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Kinetics
Reaction Rates
Kinetics• The study of reaction rates.• Thermodynamics will tell us whether or not
the reaction will occur• Spontaneous reactions (ΔG = neg) are
reactions that will happen - but we can’t tell how fast.
• Diamond will spontaneously turn to graphite – eventually.
• Kinetics is about how fast
Review- Collision Model of Chemical Reactions
Particles must collide
with enough energy
to break bonds
EFFECTIVE COLLISIONS
Factors that affect the rate of reactionIncrease number of effective collisions
increase rate of reaction
• Temperature
• Concentration
• Surface area
• Pressure
• Agitation
• Catalyst
Reaction Rate
• Rate = Conc. of A at t2 -Conc. of A at t1t2- t1
• Rate =[A]t
• Change in concentration per unit time• Consider the reaction
N2(g) + 3H2(g) 2NH3(g)
• As the reaction progresses the concentration H2 goes down
Concentration
Time
[H[H22]]
N2 + 3H2 → 2NH3
Δ[H2]/time = neg
• As the reaction progresses the concentration N2 goes down 1/3 as fast
Δ[H2]/time = neg
Concentration
Time
[H[H22]]
[N[N22]]
N2 + 3H2 → 2NH3
Δ[N2]/time = neg
Δ[H2]/time = neg
• As the reaction progresses the concentration NH3 goes up 2/3 times
Concentration
Time
[H[H22]]
[N[N22]]
[NH[NH33]]
N2 + 3H2 → 2NH3
Δ[NH3]/time = pos
Reaction Rates are always positive
• Over time – the concentration of products increases– the concentration of reactants decreases
• So Δ[P]/time is positive
• But Δ[R]/time is negative, so the rate would be expressed as - Δ[R]/time
If rate is measured in terms of concentration of Hydrogen
-Δ[H2] - mol H2
time L timeand N2 + 3H2 2NH3
Then using the stoichiometry of the reaction
-Δmol H2 1 mol N2
L time 3 mol H2
Then
-Δmol H2 2 mol NH3
L time 3 mol H2
Calculating Rates
• Average rates are taken over a period of time interval
• 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
Average Rate – Find the slope of the line between the two points
Note: slope =
Concentration
Time
[H[H22]]
tt
Note: slope = Δy / Δx
Δconcentration / Δtime
THE RATE
Instantaneous Rate – Find the slope of the line tangent to that point
Concentration
Time
[H[H22]]
ttd[Hd[H22]]
dtdt
12_291
0.000370s
O2
0.0025
0.005
0.0075
0.0100
0.0006
70s
0.0026
110 s
NO2
NO
50 100 150 200 250 300 350 400
Con
cent
ratio
ns (
mol
/L)
Time (s)
[NO2 ]
t
INSTANTANEOUS RATES
2NO2 2NO + O2
C4H9Cl + H2O C4H9OH + HCl
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.
• In addition, as the reaction progresses the concentration of reactants decrease
• This is called the initial rate method.
• Rate laws show the effect of concentration
on the rate of the reaction
• The concentration of the products do not appear in the rate law because this is an initial rate.
• The order (exponent)
– must be determined experimentally,
– cannot be obtained from the equation
Rate LawsRate Laws
• You will find that the rate will only depend on the concentration of the reactants. (Initially)
• 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
NO2 NO + ½ O2
The data for
the reaction at 300ºC
Plot the data to show concentration as a
function of time.
Time (s)[NO2]
(mol/L)
0.00 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
NO2 NO + ½ O2
To determine the rate law for the reaction, we must find the exponent in the equation.
Rate = k [NO2]x
Experiment [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040
NO2 NO + ½ O2
To determine the rate law for the reaction, we must find the exponent in the equation.
Rate = k [NO2]x
Exp [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040
2x [NO2] 4x rate
NO2 NO + ½ O2
The exponent for the concentration of NO2
must be 2 to show the effect of changing the concentration of NO2 on the rate.
Rate = k [NO2]2
Exp [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040
3x [NO2] 9x rate
NO2 NO + ½ O2
Rate = k [NO2]x
Pick two experiments to plug into the equation.
Experiment [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040
NO2 NO + ½ O2
Use the rate and one of the experiments to calculate the rate constant.
Experiment [NO2] M Rate, M s-1
1 0.060 0.0360
2 0.030 0.0090
3 0.020 0.0040
2ClO2 + 2OH- ClO3- + ClO2
- + H2O
• Determine the rate law for the reaction.• Calculate the rate constant.• What is the overall reaction order?
Experiment [ClO2-] M [OH-] M Rate, M s-1
1 0.060 0.030 0.0248
2 0.020 0.030 0.00276
3 0.020 0.090 0.00828
Types of Rate Laws• Differential Rate Law - describes how rate depends on
concentration. Method to determine- change initial concentration and measure effect on rate
• Integrated Rate Law - describes how concentration depends on time. Method to determine- measure the concentration of reactants as function of time
• For each type of differential rate law there is an integrated rate law and vice versa.
Determining Rate Laws• The first step is to determine the form of
the rate law (especially its order).• Must be determined from experimental
data.• For this reaction
2 N2O5 (aq) 4NO2 (aq) + O2(g)• The reverse reaction won’t play a role
because the gas leaves
[N[N22OO55] (mol/L) ] (mol/L) Time (s) 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
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.
[N[N22OO55] ]
(mol/L)(mol/L)
Time (s)
0
0.2
0.4
0.6
0.8
1
1.2
0 200
400
600
800
1000
1200
1400
1600
1800
2000
At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4 (200 - 600) -400
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 - .32) = 0.20 =- 2.5 x 10 -4 (1000-1800) -800
• At 0.8 M rate is 5.0 x 10-4
• At 0.4 M rate is 2.5 x 10-4
• Since the rate is twice as fast when the concentration is twice as big the rate law must be.. First power
• 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.
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.
An example• 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
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
Calculate the rate law and k
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
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[N2O5]0 = initial concentration of N2O5
[N2O5]t = concentration of N2O5 after some time (t)
ln = natural log
= - ktln ( )[N2O5]0
[N2O5]t
First Order rate = k[N2O5]• For the reaction 2N2O5 4NO2 + O2
• Rearrange this equation to y = mx + b form– Since- log of quotient equals difference of logs
• SO. ln[N2O5]t = - k t + ln[N2O5]0
- ln[N2O5]t=ln ( )[N2O5]0
[N2O5]t
= ktln ( )[N2O5]0
[N2O5]t
ln[N2O5]t
y mx b+=
First Order• Linear form integrated rate law
•ln[N2O5]t = - k t + ln[N2O5]0
= ktln ( )[N2O5]0
[N2O5]t
y mx b+=
ln[N2O5]t
t
m = slope = -k
• Graphing natural log of concentration of reactant as a function of time
• If you get a straight line, you may determine it is first order
• Integrated First Order
First Order
= - ktln
( ) [R]0
[R]t
Half Life• The time required to reach half the
original concentration.
• [R]t = [R]0/2 when t = t1/2
• If the reaction is first order
= - ktln
( ) [R]0
[R]t Since [R]t = [R]0/2
[R]t = [R]0 equals 2
ln (2) = kt1/2
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
• The highly radioactive plutonium in nuclear waste undergoes first-order decay with a half-life of approximately 24,000 years. How many years must pass before the level of radioactivity due to the plutonium falls to 1/128th (about 1%) of its original potency?
Second Order
• Rate = -[R]/t = k[R]2
• integrated rate law
• 1/[R]t = 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 t
Second Order Rate = k [R]2
integrated rate law
1/[R]t = k t + 1/[A]0
y = m x + b
t
1/[R] Slope = k
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
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
• Most often when reaction happens on a surface because the surface area stays constant.
• Also applies to enzyme chemistry.
Zero Order Rate Law
Time
Concentration
Time
Concentration
A]/t
t
k =
A]
Zero order First order Second orderRate law
Graph [A] vs time
Integrated
rate law
Zero order First order Second order
Linear graph with time
Integrated
rate law
Rearrange
to y=mx+b
Summary of Rate Laws
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.
• 2NO2 + F2 2NO2F Rate = k[NO2][F2]
• The overall equation is a summary.• In order for this reaction to occur as
written, three molecules must collide simultaneously
• An unlikely event.• So a reaction mechanism is proposed
showing a series of likely steps
Reaction Mechanisms
• 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
Reaction Mechanisms
• Each of the two reactions is called an elementary step .
• The rate for a reaction can be written from its molecularity. The reaction rate reflects the stoichimetry.
• Molecularity is the number of pieces that must come together.
• Elementary steps add up to the balanced equation
Reaction Mechanisms
• Unimolecular step involves one molecule - Rate is first order.
• Bimolecular step - requires two molecules - Rate is second order
• Termolecular step- requires three molecules - 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.
• A products Rate = k[A]
• A+A products Rate= k[A]2
• 2A products Rate= k[A]2
• A+B products Rate= 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]
Molecularity and Rate LawsFor an elementary step –
the rate law reflects the stoichiometry
Reaction Mechanisms
• Proposed series of elementary steps.– Describing each collision
• The rate of the reaction is determined by the rate limiting step.
• The overall experimentally determined rate law must be consistent with the rate law that reflects the rate limiting step.
• The proposed mechanism isNO2 + F2 NO2F + F (slow) rate = k [NO2][F2]
F + NO2 NO2F (fast) rate = k [F][NO2]
Since the first step is the slow step, it is the rate limiting step and thus the rate of the overall reaction depends upon the rate of this step. Note: The rate of this step is also the rate of the overall reaction.
2NO2 + F2 2NO2F Rate = k[NO2][F2]
The steps must add up to the overall reaction.
• 2 NO2Cl → 2 NO2 + Cl2
• Proposed steps
• NO2Cl → NO2 + Cl (slow)
NO2Cl + Cl → NO2 + Cl2 (fast)
• Write the overall rate law that would be consistent with this proposed reaction mechanism.
How to get rid of intermediates• They can’t appear in the rate law.• Slow step determines the rate and the rate
law• 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.
Formed in reversible reactions• 2 NO + O2 2 NO2
• Mechanism
• 2 NO N2O2 (fast)
• N2O2 + O2 2 NO2 (slow)
• rate = k2[N2O2][O2]
• k1[NO]2 = k1[N2O2] (equilibrium)
• Rate = k2 (k1/ k-1)[NO]2[O2]
• Rate =k [NO]2[O2]
Formed in fast reactions• 2 IBr I2+ Br2 • Mechanism• IBr I + Br (fast)• IBr + Br I + Br2 (slow)• I + I I2 (fast)• Rate = k[IBr][Br] but [Br]= [IBr] because
the first step is fast• Rate = k[IBr][IBr] = k[IBr]2
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.
Potential Energy
Reaction Coordinate
Reactants
Products
Potential Energy
Reaction Coordinate
Reactants
Products
Activation Energy Ea
Potential Energy
Reaction Coordinate
Reactants
Products
Activated complex
Potential Energy
Reaction Coordinate
Reactants
ProductsE}
Potential Energy
Reaction Coordinate
2BrNO
2NO + Br
Br---NO
Br---NO
2
Transition State
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.
Arrhenius• Said that 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.
Arrhenius• Number of collisions with the required
energy = ze-Ea/RT
• z = total collisions
• e is Euler’s number (inverse of ln)
• Ea = activation energy
• R = ideal gas constant (in J/K mol)
• T is temperature in Kelvin
Problem with this• Observed rate is too small
• Due to molecular orientation- they have to be facing the right way
• written into equation as p the steric factor.
ON
Br
ON
Br
O N Br ONBr ONBr
O NBr
O N BrONBr No Reaction
O NBr
O NBr
Arrhenius Equation
• k = zpe-Ea/RT = Ae-Ea/RT
• A is called the frequency factor = zp• k is the rate constant• ln k = -(Ea/R)(1/T) + ln A• Another line !!!!• ln k vs 1/T is a straight line• With slope Ea/R so we can find Ea
• And intercept ln A
Arrhenius Equation- Another line !!!!
ln k = -(Ea/R)(1/T) + ln A
y = m x + b
ln k
1/T
slope = -(Ea/R)
Activation Energy and Rates
The final saga
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.
• This reaction takes place in three steps
Ea
First step is fast
Low activation energy
Second step is slowHigh activation energy
Ea
Ea
Third step is fastLow activation energy
Second step is rate determining
Intermediates are present
Activated Complexes or Transition States
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.
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.
• Does not change E• Show up as a reactant in one step and a
product in a later step
Pt surface
HH
HH
HH
HH
• Hydrogen bonds to surface of metal.
• Break H-H bonds
Heterogenous Catalysts
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
• The double bond breaks and bonds to the catalyst.
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
• The hydrogen atoms bond with the carbon
Pt surface
H
Heterogenous Catalysts
C HH C
HH
H HH
Homogenous Catalysts• Chlorofluorocarbons (CFCs) catalyze the
decomposition of ozone.
• Enzymes regulating the body processes. (Protein catalysts)
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
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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