Prentice Hall © 2003Chapter 14 Chapter 14 Chemical Kinetics CHEMISTRY The Central Science 9th...

Prentice Hall © 2003 Chapter 14

Chapter 14Chapter 14Chemical KineticsChemical Kinetics

CHEMISTRY The Central Science

9th Edition

David P. White


• 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.

Factors that Affect Reaction Factors that Affect Reaction RatesRates


• 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.

Reaction RatesReaction Rates

t

B of molesin time change

B of moles ofnumber in changerate Average




– 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 0min 10mol 0 mol 26.0

min 0min 100at B of moles10at B of moles

B of molesrate Average

ttt


• 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).


t

A of molesA respect to with rate Average


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



Change of Rate with TimeC4H9Cl(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.



Reaction Rate and Stoichiometry• For the reaction

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

we know

• In general foraA + bB cC + dD


tt

OHHCClHCRate 9494

tdtctbta

D1C1B1A1Rate


• In general rates increase as concentrations increase.NH4

+(aq) + NO2-(aq) N2(g) + 2H2O(l)

Concentration and RateConcentration and Rate


• For the reactionNH4

+(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:

• The constant k is the rate constant.


]NO][NH[Rate 24k


Exponents in the Rate Law• For a general reaction with rate law

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.


nmk ]2reactant []1reactant [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 nth order if doubling the concentration causes an 2n increase in rate.

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



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.

The Change of The Change of Concentration with TimeConcentration with Time

kt

kt

kt

t

t

0

0

AA

ln

AlnAln

]A[A][Rate


First Order Reactions• 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.



First Order Reactions


0AlnAln ktt


Second 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


Second Order Reactions


0A1

A1 kt

t


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, t½ is the time taken for

[A]0 to reach ½[A]0.• Mathematically,


kk

t 693.0ln 21

21


Half-Life• For a second order reaction, half-life depends in the

initial concentration:


0A1

21 k

t


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 RateTemperature and Rate

Temperature and RateTemperature and RateThe Collision

Model• As temperature

increases, the rate increases.


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 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?• Observations: rates of reactions are affected by

concentration and temperature.



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.



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.



The Orientation Factor• Consider:

Cl + NOCl NO + Cl2

• There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not.



The Orientation Factor



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



Activation Energy• Consider the rearrangement of methyl isonitrile:

– 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


Activation 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).



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 Ea is

where R is the gas constant (8.314 J/mol·K).


RTEa

ef


Activation Energy



The 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.


RTEa

Aek


Determining the Activation Energy• If we have a lot of data, we can determine Ea and A

graphically by rearranging the Arrhenius equation:

• From the above equation, a plot of ln k versus 1/T will have slope of –Ea/R and intercept of ln A.


ARTEk a lnln



Determining the Activation Energy• If we do not have a lot of data, then we recognize


1221

2121

22

11

11ln

lnlnlnln

lnln and lnln

TTRE

kk

ARTEA

RTEkk

ARTEkA

RTEk

a

aa

aa


• 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 MechanismsReaction 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).



Multistep Mechanisms• Some reaction proceed through more than one step:

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)• Notice that if we add the above steps, we get the overall

reaction:NO2(g) + CO(g) NO(g) + CO2(g)



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.



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.



Rate Laws for Elementary Steps



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) + Br2(g) 2NOBr(g)



Mechanisms with an Initial Fast Step2NO(g) + Br2(g) 2NOBr(g)

• The experimentally determined rate law isRate = k[NO]2[Br2]

• Consider the following mechanism


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

k-1

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

Step 1:

Step 2:

(fast)

(slow)


Mechanisms 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 (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


]NO][Br[]NOBr[ 21

12

kk


Mechanisms with an Initial Fast Step• By definition of equilibrium:

• Therefore, the overall rate law becomes

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


]NOBr[]NO][Br[ 2121 kk

][BrNO][NO][]NO][Br[Rate 22

11

221

12

kkk

kkk


• 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.

CatalysisCatalysis


CatalysisCatalysis


Homogeneous Catalysis• Hydrogen peroxide decomposes very slowly:

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.

CatalysisCatalysis


Homogeneous Catalysis• Generally, catalysts operate by lowering the activation

energy for a reaction.

CatalysisCatalysis

CatalysisCatalysis


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 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.

CatalysisCatalysis


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.

CatalysisCatalysis


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.

CatalysisCatalysis


CatalysisCatalysis


Heterogeneous 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.

CatalysisCatalysis


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 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.

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

masses (10,000 to 106 amu).

CatalysisCatalysis


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.

CatalysisCatalysis


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 (103 - 107 per second).

CatalysisCatalysis


Enzymes

CatalysisCatalysis


End of Chapter 14End of Chapter 14Chemical KineticsChemical Kinetics

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Prentice Hall © 2003Chapter 14 Chapter 14 Chemical Kinetics CHEMISTRY The Central Science 9th...

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