From the Arrhenius equation we have: 301. From the Arrhenius equation we have: 302.

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From the Arrhenius equation we have:

)( /RTEAlnln a ek

)( /RTElnlnAln a ek

)(ek lnRTElnAln a

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From the Arrhenius equation we have:

)( /RTEAlnln a ek

)( /RTElnlnAln a ek

)(ek lnRTElnAln a

RTElnAln ak

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From the Arrhenius equation we have:

Recall the equation of a straight line: where y = lnk, b= ln A, m = -Ea/R, and x = T-1.

)( /RTEAlnln a ek

)( /RTElnlnAln a ek

)(ek lnRTElnAln a

RTElnAln ak

mxby

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So if we have temperature dependent data for the rate constant, we can make the following plot to obtain Ea:

T-1

lnk slope = -Ea/R

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H

6

7

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Sample problem: Some reactions double their reaction rates with every 10 oC rise in temperature. Assume a reaction to take place at 295 K and at

305 K. What must the activation energy be for the rate constant to exactly double?

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Sample problem: Some reactions double their reaction rates with every 10 oC rise in temperature. Assume a reaction to take place at 295 K and at

305 K. What must the activation energy be for the rate constant to exactly double?

Approach: Start with the Arrhenius equation applied to the two separate conditions, where T2 is the higher temperature.

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Sample problem: Some reactions double their reaction rates with every 10 oC rise in temperature. Assume a reaction to take place at 295 K and at

305 K. What must the activation energy be for the rate constant to exactly double?

Approach: Start with the Arrhenius equation applied to the two separate conditions, where T2 is the higher temperature.

1a RTEA

/

1

ek 2a RTE

A/

2

ek

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Take the ratio

1a

2a

RTEA

RTEA

/

/

1

2

e

ekk

12

Take the ratio

Now so that

12 2kk

1a

2a

RTEA

RTEA

/

/

1

2

e

ekk

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Take the ratio

Now so that

12 2kk

1a

2a

RTEA

RTEA

/

/

1

2

e

ekk

1a

2a

RTEA

RTEA

/

/

2

e

e

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Now take the natural log of both sides.

)11(2 21 TTR

Ea e

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Now take the natural log of both sides.

)11(2 21 TTR

Ea e

)11(2 21 TTR

Elnln

a e

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Now take the natural log of both sides.

That is:

)11(2 21 TTR

Ea e

elnTTREln

21a )11(2

)11(2 21 TTR

Elnln

a e

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Now take the natural log of both sides.

That is:

(recall that )

)11(2 21 TTR

Ea e

elnTTREln

21a )11(2

)11(2 21 TTR

Elnln

a e

)ln( Y )Yln( ee

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The preceding result simplifies to give:

)()11( 12

21

21

a TTln2R TT

TT

ln2RE

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The preceding result simplifies to give:

The numerical value for Ea is given as:

= 52 kJ mol-1

)()11( 12

21

21

a TTln2R TT

TT

ln2RE

K100.693mol JK8.314 K 305K 295E

1-1-

a

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The Arrhenius equation is quite useful when studying reactions involving simple species (atoms or diatomic molecules). For more complex systems, the Arrhenius equation is modified to the form:

RTEPA a / ek

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The Arrhenius equation is quite useful when studying reactions involving simple species (atoms or diatomic molecules). For more complex systems, the Arrhenius equation is modified to the form:

where P, the probability factor, accounts for the fact that in order to react, molecules must be properly oriented with respect to each other during a collision.

RTEPA a / ek

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This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well.

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This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well.

Example: NO2Cl + Cl NO2 + Cl2

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This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well.

Example: NO2Cl + Cl NO2 + Cl2

The two Cl atoms must come into “contact” for reaction to occur.

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NO2Cl + Cl NO2 + Cl2

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NO + NO3 2NO2

Red = oxygenblue = nitrogen

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For reactions involving only atoms P = 1; for reactions involving simple small molecules, P varies between approximately 0.2 and 0.001.

For reactions involving complex polyatomic molecules, P can be as small as 10-6.

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Reaction Mechanisms

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Reaction Mechanisms

Reaction Mechanism: The sequence of elementary steps that leads to product formation.

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Reaction Mechanisms

Reaction Mechanism: The sequence of elementary steps that leads to product formation.

Elementary Step: (Single step reaction) A reaction that occurs on the molecular level exactly as written.

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Reaction Mechanisms

Reaction Mechanism: The sequence of elementary steps that leads to product formation.

Elementary Step: (Single step reaction) A reaction that occurs on the molecular level exactly as written.

An overall reaction may involve one or several elementary steps.

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Example: Use of isotope labeling method.

O O CH3 C O CH3 + H2O CH3 C OH + CH3OH

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Example: Use of isotope labeling method.

O O CH3 C O CH3 + H2O CH3 C OH + CH3OH

For this reaction there are two obvious ways to arrive at CH3OH.

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Example: Use of isotope labeling method.

O O CH3 C O CH3 + H2O CH3 C OH + CH3OH

For this reaction there are two obvious ways to arrive at CH3OH.

O CH3 C O CH3

case 1

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Example: Use of isotope labeling method.

O O CH3 C O CH3 + H2O CH3 C OH + CH3OH

For this reaction there are two obvious ways to arrive at CH3OH.

O O CH3 C O CH3 or CH3 C O CH3

case 1 case 2

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The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different.

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The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different.

case 1 O O CH3 C O CH3 + H2

18O CH3 C 18OH + CH3OH

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The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different.

case 1 O O CH3 C O CH3 + H2

18O CH3 C 18OH + CH3OH

case 2 O O CH3 C O CH3 + H2

18O CH3 C OH + CH318OH

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Question: What might case 3 be?

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No CH318OH is found in the experiment, that means

case 1 is the correct bond breaking step.

O O CH3 C O CH3 + H2

18O CH3 C 18OH + CH3OH

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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps:

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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps:

N2O N2 + O rate

1k][ ON21k

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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps:

N2O N2 + O rate

(the O atom is a reaction intermediate)

1k][ ON21k

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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps:

N2O N2 + O rate

(the O atom is a reaction intermediate)

N2O + O N2 + O2 rate ]][[ OON22k2k

1k][ ON21k

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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps:

N2O N2 + O rate

(the O atom is a reaction intermediate)

N2O + O N2 + O2 rate

Overall reaction: 2 N2O 2 N2 + O2

]][[ OON22k2k

1k][ ON21k

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and are the rate constants for the two individual steps.

1k 2k

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and are the rate constants for the two individual steps.

Key Point: The exponents in the rate law for an elementary process are equal to the coefficients obtained from the chemical equation for that elementary process.

1k 2k

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and are the rate constants for the two individual steps.

Key Point: The exponents in the rate law for an elementary process are equal to the coefficients obtained from the chemical equation for that elementary process.

Important reminder: You cannot get the rate law exponents for the overall reaction by looking at the balanced equation.

1k 2k

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For the reaction: 2 N2O 2 N2 + O2

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For the reaction: 2 N2O 2 N2 + O2

the experimental rate law is:

overall rate

][ ON21k

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For the reaction: 2 N2O 2 N2 + O2

the experimental rate law is:

overall rate

Notice that this is exactly the same as the rate law for the first elementary step.

][ ON21k

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For the reaction: 2 N2O 2 N2 + O2

the experimental rate law is:

overall rate

Notice that this is exactly the same as the rate law for the first elementary step.

The observed rate can be explained by assuming that the second step is faster than the first step, i.e.

][ ON21k

12 kk

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Thus, the overall rate of decomposition is then completely controlled by the rate of the first step, which is called the rate-determining step.

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Thus, the overall rate of decomposition is then completely controlled by the rate of the first step, which is called the rate-determining step.

Rate-determining step : The slowest step in the sequence of steps leading to the formation of products.

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Second example: 2 NO2 + F2 2 NO2F

The accepted mechanism for the reaction is:

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Second example: 2 NO2 + F2 2 NO2F

The accepted mechanism for the reaction is:

NO2 + F2 NO2F + F slow step

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Second example: 2 NO2 + F2 2 NO2F

The accepted mechanism for the reaction is:

NO2 + F2 NO2F + F slow step

NO2 + F NO2F fast step

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Second example: 2 NO2 + F2 2 NO2F

The accepted mechanism for the reaction is:

NO2 + F2 NO2F + F slow step

NO2 + F NO2F fast step

Note that the two elementary steps add to the overall chemical equation.

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2 NO2 + F2 2 NO2F

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Summary comments on mechanism For a reaction mechanism to be viable, two main

conditions apply.