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.

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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.

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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

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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

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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.

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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)

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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

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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:

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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

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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.)

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( 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

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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

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How do you know which step is theHow do you know which step is the Rate Determining Step

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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.)

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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.

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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

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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

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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

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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

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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

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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

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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

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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

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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

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