WK2 C14 Kinetics
Transcript of WK2 C14 Kinetics
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Chapter 14Chapter 14--
ChemicalChemicalReactionReaction
KineticsKinetics
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14.1 Reactionrates14.1 Reactionrates
14.2 Factorsthataffectreactionrate14.2 Factorsthataffectreactionrate
14.3 Rate laws14.3 Rate laws
14.4 Theoryofchemical kinetics14.4 Theoryofchemical kinetics
14.5 Reactionmechanisms14.5 Reactionmechanisms
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14.1 Reaction rates14.1 Reaction rates
Chemical kinetics is the study of the
rate of a reaction
The reaction rate is the speed withwhich reactants disappear and
products form
Reaction rates are always reported aspositive values
The unit is mol L1 s1
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Figure 14.1 Reactionrates
The progressofthereactionA 2B, showingtheconcentration
changesofA andB overtime
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14.1 Reaction rates14.1 Reaction ratesConsiderthereaction A B
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14.1 Reaction rates14.1 Reaction rates
Calculate the average rate at which
[A] changes in the first 50 seconds
2.4 10-4 mol L-1 s-1
rate (A)t
-[A]
-([A]t=50 [A]t=0)
50 s 0 s
-(0.0629 mol L-1 0.0750 mol L-1)
50 s
-(-2.4 10-4 mol L-1 s-1)
=
=
=
=
=
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14.1 Reaction rates14.1 Reaction rates
The rate at a particular time is called
the instantaneous rate
This can be determined from theslope of a tangent to the curve
For example, figure 14.1 at t = 50 s
rate = -(slope of tangent line)dt
=d[A]
= 2.3 10-4 mol L-1 s-1
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For the general reaction:
where a to dare the stoichiometric coefficients of the reactants A and B and the
products Cand D respectively, we define the rate of reaction as
Therefore, for our reaction A 2B we would write
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Estimatingthe Initial Rateofa Reaction
The following data have been measured at 508 C for the reaction 2HI(g) C H2(g) + I2(g)
What is the initial rate of decay of HI, and what is the initial rate of reaction
at this temperature
Time (s)
[HI] (mol L
1)
Time
(s) [HI] (mol L1)
0 0.100 200 0.0387
50 0.0716 250 0.0336
100 0.0558 300 0.0296
150 0.0457 350 0.0265
The initial rate of decay of HI is the instantaneous rate of decay of HI at time zero
The slope of the line can be calculated from the coordinates of any
two points (x1, y1) and (x2, y2) using the equation:
The initial rate of reaction can be calculated using the stoichiometry of the reaction.
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14.214.2 Factors that affectFactors that affect
reaction ratereaction rate
Five principal factors influence
reaction rates:
1.Ch
emical nature of the reactants
2. Ability of the reactants to come in
contact with each others
3. Concentrations of the reactants
4. Temperature
5. Availability of rate-accelerating agents
called catalysts
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14.2 Factors that affect14.2 Factors that affect
reaction ratereaction rate
Chemical nature of the reactants
During reactions, bonds break and new
bonds form
Some reactions are fast by nature and
others are slow
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14.2 Factors that affect14.2 Factors that affect
reaction ratereaction rate
The area of contact between the phases
determines the reaction rate
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14.2 Factors that affect14.2 Factors that affect
reaction ratereaction rate
Concentrations of the reactants
The rates of bothhomogeneous and
heterogeneous reactions are affected by
the concentrations of the reactants
For example, red-hot steel wool bursts into
flame when thrust into pure oxygen
Temperature of the system
Almost all chemical reactions occur
faster at higher temperatures
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14.2 Factors that affect14.2 Factors that affect
reaction ratereaction rate
Presence of catalysts
Catalysts are substances that increase
the rate of chemical reactions withoutbeing used up
For example, enzymes that direct our
body chemistry are all catalysts
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14.3 Rate laws14.3 Rate laws
A rate law is an equation in which the
rate is given as a function of reactant
concentrations
for example, rate = k[HI]n
kis the rate constant of the reaction
n is the order of the reactant
Both rate constant and order must be
experimentally determined
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14.3 Rate laws14.3 Rate laws
The value of the rate constant
depends on the particular reaction
being studied and the temperature at
which the reaction occurs
The order can be positive or negative,
an integer or a fraction The ordercannotbe deduced from
the balanced equation
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14.3 Rate laws14.3 Rate laws
Worked example 14.2
2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)
If the butane concentration is decreasingat a rate of 0.20 mol L1 s1, at what rate
is the oxygen concentration decreasing?
Solution:
1.3 mol O2rate (O2) =
0.20 mol C4H10
L s 2 mol C4H10
13 mol O2x
L s=
Oxygen reacts at a rate of 1.3 mol L1 s1
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Solution
From our definition of rate of reaction, we can write:
We are told that:
we can therefore use this, together with our expression for the rate of reaction to
determine
Oxygen reacts at a rate of 1.3 mol L1
s1
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14.3 Rate laws14.3 Rate laws
Types of rate laws:
A differential rate law expresses the rate
as a function of concentration An integrated rate law expresses the
concentration as a function of time
Rate laws help to identify the steps bywhich a chemical reaction occurs
The sum of the individual reaction steps
is called the reaction mechanism
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14.3 Rate laws14.3 Rate laws
The differential rate law
Consider the hypothetical reaction:
A + B
products The rate law for the reaction is:
rate = k[A]n[B]m
The values of n and m can bediscovered by looking for patterns in the
rate data
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14.3 Rate laws14.3 Rate laws
Suppose that the data in the table below
has been obtained in a series of five
experiments:
A + B C
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14.3 Rate laws14.3 Rate laws
For experiments 1, 2 and 3 [B] is held
constant. Any changes in the rate must
be due to the change in [A]
The rate law says that at constant [B] therate must be proportional to [A]n
rateexp 1
rateexp 2
= [A]exp 1
[A]exp 2n
rateexp 1
rateexp 2
=
0.40 mol L-1 s-1
0.20 mol L-1 s-1 = 2
and
[A]exp 1
[A]exp 2=
0.20 mol L-1
0.10 mol L-1= 2
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14.3 Rate laws14.3 Rate laws
In the final three experiments, the
concentration of B changes while the
concentration of A is held constant
This time it is [B] that affects the rate
For experiments 3 and 4, we have:
4.0 = 2.0m
Therefore, m must equal 2
rateexp 3
rateexp 4=
[B]exp 3
[B]exp 4m
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14.3 Rate laws14.3 Rate laws
We now know that the rate law for the
reaction must be:
rate = k[A]1
[B]2
The overall order is the sum of the orders
for each reactant in the rate law
Thus, ourhypothetical reaction has an
overall order ofthree
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To calculate the value ofkfor the A + B products reaction, we need only
substitute rate and concentration data into the rate law.
rate = k[A]1[B]2
from experiment 1 in table 14-3:
After cancelling units, the value ofkwith the net units is:
Note that the unit of a third-order rate constant is concentration2 time1.
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14.3 Rate laws14.3 Rate laws
First-order rate laws Assume A products
is first order in A
Th
e differential rate law is then:
The integrated rate law is:
rate = d[A]
dt= k[A]
[A]t
[A]0 = ktln [A]0e-kt[A]t =or
ln[A
]t=
kt+ ln[
A]0
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First-orderrate laws
Let's assume that ourhypothetical reaction A products is first order in A.
The differential rate law is then:
Rearrangement gives
the concentration from the beginning of the reaction (at t= 0) until a certain
reaction time t
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The equation ln[A]t= kt+ ln[A]0 is of the form y= m x+ b,
where a plot ofyversus xgives a straight line with the slope m and
the intercept b (the intercept may also be denoted by a c).
For the reaction A products,
the plot of ln[A]tversus tis a straight line, if the reaction is first order.Conversely, if the plot doesn't give a straight line,
then the reaction is not first order.
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14.3 Rate laws14.3 Rate laws
Forfirst order reactions a plot of the
natural logarithm of concentration versus
reaction time always gives a straight line
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SolutionFirst we calculate the natural logarithm of [N2O5], which gives the
following dataTime (s) ln[N
2
O5
] Time (s) ln[N2
O5
]
0 1.609 200 2.996
50 1.956 300 3.689
100 2.303 400 4.382
The plot is a straight line, which confirms that the reaction is first order in [N2
O5
]
(we could also use a calculator to do this analysis). In this case, the slope of the
line equals k. This gives us:
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Half-lifeoffirst-orderreactions
The half-life ( ) ofareactantisaconvenientwaytodescribehowfast
itreacts, particularlyforanoverall first-orderprocess. A reactant'shalf-life,
, is the time required for half of the reactant to disappear
For a first-order reaction, the half-life of the reactant can be obtained from:
by setting [A]tequal to half of the initial concentration [A]0.
=
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14.3 Rate laws14.3 Rate laws
Second-order rate laws
Assume B products is second order
The differential rate law is:
The integrated rate law would be:
rate = d[B]
dt= k[B]2
[B]0
1= kt +
[B]t
1
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14.3 Rate laws14.3 Rate laws
When a reaction is second order, a plot
of 1/[B]t versus t should yield a straight
line with a slope k
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14.3 Rate laws14.3 Rate laws
Zero-order rate laws
The differential rate law is:
rate = k[C]0 = k(1) = k
The integrated rate law is:
[C]t = kt+ [C]0
A plot of [C] versus time gives a straight
line of slope k
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14.3 Rate laws14.3 Rate laws
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
Collision theory
One of the simplest models to explain
reaction rates is collision theory
Collision theory says that the rate of a
reaction is proportional to the number of
effective collisions per second among
the reactant molecules
An effective collision is one that actually
gives product molecules
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
Concentration can influence the number
of effective collisions per second
As reactant concentrations increase, the
number of effective collisions increases
Not every collision between reactant
molecules actually results in a chemical
change
Only a very small fraction of all the
collisions can really lead to a net change
Why is this so?
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
Molecular orientation
When two reactant molecules collide they
must be oriented correctly for a reaction to
occur (see next slide)
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
Activation energy
The activation energy (Ea) is the
minimum energy required for a reactionto occur
The activation energy determines the
rate of a reaction
At a given temperature, only a certain
fraction of the collisions possess enough
energy to be effective and form products
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14.4 Theory of chemical14.4 Theory of chemical
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
The arrangement found on the top of the
potential energy hill is called the
activated complex or transition state
Once the transition state is reached, the
reaction proceeds to give products, with
the release of energy
A reaction intermediate corresponds toan energy minimum between two
transition states
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
Temperature effects
The rate of a reaction increases with
increasing temperature
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
The activation energy is related to the
rate constant by the Arrhenius equation
k= Ae-Ea/RT
k = rate constant
Ea = activation energy
R = universal gas constant
T = temperature (in Kelvin)
A = pre-exponential or frequency factor
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14.4 Theory of chemical14.4 Theory of chemical
kineticskinetics
The activation energy can also be
obtained from two rate components
measured at different temperatures
Most reactions obey the Arrheniusequation to a good approximation
lnk
2k1
=R
Ea
1
T2
1
T1
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14.5 Reaction mechanisms14.5 Reaction mechanisms
Most reactions do not occur in a
single step
The net overall reaction is the result ofa series of simple reactions
Each of these is called an elementary
process The entire set of elementary
processes is the reaction mechanism
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14.5 Reaction mechanisms14.5 Reaction mechanisms
The rate law of an elementary process
can be written from its chemical
equation
This rule only applies to elementary
processes
The overall rate law derived from t
hemechanism must agree with the
observed rate law for the overall
reaction
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14.5 Reaction mechanisms14.5 Reaction mechanisms
The rate-determining step
Consider the gaseous reaction:
2NO2Cl 2NO2 + Cl2
The actual mechanism of the reaction is
the following two-step sequence of
elementary processes:
NO2Cl NO2 + Cl (slow)
NO2Cl + Cl NO2 + Cl2 (fast)
The Cl atom formed is an intermediate
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14.5 Reaction mechanisms14.5 Reaction mechanisms
In any multistep mechanism, one step
is usually much slower than the others
The slow step in a mechanism iscalled the rate-determining step
The rate law for the rate-determining
step is directly related to the rate lawfor the overall reaction
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14.5 Reaction mechanisms14.5 Reaction mechanisms
The steady-state approximation
In complex reaction mechanisms, the
rate-determining step may vary whenreaction conditions are changed
In such cases, the steady-state
approximation is used
This method is the assumption that the
concentration of an intermediate remains
constant as the reaction proceeds
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14.5 Reaction mechanisms14.5 Reaction mechanisms
Consider the hypothetical reaction:
2AB + C2 A2B + C2B
This reaction may proceed via the
following mechanism:2AB A2B2
A2B2 + C2 A2B + C2B
Th
e intermediate in this mec
hanism isA2B2. Thus:
d[A2B2]
dt= 0
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14.5 Reaction mechanisms14.5 Reaction mechanisms
The steady-state approximation is
applied by setting:
rate of A2B2 production = rate of A2B2 consumption
All steps that are producing and
consuming intermediate A2B2 need to be
identified The rate law for each also has to be
written
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14.5 Reaction mechanisms14.5 Reaction mechanisms
(a) Rate of production of A2B2
In this mechanism, A2B2 is only produced
in the forward reaction
The rate law for this step is:
d[A2B2]
dt =k
1[AB]2
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14.5 Reaction mechanisms14.5 Reaction mechanisms
(b) Rate of consumption of A2B2
A2B2 is consumed in the reverse reaction
of the first step and in the second step
The rate laws for these steps are:
d[A2B2]
dt
= k-1[A2B2]
d[A2B2]
dt= k2[A2B2][C2]
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14.5 Reaction mechanisms14.5 Reaction mechanisms
(c) Application of steady-state condition
Under steady-state conditions, we have
for the intermediate A2B2:
k1[AB]2 = k-1[A2B2] + k2[A2B2][C2]
If we solve for the concentration of A2B2:
Now the rate law for the overall reaction
is written
[A2B2] =
k1[AB]2
k-1 + k2[C2]
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14.6 Catalysts14.6 Catalysts
A catalyst is a substance that
changes the rate of a chemical
reaction without being used up
The action caused by a catalyst is
called catalysis
Positive catalysts speed up reactions
Negative catalysts (inhibitors) slow
reactions down
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14.6 Catalysts14.6 Catalysts
The catalyst provides a path to the
products that has a rate-determining
step with a lower activation energy
(see next slide)
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14.6 Catalysts14.6 Catalysts
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14.6 Catalysts14.6 Catalysts
Homogeneous catalysts
Exist in the same phase as the reactants
An example is found in the lead chamber
process for manufacturing sulfuric acid.
The reaction is:
S + O2 SO2
SO2 + O2 SO3
SO3 + H2O H2SO4
Unassisted, the second reaction, oxidation of
SO2 to SO3, occurs slowly.
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14.6 Catalysts14.6 Catalysts
The SO2 is combined with a mixture ofNO,
NO2, air and steam
The NO2 serves as a catalyst by being an
oxygen carrier and providing a low-energy
path for the oxidation of SO2 to SO3
The NO is then reoxidised to NO2 by oxygen
The reaction is:
NO2 + SO2 NO +SO3
NO + O2 NO2
The NO2 is regenerated
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14.6 Catalysts14.6 Catalysts
Heterogeneous catalysts
Are a separate phase from the reactants
They are commonly solids
Usually functions by promoting a
reaction on its surface
An example is the Haber-Bosch process
3H2 +N
2
2N
H3 The reaction takes place on the surface of an
iron catalyst that contains traces of
aluminium and potassium oxides.
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14.6 Catalysts14.6 Catalysts
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14.6 Catalysts14.6 Catalysts
Enzyme kinetics
Enzymes consist of proteins, and are
some of the most powerful homogenous
catalysts
Enzyme catalysis can be represented by
a series of reactions
There are two different hypotheses forhow a substrate is bound into an enzyme
The lock-and-keyhypothesis assumes that
the substrate simple fits into the active site
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14.6 Catalysts14.6 Catalysts
The induced fithypothesis assumes that theenzyme molecule changes shape as the
substrate molecule comes close
The action of an enzyme can be described
by th
e Michaelis-Menten mec
hanism
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SummarySummary
Chemical kinetics is the study of the
rate of a reaction
Reaction rates are controlled by fivefactors
The differential rate law shows how
reaction rate depend on concentration
The integrated rate law shows how
concentration depends on time
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SummarySummary
According to collision theory, the rate
of a reaction depends on the number
of effective collisions per second
Colliding molecules must possess a
minimum kinetic energy, called the
activation energy, Ea
The Arrhenius equation relates how
changes in activation energy and
temperature affect a rate constant
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SummarySummary
In more complicated reaction
mechanisms the steady-state
approximation can be used to derive
the rate law
Catalysts are substances that change
a reaction rate but are not consumed
by the reaction Catalysts in living systems are called
enzymes