of O2 DO2 / Dt
(0.0049 mol?L-1 0.0040 mol?L-1) / (400 s 300
s) 9 x 10-6 mol?(Ls)-1
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1. Average reaction rate
2. Slopes
3. Time 0

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2 N2O5 (g) ? 4 NO2 (g) O2 (g)

• The three values for rate that we calculated are
  not the same!
• Why?
• We have different molar amounts.
• But the relative rates ARE THE SAME!

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2 N2O5 (g) ? 4 NO2 (g) O2 (g)

• The relative rate of formation of O2 is
• (1/1) 9 x 10-6 mol?(Ls)-1 9 x 10-6 mol?(Ls)-1
  
• The relative rate of formation of NO2 is
• (1/4) 3.7 x 10-5 mol?(Ls)-1 9.3 x 10-6
  mol?(Ls)-1
• The relative rate of decomposition of N2O5 is
• (1/2) 1.9 x 10-5 mol?(Ls)-1 9.5 x 10-6
  mol?(Ls)-1

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Instantaneous Reaction Rates

• Whats happening at this instant in time?

We can use instantaneous reaction rates.
The initial rate is the instantaneous reaction
rate for a reaction at time zero.
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Problem

• Consider the following reaction
• 3 I- (aq) H3AsO4 (aq) 2 H (aq)
• ? I3 (aq) H3AsO3 (aq) H2O (l)
• a) If DI-/Dt 4.8 x 10-4 mol?(Ls)-1, what
  is the value of DI3-/Dt during the same time
  interval?
• b) What is the average rate of consumption of H
  during the same time interval?

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Rate Laws and Reaction Order

• The rate of a chemical reaction depends on the
  concentration of some or all of the reactants.

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Rate laws

• The rate law for a reaction
• is the equation showing the dependence of the
  reaction rate on the
• concentrations of the reactants.

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aA bB ? products

• Rate k AmBn
• k is a constant for the reaction at a given
  temperature, and is called the rate constant.

Be Careful!
m does not have to equal a n does not have to
equal b
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Sensitivity to concentration change
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Reaction order

• Reaction order
• with respect to a given reactant
• is the value of the exponent of the rate law
  equation for the specific reactant only.
• The overall reaction order is the
• sum of the reaction orders for
• all reactants.

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Reaction order example

• rate k A2B
• The reaction order with respect to A is 2
• or the reaction is second order in A
• The reaction order with respect to B is 1
• or the reaction is first order in B
• The overall reaction order is 3 (2 1 3)
• or the reaction is third order overall

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Problem

• Consider three reactions with their given rate
  laws below. What is the order of each reaction
  in the various reactants, and what is the overall
  reaction order for each reaction?

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Experimental Determination of a Rate Law

• Reaction rate laws can only be determined
  experimentally!
• We most commonly carry out a series of
  experiments in which the
• initial rate of the reaction is measured
• as a function of
• different initial concentrations of reactants

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Method of initial rates
If you see a table like this with chemical
concentrations or pressures and rate data,
chances are good the question is a method of
initial rates problem.
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Method of initial rates

• IGNORE THE REACTION with this type of problem.
  The chemicals in the TABLE are the interesting
  ones.
• You always require at least one more experimental
  reaction than your number of chemicals given in
  your table!
• Sometimes we are given a table with an extra
  experiment which we can use to check if weve
  done everything correctly.

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2 NO (g) O2 (g) ? NO2 (g)

• Since rate laws are always expressed in terms of
  reactants (and sometimes catalysts well see
  these later), lets create a general form of the
  rate law for this reaction based on what
  chemicals the TABLE tells us are involved in the
  rate of the reaction.

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2 NO (g) O2 (g) ? NO2 (g)

• rate k NOmO2n

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Method of initial rates

• For our initial reactant order determination we
  need to choose a pair of reactions where only one
  reactant concentration changes. Experiments 1
  and 2 fulfill this condition.

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Method of initial rates

• rate k NOmO2n
• Since k is a constant then
• k for experiment 1
• IS EQUAL TO
• k for experiment 2!
• k rate / NOmO2n

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Reaction order w.r.t. NO
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Reaction order w.r.t. O2
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Our rate law

• rate k NO2O21

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Rate constant using experiment 1
k rate / NO2O21
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Rate constant using experiment 2
k rate / NO2O21
The rate constant is the same, as it should be!
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Check using extra experiment
rate (1.42 x 104 M-2?s-1) NO2O21
The rate is the same as the experimentally
observed rate (within rounding errors). We MUST
have done everything right!
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Units of rate constants

• Rate always has the units
• mol?(Ls)-1
• To ensure we get the right units for rate means
  the rate constant must have different units
  depending on the overall reaction order.

Be Careful!
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Problem

• H2O2 (aq) 3 I- (aq) 2 H (aq)
• ? I3- (aq) 2 H2O (l)
• DI3-/Dt can be determined by measuring the rate
  of appearance of the colour.

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Problem

• a) What is the rate law for the formation of I3-?
  
• b) What is the value for the rate constant?
• c) What is the initial rate of formation of
  triiodide when the concentrations are H2O2
  0.300 mol?L-1 and I- 0.400 mol?L-1?

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Reaction Rates and Temperature

• Increasing the temperature increases a chemical
  reactions rate.
• In general, reaction rates approximately double
  if you increase the temperature by 10 C.

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Bumper cars
A gasp of surprise could be a reaction when
riding in a bumper car.

Reactions occur ONLY when the bumps are very
hard and occur from behind.
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Collision theory

• A BC ? AB C
• If this reaction occurs in a single step, then at
  some point in time, the B-C bond starts to break,
  while the A-B bond starts to form.
• At this point, all three nuclei are weakly linked
  together.

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Collision theory

• Molecules tend to repel each other when they get
  close.
• We must insert energy to force the molecules
  close together. This is like forcing together
  the north poles of two magnets.
• This inserted energy is the kinetic energy of the
  molecules. It becomes potential energy as the
  molecules get closer.

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Collision theory

• A---B---C
• has a higher potential energy than either
• A B-C or A-B C
• A---B---C
• is the transition state
• or the activated complex

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Figure 13.11
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Figure 13.11
There are two useful energy differences in the
Figure.
The difference in energy between products and
reactants is DH The difference in energy between
the transition state and the reactants is Ea
the activation energy
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Activation energy

• The activation energy (Ea) of a reaction is the
  will always be positive!

The energy of collision between two molecules
must be AT LEAST as big as Ea otherwise we cannot
make it to the transition state.
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• Collisions between molecules at higher
  temperatures are more likely to have collision
  energy GREATER THAN the activation energy.
• Higher temperatures mean higher rates of reaction!

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Collisions

• An individual molecule collides with other
  molecules about once every billionth of a second
  (one billion collisions per second).
• If every collision was successful in creating
  products, then every reaction would be almost
  instantaneous. This is not the case.
• Not every collision breaks the activation energy
  barrier!

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Collisions

• The fraction of collisions that have enough
  energy to break the activation barrier is given
  by
• f e-Ea/RT

e is approximately 2.7183, Ea is the activation
energy, T is the temperature in Kelvin, R is
the gas law constant (8.314 J?K-1?mol-1)
Be Careful!
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Bumper cars and energy

• bumper car - a more energetic collision is more
  likely to make us gasp (our reaction)
• molecular collisions -higher energy collisions
  are more likely to lead to reaction
• (by overcoming the activation energy)

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Bumper cars and orientation

• You are also more likely to gasp if you are hit
  from behind by another bumper car. The
  orientation of how the collision occurs is also
  important to get a reaction.
• The same is true for molecules where the fraction
  of collisions that have the right orientation is
  p. We call this fraction p the steric factor.

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Figure 13.9
Cl2 MUST collide with the N side of NO to form
the transition state ON--Cl--Cl.
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Figure 13.9
If Cl2 hits the O side, we get a different
transition state that might not give the same
products or has a higher activation energy.
(molecules bounce off each other)
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Steric factor
Our steric factor in this case would be p 0.5
since half the collisions lead to the wrong
transition state.
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General reaction A BC ? AB C

• Collision rate Z A BC
• Z is a constant related to the collision
  frequency.
• Recall only a fraction (f) of the collisions have
  a collision energy greater than or equal to the
  activation energy.
• Of those collisions, only a fraction (p) have the
  correct orientation to proceed through the
  transition state to the products.

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General reaction A BC ? AB C

• Reaction rate p x f x Collision rate
• Reaction rate pfZ A BC
• Since for our general reaction
• Reaction rate k A BC
• k pfZ pZ e-Ea/RT A e-Ea/RT
• (where A pZ)
• (frequency factor)

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Arrhenius Equation
As T increases k increases
pZ A
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Problem

• AB CD ? AC BD
• What is the value of the activation energy for
• this reaction? Is the reaction endothermic or
• exothermic?

Suggest a plausible structure for the transition
state.
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Using the Arrhenius Equation

• If we know the rate constants for a reaction at
  two different temperatures, we can then calculate
  the activation energy.
• k A e-Ea/RT
• ln k ln (A e-Ea/RT)
• ln k ln (A) ln (e-Ea/RT)

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• ln k ln (A) (Ea/RT)
• This is the equation for a straight line!

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• If we graph the natural logarithm of the rate
  constant versus inverse temperature
• ln k (y axis) vs 1/T (x axis)
• we get a straight line where the
• slope -Ea/R
• So Ea - slope x R

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ln k vs 1/T
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• ln k2 ln k1 (Ea/R) (1/T2 1/T1)
• OR
• D (ln k) (Ea/R) D(1/T)

Changes! Final minus initial
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Be Careful!

• Your textbook says
• ln (k2/k1) (Ea/R) (1/T1 1/T2)

This is absolutely correct as well! Use
whichever form of the relation that you feel more
comfortable with mathematically.
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Problem
Rate constants for the decomposition of gaseous
dinitrogen pentaoxide are 4.8 x 10-4 s-1 at 45
C and 2.8 x 10-3 s-1 at 60 C 2 N2O5 (g) ? 4
NO2 (g) O2 (g) What is the activation energy
of this reaction in kJ?mol-1? What is the rate
constant at 35?C?
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Reaction Mechanisms
A reaction mechanism is the sequence of molecular
events (elementary steps or elementary reactions)
that defines the pathway from the reactants to
the products in the overall reaction.
The elementary reactions describe the behaviour
of individual molecules while the overall
reaction tells us stoichiometry.
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NO2 (g) CO (g) ? NO (g) CO2 (g) (Overall
Reaction)

• The reaction actually takes place in two
  elementary reactions!
• ? 2 NO2 ? NO and NO3
• ?NO3 CO ?NO2 and CO2

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NO2 (g) CO (g) ? NO (g) CO2 (g) (Overall
Reaction)

• Elementary reactions must add together to give
  the overall equation!

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NO2 (g) CO (g) ? NO (g) CO2 (g) (Overall
Reaction)

• Some of the crossed-out chemicals are neither
  reactants nor products in the overall reaction.
• For example, in the above reaction NO3 is formed
  in one elementary step and consumed in a later
  elementary step.

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Reaction intermediate

• A reaction intermediate is a species that is
  formed in an elementary step reaction, that is
  consumed in a later elementary step reaction.

We never see reaction intermediates in the
overall reaction!
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Molecularity

• The molecularity of an elementary reaction is the
  number of molecules
• on the reactant side
• of the elementary step reaction.

A one molecule elementary reaction is
unimolecular. A two molecule elementary
reaction is bimolecular. A three molecule
elementary reaction is termolecular.
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Molecularity
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Chances for molecularity
The chances of a unimolecular reaction only
depend on the one molecule, and are good. A
bimolecular reaction requires that two molecules
collide with each other. This isnt difficult
and happens quite often. A termolecular reaction
requires that three molecules collide with each
other at the same time. The chances of this
happening are not very good. The chances of four
or more molecules colliding at the same time are
almost impossible.
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Bumper cars
Consider bumper cars. Very often, you will hit
one other bumper car. Every once and a while,
you and another car will hit a third car at the
same time. It is a very rare occurrence to
have a bumper car pile-up where many cars hit
a single car at exactly the same time.
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Problem

• A suggested mechanism for the reaction of
  nitrogen dioxide and molecular fluorine is

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Problem

• a) Give the chemical equation for the overall
  reaction, and identify any reaction intermediates
  
• b) What is the molecularity of each of the
  elementary reactions?

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Rate Laws and Reaction Mechanisms

• Unlike an overall reaction the
• rate law for an elementary reaction
• follows DIRECTLY from the
• molecularity of the step reaction!
• For a general elementary step reaction
• aA bB ? products
• rate k Aa Bb

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Ozone

• Unimolecular decomposition of ozone.
• O3 (g) ? O2 (g) O (g)
• The rate law will be first order with respect to
  ozone
• rate k O3

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Bimolecular reaction

• A B ? products
• Reaction depends on collisions between
• molecules A and B
• Increase A, you increase collisions
• Increase B, you increase collisions
• rate k A B

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Elementary reaction rate laws
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Mechanisms and overall rate law

• The mechanism of the overall reaction is
  predicted through the elementary reactions
  therefore
• the elementary reactions will determine the rate
  law of the overall reaction!

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Mechanisms and overall rate law

• If the overall reaction occurs in ONE elementary
  step, then the elementary reaction and the
  overall reaction ARE THE SAME.
• The rate law for the overall reaction is given by
  the rate law for the step reaction
• rate k CH3Br OH-

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Rate-determining step

• The rate-determining step
• of an overall reaction with a mechanism of two or
  more steps is the
• elementary step reaction
• which has the slowest rate.
• The overall reaction can occur NO FASTER than its
  SLOWEST elementary reaction.

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NO2 (g) CO(g) ? NO (g) CO2 (g)

• The second step has to wait for the first step to
  create the NO3, which is then used rapidly for
  the second step reaction.

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NO2 (g) CO(g) ? NO (g) CO2 (g)

• Is the proposed mechanism plausible?
• The elementary steps MUST ADD UP to give the
  overall reaction
• AND
• the mechanism rate law MUST BE CONSISTENT with
  the observed rate law.

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NO2 (g) CO(g) ? NO (g) CO2 (g)

• The elementary reactions DO add up to the overall
  reaction.
• The rate law of the rate-determining step is
• rate k1 NO22
• Since this is the same as the experimentally
  observed rate law, so this mechanism is plausible.

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Just because a mechanism is plausible doesnt
mean it is right!
Be Careful!
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Problem

• Write the rate law for each of the elementary
  reactions
• O3 (g) O (g) ? 2 O2 (g)
• Br (g) Br (g) Ar (g) ? Br2 (g) Ar (g)
• Co(CN)5(H2O)2- (aq) ? Co(CN)52- (aq) H2O (l)

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Problem

• The following substitution reaction has a first
  order rate law
• Co(CN)5(H2O)2- (aq) I- (aq) ? Co(CN)5I3- (aq)
  H2O (l)
• rate k Co(CN)5(H2O)2-
• Suggest a possible reaction mechanism, and show
  that your reaction mechanism is in accord with
  the observed rate law.

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Catalysis

• Reaction rates are not just affected by reactant
  concentrations and temperatures.

A catalyst is a substance that increases the rate
of a reaction without being consumed in the
reaction.
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How does a catalyst work?

• A catalyst makes available a different reaction
  mechanism that is more efficient than the
  uncatalyzed mechanism.

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How does a catalyst work?

• To get from one side of a mountain to the other
  we have to climb up to the top (the activation
  energy), and then down the other side of the
  mountain.
• If there is a mountain pass partway up the
  mountain then we can climb up to the pass (a
  lower activation energy) and then climb down to
  the other side.
• Going through the pass will be quicker than going
  to the top!

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2 H2O2 (aq) ? 2 H2O (l) O2 (g)

• Ea for this reaction is 76 kJ?mol-1
• At room temperature, the reaction is slow.
• In the presence of iodide ion the reaction is
  faster because a new pathway with a lower
  activation energy is made available.

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• This catalyzed overall reaction is faster than
  the uncatalyzed reaction because it has a lower
  activation energy (our mountain pass) of 19
  kJ?mol-1.
• Because the activation energy is about 3.75 times
  lower than in the uncatalyzed reaction, the
  catalyzed reaction rate will be about 40 times
  faster than the uncatalyzed rate constant.

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Figure 13.13 (O3 O is catalyzed by Cl atoms)
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Homogeneous and Heterogeneous Catalysts

• A homogeneous catalyst exists in the same phase
  as the reactants.
• A heterogeneous catalyst exists in a different
  phase (usually solid) than the reactants.

I- is a homogeneous catalyst here Pt is a
heterogeneous catalyst here
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Figure 13.15
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Heterogeneous catalysts

• Most catalysts used in industry are heterogeneous
• It is much easier to separate a solid from a gas
  or liquid (for example) than two liquids or
  gases).

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Enzymes are catalysts

• In living beings catalysts are usually called
  enzymes
• Carbonic anhydrase catalyzes the reaction of
  carbon dioxide with water
• CO2 (g) H2O (l) ? H (aq) HCO3- (aq)
• The enzyme increases the rate of this reaction by
  a factor of 106. Equivalent to about a 200 K
  increase

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Enzymes
Lock-and-key model