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Chapter 14 Chemical Kinetics

Factors that Affect Reaction Rates

• Kinetics is the study of how fast chemical reactions occur.

• There are 4 important factors which affect rates of reactions: – reactant concentration, – temperature, – action of catalysts, and – surface area.

• Goal: to understand chemical reactions at the molecular level.

Reaction Rates

• Speed of a reaction is measured by the change in concentration with time.

• For a reaction A  B Average rate  change in number of moles of B    moles change in time of B  

t

• Suppose A reacts to form B. Let us begin with 1.00 mol A.

Reaction Rates

Reaction Rates

– At

t

= 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present.

– At

t

= 20 min, there is 0.54 mol A and 0.46 mol B.

– At

t

= 40 min, there is 0.30 mol A and 0.70 mol B.

– Calculating, Average rate      moles of  moles 

t

of B  B at

t

 10   moles of 0 .

26 mol 10 min   0 mol 0 10 min min  0  0 .

026 min mol/min B at

t

 0 

Reaction Rates

• For the reaction A  B there are two ways of measuring rate: – the speed at which the products appear (i.e. change in moles of B per unit time), or – the speed at which the reactants disappear (i.e. the change in moles of A per unit time).

Average rate with respect to A

   

moles of

t

A

Reaction Rates

Change of Rate with Time

• For the reaction A  B there are two ways of • Most useful units for rates are to look at molarity. Since volume is constant, molarity and moles are directly proportional.

• Consider: C 4 H 9 Cl(

aq

) + H 2 O(

l

)  C 4 H 9 OH(

aq

) + HCl(

aq

)

Reaction Rates

Change of Rate with Time

C 4 H 9 Cl(

aq

) + H 2 O(

l

)  C 4 H 9 OH(

aq

) + HCl(

aq

) – We can calculate the average rate in terms of the disappearance of C 4 H 9 Cl.

– The units for average rate are mol/L·s or

M/s

.

– The average rate decreases with time.

– We plot [C 4 H 9 Cl] versus time.

– The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve.

– Instantaneous rate is different from average rate.

– We usually call the instantaneous rate the rate.

Reaction Rates

Reaction Rate and Stoichiometry

• For the reaction C 4 H 9 Cl(

aq

) + H 2 O(

l

)  we know

Rate

• In general for    

C

4

H

9 

t

C 4 H 9 OH(

aq

) + HCl(

aq

)

Cl

   

C

4

H

9

OH

t

Rate

 

1

a a

A    

t + b

B  

1

b

 

c

C   

t

+ d

D

1

c

   

t

1

d

   

t

Concentration and Rate

• In general rates increase as concentrations increase.

NH 4 + (

aq

) + NO 2 (

aq

)  N 2 (

g

) + 2H 2 O(

l

)

Concentration and Rate

• For the reaction NH 4 + (

aq

) + NO 2 (

aq

)  N 2 (

g

) + 2H 2 O(

l

) we note – as [NH 4 + ] doubles with [NO 2 ] constant the rate doubles, – as [NO 2 ] doubles with [NH 4 + ] constant, the rate doubles, – We conclude rate  [NH 4 + ][NO 2 ].

• Rate law: • The constant

k

Rate

k

[ NH

 4

][ NO

 2

]

is the rate constant.

Concentration and Rate

Exponents in the Rate Law

• For a general reaction with rate law

Rate

k

[ reactant 1 ]

m

[ reactant 2 ]

n

we say the reaction is order in reactant 2.

m

th order in reactant 1 and • The overall order of reaction is

m + n +

….

• A reaction can be zeroth order if

m

,

n

, … are zero.

n

• Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry.

th

Concentration and Rate

Using Initial Rates to Determines Rate Laws

• A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect.

• A reaction is first order if doubling the concentration causes the rate to double.

• A reacting is

n

th order if doubling the concentration causes an 2

n

increase in rate.

• Note that the rate constant does not depend on concentration.

The Change of Concentration with Time

First Order Reactions

• Goal: convert rate law into a convenient equation to give concentrations as a function of time.

• For a first order reaction, the rate doubles as the concentration of a reactant doubles.

Rate    [  A]

t

k

[ ln  

t

 ln   0   A

kt

] ln      

t

0    

kt

The Change of Concentration with Time

First Order Reactions

• A plot of ln[A]

t

versus t is a straight line with slope and intercept ln[A] 0 .

k

• In the above we use the natural logarithm, ln, which is log to the base

e

.

The Change of Concentration with Time

ln  

t

 

kt

First Order Reactions

 ln   0

The Change of Concentration with Time

Second Order Reactions

• For a second order reaction with just one reactant 1  

t

kt

 1   0 • A plot of 1/[A]

t

intercept 1/[A] 0 versus t is a straight line with slope

k

• For a second order reaction, a plot of ln[A]

t

linear.

vs.

t

and is not

The Change of Concentration with Time

Second Order Reactions

1  

t

kt

 1   0

The Change of Concentration with Time

Half-Life

• Half-life is the time taken for the concentration of a reactant to drop to half its original value.

• For a first order process, half life,

t

½ [A] 0 to reach ½[A] 0 .

• Mathematically,

t

1 2   ln   2

k

k

is the time taken for 0 .

693

The Change of Concentration with Time

Half-Life

• For a second order reaction, half-life depends in the initial concentration:

t

1 2  

k

1   0

Temperature and Rate

The Collision Model

• Most reactions speed up as temperature increases. (E.g. food spoils when not refrigerated.) • When two light sticks are placed in water: one at room temperature and one in ice, the one at room temperature is brighter than the one in ice.

• The chemical reaction responsible for chemiluminescence is dependent on temperature: the higher the temperature, the faster the reaction and the brighter the light.

Temperature and Rate

The Collision Model

• As temperature increases, the rate increases.

Temperature and Rate

The Collision Model

• Since the rate law has no temperature term in it, the rate constant must depend on temperature.

• Consider the first order reaction CH 3 NC  CH 3 CN. – As temperature increases from 190  C to 250  C the rate constant increases from 2.52  10 -5 s -1 to 3.16  10 -3 s -1 .

• The temperature effect is quite dramatic. Why?

• Observations: rates of reactions are affected by concentration and temperature.

Temperature and Rate

The Collision Model

• Goal: develop a model that explains why rates of reactions increase as concentration and temperature increases.

• The collision model: in order for molecules to react they must collide.

• The greater the number of collisions the faster the rate.

• The more molecules present, the greater the probability of collision and the faster the rate.

Temperature and Rate

The Collision Model

• The higher the temperature, the more energy available to the molecules and the faster the rate.

• Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

The Orientation Factor

• In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.

Temperature and Rate

The Orientation Factor

• Consider: Cl + NOCl  NO + Cl 2 • There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not.

Temperature and Rate

The Orientation Factor

Temperature and Rate

Activation Energy

• Arrhenius: molecules must posses a minimum amount of energy to react. Why?

– In order to form products, bonds must be broken in the reactants.

– Bond breakage requires energy.

• Activation energy,

E a

, is the minimum energy required to initiate a chemical reaction.

Temperature and Rate

Activation Energy

• Consider the rearrangement of methyl isonitrile:

N H 3 C N C H 3 C H 3 C C N C

– In H 3 C-N  C, the C-N  C bond bends until the C-N bond breaks and the N  C portion is perpendicular to the H 3 C portion. This structure is called the activated complex or transition state.

– The energy required for the above twist and break is the activation energy,

E a

.

– Once the C-N bond is broken, the N  C portion can continue to rotate forming a C-C  N bond.

Temperature and Rate

Activation Energy

• The change in energy for the reaction is the difference in energy between CH 3 NC and CH 3 CN.

• The activation energy is the difference in energy between reactants, CH 3 NC and transition state.

• The rate depends on

E a

.

• Notice that if a forward reaction is exothermic (CH 3 NC  CH 3 CN), then the reverse reaction is endothermic (CH 3 CN  CH 3 NC).

Temperature and Rate

Activation Energy

• How does a methyl isonitrile molecule gain enough energy to overcome the activation energy barrier?

• From kinetic molecular theory, we know that as temperature increases, the total kinetic energy increases.

• We can show the fraction of molecules,

f

, with energy equal to or greater than

E a

is

f

e

E a RT

where

R

is the gas constant (8.314 J/mol·K).

Temperature and Rate

Activation Energy

Temperature and Rate

The Arrhenius Equation

• Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation:

k

Ae

E a RT

k

is the rate constant,

E a

is the activation energy, constant (8.314 J/K-mol) and

T R

is the gas is the temperature in K.

– –

A A

is called the frequency factor.

is a measure of the probability of a favorable collision.

– Both

A

and

E a

are specific to a given reaction.

Temperature and Rate

Determining the Activation Energy

• If we have a lot of data, we can determine

E a

and graphically by rearranging the Arrhenius equation:

A

ln

k

 

E a RT

ln

A

• From the above equation, a plot of ln

k

versus 1/

T

will have slope of –

E a /R

and intercept of ln

A

.

Temperature and Rate

Temperature and Rate

Determining the Activation Energy

• If we do not have a lot of data, then we recognize ln

k

1  

E a RT

1  ln

A

and ln

k

2  

E a RT

2  ln

A

ln

k

1  ln

k

2    

E a RT

1  ln

A

     

E a RT

2  ln

A

  ln

k

1

k

2 

E a R

  1

T

2  1

T

1  

Reaction Mechanisms

• The balanced chemical equation provides information about the beginning and end of reaction.

• The reaction mechanism gives the path of the reaction.

• Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction.

Elementary Steps

• Elementary step: any process that occurs in a single step.

Reaction Mechanisms

Elementary Steps

• Molecularity: the number of molecules present in an elementary step.

– Unimolecular: one molecule in the elementary step, – Bimolecular: two molecules in the elementary step, and – Termolecular: three molecules in the elementary step.

• It is not common to see termolecular processes (statistically improbable).

Reaction Mechanisms

Multistep Mechanisms

• Some reaction proceed through more than one step: NO 2 (

g

) + NO 2 (

g

)  NO 3 (

g

) + CO(

g

)  NO 3 (

g

) + NO(

g

NO 2 (

g

) + CO 2 (

g

) ) • Notice that if we add the above steps, we get the overall reaction: NO 2 (

g

) + CO(

g

)  NO(

g

) + CO 2 (

g

)

Reaction Mechanisms

Multistep Mechanisms

• If a reaction proceeds via several elementary steps, then the elementary steps must add to give the balanced chemical equation.

• Intermediate: a species which appears in an elementary step which is not a reactant or product.

Reaction Mechanisms

Rate Laws for Elementary Steps

• The rate law of an elementary step is determined by its molecularity: – Unimolecular processes are first order, – Bimolecular processes are second order, and – Termolecular processes are third order.

Rate Laws for Multistep Mechanisms

• Rate-determining step: is the slowest of the elementary steps.

Reaction Mechanisms

Rate Laws for Elementary Steps

Reaction Mechanisms

Rate Laws for Multistep Mechanisms

• Therefore, the rate-determining step governs the overall rate law for the reaction.

Mechanisms with an Initial Fast Step

• It is possible for an intermediate to be a reactant.

• Consider 2NO(

g

) + Br 2 (

g

)  2NOBr(

g

)

Reaction Mechanisms

Mechanisms with an Initial Fast Step

2NO(

g

) + Br 2 (

g

)  2NOBr(

g

• The experimentally determined rate law is ) Rate =

k

[NO] 2 [Br 2 ] • Consider the following mechanism

Step 1: NO(g) + Br 2

k

(g) NOBr

k

-1 2 (g) (fast) Step 2: NOBr 2

k

(g) + NO(g) 2NOBr(g) (slow)

Reaction Mechanisms

Mechanisms with an Initial Fast Step

• The rate law is (based on Step 2 ): Rate =

k

2 [NOBr 2 ][NO] • The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable).

• Assume NOBr 2 is unstable, so we express the concentration of NOBr 2 in terms of NOBr and Br 2 assuming there is an equilibrium in step 1 [ NOBr 2 ] 

k

1 [ NO][Br 2 ]

k

 1 we have

Reaction Mechanisms

Mechanisms with an Initial Fast Step

• By definition of equilibrium:

k

1

[ NO][Br

2

]

k

 1

[ NOBr

2

]

• Therefore, the overall rate law becomes Rate 

k

2

k k

1  1 [ NO][Br 2 ] [ NO] 

k

2

k k

1  1 [ NO] 2 • Note the final rate law is consistent with the [Br 2 ] experimentally observed rate law.

Catalysis

• A catalyst changes the rate of a chemical reaction.

• There are two types of catalyst: – homogeneous, and – heterogeneous.

• Chlorine atoms are catalysts for the destruction of ozone.

Homogeneous Catalysis

• The catalyst and reaction is in one phase.

Catalysis

Catalysis

Homogeneous Catalysis

• Hydrogen peroxide decomposes very slowly: 2H 2 O 2 (

aq

)  2H 2 O(

l

) + O 2 (

g

) • In the presence of the bromide ion, the decomposition occurs rapidly: – 2Br (

aq

) + H 2 O 2 (

aq

) + 2H + (

aq

)  – Br 2 (

aq

) is brown.

– Br 2 (

aq

) + H 2 O 2 (

aq

)  2Br (

aq

Br ) + 2H 2 + ( (

aq aq

) + 2H ) + O 2 ( 2

g

O( ).

l

).

– Br is a catalyst because it can be recovered at the end of the reaction.

Catalysis

Homogeneous Catalysis

• Generally, catalysts operate by lowering the activation energy for a reaction.

Catalysis

Catalysis

Homogeneous Catalysis

• Catalysts can operate by increasing the number of effective collisions.

• That is, from the Arrhenius equation: catalysts increase

k

be increasing

A

or decreasing

E a

.

• A catalyst may add intermediates to the reaction.

• Example: In the presence of Br , Br 2 (

aq

) is generated as an intermediate in the decomposition of H 2 O 2 .

Catalysis

Homogeneous Catalysis

• When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction. The catalyst is in a different phase than the reactants and products.

Heterogeneous Catalysis

• Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars).

• Most industrial catalysts are heterogeneous.

Catalysis

Heterogeneous Catalysis

• First step is adsorption (the binding of reactant molecules to the catalyst surface).

• Adsorbed species (atoms or ions) are very reactive.

• Molecules are adsorbed onto active sites on the catalyst surface.

Catalysis

Catalysis

Heterogeneous Catalysis

• Consider the hydrogenation of ethylene: C 2 H 4 (

g

) + H 2 (

g

)  C 2 H 6 (

g

), 

H

= -136 kJ/mol.

– The reaction is slow in the absence of a catalyst.

– In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature.

– First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface.

– The H-H bond breaks and the H atoms migrate about the metal surface.

Catalysis

Heterogeneous Catalysis

– When an H atom collides with an ethylene molecule on the surface, the C-C  bond breaks and a C-H  bond forms.

– When C 2 H 6 forms it desorbs from the surface.

– When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered.

Enzymes

• Enzymes are biological catalysts.

• Most enzymes are protein molecules with large molecular masses (10,000 to 10 6 amu).

Catalysis

Enzymes

• Enzymes have very specific shapes.

• Most enzymes catalyze very specific reactions.

• Substrates undergo reaction at the active site of an enzyme.

• A substrate locks into an enzyme and a fast reaction occurs.

• The products then move away from the enzyme.

Catalysis

Enzymes

• Only substrates that fit into the enzyme lock can be involved in the reaction.

• If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).

• The number of events (turnover number) catalyzed is large for enzymes (10 3 - 10 7 per second).

Catalysis

Enzymes

End of Chapter 14 Chemical Kinetics