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Chapter 13

“ChemicalKinetics”

Kinetics

• Kinetics is– the study of the factors that affect the speed of a

reaction– the mechanism by which a reaction proceeds.

• Experimentally it is shown that there are 4factors that influence the speed of a reaction:– nature of the reactants– temperature– catalyst– concentration

Rate

• Rate is:– how much a quantity changes in a given period of

time• The speed you drive your car is a rate

– the distance your car travels (miles) in a givenperiod of time (1 hour)

– so the rate of your car has units of mi/hr

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Defining Reaction Rate

• The rate of a chemical reaction isgenerally measured in terms of– how much the concentration of a reactant

decreases in a given period of time– or product concentration increases– for reactants, a negative sign is placed in

front of the definition

Reaction Rate Changes Over Time

• As time goes on, the rate of a reactiongenerally slows down– because the concentration of the reactants

decreases.• At a given time the reaction will stop,

either because the reactants run out orbecause the system has reachedequilibrium.

Fast versus Slow

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Fast versus Slow

Fast versus Slow

Reaction Rate and Stoichiometry• In most reactions, the coefficients of the balanced

equation are not all the sameH2 (g) + I2 (g) → 2 HI(g)

• In these reactions, the change in the number ofmolecules of one substance is a multiple of thechange in the number of molecules of another– for the above reaction, for every 1 mole of H2 used, 1

mole of I2 will also be used and 2 moles of HI made– therefore the rate of change will be different

• In order to be consistent, the change in theconcentration of each substance is multiplied by1/coefficient

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Average Rate

• The average rate is the change inmeasured concentrations in anyparticular time period– linear approximation of a curve

• The larger the time interval, the morethe average rate deviates from theinstantaneous rate

Instantaneous Rate

• The instantaneous rate is the change inconcentration at any one particular time– slope at one point of a curve

• Determined by taking the slope of a linetangent to the curve at that particularpoint– first derivative of the function

H2 (g) + I2 (g) → 2 HI (g)

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Example

• For the reaction given, the [I−] changes from 1.000 Mto 0.868 M in the first 10 s. Calculate the averagerate in the first 10 s and the Δ[H+].H2O2 (aq) + 3 I−(aq) + 2 H+

(aq) → I3−(aq) + 2 H2O(l)

Measuring Reaction Rate• In order to measure the reaction rate you

need to be able to measure the concentrationof at least one component in the mixture atmany points in time

• There are two ways of approaching thisproblem– (1) for reactions that are complete in less than 1

hour, it is best to use continuous monitoring of theconcentration

– (2) for reactions that happen over a very long time,sampling of the mixture at various times can beused

Continuous Monitoring• Polarimetry – measuring the change in the degree of

rotation of plane-polarized light caused by one of thecomponents over time

• Spectrophotometry – measuring the amount of lightof a particular wavelength absorbed by onecomponent over time– the component absorbs its complimentary color

• Total pressure – the total pressure of a gas mixtureis stoichiometrically related to partial pressures of thegases in the reaction

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Sampling

• Gas chromatography can measure theconcentrations of various components in amixture– for samples that have volatile components– separates mixture by adherence to a surface

• Drawing off periodic aliquots from themixture and doing quantitative analysis– titration for one of the components– gravimetric analysis

Factors Affecting Reaction RateNature of the Reactants

• Nature of the reactants means whatkind of reactant molecules and whatphysical condition they are in.– small molecules tend to react faster than large molecules;– gases tend to react faster than liquids which react faster

than solids;– powdered solids are more reactive than “blocks”– certain types of chemicals are more reactive than others– ions react faster than molecules

Factors Affecting Reaction RateTemperature

• Increasing temperature increases reactionrate– chemist’s rule of thumb - for each 10°C rise in

temperature, the speed of the reaction doubles• There is a mathematical relationship between

the absolute temperature and the speed of areaction discovered by Svante Arrheniuswhich will be examined later

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Factors Affecting Reaction RateCatalysts

• Catalysts are substances which affect thespeed of a reaction without beingconsumed.

• Most catalysts are used to speed up areaction, these are called positive catalysts

• Homogeneous = present in same phase• Heterogeneous = present in different phase

Factors Affecting Reaction RateReactant Concentration

• Generally, the larger the concentration ofreactant molecules, the faster the reaction– increases the frequency of reactant

molecule contact– concentration of gases depends on the partial

pressure of the gas• higher pressure = higher concentration

• Concentration of solutions depends on thesolute to solution ratio (molarity)

The Rate Law

• The Rate Law of a reaction is the mathematicalrelationship between the rate of the reaction andthe concentrations of the reactants– and homogeneous catalysts as well

• The rate of a reaction is directly proportional tothe concentration of each reactant raised to apower

• For the reaction aA + bB → products therate law would have the form given below– n and m are called the orders for each reactant– k is called the rate constant

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Reaction Order

• The exponent on each reactant in the rate lawis called the order with respect to thatreactant

• The sum of the exponents on the reactants iscalled the order of the reaction

• The rate law for the reaction:2 NO(g) + O2(g) → 2 NO2(g)

is Rate = k[NO]2[O2]

Reactant Concentration vs. TimeA → Products

Half-Life

• The half-life, t1/2, of areaction is the length oftime it takes for theconcentration of thereactants to fall to ½ itsinitial value

• The half-life of thereaction depends on theorder of the reaction

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Zero Order Reactions• Rate = k[A]0 = k

– constant rate reactions• [A] = -kt + [A]0• Graph of [A] vs. time is

straight line with slope = -kand y-intercept = [A]0

• t ½ = [A0]/2k• when Rate = M/sec, k =

M/sec

[A]0

[A]

time

slope = - k

First Order Reactions• Rate = k[A]• ln[A] = -kt + ln[A]0• graph ln[A] vs. time gives

straight line with slope = -kand y-intercept = ln[A]0– used to determine the rate

constant• t½ = 0.693/k• the half-life of a first order

reaction is constant• the when Rate = M/sec, k =

sec-1

ln[A]0

ln[A]

time

slope = −k

Half-Life of a First-Order ReactionIs Constant

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Second Order Reactions

• Rate = k[A]2• 1/[A] = kt + 1/[A]0• graph 1/[A] vs. time gives

straight line with slope = kand y-intercept = 1/[A]0– used to determine the rate

constant• t½ = 1/(k[A0])• when Rate = M/sec, k = M-

1·sec-1

l/[A]0

1/[A]

time

slope = k

Determining the Rate Law• Can only be determined experimentally• Graphically

– rate = slope of curve [A] vs. time– if graph [A] vs time is straight line, then exponent

on A in rate law is 0, rate constant = -slope– if graph ln[A] vs time is straight line, then exponent

on A in rate law is 1, rate constant = -slope– if graph 1/[A] vs time is straight line, exponent on

A in rate law is 2, rate constant = slope• Initial rates

– by comparing effect on the rate of changing theinitial concentration of reactants one at a time

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Example• The reaction SO2Cl2(g) → SO2(g) + Cl2(g) is first order

with a rate constant of 2.90 x 10-4 s-1 at a given set ofconditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]0= 0.0225 M

Initial Rate Method• Another method for determining the order of a

reactant is to see the effect on the initial rateof the reaction when the initial concentrationof that reactant is changed– for multiple reactants, keep initial

concentration of all reactants constant exceptone

– zero order = changing the concentration has noeffect on the rate

– first order = the rate changes by the same factoras the concentration

• doubling the initial concentration will double the rate– second order = the rate changes by the square of

the factor the concentration changes• doubling the initial concentration will quadruple the rate

Example• Determine the rate law and rate constant for the reaction

NO2(g) + CO(g) → NO(g) + CO2(g)given the data below.

4.3.2.1.

Expt.Number

0.0330.100.400.00830.200.200.00820.100.200.00210.100.10

InitialRate(M/s)

Initial[CO],(M)

Initial[NO2],

(M)

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Example• Determine the rate law and rate constant for the reaction

NH4+1 + NO2

-1 → Ν2 + 2 Η2Οgiven the data below.

21.60.04040.200410.80.02020.200332.30.2000.0600210.80.2000.02001

InitialRate,(x 10-7),M/s

Initial[NO2

-], MInitial[NH4

+], MExpt.No.

The Effect of Temperature on Rate• Changing the temperature changes the

rate constant of the rate law• Svante Arrhenius investigated this

relationship and showed that:

!!"

#$$%

&=

'

RT

Ea

eAk

where T is the temperature in kelvinsR is the gas constant in energy units, 8.314 J/(mol·K)A is a factor called the frequency factorEa is the activation energy, the extra energy needed to startthe molecules reacting

Activation Energy

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Activation Energy and theActivated Complex

• Activation Energy is a barrier to the reaction• It is the amount of energy needed to convert

reactants into the activated complex– aka transition state

• The activated complex is a chemical specieswith partially broken and partially formedbonds

The Arrhenius Equation:The Exponential Factor

• The exponential factor in the Arrheniusequation is a number between 0 and 1

• It represents the fraction of reactantmolecules with sufficient energy so they canmake it over the energy barrier

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Arrhenius Plots• the Arrhenius Equation can be algebraically

solved to give the following form:

( )AR

Ek

a lnT

1)ln( +!

"

#$%

&'=

this equation is in the form y = mx + bwhere y = ln(k) and x = (1/T)a graph of ln(k) vs. (1/T) is a straight line

(-8.314 J/mol·K)(slope of the line) = Ea, (in Joules)

ey-intercept = A, (unit is the same as k)

Determine the activation energy and frequencyfactor for the reaction O3(g) → O2(g) + O(g) given

the following data:

1.19 x 10919003.81 x 1071200

8.55 x 10818001.63 x 1071100

5.93 x 10817005.90 x 1061000

3.93 x 10816001.70 x 106900

2.46 x 10815003.58 x 105800

1.45 x 10814004.83 x 104700

7.83 x 10713003.37 x 103600

k, M-1·s-1Temp, Kk, M-1·s-1Temp, K

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Collision Theory of Kinetics

• For most reactions, in order for a reaction totake place, the reacting molecules mustcollide into each other.

• Once molecules collide they may reacttogether or they may not, depending on twofactors -

whether the collision has enough energy to "break thebonds holding reactant molecules together";

whether the reacting molecules collide in the properorientation for new bonds to form.

Effective Collisions

• Collisions in which these two conditions aremet (and therefore lead to reaction) are calledeffective collisions

• The higher the frequency of effectivecollisions, the faster the reaction rate

• When two molecules have an effectivecollision, a temporary, high energy (unstable)chemical species is formed - called anactivated complex or transition state

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Effective CollisionsOrientation Effect

Collision Theory andthe Arrhenius Equation

• A is the factor called the frequency factorand is the number of molecules that canapproach overcoming the energy barrier

• there are two factors that make up thefrequency factor – the orientation factor (p)and the collision frequency factor (z)

RT

E

RT

E aa

pzeeAk

!!

=""#

$%%&

'=

Orientation Factor• The proper orientation results when the

atoms are aligned in such a way that the oldbonds can break and the new bonds can form

• The more complex the reactant molecules,the less frequently they will collide with theproper orientation

• For most reactions, the orientation factor isless than 1

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Reaction Mechanisms• Generally a chemical reactions with an

equation listing all the reactant molecules andproduct molecules

• The probability of more than 3 moleculescolliding at the same instant with the properorientation and sufficient energy to overcomethe energy barrier is negligible– Therefore most reactions occur in a series of small

reactions involving 1, 2, or at most 3 molecules– To describe the series of steps that occur to

produce the overall observed reaction is called areaction mechanism

– Knowing the rate law helps us understand thesequence of steps in the mechanism

Example

H2(g) + 2 ICl(g) → 2 HCl(g) + I2(g)1) H2(g) + ICl(g) → HCl(g) + HI(g)2) HI(g) + ICl(g) → HCl(g) + I2(g)

Rate Determining Step• In most mechanisms, one step occurs slower than

the other steps• The result is that product production cannot occur

any faster than the slowest step – the stepdetermines the rate of the overall reaction

• We call the slowest step in the mechanism the ratedetermining step

• The rate law of the rate determining step determinesthe rate law of the overall reaction

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Reaction Mechanism ExampleNO2(g) + CO(g) → NO(g) + CO2(g) Rateobs = k[NO2]2

1) NO2(g) + NO2(g) → NO3(g) + NO(g)Rate = k1[NO2]2 slow

2) NO3(g) + CO(g) → NO2(g) + CO2(g) Rate = k2[NO3][CO] fast

Catalyst• Catalysts are substances that affect the rate

of a reaction without being consumed• Catalysts work by providing an alternative

mechanism for the reaction– with a lower activation energy

• Some catalysts are consumed in an earlymechanism step, then made in a later step

Catalysts

• Homogeneous catalysts are in the samephase as the reactant particles– Cl(g) in the destruction of O3(g)

• Heterogeneous catalysts are in a differentphase than the reactant particles– solid catalytic converter in a car’s exhaust system

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Catalysts

Catalytic HydrogenationH2C=CH2 + H2 → CH3CH3

Enzymes

• Because many of the molecules are large andcomplex, most biological reactions require acatalyst to proceed at a reasonable rate

• Protein molecules that catalyze biologicalreactions are called enzymes

• Enzymes work by adsorbing the substratereactant onto an active site that orients it forreaction

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Enzyme-Substrate Binding

Example