kinetics
DESCRIPTION
KINETICS. The speed of a reaction. Kinetics. The study of reaction rates. Spontaneous reactions are reactions that will happen - but we can’t tell how fast. Diamond will spontaneously turn to graphite – eventually. Reaction mechanism- the steps by which a reaction takes place. Reaction Rate. - PowerPoint PPT PresentationTRANSCRIPT
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KINETICSThe speed of a reaction
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Kinetics• The study of reaction rates.
• Spontaneous reactions are reactions that will happen - but we can’t tell how fast.
• Diamond will spontaneously turn to graphite – eventually.
• Reaction mechanism- the steps by which a reaction takes place.
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Reaction Rate
• Rate = Conc. of A at t2 -Conc. of A at t1t2- t1
• Rate =[A]t
• Change in concentration per unit time
• For this reaction
• N2 + 3H2 2NH3
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• As the reaction progresses the concentration of H2 goes down
Concentration
Time
[H[H22]]
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• As the reaction progresses the concentration of N2 goes down 1/3 as fast
Concentration
Time
[H[H22]]
[N[N22]]
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• As the reaction progresses the concentration of NH3 goes up.
Concentration
Time
[H[H22]]
[N[N22]]
[NH[NH33]]
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Calculating Rates• Average rates are taken over long
intervals
• Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time
• Derivative.
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• Average slope method
Concentration
Time
[H[H22]]
tt
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• Instantaneous slope method.
Concentration
Time
[H[H22]]
tt
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Defining Rate
• We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.
• In our example: N2 + 3H2 2NH3
• -[N2] = -3 [H2] = 2 [NH3] t t t
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Rate Laws
• Reactions are reversible.
• As products accumulate they can begin to turn back into reactants.
• Early on the rate will depend on only the amount of reactants present.
• We want to measure the reactants as soon as they are mixed.
• This is called the Initial rate method and time = 0.
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• Two key points
• The concentration of the products do not appear in the rate law because this is an initial rate.
• The order must be determined experimentally because it can’t be obtained from the equation
Rate LawsRate Laws
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• You will find that the rate will only depend on the concentration of the reactants.
• Rate = k[NO2]n
• This is called a rate law expression.• k is called the rate constant.• n is the order of the reactant - usually a
positive integer.
2 NO2 2 NO + O2
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• The rate of appearance of O2 can be said to be.
• Rate' = [O2] = k'[NO2]2
t• Because there are 2 NO2 for each O2
• Rate = 2 x Rate'
• So k[NO2]n = 2 x k'[NO2]
n
• So k = 2 x k'
2 NO2 2 NO + O2
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Types of Rate Laws
• Differential Rate law - describes how rate depends on concentration.
• Integrated Rate Law - Describes how concentration depends on time.
• For each type of differential rate law there is an integrated rate law and vice versa.
• Rate laws can help us better understand reaction mechanisms.
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Determining Rate Laws
• The first step is to determine the form of the rate law (especially its order).
• Must be determined from experimental data which is usually given.
• For this reaction 2 N2O5 (aq) 4NO2 (aq) + O2(g)
• The reverse reaction won’t play a role
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[N[N22OO55] (mol/L) Time (s)] (mol/L) Time (s)
1.001.00 00
0.880.88 200200
0.780.78 400400
0.690.69 600600
0.610.61 800800
0.540.54 10001000
0.480.48 12001200
0.430.43 14001400
0.380.38 16001600
0.340.34 18001800
0.300.30 20002000
Now graph the data
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0
0.2
0.4
0.6
0.8
1
1.2
0 200
400
600
800
1000
1200
1400
1600
1800
2000• To find rate we have to find the
slope at two points• We will use the tangent method.
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0
0.2
0.4
0.6
0.8
1
1.2
0 200
400
600
800
1000
1200
1400
1600
1800
2000
At .90 M the rate is (.98 - .76) = 0.22 =- 5.5x 10 -4 (0-400) -400
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0
0.2
0.4
0.6
0.8
1
1.2
0 200
400
600
800
1000
1200
1400
1600
1800
2000
At .40 M the rate is (.52 - .31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800
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• Since the rate at twice the concentration is twice as fast the rate law must be..
• Rate = -[N2O5] = k[N2O5]1 = k[N2O5] t
• We say this reaction is first order in N2O5
• The only way to determine order is to run the experiment.
• So the Rate Law is Rate = k[N2O5]
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The Method of Initial Rates• This method requires that a reaction be run
several times.
• The initial concentrations of the reactants are varied.
• The reaction rate is measured just after the reactants are mixed.
• Eliminates the effect of the reverse reaction.
• In our problems, this isn’t a concern.
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An Example of a Rate Law• For the reaction
BrO3- + 5 Br- + 6H+ 3Br2 + 3 H2O
• The general form of the Rate Law is:
• Rate = k[BrO3-]n[Br-]m[H+]p
• We use experimental data to determine the values of n, m, and p
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Initial concentrations (M)
Rate (M/s)
BrOBrO33-- BrBr--
HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--
44
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--
33
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--
33
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--
33
Now we have to see how the rate changes with concentration
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Remember our Formula ! - Δ Concentration order = Δ Rate
In words, the change in concentration raised to the order of the exponent = the change in the rate.
- Fortunately, most of the time, these can be solved on inspection. Eg. If the concentration is tripled and the rate is tripled, then 3x = 3 so x = 1.
HOWEVER…
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Sometimes more drastic measures are needed.
• Namely the law of logarithms.
• Δ Concentration order = Δ Rate becomes
Order x (Log (conc.)) = Log (rate)
• It is simple and works every time.
• 23 = 8 3 log 2 = log 8 3 x .301 = .903
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Integrated Rate Law• Expresses the reaction concentration as a
function of time.
• Form of the equation depends on the order of the rate law (differential).
• Changes Rate = [A]n t
• We will only work with n=0, 1, and 2
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First Order
• For the reaction 2N2O5 4NO2 + O2
• We found the Rate = k[N2O5]1
• If concentration doubles rate doubles.
• If we integrate this equation with respect to time we get the Integrated Rate Law
• ln[N2O5] = - kt + ln[N2O5]0
• ln is the natural log
• [N2O5]0 is the initial concentration.
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• General form Rate = [A] / t = k[A]
• ln[A] = - kt + ln[A]0
• In the form y = mx + b
• y = ln[A] m = - k
• x = t b = ln[A]0
• A graph of ln[A] vs time is a straight line.
First Order
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• By getting the straight line you can prove it is first order.
• Often expressed in a ratio.
First Order Rates
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• By getting the straight line you can prove it is first order.
• Often expressed in a ratio
First Order
lnA
A = kt0
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Half Life• The time required to reach half the
original concentration.
• If the reaction is first order
• [A] = [A]0/2 when t = t1/2
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Half Life• The time required to reach half the
original concentration.
• If the reaction is first order
• [A] = [A]0/2 when t = t1/2
ln
A
A = kt0
01 2
2
• ln(2) = kt1/2
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Half Life• t1/2 = 0.693/k
• The time to reach half the original concentration does not depend on the starting concentration.
• An easy way to find k.
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Second Order
• Rate = -[A] / t = k[A]2
• integrated rate law
• 1/[A] = kt + 1/[A]0
• y= 1/[A] m = k
• x= t b = 1/[A]0
• A straight line if 1/[A] vs t is graphed
• Knowing k and [A]0 you can calculate [A] at any time time.
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Second Order Half Life• [A] = [A]0 /2 at t = t1/2
1
20
2[ ]A = kt +
1
[A]10
22[ [A]
- 1
A] = kt
0 01
tk[A]1 =
1
02
1
[A] = kt
01 2
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Zero Order Rate Law
• Rate = k[A]0 = k
• Rate does not change with concentration.
• Integrated [A] = -kt + [A]0
• When [A] = [A]0 /2 t = t1/2
• t1/2 = [A]0 / 2k
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• Most often when reaction happens on a surface because the surface area stays constant.
• Also applies to enzyme chemistry.
• The material is said to be at the zeroeth order.
Zero Order Rate Law
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Time
Concentration
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Time
Concentration
A]/t
t
k =
A]
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More Complicated Reactions
• BrOBrO33-- + 5 Br + 5 Br -- + 6H + 6H++ 3Br 3Br22 + 3 H + 3 H22OO
• For this reaction we found the rate law to For this reaction we found the rate law to be:be:
• Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22
• To investigate this reaction rate we need To investigate this reaction rate we need to control the conditions. to control the conditions.
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Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22
• We set up the experiment so that two of the reactants are in large excess.
• [BrO3-]0= 1.0 x 10-3 M
• [Br-]0 = 1.0 M
• [H+]0 = 1.0 M
• As the reaction proceeds [BrO3-]
changes noticably • [Br-] and [H+] don’t
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• This rate law can be rewritten
• Rate = k[BrORate = k[BrO33--][Br][Br--]]00[H[H
++]]0022
• Rate = k[BrRate = k[Br--]]00[H[H++]]00
22[BrO[BrO33--]]
• Rate = k’[BrORate = k’[BrO33--]]
• This is called a pseudo first order rate This is called a pseudo first order rate law.law.
• k =k = k’ k’
[Br[Br--]]00[H[H++]]00
22
Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22
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Summary of Rate Laws• Know your integrated rate laws.
• Know how to find the order for the reaction
• Know how to find k after you determine the orders.
• As always, read the question carefully to make sure you are using all the given information.
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Reaction Mechanisms
• The series of steps that actually occur in a chemical reaction.
• Kinetics can tell us something about the mechanism.
• A balanced equation does not tell us how the reactants become products.
• We will always want to find the rate determining step, which will usually be the slowest step in the mechanism.
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• 2NO2 + F2 2NO2F • Rate = k[NO2][F2]• The proposed mechanism is• NO2 + F2 NO2F + F (slow)• F + NO2 NO2F (fast) • F is called an intermediate It is formed
then consumed in the reaction, so it isn’t present in the given equation.
Reaction Mechanisms
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• Each of the two reactions is called an elementary step .
• The rate for a reaction can be written from its molecularity .
• Molecularity is the number of pieces that must come together.
Reaction Mechanisms
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• Unimolecular step requires one molecule so the rate is first order.
• Bimolecular step - requires two molecules so the rate is second order.
• Termolecular step - requires three molecules so the rate is third order.
• Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.
Molecularity
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• A Products Rate = k[A]
• A + A Products Rate= k[A]2
• 2A Products Rate= k[A]2
• A + B ProductsRate= k[A][B]
• A + A + B Products Rate= k[A]2[B]
• 2A + B Products Rate= k[A]2[B]
• A + B + C Products Rate= k[A][B][C]
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How to get rid of intermediates
• Use the reactions that form them
• If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry.
• If it is formed by a reversible reaction set the rates equal to each other.
• A picture is worth 1,000 words here, so let’s check out the next slide.
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Formed in reversible reactions• 2 NO + O2 2 NO2
• Mechanism
• 2 NO N2O2 (fast) Step 1
• N2O2 + O2 2 NO2 (slow) Step 2
• rate = k2[N2O2][O2]
• k1[NO]2 = k-1[N2O2]
• rate = k2 (k1/ k-1)[NO]2[O2]=k[NO]2[O2]
• NOTICE: Step 1 is in equilibrium so the rate of formation of the N2O2 is equal to the rate of formation of the 2 NO.
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Formed in fast reactions• 2 IBr I2+ Br2
• Mechanism• IBr I + Br (fast)
• IBr + Br I + Br2 (slow)
• I + I I2 (fast)
• What is the rate determining step ?
• Rate = k[IBr][Br] but [Br]= [IBr]
• Rate = k[IBr][IBr] = k[IBr]2
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Collision theory• Molecules must collide to react.
• Concentration affects rates because collisions are more likely.
• Must collide hard enough.
• Temperature and rate are related.
• Only a small number of collisions produce reactions.
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Potential Energy
Reaction Coordinate
Reactants
Products
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Potential Energy
Reaction Coordinate
Reactants
Products
Activation Energy Ea
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Potential Energy
Reaction Coordinate
Reactants
Products
Activated complex
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Potential Energy
Reaction Coordinate
Reactants
ProductsE}
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Potential Energy
Reaction Coordinate
2BrNO
2NO + Br
Br---NO
Br---NO
2
Transition State
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Terms• Activation energy - the minimum energy
needed to make a reaction happen.
• Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.
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Arrhenius• Said the at reaction rate should increase
with temperature.
• At high temperature more molecules have the energy required to get over the barrier.
• The number of collisions with the necessary energy increases exponentially.
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Arrhenius• Number of collisions with the required
energy = ze-Ea/RT
• z = total collisions
• e is Euler’s number (opposite of ln)
• Ea = activation energy
• R = ideal gas constant
• T is temperature in Kelvin
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Problems• Observed rate is less than the number of
collisions that have the minimum energy.
• Due to Molecular orientation
• written into equation as p the steric factor.
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ON
Br
ON
Br
ON
Br
ON
Br
O N Br ONBr ONBr
O NBr
O N BrONBr No Reaction
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Arrhenius Equation
• k = zpe-Ea/RT = Ae-Ea/RT
• A is called the frequency factor = zp
• ln k = -(Ea/R)(1/T) + ln A
• Another line !!!!
• ln k vs t is a straight line
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Activation Energy and Rates
The final saga
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Mechanisms and rates • There is an activation energy for each
elementary step.
• Activation energy determines k.
• k = Ae- (Ea/RT)
• k determines rate
• Slowest step (rate determining) must have the highest activation energy.
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This reaction takes place in three steps
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Ea
First step is fast
Low activation energy
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Second step is slowHigh activation energy
Ea
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Ea
Third step is fastLow activation energy
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Second step is rate determining
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Intermediates are present
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Activated Complexes or Transition States
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Catalysts• Speed up a reaction without being used up
in the reaction.
• Enzymes are biological catalysts.
• Homogenous Catalysts are in the same phase as the reactants.
• Heterogeneous Catalysts are in a different phase as the reactants.
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How Catalysts Work
• Catalysts allow reactions to proceed by a different mechanism - a new pathway.
• New pathway has a lower activation energy.
• More molecules will have this activation energy.
• Do not change E
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Pt surface
HH
HH
HH
HH
• Hydrogen bonds to surface of metal.
• Break H-H bonds
Heterogenous Catalysts
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Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
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Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
• The double bond breaks and bonds to the catalyst.
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Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
• The hydrogen atoms bond with the carbon
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Pt surface
H
Heterogenous Catalysts
C HH C
HH
H HH
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Homogenous Catalysts• Chlorofluorocarbons catalyze the
decomposition of ozone.
• Enzymes regulating the body processes. (Protein catalysts)
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Catalysts and rate• Catalysts will speed up a reaction but only
to a certain point.
• Past a certain point adding more reactants won’t change the rate.
• Zero Order
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Catalysts and rate.
Concentration of reactants
Rate
• Rate increases until the active sites of catalyst are filled.
• Then rate is independent of concentration
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