Chemical Kinetics

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CHAPTER 14 Chemical Kinetics
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Chemical Reactions There are two things that
we are interested in concerning chemical
reactions 1) Where is the system going
(chemical equilibrium, covered in chapter
15). 2) How long will it take for the system to
get to where it is going (chemical kinetics,
covered in chapter 14). Kinetics is the study of
the rate at which a chemical reaction takes
place, and the mechanism by which reactants are
converted into products.
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Example Consider the following irreversible
chemical reaction A(g) ? B(g)
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Example (continued) We may follow the rate of
the reaction by either observing the
disappearance of reactant A or the appearance of
product B. The average rate of the reaction over
some time period from t1 to t2 is Ave. rate
- ( A2 - A1 ) - ?A
t2 - t1 ?t ( B2 -
B1 ) ?B t2 - t1 ?
?t where ?A A2 - A1 ?B B2 - B1
?t t2 - t1 We insert a negative sign when
find the rate of disappearance of a reactant to
make the rate of reaction positive.
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Example (continued)
Average rate between 10 s and 20 s is (looking at
disappearance of A) Ave. rate - (0.022 M -
0.030 M) 0.00080 mol/L.s
(20. s - 10. s)
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Instantaneous Rate of Reaction
The instantaneous rate of reaction at time t is
equal to the value for the slope of the tangent
line in a plot of concentration vs time (negative
if a reactant, positive if a product).
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Stoichiometry and Reaction Rate Different
reactants (and products) may disappear (or
appear) at different rates depending on the
stoichiometry of the reaction. For example
H2(g) 2 ICl(g) ? 2 HCl(g) I2(g) In the
above reaction ICl disappears twice as fast as
H2, and HCl appears twice as fast as I2. To take
this into account we can define the average rate
of reaction as the change in concentration of a
reactant or product divided by the stoichiometric
coefficient used to balance the reaction. Ave.
rate - ?H2 - 1 ?ICl 1 ?HCl
?I2 ?t 2 ?t
2 ?t ?t Note we still insert a
negative sign when looking at the disappearance
of a reactant. By this method we will obtain the
same value for average rate no matter which
reactant or product we observe.
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Rate Law A rate law is an expression that gives
the rate of a chemical reaction in terms of the
concentrations of the reactants (and occasionally
other concentrations as well). For example,
consider the following general chemical
reaction a A b B ? products where a and b
are stoichiometric coefficients, and we have
assumed the reaction is irreversible (proceeds
only in the forward direction). The rate law for
reactions of this type can often be written
as rate k Am Bn m order of reaction with
respect to A n order of reaction with
respect to B m n overall reaction
order k rate constant, value depends only on
temperature. Units determined by dimensional
analysis.
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For example, for the reaction 2 NO2(g)
F2(g) ? 2 NO2F(g) rate - ?F2 k NO2
F2 ?t Reaction is 1st order in NO2, 1st
order in F2, and 2nd order overall. Note the
following 1) The reaction orders are usually
small whole numbers (0, 1, or 2 occasionally 1/2
or -1 rarely any other values). 2) There is no
general relationship between the reaction orders
and the stoichiometric coefficients for the
reaction. For example, in the above reaction the
stoichiometric coefficients for NO2 and F2 are 2
and 1, but the reaction orders (determined by
experiment) are 1 and 1.
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Experimental Determination of the Rate Law There
are several methods that have been developed for
finding the rate law for a chemical reaction.
The easiest and most common method used is the
initial rates method. Consider a general
reaction of the form a A b B ?
products where we assume the rate law for the
reaction has the form rate k Am Bn For
a particular set of initial concentrations, the
initial rate of reaction is the rate of the
reaction measured before the initial
concentrations of reactants have had a chance to
change significantly. When we measure the
initial rate of reaction, we can then use the
initial concentrations of our reactants in the
rate law.
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Consider the following two experiments carried
out at the same temperature. trial 1, initial
concentrations of A and B are A1 and B1,
initial rate of reaction is R1. trial 2, initial
concentrations of A and B are A2 and B2,
initial rate of reaction is R2. R1 k A1m
B1n R2 k A2m B2n Then R2 k A2m
B2n ( A2/A1 )m ( B2/B1 )n R1
k A1m B1n The above is true in general.
Now, consider the case where the initial
concentration of B is the same in both trials.
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If B2 B1, then ( B2/B1 )n 1n
1 and R2 ( A2/A1 )m R1 We can
usually find the value for m by inspection, but
for a rigorous method for determining m we can
simply take the natural logarithm of both sides
of the above equation ln (R2/R1) ln (
A2/A1 )m m ln (A2/A1 ) m ln
(R2/R1) ln (A2/A1 ) The
value for n (order of reaction with respect to B)
can be found by comparing trials where the
initial concentration of A is the same in both
trials.
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Example The following data were obtained for the
reaction A B ? products trial initial
A initial B initial rate (mol/L) (mol/L) (mol/L.
s) 1 0.100 0.100 3.1 x 10-5
2 0.100 0.200 3.0 x 10-5 3 0.200 0.200 1.2
x 10-4 We may assume the rate of reaction is
given by the expression rate k Am Bn Based
on the above data, find m, n, and k.
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Finding m. Compare trial 3 and trial 2. R3
k A3m B3n ( A3/A2 )m (B3/B2 )n
R2 k A2m B2n 1.2 x 10-4
(0.200/0.100)m (0.200/0.200)n 3.0 x 10-5
4.0 2m , and so m 2 Finding n.
Compare trial 2 and trial 1. R2 k A2m
B2n ( A2/A1 )m (B2/B1 )n R1
k A1m B1n 3.0 x 10-5 (0.100/0.100)m
(0.200/0.100)n 3.1 x 10-5 0.97 2n ,
and so n 0 So the rate law is R k A2
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To find k we can now use any of the trials. If
we use trial 1, then R1 k A12 k
R1/A12 (3.1 x 10-5 mol/L.s) 3.1 x 10-3
L/mol.s
(0.100 mol/L)2 Notice that the units for k are
determined by dimensional analysis. For a real
set of experimental data we would find a value
for k from each data set, and then average to
find the best value for k. Also note for real
data the values for the reaction orders would
likely not work out to be exactly integers due to
experimental error.
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For example Experimental value Reaction
order 1.94 2n n 1 9.4
3m m 2 1.45 2p
p 1/2 We have assumed in the
above that the true values for the reaction
orders should be integer or half-integer values,
and that the small differences we observe are due
to random error in the experimental data.
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Typical Types of Rate Laws As previously
discussed, the rate law for a reaction can often
be written as rate - ?A k Am Bn
?t Common rate laws. Zero
order rate k First order homogeneous rate
k A Second order homogeneous rate k A2
Second order heterogeneous rate k A B
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First Order Homogeneous Rate Law For a first
order homogeneous rate law rate k A We may
show that concentration vs time is given by the
expression At A0 e-kt
where At concentration of A at time t A0
concentration of A at t 0 k rate constant
(units of 1/time) A plot of concentration vs
time will exhibit what is called an exponential
decay in the concentration of A.
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Finding the Rate Constant It is difficult to
find the value for k (rate constant) from a plot
of concentration vs time since we get nonlinear
behavior. We can find a linear relationship as
follows At A0 e-kt Take the ln of both
sides ln At ln A0 - kt (y) b
m(x) y lnAt x t This predicts
that for a first order reaction a plot of ln At
vs time will give a straight line, with slope
-k (and intercept ln A0)
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One way to test whether or not a reaction is
first order is to plot the logarithm of
concentration vs time. If you get a linear
result in the plot, then you know the reaction is
first order. If you do not get a linear result
in the plot, then you know the reaction is not
first order.
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Half-life By definition, the half-life for a
chemical reaction, t1/2, is the time it takes for
the concentration of a reactant to decrease to
1/2 of its initial value. For a first order
reaction At A0 e-kt At t t1/2, At
A0/2, so A0/2 A0 e-kt½ Divide both
sides by A0 then 1/2 e-kt½ Take the ln
of both sides then ln(1/2) -kt1/2 Divide by
-k then t1/2 - ln(1/2) k But
- ln(1/2) ln(2), so t1/2 ln(2) ?
0.693 k k
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Note the following 1) Since t1/2 ln(2)/k for
a first order reaction, the value for the
half-life is independent of the initial
concentration of the reactant. 2) First order
reactions are the only reactions where the
half-life is independent of concentration. 3) If
we wait two half-lives, the concentration will
decrease to 1/4 of the initial value. For three
half-lives the concentration will decrease to 1/8
of the initial value, and so forth.
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Second Order Homogeneous Rate Law For a second
order homogeneous rate law rate k A2 we may
show that concentration vs time is given by the
expression At A0 (1
ktA0)
where At concentration of A at time t A0
concentration of A at t 0 k rate constant
(units of
1/(concentration)(time)) Concentration will not
decrease as quickly in a second order reaction as
it does for a first order reaction.
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Finding the Rate Constant Since At
A0 (1 ktA0) then if we
invert both sides of this equation we get 1
(1 ktA0) 1 kt At
A0 A0 (y)
b m(x) y 1/At x
t This predicts that for a second order reaction
a plot of 1/At vs time will give a straight
line, with slope k (and intercept 1/A0).
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One way to test whether or not a reaction is
second order is to plot 1/concentration vs time.
If you get a linear result in the plot, then you
know the reaction is second order. If you do not
get a linear result in the plot, you know the
reaction is not second order.
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Half-life We may use the definition of half-life
to find an expression for t1/2 for a second order
reaction. It is easiest to do this starting with
the expression 1 1 kt At
A0 If we substitute At A0/2 at t
t1/2, we get (after some algebra) t1/2 1
k A0 Unlike a first
order reaction, the half-life for a second order
reaction depends on concentration. As the
initial concentration decreases the half life
becomes longer.
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Zero Order Rate Law For a zero order rate
law rate k A0 k we may show that
concentration vs time is given by the
expression At A0 - kt , t lt A0/k
0 , t ? A0/k
where At concentration of A at time t A0
concentration of A at t 0 k rate constant
(units of
concentration/time) Concentration decreases at a
constant rate until all of the reactant has
disappeared.
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Finding the Rate Constant and Half-Life Since
At A0 - kt , t lt A0/k a plot of
concentration vs time will be a straight line,
with slope -k, (and intercept A0). The
half life for a zero order reaction will be t1/2
A0/2k
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Summary of Results
First order At A0 e-kt Second order At
A0
(1 ktA0)
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Sample Problem Consider the reaction A ?
products The reaction obeys first order
homogeneous kinetics. The initial concentration
of A in the system is A0 0.418 M. After 100. s
the concentration of A is 0.322 M. Find k (the
rate constant), and the concentration of A after
500. s.
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Consider the reaction A ? products The
reaction obeys first order homogeneous kinetics.
The initial concentration of A in the system is
A0 0.418 M. After 100. s the concentration of
A is 0.322 M. Find k (the rate constant), and
the concentration of A after 500. s. At A0
e-kt ln(At) ln(A0) - kt ln(At/A0)
- kt k - (1/t) ln(At/A0) (1/t)
ln(A0/At) So k (1/100. s) ln(.418/.322)
2.61 x 10-3 s-1. At 500. s, A500 (0.418 M)
exp-(2.61 x 10-3 s-1)(500. s) 0.1134 M
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Temperature Dependence of the Rate Constant For
a reaction that obeys the rate law rate k
Am Bn the rate, and therefore the rate
constant, for the reaction usually increases as
temperature increases. Experimentally it is
found that the temperature dependence of the rate
constant for the reaction often follows a simple
expression called the Arrhenius equation k A
e-Ea/RT where k - the rate constant for the
reaction A - pre-exponential factor Ea -
activation energy for the reaction If we take the
ln of both sides of the Arrhenius equation, we
get ln(k) ln(A) - (Ea/R)(1/T)
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Example 2 HI(g) ? H2(g) I2(g)
If we assume the data fit the Arrhenius equation,
then ln k ln A - (Ea/R) (1/T) and so for each
experimental value for k and T we must 1)
Convert T into units of K, then find 1/T 2) Find
the value for ln k and then plot the results.
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The value of Ea is found from the slope of the
above plot. The value for A is then found by
substituting one of the data points into the
Arrhenius equation.
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x y 0.0013 K-1 -
4.0 0.0018 K-1 - 15.0 At T 556. K we have
k 3.52 x 10-7 L/mol.s
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x y 0.0013 K-1 -
4.0 0.0018 K-1 - 15.0 slope ?y (-
15.0) - (-4.0) - 22000. K
?x 0.0018 - 0.0013K-1 So
Ea - R (slope) - (8.314 x 10-3 kJ/mol.K) (-
22000. K)
182.9 kJ/mol
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Since k A e-Ea/RT, then A k eEa/RT At T
556. K we have k 3.52 x 10-7 L/mol.s, and
so A (3.52 x 10-7 L/mol.s) e(182900.
J/mol)/(8.314 J/mol.K)(556.K) 5.37 x 1010
L/mol.s
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Other Forms of the Arrhenius Equation Beginning
with the Arrhenius equation k A e-Ea/RT we can
derive the following equation ln (k2/k1) -
(Ea/R) 1 1
T2 T1 In this
equation k1 is the rate constant at T1 and k2 is
the rate constant at T2. By knowing the value
for the rate constant at two different
temperatures we may use this equation to find Ea,
and then use the value of k at either temperature
to find A.
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Collision Theory The theoretical model on which
the Arrhenius equation is based is called
collision theory. The theory makes the following
assumptions 1) For a reaction to take place a
collision between reactant molecules must
occur. 2) The reactants must collide with
sufficient kinetic energy to overcome the energy
barrier separating reactants and products. 3)
The reactants must have a favorable orientation
for reaction to occur. For example, consider the
reaction AB C ? A BC
Transition state - The species that forms as
reactants are converted into products. It has a
structure intermediate between that of the
reactants and that of the products of the
reaction.
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How will the energy of the system change as we
proceed from reactants to products? We can
represent this by an energy diagram.
In this diagram Ea, the activation energy, is the
height of the barrier separating the reactants
(AB C) and products (A BC). ?E represents the
change in energy in going from reactants to
products. This reaction is exothermic (?E lt 0).
A---B---C is the transition state. Note that it
is a good approximation to say ?H ? ?E.
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Collision Theory and the Arrhenius Equation How
does collision theory relate to the Arrhenius
equation? We can use the theory to give a
physical interpretation to the constants that
appear in the equation. k A e-Ea/RT pz
e-Ea/RT A depends on the collision frequency
(z, the number of collisions between reactant
molecules per unit time) multiplied by an
orientation factor (p, the fraction of collisions
that have the correct orientation of reactant
molecules). e-Ea/RT represents the fraction of
collisions that have sufficient kinetic energy to
pass over the barrier separating reactants from
products. Notice that as T becomes larger, this
term also becomes larger, which makes sense, as
the molecules move faster and have higher kinetic
energy at high temperature than they do at low
temperature. Note As previously discussed in
CHM 1045, vave (3RT/M)1/2
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Orientation The pre-exponential factor A in the
Arrhenius equation depends on two factors - the
rate at which collisions take place and the
fraction of collisions that have a favorable
orientation for a reaction to occur. k A
e-Ea/RT Example Cl(g) NOCl(g) ? Cl2(g)
NO(g) Favorable orientation Unfavorable
orientation
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Reaction Mechanism A reaction mechanism is a
sequence of elementary reactions that take place
on a molecular level and lead from reactants to
products. For example, consider the
stoichiometric reaction NO2(g) CO(g) ?
NO(g) CO2(g) One possible mechanism for this
reaction is step 1 NO2 NO2 ? NO3 NO step
2 NO3 CO ? NO2 CO2 In this mechanism NO3
is a reaction intermediate, a substance that is
neither a reactant nor a product, but which is
produced and consumed as reactants are converted
into products.
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Types of Elementary Reactions There are three
common types of elementary reactions that appear
in reaction mechanisms. Unimolecular A ?
products rate k A Bimolecular A A
? products rate k A2 A B ?
products rate k A B Termolecular A
A A ? products rate k A3 A A B
? products rate k A2 B A B C ?
products rate k A B C Note that the
rate for a particular elementary reaction is a
rate constant multiplied by the concentrations of
the reactants.
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Requirements For a Reaction Mechanism An
acceptable reaction mechanism must satisfy two
requirements. 1) The individual elementary steps
in the mechanism must add up to the overall
reaction. 2) The rate law predicted by the
mechanism must agree with the experimentally
determined rate law. If the above two
requirements are not met, then the reaction
mechanism is not acceptable. Unfortunately, the
reverse is not true. In principle all we can do
with a mechanism is show it is consistent with
experiment. Also note that the final expression
for the rate law should not involve any reaction
intermediates.
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Finding the Rate Law From the Reaction
Mechanism In general, it is difficult to obtain
the rate law for a reaction from the reaction
mechanism (in fact, for complicated systems one
often uses a computer to model the
reaction). There are a few simple cases where we
can obtain a rate law from a mechanism 1) One
step mechanism. Examples O(g) HBr(g) ?
OH(g) Br(g) rate k O HBr H(aq)
OH-(aq) ? H2O(?) rate k H OH-
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2) Mechanisms with a single slow step. For a
multistep mechanism with one slow step, the
overall rate of the reaction will be the rate of
the slow step. This makes it possible to obtain
a rate law as long as there are no reaction
intermediates involved in the slow
step. Example step 1 H2(g) ICl(g) ? HI(g)
HCl(g) slow step 2 HI(g) ICl(g) ? I2(g)
HCl(g) fast overall H2(g) 2 ICl ?
I2(g) 2 HCl(g) (HI is an
intermediate) Since the overall rate of reaction
is equal to the rate of the slow step, we may
say rate k1 H2 ICl 1st order in H2, 1st
order in ICl 2nd order overall
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We can often use experimental data to decide
whether a particular reaction mechanism is
possible or not. Example Consider the
following two mechanisms for the
reaction NO2(g) CO(g) ? NO(g) CO2(g) One
step mechanism NO2(g) CO(g) ? NO(g)
CO2(g) predicted rate law rate k1
NO2CO Two step mechanism step 1 NO2(g)
NO2(g) ? NO3( g) NO(g) slow step 2 NO3(g)
CO(g) ? NO2(g) CO2(g) fast predicted rate
law rate k1 NO22 Experimentally it is found
that this reaction is 0th order in CO and 2nd
order in NO2. That means the first mechanism is
not correct, and that the second mechanism is
consistent with the observed rate law.
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Fast and Reversible Elementary Reactions In some
reaction mechanisms there will be a step that is
both fast and which goes in both directions.

k1 A B C
D fast, reversible
k-1
Since the reaction is fast in both directions we
can assume that equilibrium is rapidly achieved.
At that point, the rate of the forward and
reverse reactions must be the same, so k1 A
B k-1 C D We can solve the above
expression for the concentration of one reactant
in terms of other reactants and the rate
constants. This is often useful in finding the
rate law from a mechanism
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Example Consider the following three step
mechanism for the stoichio-metric reaction 2 H2
2 NO ? 2 H2O N2 overall reaction
step 1 2 NO N2O2 fast, reversible
step 2 H2 N2O2 ? H2O N2O slow step
3 N2O H2 ? N2 H2O fast (N2O2, N2O are
intermediates) What is the rate law predicted
for the above mechanism?
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step 1 2 NO N2O2 fast,
reversible step 2 H2 N2O2 ? H2O
N2O slow step 3 N2O H2 ? N2
H2O fast Overall rate rate of slow step k2
H2N2O2 N2O2 is a reaction intermediate and
so should not appear in the final rate
expression. But from step 1 we may say k1 NO2
k-1 N2O2 so N2O2 (k1/k-1) NO2 So
rate k2 H2 (k1/k-1) NO2 (k1k2/k-1)
H2 NO2 koverall H2 NO2
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Catalyst A catalyst is a substance that changes
the rate of a chemical reaction without itself
being produced or consumed by the
reaction. Example The reaction 2 KClO3(s) ?
2 KCl(s) 3 O2(g) is slow even at high
temperatures. However, if a small amount of
solid MnO2 is added to the reaction mixture the
reaction will become fast. The MnO2 is not
consumed in the reaction and can be recovered at
the end of the reaction. Homogeneous catalyst -
Appears in the same phase as the
reactants. Heterogeneous catalyst - Appears in a
different phase than the reactants.
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Many catalysts work by lowering the activation
energy for a reaction.
Catalysts can also affect the rate of reaction
by providing new pathways for converting
reactants to products, or by changing the value
for the pre-exponential factor (A), usually by
forcing the reactants into a favorable
orientation. Biological catalysts (enzymes)
often work in this way.
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Example Hydrogenation of ethene. C2H4(g)
H2(g) ? C2H6(g)
55
Enzyme - A biological catalyst. Most biological
enzymes are proteins, composed of chains of amino
acids, that take on a particular shape, making it
possible to catalyze a specific biological
reaction.
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End of Chapter 14
He (van't Hoff) was a modest and unassuming
man who never made priority claims indeed, in
his later books he sometimes gave credit to
others for things - like the Arrhenius equation
and the Le Chatlier principle - that he had
first discovered himself. - K. J. Laidler, The
World of Physical Chemistry Nothing has such
power to broaden the mind as the ability to
investigate systematically and truly all that
comes under your observa-tion in life. - Marcus
Aurelius