chemical kinetics ppt
TRANSCRIPT
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Chemical Kinetics
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We can use thermodynamics to tell if a reaction isproduct- or reactant-favored.
But this gives us no info on HOW FAST reaction goes from
reactants to products. KINETICS the study ofREACTION RATES and their
relation to the way the reaction proceeds, i.e., itsMECHANISM.
Chemical Kinetics
The reactionmechanismis our goal!
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What we have to study ?
Reaction Rates How we measure rates.
Rate LawsHow the rate depends on amounts
of reactants.
Integrated Rate LawsHow to calc amount left or time toreach a given amount.
Half-lifeHow long it takes to react 50% ofreactants.
Arrhenius Equation How rate constant changes with T.
MechanismsLink between rate and molecularscale processes.
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Rate of reaction is typically measured as the change in
concentration with time
This change may be a decrease or an increase
Likewise the concentration change may be of reactants or
products
Rate = ______________ = ______________
change in timechange in time
in [products] in [reactants]
Rate = concentration changetime change
Reaction Rates
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A B
rate = -D[A]
Dt
rate =D[B]
Dt
time
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Note the use of the negative sign
- rate is defined as a positive quantity
- rate of disappearance of a reactant is negative
t
[N2O5]Rate ofreaction=
12 t
[NO2]14 t
[O2]= =-
2N2O5(g) 4NO2(g) + O2(g)For Example
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1. Average reaction rate: a measure of the change in
concentration with time
2. Instantaneous rate: rate of change of concentration at any
particular instant during the reaction
3. Initial rate: instantaneous rate at t = 0- that is, when the reactants are first mixed
Rate may be expressed in
three main ways
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Average reaction rate
In this reaction, theconcentration of butylchloride, C4H9Cl, wasmeasured at various
times, t.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)Consider the following reaction
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The average rate of thereaction over eachinterval is the change inconcentration divided bythe change in time:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Average Rate, M/s
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Note that the average ratedecreases as the reactionproceeds.
This is because as thereaction goes forward,there are fewer collisionsbetween reactant
molecules.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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Instantaneous rate
A plot of concentrationvs. time for this reaction
yields a curve like this. The slope of a line
tangent to the curve atany point is the
instantaneous rate atthat time.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Consider the following reaction
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The reaction slows down
with time because theconcentration of thereactants decreases.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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In this reaction, the ratioof C4H9Cl to C4H9OH is1:1.
Thus, the rate ofdisappearance of C4H9Clis the same as the rate ofappearance of C4H9OH.
Rate =-[C4H9Cl]
t=[C4H9OH]
t
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = -
[CH4]t
= - [O2]t
12
= [H2O]t
12
=
[CO2]t
Example Problem
You should
careful aboutpositive or
negative sign !
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I. Rate is not Constant
Throughout Reaction
II. Reaction Rate isHigher with Higher
Concentration
(-) slope means concentrationis decreasing.
Curved Rather
than Straight LineTell Us That.....
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1. Rate = - A/ t =; + B/ t
Determined from stoichiometry
Uses both reactants & products
2. Rate Law; rate =k[A]x
Determined by experimentaldata- Stoichiometry of equationis irrelevant
Only reactants in rate law
Two MathematicalExpressions toDescribe Reaction
Rate
A B
Consider the reaction
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The Rate Law
The rate law expresses the relationship of the rate of a reactionto the rate constant and the concentrations of the reactantsraised to some powers.
aA + bB cC + dD
Rate = k[A]x[B]y
Where , k is the Rate Law Constant
Consider the following reaction
x and y are determinedexperimentally, and do notdepend on stoichiometriccoefficients from balancedequation
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reaction isxth order in A
reaction isyth order in B
reaction is (x +y)th order overall
Reaction Order
Reaction order tells how quickly rate will
increase when concentration increases
The Rate Law
The rate law expresses the relationship of the rate of a reactionto the rate constant and the concentrations of the reactantsraised to some powers.
aA + bB cC + dD
Rate = k[A]x[B]y
Consider the following reaction
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For a general reaction with rate law
we say the reaction is mth order in reactant 1 and nth order inreactant 2.
The overall order of reaction is m + n + .
A reaction can be zeroth order ifm, n, are zero.
Note the values of the exponents (orders) have to bedetermined experimentally. They are not simply related tostoichiometry.
nm2][reactant1]k[reactantRate
Exponents in the Rate Law
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F2(g) + 2ClO2(g) 2FClO2(g)
rate = k[F2][ClO2]1
For the above
reaction , rate isgiven as
Consider the following reaction
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F2(g) + 2ClO2(g) 2FClO2(g)
rate = k[F2]x[ClO2]
y
Double [F2] with [ClO2] constant
Rate doubles x= 1
Quadruple [ClO2
] with [F2
] constant
Rate quadruples
y= 1
rate = k[F2][ClO2]
Experimental Calculation
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Determine the rate law and calculate the rate constantfor the following reaction from the following data:S2O8
2- (aq) + 3I- (aq) 2SO42- (aq) + I3
- (aq)
Experiment [S2O82-] [I-]
Initial Rate
(M/s)1 0.08 0.034 2.2 x 10-4
2 0.08 0.017 1.1 x 10-4
3 0.16 0.017 2.2 x 10-4
Example Problem
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rate = k[S2O82-]x[I-]y
y = 1
x = 1rate = k[S2O8
2-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
k=rate
[S2O82-][I-]
=2.2 x 10-4M/s
(0.08M)(0.034M)= 0.08/Ms
Thinkabout thisproblem as
before
Solution
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Molecularity of a reaction
Order of a reaction
Half life
Two Important
terms inkinetics
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Molecularity
Molecularityis the number of molecules coming together toreact in an elementary step.
Elementary reactions are simple reactions (described bymolecularity)
(a) A Products UNI-molecular reaction
e.g. H2C
H2C CH2
CH3 CH
CH2
(b) A + A Products or A + B Products BI-molecular
e.g. CH3I + CH3CH2O- CH3OCH2CH3 + I
-
(c) 2A + B P or A + B + C P Ter-molecular
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Order of a reaction
The order of reaction is determined with respect to eachreactant in the reaction.The order of a reaction is the power to which the
concentration of a reactant is raised.This is the order with respect to one reactant only.
For the reaction
where there are more than one reactants, the order of thereaction is determined with A and then with B.
aA + bB cC + dD
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As mentioned earlier in equation the rate law which is
expressed as rate (r) = k [A]m
[B]n
The exponent 'm' denotes the order of the reaction withrespect to reactant A and the exponent 'n' denotes the orderof the reaction with respect to B.
The overall order of the reaction is then (m + n).
It should be noted that 'm' and 'n' do not necessarily represent
thestoichiometric coefficients 'a' and 'b' of the reaction.
Order of a reaction is experimentally determined.
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Differences between Molecularity and
Order of a Reaction
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Half-Life
Half-life is defined asthe time required forone-half of a reactant
to react.
Because [A] at t1/2 isone-half of the original
[A], [A]t = 0.5 [A]0.
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Types of order of a reaction
0th Order1st Order2nd Order
Pseudo First Order Reaction
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(Order = 0) have a constant rate.
This rate is independent of the conc of the reactants.
The rate law is: k, with k having the units of M/sec.
Zero-Order Reactions
0[A]kt[A]
IntegratedRate Law
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Zero OrderRate Law k
RateConstant
Slope = - kIntegratedRate Law
[A] = -kt + [A]0
Graph [A] versus t Life t =[A]0/2k
Concentration versus time profilefor a zero order reaction
Rate of reaction versus time for azero order reaction
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Ammonia (NH3) gas decomposes over platinum catalyst tonitrogen gas (N2) and hydrogen gas (H2).
The chemical reaction is as follows:
The reaction follows zero order kinetics. Therefore the ratelaw is rate r = k[NH3]
o
For a zero order reaction the concentration versus timeprofile is linear and the rate of reaction versus time hasthe profile.
Example Problem
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k
][][
Ak
dt
AdR
Rearranging gives:
kdtA
Ad
][
][
At time t = 0, [A] = [A]0And when t = t, [A] = [A]t
First-order reactions
Consider the following reaction
Then the rate of disappearance of A is:
A B
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Integrating:
tA
A
dtkA
Adt
0
][
][ 0 ][
][
ktAtA
A ]][ln[
][
][ 0
lnxdxx1
thatRecall
ktAAt
)]ln[](ln[ 0
ln[A]t = ln[A]0 -kt
Integrated form of the
1st order rate expression
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ln[A]t
t / s
-slope = -k
Intercept = ln[A]0kt
A
At
0][
][ln
-slope = -k
t / s
ln([A]t/[A]0)
Recall ln[A]t = ln[A]o - kt
[A]t = [A]0 e-kt
Antilog gives: [A]t
t / s
Intercept = [A]0
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The first order integrated rate law can be used to determine
the concentration of [A] at any time. It can be determined graphically
Where y = ln[A]
x = time m = -k b = ln[A] 0
Integrated Rate Law: First Order
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First-Order Processes Example
Consider the process in whichmethyl isonitrile is convertedto acetonitrile.
CH3NC CH3CN
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This data was collectedfor this reaction at198.9C.
CH3NC CH3CN
First-Order Processes Example
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When ln Pis plotted as a function of time, a
straight line results. Therefore,
The process is first-order.
kis the negative slope: 5.1 10-5 s1.
First-Order Processes Example
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Half-Life for first order reaction
For a first-order process, this becomes
0.5 [A]0
[A]0ln = kt1/2
ln 0.5 = kt1/2
0.693 = kt1/2
= t1/20.693
k
NOTE: For a first-order process, the half-life does not dependon [A]0.
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Second-order reactionsTwo possible cases:
Case I : A + A Products
OR 2A Products
Case II : A + B Products
2][
][
2
1Ak
dt
Adr
Rearranging gives:kdt
A
Ad2
][
][2
At time t = 0, [A] = [A]0
And when t = t, [A] = [A]t
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Integrating:
tA
Adtk
A
Adt
0
][
][ 2 2][
][
0
xx
xdxxdx
x
1
12
1 112
2
2
kt
A
tA
A
2
][
1][
][ 0
ORkt
A
tA
A
2
][
1][
][ 0
ktAA
t
2][
1
][
1
0
Integrated form of the
2nd
order rate expressiony = c + mx
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(1/[A]t) / dm3 mol-1
t / s
slope = 2k
Intercept = 1/[A]0
ktAA
t
2][
1
][
1
0
y = c + mx
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Half-Life for second order reaction
For a second-order process,
1
0.5 [A]0= kt1/2 +
1
[A]0
2
[A]0= kt1/2 +
1
[A]0
2
1[A]0
= kt1/21
[A]0=
= t1/21
k[A]0
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15 Chemical Kinetics
Variation of Reaction rates and Order
First order, rate= k [A]
k = rate, 0th
order
[A]
rate
2nd order, rate= k [A]2
The variation ofreaction rates as functions of concentration for
various order is interesting.
Mathematical analysis is an important scientific tool, worth noticing.
[A] = ___?
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For a reaction A + B C + D
The rate law for the reaction is expressed as rate = k [A]m [B]n
If the concentration of B is kept large, then [B] essentiallydoes not change over time and if the rate of reaction isdetermined as rate = k [A] then this reaction is a pseudo firstorder reaction.
Acidic hydrolysis of ethyl acetate is an example of a first orderreaction, where the acid concentration is kept in excess in thesolvent, water.
Pseudo First Order Reaction
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Methods to Determine
Order of a Reaction
For determination of the order of a reaction, following
methods are usually employed.
a) Graphical Methodb) Initial Rate Methodc) Isolation method
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Graphical Method of
Order DeterminationSteps:
First, the data of concentrations versus time is obtained by asuitable method.
Then, the data is plotted as concentration versus time.
From the resulting plot, the instantaneous rates aredetermined by drawing tangents to curve and thencalculating their slopes.
The reaction rate so obtained are plotted againstconcentrations raised to various powers.
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Decomposition rates of N2O5 g at different values of concentrationsof N2O5 as given in the data below:
Example Problem
Solution
These values are plotted as rate versus concentration and
rate versus (concentration)2
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Rate of decomposition of [N2O5]versus concentration of [N2O5]
Rate of decomposition of [N2O5]
versus concentration of [N2O5]2
It is seen that the rate is proportional to the concentration of N2O5 raised
to the power one. Therefore, dinitrogen pentoxide decomposes in a firstorder manner. The value of the rate constant k, as evaluated from thedata, is
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The rate is measured at the beginning of the reaction for severaldifferent initial concentrations of reactants.
[A]t
t / s
Initial rate
Follow reaction to ~ 10%
completion
Initial Rate Method
Important Method inchemical Kinetics
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RecallA + B P, Rate0 = k[A]0
a[B]0b
Taking logs
log Rate0 = log k+ a log [A]0 + b log[B]0
ym xc
** Keep [A]0 constant for varying values of [B]0 to find b
Log Ro
log[B]0
slope = b
Intercept = log k+ a log[A]0
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** Keep [B]0 constant for varying values of [A]0 to find a
from the slope of the graph, log R0 vs log [A]0
** Substitute values ofa, b, [A]0, [B]0 to find k.
However, in some cases, there may be no need to use
the plots as shown previously.
EXAMPLE
R1 = k[A]a[B]b
R2 = k[nA]a[B]b
Dividing R2 by R1
For these experiments, B is kept constant
while A is varied and R1 and R2 are known.
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ba
ba
BAk
BnAk
R
R
][][
][][
1
2
a
a
A
nA
][
][
a
aa
A
An
][
][
a
n
na
R
Rloglog
1
2
n
R
R
a
log
log1
2
(a) IfR2 = 2R1, and n=2, then a = 1, so 1st order with respect to A
(b) IfR2 = 4R1, and n=2, then a = 2, so 2nd order with respect to A
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Concluding:
if n=2,and Rate doubles 1st order
Rate increases by a factor of 4 2nd order
Rate increases by a factor of 9 3rd order
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This technique simplifies the rate law by making all the
reactants except one, in large excess.Therefore,
The dependence of the rate on each reactant can be foundbyisolating each reactant in turn and keeping all other
substances (reactants) in large excess.
Using as example: r= k[A]tm [B]t
n
Make B in excess, so [B]>>[A].
Hence, by the end of the reaction [B] would not have
changed that much, although all of A has been used up
And we can say, [B] [B]0
Isolation Method:
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r= k[A]tm , where k = k[B]0
n
Since A is the reactant that changes, then the rate
becomes dependent on A, and we can say
Created a false first-order (imitating first-order)
PSEUDO-FIRST-ORDER,
Logging both sides gives:
log r= log k + m log [A]ty = c + m x
A plot of log rvs log [A]t gives a straight line with slope = m,and intercept log k
where k is the pseudo-first-order rate constant
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If m = 1, the reaction is said to be pseudo-first-order
With the roles of A and B reversed, n can be found in asimilar manner
kcan then be evaluated using any data set along with the
known values ofm and n
Remember
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Estimate the orders andrate constant k from the results
observed for the reaction? What is the rate when [H2O2] = [I-]= [H+] = 1.0 M?
H2O2 + 3 I- + 2 H+ I3
- + 2 H2O
Exprmt [H2O2] [I-
] [H+
] Initial rate M s-1
1 0.010 0.010 0.0050 1.15e-6
2 0.020 0.010 0.0050 2.30e-6
3 0.010 0.020 0.0050 2.30e-6
4 0.010 0.010 0.0100 1.15e-6
Learn the strategy to determine the rate law from this example.
Figure out the answer without writing down anything.
Example Problem
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Estimate the orders from the results observed for the reaction
H2O2 + 3 I- + 2 H+ I3
- + 2 H2O
Exprmt [H2O2] [I-] [H+] Initial rate M s-1
1 0.010 0.010 0.0050 1.15e-6
2 0.020 0.010 0.0050 2.30e-6 1 for H2O23 0.010 0.020 0.0050 2.30e-6 1 for I-
4 0.010 0.010 0.0100 1.15e-6 0 for H+
1.15e-6 = k[H2O2]x [I-]y [H+]z
1.15e-6 k(0.010)x(0.010)y(0.0050)z exprmt 1 1 1----------- = ------------------------------------- ---- = ---
2.30e-6 k(0.020)x(0.010)y(0.0050)z exprmt 2 2 2
x = 1
( )x
Solution
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Other orders are determined in a similar way as shown
before.Now, lets findk and therateThurs, rate = 1.15e-6 = k(0.010)(0.010) from exprmt 1
k= 1.15e-6 M s-1 / (0.010)(0.010) M3 = 0.0115 M-1 s-1
And the rate law is therefore,d[H2O2] k
rate == 0.0115 [H2O2] [I-]
d t total order 2
Therate when [H2O2] = [I-] = [H+] = 1.0 M
The rate is the same as the rate constant k, when
concentrations of reactants are all unity (exactly 1), doesnt
matter what the orders are.
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N2O5 decomposes according to 1st
order kinetics, and 10% of itdecomposed in 30 s. Estimatek, tand percent decomposed in
500 s.
Example Problem
If the rate-law isknown, what arethe keyparameters?
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15 Chemical Kinetics
N2O5 decomposes according to 1st order kinetics, and 10% of it
decomposed in 30 s. Estimate k, tand percent decomposed in 500 s.
Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t= 30 s or
0.9 = 1.0 e k t apply [A]o = [A] ek t
ln 0.9 = ln 1.0k30 s
0.1054 = 0k* 30k = 0.00351 s 1
t=0.693 /k= 197 s apply k t = ln 2
[A] = 1.0 e 0.00351*500 = 0.173
Percent decomposed: 1.00.173 = 0.827 or 82.7 %
After 2 t (2*197=394 s), [A] = ()2 =, 75% decomposed.
After 3 t (3*197=591 s), [A] = ()3 =1/8, 87.5% decomposed.
Apply integrated rate law to solve problems
Solution
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What is the half-life of N2O5 if it decomposes with a
rate constant of 5.7 x 10-4 s-1?
13.3
tln2
k=
0.693
5.7 x 10-4 s-1= = 1200 s = 20 minutes
How do you know decomposition is first order?
units ofk(s-1)
Example Problem
Solution
Solution
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The reaction 2A B is first order in A with a rate constant
of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decreasefrom 0.88Mto 0.14M?
ln[A] = ln[A]0 - kt
kt = ln[A]0 ln[A]
t =ln[A]0 ln[A]
k= 66 s
[A]0
= 0.88M
[A] = 0.14M
ln[A]0
[A]
k=
ln0.88M
0.14M
2.8 x 10-2 s-1=
Example Problem
Solution
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Solution
[A] / [A]0 = fraction remaining
when t = t1/2 then fraction remaining = _________Therefore, ln (1/2) = - k t1/2
- 0.693 = - k t1/2
t1/2 = 0.693 / kSo, for sugar,
t1/2 = 0.693 / k = 2100 sec = 35 min
Rate = k[sugar] and k = 3.3 x 10
-4
sec
-1
. What is the half-life ofthis reaction?
Example Problem
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Solution
2 hr and 20 min = 4 half-livesHalf-life Time Elapsed Mass Left
1st 35 min 2.50 g
2nd 70 1.25 g3rd 105 0.625 g
4th 140 0.313 g
Rate = k[sugar] and k = 3.3 x 10-4
sec-1
. Half-life is 35min. Start with 5.00 g sugar. How much is left after 2hr and 20 min (140 min)?
Example Problem
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Radioactive decay is a first order process.Tritium f electron + helium3H 0-1e
3He
t1/2 = 12.3 years
If you have 1.50 mg of tritium, how much is left after
49.2 years?
Example Problem
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Solution
ln [A] / [A]0 = -kt
[A] = ? [A]0 = 1.50 mg t = 49.2 y
Need k, so we calc k from: k = 0.693 / t1/2
Obtain k = 0.0564 y-1
Now ln [A] / [A]0 = -kt = - (0.0564 y-1)(49.2 y)
= - 2.77Take antilog: [A] / [A]0 = e
-2.77 = 0.0627
0.0627 = fraction remaining
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
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Solution
[A] / [A]0 = 0.0627
0.0627 is the fraction remaining!
Because [A]0 = 1.50 mg, [A] = 0.094 mg
But notice that 49.2 y = 4.00 half-lives
1.50 mg f 0.750 mg after 1 half-life
f0.375 mg after 2f0.188 mg after 3
f0.094 mg after 4
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
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Determine the rate law for the following reactiongiven the data below.
H2O2(aq) + 3 I-(aq) + 2H+(aq) I3
-(aq) + H2O (l)
[H2O2] [I-] [H+] InitialExpt # (M) (M) (M) Rate (M /s)
1 0.010 0.010 0.00050 1.15 x 10-6
2 0.020 0.010 0.00050 2.30 x 10-6
3 0.010 0.020 0.00050 2.30 x 10-64 0.010 0.010 0.00100 1.15 x 10-6
Example Problem
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Rate = k[H2O2]m[I-]n[H+]pCompare Expts. 1 and 2:
Rate 2 = 2.30 x 10-6 M/s = 2.00
Rate 1 = 1.15 x 10-6 M/s
Rate 2 = k (0.020 M)m(0.010 M)n(.00050 M)p = 2.00
Rate 1 k (0.010 M)
m
(0.010 M)
n
(.00050 M)
p
(0.020)m = 2.0m = 2.00
(0.010)m
2.0m = 2.00 only if m = 1
Rate = k[H2O2]1[I-]n[H+]p
Solution
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Rate = k[H2O2]1[I-]1[H+]pCompare Expts. 1 and 4:
Rate 4 = 1.15 x 10-6 M/s = 1.00
Rate 1 = 1.15 x 10-6 M/s
Rate 4 = k (0.010 M)1(0.010 M)1(.00100 M)p = 1.00
Rate 1 k (0.010 M)
1
(0.010 M)
1
(.00050 M)
p
(0.00100)p = 2.0p = 1.00
(0.00050)p
2p = 1.00 only if p = 0
Rate = k[H2O2]1[I-]1[H+]0
Rate = k[H2
O2
] [I-]
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The following data was used to determine the ratelaw for the reaction:
A + B C
Calculate the value of the rate constant if the rate lawis:
Expt # [A] (M) [B] (M) Initial rate (M /s)
1 0.100 0.100 4.0 x 10-5
2 0.100 0.200 8.0 x 10-5
3 0.200 0.100 16.0 x 10-5
Rate = k [A]2[B]
Example Problem
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Select a set of conditions:Experiment 1: Rate = 4.0 x 10-5 M /s
[A] = 0.100 M
[B] = 0.100 M
Substitute data into the rate law:
4.0 x 10
-5
M = k (0.100 M)
2
(0.100 M)s
Solve for k
k = 4.0 x 10-5 M/s = 0.040 = 0.040 M-2s-1
(0.100 M)2 (0.100M) M2 . s
Rate = k [A]2[B]
Solution
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A certain pesticide decomposes in water via a first orderreaction with a rate constant of 1.45 yr-1. What will theconcentration of the pesticide be after 0.50 years for asolution whose initial concentration was 5.0 x 10-4 g/mL?
ln[A]t= -kt + ln[A]
0
Given: k = 1.45 yr-1
t = 0.50 yr[A]o = 5.0 x 10
-4 g/mLFind: [A]t=0.5 yr
Example Problem
Solution
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ln[A]0.5 yr = -kt + ln[A]0
ln[A]0.5 yr = - (1.45 x 0.50 yr) + ln (5.0 x 10-4)
yr
ln[A]0.5 yr = - 0.725 + -7.601 = - 8.326
To find the value of [A]0.5 yr
, use the inverse naturallogarithm, ex.
[A]0.5 yr = e-8.326 = 2.42 x 10-4 = 2.4 x 10-4 g/mL
l bl
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A certain pesticide has a half life of 0.500 yr. If the initialconcentration of a solution of the pesticide is 2.5 x 10-4g/mL, what will its concentration be after 1.5 years?
Given: [A]0 = 2.5 x 10-4 g/mL
t1/2 = 0.500 yr
t = 1.5 yrFind: [A]1. 5 yr
First, find the
value for k.
Example Problem
Solution
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Find the rate constant, k:k = 0.693/t1/2
k = 0.693 = 1.386 yr-1
0.500 yr
Substitute data into
ln[A]1.5 yr = - (1.386 x 1.5 yr) + ln(2.5 x 10-4)yr
= - 10.373
ln[A]t = -kt + ln[A]0
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Solve for [A]1. 5 yr using the inverse natural logarithm:
[A]1. 5 yr = e-10.373 = 3.1 x 10-5 g/mL
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Concentrations
physical state of reactants and
products
Temperature
Catalysts
Factors Affecting Rates
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Factors That Affect Reaction Rates
Physical State of the Reactants
In order to react, molecules must come in contactwith each other.
The more homogeneous the mixture of reactants,the faster the molecules can react.
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Concentration of Reactants
As the concentration of reactants increases, so doesthe likelihood that reactant molecules will collide.
Factors That Affect Reaction Rates
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Temperature
At higher temperatures, reactant molecules havemore kinetic energy, move faster, and collide more
often and with greater energy.
Factors That Affect Reaction Rates
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Presence of a Catalyst
Catalysts speed up reactions by changing themechanism of the reaction.
Catalysts are not consumed during the course ofthe reaction.
Factors That Affect Reaction Rates
T t d R t
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The Collision Model Most reactions speed up as temperature increases. (E.g. food
spoils when not refrigerated.)
When two light sticks are placed in water: one at room temperature
and one in ice, the one at room temperature is brighter than the one
in ice.
The chemical reaction responsible for chemiluminescence is
dependent on temperature: the higher the temperature, the faster
the reaction and the brighter the light.
Temperature and Rate
T t d R t
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The CollisionModel
As temperature
increases, the rateincreases.
Temperature and Rate
T t d R t
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The Collision Model
Since the rate law has no temperature term in it, the rate
constant must depend on temperature.
Consider the first order reaction CH3NC CH3CN. As temperature increases from 190 C to 250 C the rate
constant increases from 2.52 10-5 s-1 to 3.16 10-3 s-1.
The temperature effect is quite dramatic. Why?
Observations: rates of reactions are affected by
concentration and temperature.
Temperature and Rate
T t d R t
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The Collision Model
Goal: develop a model that explains why rates of
reactions increase as concentration and temperature
increases. The collision model: in order for molecules to react they
must collide.
The greater the number of collisions the faster the rate.
The more molecules present, the greater the probability
of collision and the faster the rate.
Temperature and Rate
T t d R t
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The Collision Model
The higher the temperature, the more energy available to
the molecules and the faster the rate.
Complication: not all collisions lead to products. In fact,only a small fraction of collisions lead to product.
The Orientation Factor
In order for reaction to occur the reactant molecules must
collide in the correct orientation and with enough energy
to form products.
Temperature and Rate
T t d R t
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The Orientation Factor
Consider:
Cl + NOCl NO + Cl2
There are two possible ways that Cl atoms and NOClmolecules can collide; one is effective and one is not.
Temperature and Rate
T t d R t
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The Orientation Factor
Temperature and Rate
T t d R t
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Activation Energy Arrhenius: molecules must posses a minimum amount of
energy to react. Why?
In order to form products, bonds must be broken in thereactants.
Bond breakage requires energy.
Activation energy,Ea, is the minimum energy required to
initiate a chemical reaction.
Temperature and Rate
Temperat re and Rate
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Activation Energy
Consider the rearrangement of methyl isonitrile:
In H3C-NC, the C-NC bond bends until the C-N bond breaks
and the NC portion is perpendicular to the H3C portion. This
structure is called the activated complex or transition state.
The energy required for the above twist and break is theactivation energy,Ea.
Once the C-N bond is broken, the NC portion can continue to
rotate forming a C-CN bond.
H3C N C
C
NH3C H3C C N
Temperature and Rate
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Temperature and Rate
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Activation Energy
The change in energy for the reaction is the difference in
energy between CH3NC and CH3CN.
The activation energy is the difference in energy betweenreactants, CH3NC and transition state.
The rate depends onEa.
Notice that if a forward reaction is exothermic (CH3
NC
CH3CN), then the reverse reaction is endothermic
(CH3CN CH3NC).
Temperature and Rate
Temperature and Rate
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Activation Energy
How does a methyl isonitrile molecule gain enough
energy to overcome the activation energy barrier?
From kinetic molecular theory, we know that astemperature increases, the total kinetic energy increases.
We can show the fraction of molecules,f, with energy
equal to or greater thanEa is
whereR is the gas constant (8.314 J/molK).
RTEa
ef
Temperature and Rate
Temperature and Rate
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Activation Energy
Temperature and Rate
Example Problem
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For each reaction energy diagram below, mark the
location of the reactants, products and transition state.Identify the magnitude of Ea and DHrxn. Is each reactionendothermic or exothermic?
Example Problem
Temperature and Rate
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The Arrhenius Equation
Arrhenius discovered most reaction-rate data obeyed
the Arrhenius equation:
kis the rate constant, Eais the activation energy,R is the gas
constant (8.314 J/K-mol) and Tis the temperature in K.
A is called the frequency factor.
A is a measure of the probability of a favorable collision.
BothA andEa are specific to a given reaction.
RTEa
Aek
Temperature and Rate
Temperature and Rate
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Determining the Activation Energy If we have a lot of data, we can determineEa andA
graphically by rearranging the Arrhenius equation:
From the above equation, a plot of ln kversus 1/Twill
have slope ofEa
/R and intercept of ln A.
ARTEk a lnln
Temperature and Rate
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Can be arranged in the form of a straight line
ln k = (-Ea/R)(1/T) + ln A
Plot ln k vs. 1/T slope = -Ea/R
Ea/RTdecreases
-Ea/RTincreases
e-Ea/RT
increasesk
increasesREACTIONSPEEDS UP
If Tincreases
Temperature and Rate
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Temperature and Rate
Temperature and Rate
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Determining the Activation Energy If we do not have a lot of data, then we recognize
122
1
2121
22
11
11ln
lnlnlnln
lnlnandlnln
TTR
E
k
k
ART
EA
RT
Ekk
A
RT
EkA
RT
Ek
a
aa
aa
Temperature and Rate
Example Problem
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Calculate the activation energy for the rearrangement ofmethyl isonitrile to acetonitrile using the following data.
You can solve this in oneof several ways:
Graph ln k vs. 1/T and
determine the slope;
Find ln k and 1/T and
determine the slope using
two well spaced points
using D(ln k)/D(1/T);
Example Problem
Solution
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Personally, prefer the non-graphical method:
Use two points that are well separated and convert T toKelvin (K = oC + 273.15)
Temp
(K) 1/T (K-1) k
462.9 2.160 x 10-3
2.52 x 10-5
524.4 1.907 x 10-3
3.16 x 10-3
122
1 11lnTTR
E
k
ka
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Plug the values of k and T into the equation. Be careful toput k1 and T1 in the appropriate order:
ln 2.52 x 10-5 = Ea 1 - __1__
3.16 x 10-3 (8.314 J/mol K) 524.4 K 462.9 K
Solve for Ea.
122
1 11lnTTR
E
k
ka
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ln 2.52 x 10-5 = Ea 1 - __1__3.16 x 10-3 (8.314 J/mol K) 524.4 K 462.9 K
Solve for Ea
.
- 4.8315 = (- 3.047 x 10-5 mol/J) Ea
Ea = 1.59 x 105 J/mol
R ti M h i
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The balanced chemical equation provides information
about the beginning and end of reaction.
The reaction mechanism gives the path of the reaction. Mechanisms provide a very detailed picture of which
bonds are broken and formed during the course of a
reaction.
Elementary Steps
Elementary step: any process that occurs in a single step.
Reaction Mechanisms
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Elementary Steps Molecularity: the number of molecules present in an
elementary step.
Unimolecular: one molecule in the elementary step, Bimolecular: two molecules in the elementary step, and
Termolecular: three molecules in the elementary step.
It is not common to see termolecular processes
(statistically improbable).
Reaction Mechanisms
R ti M h i
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Multistep Mechanisms Some reaction proceed through more than one step:
NO2(g) + NO2(g) NO3(g) + NO(g)
NO3(g) + CO(g) NO2(g) + CO2(g) Notice that if we add the above steps, we get the overall
reaction:
NO2(g) + CO(g)
NO(g) + CO2(g)
Reaction Mechanisms
Reaction Mechanisms
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Multistep Mechanisms If a reaction proceeds via several elementary steps, then
the elementary steps must add to give the balanced
chemical equation.
Intermediate: a species which appears in an elementary
step which is not a reactant or product.
Reaction Mechanisms
Reaction Mechanisms
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Rate Laws for Elementary Steps The rate law of an elementary step is determined by its
molecularity:
Unimolecular processes are first order, Bimolecular processes are second order, and
Termolecular processes are third order.
Rate Laws for Multistep Mechanisms
Rate-determining step: is the slowest of the elementary
steps.
Reaction Mechanisms
Reaction Mechanisms
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Rate Laws for Elementary Steps
Reaction Mechanisms
Reaction Mechanisms
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Rate Laws for Multistep Mechanisms Therefore, the rate-determining step governs the overall
rate law for the reaction.
Mechanisms with an Initial Fast Step It is possible for an intermediate to be a reactant.
Consider
2NO(g) + Br2(g)
2NOBr(g)
Reaction Mechanisms
Reaction Mechanisms
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Mechanisms with an Initial Fast Step2NO(g) + Br2(g) 2NOBr(g)
The experimentally determined rate law is
Rate = k[NO]2[Br2] Consider the following mechanism
NO(g) + Br2(g) NOBr2(g)k1
k-1
NOBr2(g) + NO(g) 2NOBr(g)k2
Step 1:
Step 2:
(fast)
(slow)
Reaction Mechanisms
Reaction Mechanisms
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Mechanisms with an Initial Fast Step The rate law is (based on Step 2):
Rate = k2[NOBr2][NO]
The rate law should not depend on the concentration ofan intermediate (intermediates are usually unstable).
Assume NOBr2 is unstable, so we express the
concentration of NOBr2 in terms of NOBr and Br2
assuming there is an equilibrium in step 1 we have
]NO][Br[]NOBr[ 2
1
12
k
k
Reaction Mechanisms
Reaction Mechanisms
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Mechanisms with an Initial Fast Step By definition of equilibrium:
Therefore, the overall rate law becomes
Note the final rate law is consistent with theexperimentally observed rate law.
]NOBr[]NO][Br[ 2121 kk
][BrNO][NO][]NO][Br[Rate 22
1
122
1
12
k
kk
k
kk
Reaction Mechanisms
Deriving a Rate Law From a Mechanism
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Deriving a Rate Law From a Mechanism
The decomposition of H2O2 in the presence of Ifollow this
mechanism,
i H2O2 + I k1 H2O + IO
slow
ii H2O2 + IO k2 H2O + O2 + I
fast
What is the rate law? Energy
Eai
Eaii
reaction
Deriving a Rate Law From a Mechanism
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The decomposition of H2O2 in the presence of Ifollow this
mechanism,
i H2O2 + I k1 H2O + IO
slow
ii H2O2 + IO k2 H2O + O2 + I
fast
What is the rate law?
Solution
The slow step determines the rate, and the rate law is:
rate = k1 [H2O2] [I]
Since both [H2O2] and [I]are measurable in the system, this
is the rate law.
g
Deriving a rate law from a mechanism - 2
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Derive the rate law for the reaction, H2 + Br2 = 2 HBr,
from the proposed mechanism:
i Br2 2 Br fast equilibrium (k1, k-1)
ii H2 + Br k2 HBr + H slow
iii H + Br k3 HBr fast
Solution:The fast equilibrium condition simply says that
k1 [Br2] = k-1[Br]2
and [Br] = (k1/k-1 [Br2])
The slow step determines the rate law,
rate = k2 [H2] [Br] Br is an intermediate= k2 [H2] (k1/k-1 [Br2])
= k[H2] [Br2]; k = k2 (k1/k-1)
M- s -1
total order 1.5
explain
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A catalyst changes the rate of a chemical reaction.
There are two types of catalyst:
homogeneous, and heterogeneous.
Chlorine atoms are catalysts for the destruction of ozone.
Homogeneous Catalysis
The catalyst and reaction is in one phase.
Catalysis
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Homogeneous Catalysis Hydrogen peroxide decomposes very slowly:
2H2O2(aq) 2H2O(l) + O2(g)
In the presence of the bromide ion, the decompositionoccurs rapidly:
2Br-(aq) + H2O2(aq) + 2H+(aq) Br2(aq) + 2H2O(l).
Br2(aq) is brown.
Br2(aq) + H2O2(aq) 2Br-(aq) + 2H+(aq) + O2(g).
Br- is a catalyst because it can be recovered at the end of the
reaction.
Catalysis
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Homogeneous Catalysis Generally, catalysts operate by lowering the activation
energy for a reaction.
Catalysis
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Catalysis
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Homogeneous Catalysis Catalysts can operate by increasing the number of
effective collisions.
That is, from the Arrhenius equation: catalysts increase kbe increasingA or decreasingEa.
A catalyst may add intermediates to the reaction.
Example: In the presence of Br-, Br2(aq) is generated as
an intermediate in the decomposition of H2O2.
Catalysis
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Homogeneous Catalysis When a catalyst adds an intermediate, the activation
energies for both steps must be lower than the activation
energy for the uncatalyzed reaction. The catalyst is in a
different phase than the reactants and products.
Heterogeneous Catalysis
Typical example: solid catalyst, gaseous reactants and
products (catalytic converters in cars).
Most industrial catalysts are heterogeneous.
Catalysis
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Heterogeneous Catalysis First step is adsorption (the binding of reactant molecules
to the catalyst surface).
Adsorbed species (atoms or ions) are very reactive.
Molecules are adsorbed onto active sites on the catalyst
surface.
Catalysis
Catal sis
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Catalysis
Catalysis
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Heterogeneous Catalysis Consider the hydrogenation of ethylene:
C2H4(g) + H2(g) C2H6(g), H= -136 kJ/mol.
The reaction is slow in the absence of a catalyst.
In the presence of a metal catalyst (Ni, Pt or Pd) the reaction
occurs quickly at room temperature.
First the ethylene and hydrogen molecules are adsorbed onto
active sites on the metal surface. The H-H bond breaks and the H atoms migrate about the metal
surface.
Catalysis
Catalysis
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Heterogeneous Catalysis When an H atom collides with an ethylene molecule on the
surface, the C-C bond breaks and a C-H bond forms.
When C2H6 forms it desorbs from the surface.
When ethylene and hydrogen are adsorbed onto a surface, less
energy is required to break the bonds and the activation energy
for the reaction is lowered.
Catalysis