Copyright 1999, PRENTICE HALLChapter 141 Chemical Kinetics Chapter 14 David P. White University of North Carolina, Wilmington.

Copyright 1999, PRENTICE HALL Chapter 14 1

Chemical KineticsChemical Kinetics

Chapter 14Chapter 14

David P. WhiteDavid P. White

University of North Carolina, WilmingtonUniversity of North Carolina, Wilmington


Chemical KineticsChemical Kinetics• 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.


Chemical KineticsChemical KineticsReaction RatesReaction Rates• Speed of a reaction is measured by the change in

concentration with time.• For a reaction A B

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

t

B of moles

time in changeB of moles ofnumber the in change

rate Average


Chemical KineticsChemical KineticsReaction RatesReaction Rates


Chemical KineticsChemical KineticsReaction RatesReaction 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,

mol/min 026.0min 0 - min 10mol 0 - mol 26.0

min 0 - min 100at B of moles10at B of moles

B of molesrate Average

ttt


Chemical KineticsChemical KineticsReaction RatesReaction 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).

• A plot of number of moles versus time shows that as the reactants (red A) disappear, the products (blue B) appear.


Chemical KineticsChemical KineticsReaction RatesReaction Rates


Chemical KineticsChemical KineticsRates in Terms of ConcentrationsRates in Terms of Concentrations• 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:

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)


Chemical KineticsChemical KineticsRates in Terms of ConcentrationsRates in Terms of Concentrations




C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)– We can calculate the average rate in terms of the

disappearance of C4H9Cl.

– The units for average rate are mol/L•s or M/s.

– The average rate decreases with time.

– We plot [C4H9Cl] 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.




Chemical KineticsChemical KineticsReaction Rates and StoichiometryReaction Rates and Stoichiometry• For the reaction


we know

• In general for

aA + bB cC + dD

t

OHHC

t

ClHCRate 9494

tD1

tC1

tB1

tA1

Rate

dcba


The Dependence of Rate on ConcentrationThe Dependence of Rate on Concentration• In general rates increase as concentrations increase.

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)


The Dependence of Rate on ConcentrationThe Dependence of Rate on Concentration• For the reaction

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

we note – as [NH4

+] doubles with [NO2-] constant the rate doubles,

– as [NO2-] doubles with [NH4

+] constant, the rate doubles,

– We conclude rate [NH4+][NO2

-].

• Rate law:

Rate = k[NH4+][ NO2

-].

• The constant k is the rate constant.


The Dependence of Rate on ConcentrationThe Dependence of Rate on Concentration• For a general reaction with rate law

Rate = k[reactant 1]m[reactant 2]n,

we say the reaction is mth order in reactant 1 and nth order in reactant 2.

• The overall order of reaction is m + n + ….• A reaction can be zeroth order if m, n, … are zero.• Note the values of the exponents (orders) have to be

determined experimentally. They are not simply related to stoichiometry.


The Dependence of Rate on ConcentrationThe Dependence of Rate on ConcentrationUsing Initial Rates to Determine Rate LawsUsing Initial Rates to Determine 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 reaction is second order if doubling the

concentration results in a 22 increase in rate. • A reacting is nth order if doubling the concentration

causes an 2n increase in rate.• Note that the rate constant does not depend on

concentration.


The Change of Concentration with TimeThe Change of Concentration with TimeFirst-Order ReactionsFirst-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.• We can show that

• A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0.

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

0AlnAln ktt


The Change of Concentration with TimeThe Change of Concentration with TimeFirst-Order ReactionsFirst-Order Reactions

0AlnAln ktt


The Change of Concentration with TimeThe Change of Concentration with TimeHalf-LifeHalf-Life• Half-life is the time taken for the concentration of a

reactant to drop to half its original value.

• That is, half life, t1/2 is the time taken for [A]0 to reach ½[A]0.

• Mathematically,

kkt

693.0ln21

21


The Change of Concentration with TimeThe Change of Concentration with TimeHalf-LifeHalf-Life


The Change of Concentration with TimeThe Change of Concentration with TimeSecond-Order ReactionsSecond-Order Reactions• For a second order reaction with just one reactant

• A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0

• For a second order reaction, a plot of ln[A]t vs. t is not linear.

0A1

A1 kt

t


The Change of Concentration with TimeThe Change of Concentration with TimeSecond-Order ReactionsSecond-Order Reactions


The Change of Concentration with TimeThe Change of Concentration with TimeSecond-Order ReactionsSecond-Order Reactions• We can show that the half life

• A reaction can have rate constant expression of the form

rate = k[A][B],

i.e., is second order overall, but has first order dependence on A and B.

0A1

21

kt


Temperature and RateTemperature and Rate• 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.

• As temperature increases, the rate increases.


Temperature and RateTemperature and Rate


Temperature and RateTemperature and Rate• As temperature increases, the rate increases.• Since the rate law has no temperature term in it, the

rate constant must depend on temperature.

• Consider the first order reaction CH3NC CH3CN. – 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?


Temperature and RateTemperature and RateThe Collision ModelThe Collision Model• Observations: rates of reactions are affected by

concentration and temperature.• 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.


Temperature and RateTemperature and RateThe Collision ModelThe Collision Model


Temperature and RateTemperature and RateThe Collision ModelThe Collision Model• The more molecules present, the greater the

probability of collision and the faster the rate.• 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.• In order for reaction to occur the reactant molecules

must collide in the correct orientation and with enough energy to form products.


Temperature and RateTemperature and RateActivation EnergyActivation 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, Ea, is the minimum energy required to initiate a chemical reaction.


Temperature and RateTemperature and RateActivation EnergyActivation Energy


Temperature and RateTemperature and RateActivation EnergyActivation Energy• Consider the rearrangement of acetonitrile:

– In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.

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

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

H3C N CC

NH3C H3C C N




Temperature and RateTemperature and RateActivation EnergyActivation Energy• The change in energy for the reaction is the difference

in energy between CH3NC and CH3CN.

• The activation energy is the difference in energy between reactants, CH3NC and transition state.

• The rate depends on Ea.

• Notice that if a forward reaction is exothermic (CH3NC CH3CN), then the reverse reaction is endothermic (CH3CN CH3NC).


Temperature and RateTemperature and RateActivation EnergyActivation Energy• Consider the reaction between Cl and NOCl:– If the Cl collides with the Cl of NOCl then the products are

Cl2 and NO.

– If the Cl collided with the O of NOCl then no products are formed.

• We need to quantify this effect.




Temperature and RateTemperature and RateThe Arrhenius EquationThe Arrhenius Equation• Arrhenius discovered most reaction-rate data obeyed the Arrhenius

equation:

– k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.

– A is called the frequency factor.

– A is a measure of the probability of a favorable collision.

– Both A and Ea are specific to a given reaction.

• If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation:

• If we do not have a lot of data, then we can use

RTaE

Aek


Temperature and RateTemperature and RateThe Arrhenius EquationThe Arrhenius Equation• If we have a lot of data, we can determine Ea and A

graphically by rearranging the Arrhenius equation:

• If we do not have a lot of data, then we can use

ART

Ek a lnln

122

1 11ln

TTR

E

kk a


Reaction MechanismsReaction 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.


Reaction MechanismsReaction MechanismsElementary StepsElementary Steps• Elementary step: any process that occurs in a single

step.• 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 MechanismsReaction MechanismsElementary StepsElementary Steps• 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.

Rate Laws of Elementary StepsRate Laws of 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.


Reaction MechanismsReaction MechanismsRate Laws of Multistep MechanismsRate Laws of Multistep Mechanisms• Rate-determining step: is the slowest of the

elementary steps.• Therefore, the rate-determining step governs the

overall rate law for the reaction.

Mechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• It is possible for an intermediate to be a reactant.• Consider

2NO(g) + Br2(g) 2NOBr(g)


Reaction MechanismsReaction MechanismsMechanisms with an Initial Fast StepMechanisms with an Initial Fast Step

2NO(g) + Br2(g) 2NOBr(g)

• The experimentally determined rate law is

Rate = k[NO]2[Br2]

• Consider the following mechanism

for which the rate law is (based on Step 2):

NO(g) + Br2(g) NOBr2(g)k1

k-1

NOBr2(g) + NO(g) 2NOBr(g)k2

Step 1:

Step 2:

(fast)

(slow)


Reaction MechanismsReaction MechanismsMechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• The rate law is (based on Step 2):

Rate = k2[NOBr2][NO]

• The rate law should not depend on the concentration of an intermediate because intermediates are usually unstable.

• Assume NOBr2 is unstable, so we express the concentration of NOBr2 in terms of NOBr and Br2 assuming there is an equilibrium in step 1 we have

]Br][NO[NOBr 21

12

kk


Reaction MechanismsReaction MechanismsMechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• By definition of equilibrium:

k1[NO][Br2] = k-1[NOBr2]

• Therefore, the overall rate law becomes

• Note the final rate law is consistent with the experimentally observed rate law.

]Br[]NO[Rate 22

1

12

kk

k


CatalysisCatalysis• 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 CatalysisHomogeneous Catalysis• The catalyst and reaction is in one phase.• Hydrogen peroxide decomposes very slowly:

2H2O2(aq) 2H2O(l) + O2(g).


CatalysisCatalysisHomogeneous CatalysisHomogeneous Catalysis

2H2O2(aq) 2H2O(l) + O2(g).

• In the presence of the bromide ion, the decomposition occurs rapidly:– 2Br-(aq) + H2O2(aq) + 2H+(aq) Br2(aq) + 2H2O(l).

– Br2(aq) is brown.

– Br2(aq) + H2O2(aq) 2Br-(aq) + 2H+(aq) + O2(g).

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

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


CatalysisCatalysisHomogeneous CatalysisHomogeneous Catalysis


CatalysisCatalysisHomogeneous CatalysisHomogeneous 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 Ea.

• A catalyst may add intermediates to the reaction.

• Example: In the presence of Br-, Br2(aq) is generated as an intermediate in the decomposition of H2O2.

• When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction.


CatalysisCatalysisHeterogeneous CatalysisHeterogeneous Catalysis• The catalyst is in a different phase than the reactants

and products.• Typical example: solid catalyst, gaseous reactants and

products (catalytic converters in cars).• Most industrial catalysts are heterogeneous.• 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 (red spheres)

on the catalyst surface.


CatalysisCatalysisHeterogeneous CatalysisHeterogeneous Catalysis


CatalysisCatalysisHeterogeneous CatalysisHeterogeneous Catalysis• Consider the hydrogenation of ethylene:

C2H4(g) + H2(g) C2H6(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.


CatalysisCatalysisHeterogeneous CatalysisHeterogeneous Catalysis• Consider the hydrogenation of ethylene:

C2H4(g) + H2(g) C2H6(g), H = -136 kJ/mol.– When an H atom collides with an ethylene molecule on the

surface, the C-C bond breaks and a C-H bond forms.

– When C2H6 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.


CatalysisCatalysisEnzymesEnzymes• Enzymes are biological catalysts.• Most enzymes are protein molecules with large

molecular masses (10,000 to 106 amu).• 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.


CatalysisCatalysisEnzymesEnzymes


CatalysisCatalysisEnzymesEnzymes• The products then move away from the enzyme.• 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 (103 - 107 per second).


CatalysisCatalysisNitrogen Fixation and NitrogenaseNitrogen Fixation and Nitrogenase• Nitrogen gas cannot be used in the soil for plants or

animals.

• Nitrogen compounds, NO3, NO2-, and NO3

- are used in the soil.

• The conversion between N2 and NH3 is a process with a high activation energy (the NN triple bond needs to be broken).

• An enzyme, nitrogenase, in bacteria which live in root nodules of legumes, clover and alfalfa, catalyses the reduction of nitrogen to ammonia.


CatalysisCatalysisNitrogen Fixation and NitrogenaseNitrogen Fixation and Nitrogenase


CatalysisCatalysisNitrogen Fixation and NitrogenaseNitrogen Fixation and Nitrogenase• The fixed nitrogen (NO3, NO2

-, and NO3-) is consumed

by plants and then eaten by animals.• Animal waste and dead plants are attacked by

bacteria that break down the fixed nitrogen and produce N2 gas for the atmosphere.


End of Chapter 14End of Chapter 14

Chemical KineticsChemical Kinetics

Copyright 1999, PRENTICE HALLChapter 141 Chemical Kinetics Chapter 14 David P. White University of North Carolina, Wilmington.

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