CHEMICAL KINETICS

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CHEMICAL KINETICS In thermochemistry we studied
the question of whether or not a reaction can
occur under given conditions. is ?G ve or
-ve In chemical kinetics we are concerned with
two primary characteristics of the
reaction. What are these properties?
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A) The rate at which a reaction occurs How can
this be measured? The increase in concentration
of a product or the decrease in concentration of
a reactant per unit time. Properties to be
monitored can be conductivity, absorbance, pH or
any other physical property which is convenient.
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B) The mechanism by which a reaction occurs
the pathway or series of steps involved. Why
would the pathway or series of steps effect the
rate? There are many cases in which ?G? is ve
but the reactions occur so slowly that the rate
is immeasurable at room temperature.
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Example Diamond in air C (diamond) O2 (g)
? CO2 (g) ?G? -396 kJ Why does
diamond not oxidise to cardon dioxide in
air? This can be explained by kinetics.
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What factors will affect the rate of a chemical
reaction? (1) nature of the reactants (2) concent
ration of the reactants (3) temperature (4) cataly
sis
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What must atoms do before they can react with one
another? Do all collisions result in
reactions? WHY? What factors are involved?
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Collision theory of reaction rates For a
collision to be effective the reacting species
must (a) possess a minimum (kinetic energy) KE
to overcome repulsion between their electron
clouds and break existing bonds (b) have the
proper orientations toward each other
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Example Consider the reaction (Whitten) NO N2O
? NO2 N2
Bond breaking
N
N
N
N
N
N
O
O
O
N
N
O
O
N
O
Bond forming
EFFECTIVE
Collision
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N
N
N
N
N
O
N
O
O
O
O
N
N
N
O
Collision
INEFFECTIVE
Unfavourably placed for NO bond formation.
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BE is Potential energy What property involves
the measure of potential energy? manifests
itself as ?H.
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Transition state
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The first section of the curve shows... Reactants
well separated and having no effect on each
other.
As they approach each other there is an increase
in PE due to ... Electrostatic repulsion between
their electron clouds, including those of the
bonds which are to be broken.
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If the reactants have enough KE to overcome this
repulsion they form a high-energy, short lived
transition complex which then breaks up into the
products. If ?H is -ve the reaction is said to
be exothermic (heat released) if ?H is ve the
reaction is endothermic (heat absorbed).
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The amount of KE that must be converted to PE for
the transition state to be reached is called the
activation energy, Ea. Ea is then released as
the products move apart.
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Reaction rate is defined as the change in
concentration of a reactant or product per unit
time.
If a moles of A that disappears, b moles of B
appears. If a 1 say and b 3, then 3 mol of
B disappear for every 1 mol of A.
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Determination of reaction rate This is done by
measuring at t0, t1, t2, etc. and
plotting as follows.
The slope of the tangent to the curve at any
given time gives the reaction rate at that time.
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Rate decreases with time due to fewer collisions
between reacting species. Concentration of
reactants is decreasing by convention the rate
is ve because of the method used for the
determination of the slope of the tangent.
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That is tangent equals at t2 - at
t1 t2 - t1 Since at t2 lt
at t1, the value of the tangent is ve and hence
Tangent -(- ?conc./?t) ?conc./?t
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Rate equations and reaction order These
instantaneous rates can be used to find the
mathematical relationship between reaction rate
and concentration of each reactant, that is the
rate equation. Rate k Aa Bb Cc
for the reaction aA bB cC ? k rate
constant, which is characteristic of a particular
reaction.
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k reflects the fraction of collisions that are
effective in producing a reaction at a given
temperature. For example for Rate k A2
B A B ? no dependence on
products The reaction is described as second
order in A and first order in B. The
overall reaction order is the sum of the
exponents, for example in the above reaction,
overall reaction order 3
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Example If one reactant is present in large
excess, so that its concentration does not change
appreciably as the reaction proceeds HCl
H2O ? H3O Cl- (solvent) That is a one-step
reaction but H2O not significant HCl dissolved
in H2O
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Determination of rate equations For the simple
case Reactions that are known to be elementary
(ie occur in one step) and do not involve the
solvent rate equations can be written based
solely on the chemical equation for the
reaction. eg A 2B ? C Rate k A B2
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What if a two step reaction process is
involved? Whats the so called rate determining
step? The rate determining step will be the
slower of the two steps. eg For the overall
reaction 2NO2 (g) F2 (g) ? 2NO2F
(g) Experimentally determined rate equation is
Rate k NO2 F2
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This is because the reaction takes place in 2
steps as follows. NO2 (g) F2 (g) ? 2NO2F
F NO2 (g) F (g) ? 2NO2F (g) Rate
depends on the rate at which F is produced in the
first reaction since it is used in the second
reaction immediately as it is formed.
Slow
Fast
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INITIAL RATES FOR FINDING RATE EQUATIONS Values
of concentration vs time data from a series of
separate experiments. Each must have a different
initial concentration of one or more
reactants A 2B ? AB2 Trial run Initial
A Initial B Initial rate of AB2 formation
(mol L-1) (mol L-1 s-1 ) 1 1.0 x
10-2 1.0 x 10-2 1.5 x 10-4 2 1.0 x
10-2 2.0 x 10-2 1.5 x 10-4 3 2.0 x
10-2 3.0 x 10-2 6.0 x 10-4
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For trial 1,
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Equations for first order reactions Rate k
A
Integrating gives k ?t - ?A/A log A
log A0 - kt/2.303 where A0
concentration at t 0 A concentration at
time t
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This equation gives us a lot of information on
the progress of a first order reaction. If k is
known, the time needed to reach a given
concentration of reactant can be calculated, or
can calculate the conc of the reactant after a
given amount of time.
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Example For the decomposition of N2O5 in
CHCl3 2N2O5 ? 4NO2 O2 Reaction is 1st
order, k 6.32 x 10-4 s-1 Initial N2O5
0.40 mol L-1 Find N2O5 after 1 hr (3600 s)
Use log A log A0 - kt/2.303
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log N2O5 log 0.40 - (6.32x10-4)(3600)/2.303
-1.39 Taking antilog N2O5 0.041
mol L-1
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Example The reaction N2O5 (g) ?N2O4 (g) ½
O2 (g) obeys the rate law Rate k
N2O5 where k 1.68 x 10-2 s-1 Initial
N2O5 2.50 moles container volume 5.0 L How
many moles of N2O5 would remain after 1 minute?
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Taking antilog A 1.82 x 10-1 mol
L-1 Moles A remaining 0.182 mol L-1 x
5L 0.910 mol
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Half-life of first order reactions The Half-life
or t ½ of a reactant is the time it takes for
one-half of a reactant to be converted into
product. Same as that calculated for nuclear
decay. Solve previous equation for t Since A
½ A0 at t t ½ (by definition)
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Example For a solution originally containing 1.30
x 10-6 mol L-1 of AmCl3. Only 1.27 x 10-6 mol
L-1 of this radioactive substance remained after
2 hrs. Given that radioactive decay is a
first-order process, determine the half life of
240Am.
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What happens if our reaction is second order or
greater?
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Rate equations for second order reactions The
rate of a second order reaction can be dependent
upon the concentration of either one or two
reactants Therefore Rate kA2 or Rate
kAB If Rate k A2 Integrate 1/A
1/A0 kt
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The reaction A B C D is second
order in A and second order overall. And if k
0.622 L mol-1 min-1 What is the half life of A
if Ao4.10 x 10-2 mol L-1? What is the
equation for the half-life of a second order
reaction?
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For t½ A ½A0 Thus 1/A 1/A0
kt becomes 2/A - 1/A0 kt½ t½
1/kA0 for the question then where k 0.622 L
mol-1 min-1 and Ao4.10 x 10-2 mol L-1. t½
1/(0.622 )(4.10 x 10-2 ) 39.2 min
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Rate equations for zero-order reactions We have
looked at first and second order reactions, what
is a zero order? What controls the rate of this
reaction? The rate is not dependent on any
reactant conc. i.e. Rate k A0B0 k i.e.
rate is constant. That means that the rate is
controlled by something other than collisions
Like What?
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The amount of Light Photochemical reactions. Rate
k
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Graphic method for finding rate equations
Type of reaction Eqn for straight line
linear plot slope Zero order A -kt
A0 A vs t -k (rate k) first
order logA -kt/2.303 logA0 logA vs
t -k/2.303 (rate kA ) second order
1/A 1/A0 kt 1/A vs t k (rate
kA2)
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Reaction mechanisms How the reaction occurs.
Example The overall third order reaction of NO
with H2 2NO(g) 2H2(g) N2 (g)
2H2O(g) Rate kNO2H2 (experimentally
determined)
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How could we envisage the reaction occurring?
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The Arrhenius eqn. from this equation

• Reactions with large activation energies have
  smaller values of k and are slower.
• For a given Ea, k will increase as temperature
  increases, reaction proceeds at a fast rate.,

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Example The rate constant for the decomposition
of N2O5 in CHCl3 (chloroform) 2N2O5
4NO2 O2 was measured at T1 25C
(k15.54 x10-5 S-1) and T2 67C (k2 9.30 x
10-3 S-1). Find Ea for this reaction. Ans
1.0x10-5 J mol-1
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Homogeneous and heterogeneous reactions Dependent
on the opportunities for effective contact
between reactants. Liq Liq good contact Liq
Solid good contact Solid Solid difficult
contact
Catalysis A catalyst is a substance which
increases the rate of a chemical reaction and is
not itself consumed in that reaction. A catalyst
works by providing an alternative and easier
pathway from reactants to products.
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e.g
. Consider the reaction A A 2B

2AB
A A bond breaks as A is attached to the active
site.
B
B
A
A
A
A
Active sites on catalyst
surface
B
B
A
A
A B molecules
break away
(change in
electronic
arrangement
less attraction for
surface)