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KineticsKinetics
D d f t C t ti (RATE LAW)Dependence of rate on Concentration (RATE LAW)
Reaction Mechanisms
Dependence of rate on temperature
Activation Energy Ea
Activated Complex
Arrhenius Equation
Mary J. Bojan Chem 112 1
A MECHANISM is a description of what is happening on a molecular level.
REACTION MECHANISMThe process or series of elementary steps by which aThe process or series of elementary steps by which a reaction occurs
Postulate a MECHANISM for the IN AN conversion
Elementary step: process that occurs as written.
Postulate a MECHANISM for the IN → AN conversion
Chem 112 2MJ Bojan
To fully understand the mechanism of a reaction, t E P fil f th tigenerate an Energy Profile for the reaction.
Energy Profile for the proposed mechanism
1. Reactants must “approach” the barrier. There is a probability (A) associated with this.
MJ Bojan Chem 112 3
Costs energy to break the C−N bondThis is the “energy barrier”.
Ea is energy need to get to the transition state.
To get over the hill need to put energy in
2. How does a molecule get over the barrier?
Wh d th f ?MJ Bojan Chem 112 4
Where does the energy come from?
The average kinetic energy of a collection of molecules is proportional to the temperature Recall KMTproportional to the temperature. Recall KMT
All molecules in a gas sample are in motion.g p
The speeds of the molecules are related to their kinetic energykinetic energy.
Chem 112 5MJ Bojan
Kinetic Molecular Theory can be used to determine the % of molecules with sufficient energy to react.gy
ule
s T1
on
of
mo
lecu
T2
Fra
ctio
Average Kinetic Energy
2
Average Kinetic Energy
fraction of molecules (f) with energy E f α e−E/RT
As T increases the fraction of molecules with energy greater than Ea
increases.
(The number of molecules with sufficient energy to get over the barrier
Mary J. Bojan Chem 112 6
increases.)
Collision Theory
Most molecules do NOT have sufficient energy to react.
No reaction
How do they get the energy to react if they d ’ l d h i ?don’t already have it?
Energy is transferred upon collision.Reaction rate is proportional to # of collisionsReaction rate is proportional to # of collisions
MJ Bojan Chem 112 7
A successful collision must have the
Even if there is sufficient energy in a collision to overcome
proper orientation.
gythe activation energy, a collision might not be successful!
Example: Bimolecular reactionNOCl + Cl → NO + Cl2
Chem 112 8MJ Bojan
The reaction profile can be used to i th ltsummarize these results.
Activated complex(transition state)
E ti tiEa = activation energy
reactants
products
ΔE: reaction energy
If the collision energy is < Ea or orientation is incorrect: no reaction
If th lli i i E d i t ti i i t REACTION
Mary J. Bojan Chem 112 9
If the collision energy is > Ea and orientation is incorrect: REACTION
Using collision theory we learned:
R ti t d d
time
collisionssuccessfulofrate
#
Reaction rates depend on
1. # of Collisions
2 Activation Energy (E )2. Activation Energy (Ea)
3. Frequency factor and orientation (A) this is essentially a probability that reactants will collide AND have the correct orientation for reaction to occur. (How often reactants approach reaction per unit time.)
If we hold temperature constant, what can we do to change the rate?
Chem 112 10MJ Bojan
If we hold temperature constant, what can d t h th t ?we do to change the rate?
collisionssuccessfulofrate
#
If we increase # collisions
timerate
e c ease # co s o s
Ea
A
At constant T, rate concentration. The constant of proportionality = k, called the rate constant.
Mary J. Bojan Chem 112 11
y
Rate Law: relationship between the rate of a reaction and concentration of its reactants.
We can measure concentrations of reactantsWe can measure concentrations of reactants…
If a step is elementary, then the rate of that reaction is proportional to the concentration of the reactantsproportional to the concentration of the reactants.
Examples: Rate Law
UnimolecularUnimolecular
Bimolecular
Termolecular
The Rate Law for an elementary process can be determined from its molec larit (ho man molec les are in ol ed in a collision)
Mary J. Bojan Chem 112 12
its molecularity (how many molecules are involved in a collision).
Rate Law: relationship between the rate of a reaction and concentration of its reactantsreaction and concentration of its reactants.
We can measure concentrations of reactants…
Th t l iThe rate law is:
– rate = k[A]x[B]y
k is the rate constantk is the rate constant
– independent of [A] and [B]; varies with T
x and y are exponentsy p
– ≥ 0, not necessarily integers
Order of reaction = sum of exponents
– Overall order of reaction = x + y
Mary J. Bojan Chem 112 13
We want to use this theory to predict the rate law for a reactionrate law for a reaction.
What is the rate law for the following reaction? NO2 +CO → NO + CO2
Assuming this is an elementary process:
Experimental result:
Mary J. Bojan Chem 112 14
If a step is elementary, then the rate of that reaction is proportional to the concentration of the reactants.p p
We can’t tell if a reaction is an elementary process or the result of a mechanism unless we do an experiment.
RECALL: An elementary step is a reaction that proceeds as written.
PROBLEM: Most reactions do not occur in a single step They occurPROBLEM: Most reactions do not occur in a single step. They occur as the result of several elementary steps.
If the experimental rate law does not match the rate law predicted by the reaction, then we must try to elucidate the MECHANISM.
St d f ki ti h l l id t tiSt d f ki ti h l l id t tiStudy of kinetics can help us elucidate reaction Study of kinetics can help us elucidate reaction MECHANISMSMECHANISMS
MJ Bojan Chem 112 15
What is the Mechanism for the following reaction?
NO CO NO CONO2 +CO → NO + CO2
REACTION MECHANISM
Propose a mechanism:
The process or series of elementary steps by which a reaction occurs
NOTE: Elementary steps in a mechanism must add up to
NO2 + NO2 → NO3 + NO 1
NO3 + CO → NO2 + CO2 2
mechanism must add up to give the balanced equation for the overall process.
NO2 + CO → NO + CO2
NO3 is produced in step 1 and consumed in step 2Intermediate: a stable molecule
• Intermediates do not (should not) appear in the rate law. (Concentration dependence of intermediates cannot be measured.)
MJ Bojan Chem 112 16
( p )
The schematic reaction profile for a mechanism can be drawnbe drawn.
transition statesProposed mechanism:
y
NO2 + NO2 → NO3 + NO 1
NO3 + CO → NO2 + CO2 2
intermediate
Ene
rgy
NO2 + CO → NO + CO2
R ti di t
reactant
product
Reaction coordinate
MJ Bojan Chem 112 17
Find the Rate Law for each step in a multi-step mechanism
Propose a mechanism:
Experimental Rate Law:
Propose a mechanism:
NO2 + NO2 → NO3 + NO 1
NO + CO → NO + CO 2NO3 + CO → NO2 + CO2 2
NO2 + CO → NO + CO2
Then compare them with the rate 1
NO2 t
k NO2 2
experimental result.t
CONOkCO
rate 3 2
t 3
MJ Bojan Chem 112 18
How do you know which step is theHow do you know which step is the Rate Determining Step
MJ Bojan Chem 112 19
Summarize points related to mechanisms
transition states
• Elementary steps in a mechanism must add up to give the balanced equation for the overall process.
intermediatenerg
y
• NO3 is produced in step 1 and consumed in step 2Intermediate: a stable molecule intermediate
En
reactant
Intermediate: a stable moleculeNote: it is NOT the same as the
transtion state (or activated complex)
Reaction coordinate
product• Rate determining step = slow step
• Intermediates do not (should not) appear in the rate law. (Concentration dependence of intermediates cannot be measured.)
MJ Bojan Chem 112 20
To find mechanisms
1. Find the experimental rate law
2. Postulate elementary steps
3. Find the rate law predicted by the mechanism and compare to experiment.
No rate can be written in terms of intermediates
Experiments can be used to support a proposed reaction mechanism or be proof that a proposed mechanism is incorrect.
MJ Bojan Chem 112 21
p oposed ec a s s co ect
Mechanism Example Problem
Cl CHCl HCl CClCl2 + CHCl3 → HCl + CCl4
Rate = kobs [Cl2]1/2 [CHCl3]obs [ 2] [ 3]
Postulate the following mechanism: is it consistent with the experimental rate law?is it consistent with the experimental rate law?
Cl2 2Cl fast
Cl + CHCl3 → HCl + CCl3 slow
Cl + CCl3→ CCl4 fastCl CCl3 CCl4 fast
MJ Bojan Chem 11222
We have focused on reaction rates at constant T.
Reaction rates depend on
1. # of Collisions
R l t d t t ti Related to concentration
Can be used to help elucidate reaction mechanisms
2. Activation Energy (Ea)2. Activation Energy (Ea)
3. Frequency factor and orientation (A) this is essentially a probability that reactants will collide AND have the correct orientation for reaction to occur (How often reactantsorientation for reaction to occur. (How often reactants approach reaction per unit time.)
What happens when temperature changes?pp p g
Chem 112 23MJ Bojan
As T increases, the fraction of molecules with energy greater than E increasesenergy greater than Ea increases.
ole
cule
s T1
ract
ion
of
mo
T2
Fr
Average Kinetic Energy
fraction of molecules (f) with energy E RTEf a
Mary J. Bojan Chem 112 24
Temperature dependence shows up in the rate constant: krate constant: k
Measure the rate of the same reaction (samesame reaction (same conditions) at different temperatures
Plot rate constant vs. T
Chem 112 25MJ Bojan
Temperature dependence of the rate constant is p pgiven by the Arrhenius Equation
k A eEa
RT
k = rate constant is temperature dependent
A = frequency factorq yRelated to reaction frequency and orientation
Ea = Activation energya gy
R = gas constant (usually 8.314 J/mol-K)T = temperature in Kp
Chem 112 26MJ Bojan
Use the linear form of the equation to find the activation energy for a reactionactivation energy for a reaction.
ln k ln A EaRT
Arrhenius plotArrhenius plot
plot of ln k vs 1/T is a straight lineline
slope = −Ea/Rintercept = ln A
Chem 112 27MJ Bojan
It is possible to find Ea when the frequency factor A is not known
SAMPLE PROBLEM
known.
SAMPLE PROBLEM
Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines.
2NO( ) N ( )+ O ( )2NO(g) N2 (g)+ O2(g)
The decomposition of nitric oxide (NO) to N2 and O2
is second order with a rate constant of
0.0796 M−1s−1 at 737°C and
0.0815 M−1s−1 at 947°C.
Calculate the activation energy for the reaction 11EkCalculate the activation energy for the reaction.
122
1 11
TTR
E
k
k aln
Chem 112 28MJ Bojan
FYIWhere does this equation come from?
1 11Ek aln
We want to eliminate A from the Arrhenius equation.
Where does this equation come from?
122 TTRk
ln
q
This can be done if we know the reaction rate at two different temperatures:
Eliminate A by subtracting equation two from equation
one
ln k1 Ea
RT1
ln AAt T1
one.
ln k2 Ea
RT2
ln Aln k1 ln k2
Ea
RT1
Ea
RT2
At T2
R = 8.314 J/mol-Kln
k1
k2
Ea
R
1
T2
1
T1
MJ Bojan Chem 112 29
RATE VS TEMPERATUREH d t 10° t t ?
Most reactions have E = 20 – 200 kJ/mol
How does rate vary over a 10° temperature range? (e.g. from T1 = 300 K to T2 = 310 K)
Most reactions have Ea = 20 – 200 kJ/molA “typical” Ea might be 50 kJ/mol
Use this eq ation to find the ratio of the ratesUse this equation to find the ratio of the rates (= ratio of the rate constants).
k /k 0 65 1 9
1 11
TTR
E
k
k aln
k2/k1 = e0.65 = 1 9
k2/k1 = e0.65 = 1.9
Rate at T2 (=310K) is twice as fast as rate at T1 (= 300 K)
122 TTRk
rule of thumb: reaction rates double for every 10° rise in temperature
k2/k1 e 1.9
(assumes Ea ≈ 50 kJ/mol)MJ Bojan Chem 112 30
Using collision theory we learned:Reaction rates depend onReaction rates depend on
1. # of Collisions2. Activation Energy (Ea)3. Frequency factor and orientation (A) this is essentially a3. Frequency factor and orientation (A) this is essentially a
probability that reactants will collide AND have the correct orientation for reaction to occur. (How often reactants approach reaction per unit time.)
At constant temperature:Collisions are related to concentration.Rate law is the relationship between rate and concentration)Rate law is the relationship between rate and concentration).
Rate = k [A]x [B]y
Activation energy (Ea) and A are related to the rate constant: k. gy ( a)which varies with temperature.
ln k ln A EaRT
Chem 112 31MJ Bojan