Chapter 13. Chemical Kinetics · 2013-02-05 · Chapter 13. Chemical Kinetics The Rate of Reaction (Section 13.1) Rate Laws (Section 13.2) The Relation Between Reactant Concentration

Chapter 13. Chemical Kinetics

The Rate of Reaction (Section 13.1)Rate Laws (Section 13.2)The Relation Between Reactant Concentration and Time (Section 13.3)Activation Energy and Temperature Dep~ence of Reaction Rates (Section 13,4)Reaction Mechanisms (Section 13,5)Catalysis (Section 13.6)

SUMMA~_~_~

The Rate of Reaction (~Section 13.1)

Expressing the Rate of Reaction. Chemical kinetics is the area of chemistryconcerned w~th the stu.s~y- of the rates of ct~em cal reacbons. The uro~f k net c stud es

are t~the factors that affect reaction rates and determine the reaction mechanism.Knowledge of the~ctgrs that affect reaction rat~s en~l~ chemists to control rates. Findingthe reaction mechanism means we can identify the intermediate steps by which~ctants

~e in s~)me quantity with time. The rate of population growth is thecha~ge in population per cha--~g-~ {~ time. The ch-~nge in a quantity such as population isalways equal to the difference, population (after) minus population (befoi:e). The symbol for"the change in" is

z~,(poputation) = populationfl~al - populationinitial

The change in time is some appropriate time interval, ~,t, where

At = tfinal - tinitia/

The rate of population change is

rate- A(population)&t

As a chemical reaction .proceeds the_~p_oAc._&n_tr_ation~s Of reactants and products changewith time. For instance, as [he reaction A + B -+ C progresses, the concentration of~

~~he rate is expressed as the c~n the molar concentration of C~uringthe time interval At. \.~

_:~ ’rate-

For a specific reactior~, we need to take into acco~z~t t.b_he stoichiometry; that is, we ne6d.~~or example, let’s express the rate ~reaction in termsof the concentrations of the individua! reactants and pr0~

NO(g) + O2(g) ~ 2 NO2(g)

25O

Chemistry, Ch. 13: Chemical Kinetics 25!

These concentrations can be monitored experimentally as a function of time, Notice ~balanced eq_uation, that 2 tool of NQ reacts with 1 tool of Q2; therefore, the concentration ofNO will decrease twice as ~O~-..~ --

~~ is inserted before terms involving reactants~.The change in NOconcen~Fration, A[NO], is neg~i~-~ because the concentration of NO decreases with time.In~_serting a negative sign in the ex~kes the rate of’reaction a positive quantity.

For--~~ ~ ~

aA + bB --> cC

the rate can be expressed in terms of~individual reactant or product.

A[A] _ A[B]-rate= aAt bat cat

~ No matter which reactant or product we use, the reaction rate will be positive and have the

~ -~-~al~ula~ng ~nA~er~geR_R__~e. ~-~e average rate of reaction over any t~ent-erval is e’ ’~qual to the c~hange In the concentration of a reactant ~[A], or ot a Examples

product A[C] divided by the time interval, At, during which the change occurred.

change in the concentration of Aaverage rate = -length of time interval

;i3.1 - 13.3

Exercises13-1 - 13-3

[A]2 -JAIl~ average rate-At t2 - t~

The concentration term [A]2 is the concentration of A at time t2, and [A]I is the concentrationof A at time h.

Rate Laws (Section 13.2}

Effect of Concentration. The rate of a reaction is proportional to the reactantconcentrations. For the reaction

~- NO + _~O2 ~ NO2

the rate is pro.~________~portional to the concentrations of NO and 02.

---~-2~2 Chemistry, Ch. 13: Chemical Kinetics

~ The rate law (or rate equation) for the reaction is:

’~ rate = k[NO]X[O2]Y

The rELq.portionality constant k is called the rate constant. The value of k depends on thereaction and the temperatu~, x~_and y are exponents_ which are oft....,~en but not alway~s 1 or 2.

The expon~ents x~.and y~determine how stron.g~ the concentrations affect the ~ra~te. The exponent x is called the order w~ect to NO, and ~is the order with ~,02. The sum x + y is the overall order. The V"~Tues ofx and y must be determ~~ an C~nno e enve y any other means. ~-~-w~how to determinethe order of reaction in the next section. For now we wil! just use the results. For the NOreaction with ~ experiments show that x = 2 and y = 1. Therefore, the rate law for thisreaction is:

~ rate = k[NO]2[O2]

This reaction is second order in nitric oxide, and first order in oxygen. It is third order overallThe- f~t that the~reac ~on ~s~rs or er ~n O_~ means that the rate isdirectly proportional to

the O~ concentration. If [O~] doubles or triples, the radiate will double or triple also. We canshow this mathematically. Consider two experiments. In expt 1 the concentration of O2 is c.In expt 2 the concentration of 02 is doubled from c to 2c. If the concentration of NO is thesame in both experiments, it will have no effect on the rate: Use of the rate law allows us towrite the ratio of the two rates:

rate(expt 2)_ k[NO]2(2c) = 2rate(expt 1) k[NO]~ (c)

As discussed, we see that doubling the concentration of a reactant that is first order willcause the rate to double.

If the concentration of 02 is held constant in two experiments and the concentration ofNQ doubles (from c to 2c), the rate law predicts that the rate wilt quadruple.

rate(expt 2) = k[O2](2c)2 = 22 = 4rate(expt 1) k[O2](c)2

The fact that ~ction is second order in NO means that the rate is pr_Lg_portional to thes.&g_uare of the concentration of NO...,~__~: Doubling or tripling of [NO] causes the rate to increasefour- or nine-fold, respectively.

In egAg.eral, if the concentration of one reactant is doubled whi~e the other reactantcon~tion fS unchange , an t e rate ~s:

~1. unchanged, the order of the reaction is zero.order___._,with ~:~espect to the chanqin.q ~nt;~2. doubled the order of the reaction is first order with ~_ct to th~gzEg~reactant.~,,"~L-~,~3. quadrupled~ react!on is seco~d order w~th ~ect to the changlng~2.~:~

The Isolation Method, One procedure used to determine the rate law for a reactioninvolves the isoiationmethod. In this method the concentration of all l ut one reactant isfixed, and the rate of reaction is measured as a function of the concentration of the onereactant whose concentration is varied. Any variation in the rate is due to the variation of this

Chemistry, Ch. 13: Chemical Kinetics 253

reactant’s concentration. In practice the experim:mter obse~es the dependence of the initialrate on the concentration of the reactant.

To determine the order with respect to A in the following chemical reaction

2A+ B-~C

the initial rate would be measured in several ~:~.t:~e~riments in which theconcentration of A is varied and the concentration of B is held constant. Todetermine the order with respect to B, the concentration of A must be held constantand the concentration of B is varied in several experiments. The application of thismethod is illustrated in Example 13.5.

Examples13.4, 13.5

Exercises13-4, 13-5

(~-"l~h# Relation Between Reactant Concent~:ation and Time (Section 13.3)

~’~r&~"rst-Order Reactions. One of the most widely encountered kinetic forms is the firgt-

ord’~ rate equation. In this case the exponent of [A] in the rate law is 1~

A --> products

rate = A[A] = k[A]

For a first-order reaction the unit of the rate constant is reciprocal time, 1/t. Convenient units~h, etc.

The equation that relates the concentration of A remaining to the time, since the reactionstaP~ed~

f"~ In[A]t = kt~ /"’~ "=.._,--..-..A-. ~ ’- ’

[This is a ve~ useful equation called the ~irst-order equal. Here [A]0 is theconcentration of A at time = 0, and [A]t is the concentration of A at time = t. T~ate

~constantk is the first-order rat~ constant. The concentration [A]t decrease~ ~as the ti~e1~ This equation allows th~lation or [he rate ~ is known,

~<and [A]I i~ measured at time t. Also, once k ~s known, [A]t can ~e calculated for any future

time.

~ ~T° determin~w~2ther a reaction is firs~ order, we r~arranged the fizst~~thef~[~:~ __ ~ ~ ~

~ln [A]t =- kt + In [A]o ~ ~, " ~’ ’ , ~ ,

corresponding to the linear equation ~ ~ ~~ ~ ~ ~

Here m is the alo~ofthe ~ne and b is the inteme~on the y axis, Gomoaring the last twoequations, we oan equate y and ~to experimental quantities. ~

~ y = IMA]t and

254 Chemistry, Ch. 13: Chemical Kinetics

Therefore,~ 2 ~ and the~~, Thusversus t for a first-order reaction gives a straight line with a

~ e~;zg.]~_. If Iotof~l versus tields a curved line rather than a stra~reaction is not a first-order reaction. This r hical rocedure is the method usedchemists to determine whether or not a given reaction is first

In

where 0.693 is a con~qp~ta,~Lan~L~is the rate constant. Knowledge of the half-life allows thecalcul~ rate constant k. The half-life of a r---~action is ~&e~e~Jr~;L~r J:be~o~ntration of a reactant to decre~L~&ase to half of the initial value. After one half-life, the ratio[A] t/[A]o is equal to 0.5.

If the reaction continues, then [A]t will drop by 1/2 again during the second half-lifeperiod as shown in Figure 13.2. After two half-life periods the fraction of the originalconcentration of A remaining, [A]t/[A]o, will be 1/2 of the concentration remaining after thefirst halfqife, so [A]t/[A]o = 0.5 x 1/2 = 0.25.


Number of half-lives elapsed

curved (exponenti"~l) h’-=ne._~Over e._._~ach halfClife period, [A]t drops in half~. £5,5’~-~’-,<° ~

qd-Order Reactions. In a second-order reaction, the~rfional either (1)~. "~.~o the square of the Concentr~]on of one reactant,

A --~ product

o~, (2) to the product of the concentrations of two reactants, each raised to the fi~’st power:

A + B --> product

_ A~.~] k[AI[B]

T~hi~hj_~.s reaction is first order in A and first order in B and so it is se_~ond or_ d~eJr~ooverall. For a, acond-order reactio ate constant has units _l/(mo!~Eity ×~

~-~ important equation calle~ integrated second-order equation is ,

.here_[A~, [A~£, and t have their usual meanin and’ _: - onst nt.’-his equation allows the calculation of the concentration of A at any time t after the reactionI~as begun. Alternatively, if [A]o, [Air, and t are known, the rate constant can be calculated.F~ample 13.6 in the text shows an example calculation using this equation.~he half-life for a second-order reaction is given by


Nere we see that the half-life is inversely proportional to the initial concentration, [A]0. This/~_ si.tuation is different from that for a first-order reaction, where t~2 is independent of [A]0.

~ t~mr .=,.’ A~-order react,on is one in which the rate does not depe~

A -~, products

:~ ra..r..r..r..r..r.~te- -[A] = k[A],At

~fime is a straight line for a zero order reaction. The rate does not slow

Summary. The integrated rate equations and graphical methods allow us to distinguishbetween the various overall orders of reaction.

1. ~A reaction is first order When a plot of~n [A_]t~versus t is a straight line. A plot of [A]tversus time wil~ be cur~e@_. ~ ~

-:~.~.~.~.~.~.~.~.~e2. A.~[ea~ction is zero order when a plot of [A]t versus t is a stra~ line.~ 3. A reaction is second or~ versus t is a straight line, and ~

~ersus t is cu~ed.~ ~ Examples

~ 4. The half-life ~uations also provide a wa~ to distinguish be~een fiEs~-#n_ d .13.6, 13.7

~on~ re~t~ons. The half-life of a first-order reaction is indCp~odent#LExercises

~~c~Lr~, whereas the half-life of a second-order reaction is~nversely propo~ona~ to the ~n~t~al concentrat~on.~ ,

~--~ ~ _ -~_

13-6- 13.

@~ ACt~ffat~on Energy and Temperature Dependence of Reaction Rates (Section 13.4}

--_ ~ollision rhOo~. The collision theo~ of chemica~ reac~r~vides a ~eneral~!~afion of how reactbn rate~d~ reactant conc~ ~ temp~ature.The~dea~ofthetheo are: ~

,~") In order for ~les. or ions to react tb_ev must first col~. Theco,,,s,on fre .more concentrated th@c ants, the greater the collision frequency, and the, eaction

~.2"~ For molecules to react the~_ must come together in the proper orientation. Molecules can"" be com~plex ~nd often it is just on----~ ~tom in a molecule iha{ reacts upon col sion with

~ When reactant molecules collide,~th~ossess a minimum amou~nt.of kinetics.en....&QgZg£ in order for an effective collisioq--a reaction-- o~ Wit~ t ’necessaryen.~.r two molecules will just bump each other and bounce back without reacting. The_rn~mmum amount of en~~ a chemical reaction Is called the

Effect of Ternperatur_~e. ~of a reaction syslem is an important variablebecause of its ~reactioq. rates. As a~, reactio~n rate~ss a_&p_proximatelydouble with a 10 °C rise in temperature. In general, tb_e rate _e~n_ is_ ra~te_=-- k[A]x[B]y. ,

~-~Since the concentration~s [A] and [B] are unaffected~by temperature, it is the rate constant

Chemistt’y, Ch. 13." Chemical Kinetics 257

t~ temperature. In 1889, S. Arrhenius found that a plot of theIo~rithm of the rate constant (In k) versu~ the rec~..~_of the absokJte temperature 1___/~gave a straight line, See Figure~ (text). ~rrhenius ~h~ ~as

~g related to an energ~

where ~ is the id~ (in ~s) and Ea is ~ion e~. TheI~he Arrhenius equation is

where In A is the interc~Both the Arrhenius ~and ~ar~_constants for a ~a~iculprreaction. Takin~~ives the Arrhenius e~which relates the rate~¢mperature.

The MeaM~ofA and Ea. The parameter A is called the ffeq~ It is relatedto the frequency of molecular collisions, and the fr tio~[~ at ~ the c~rector~~iSCUSSed above, ~ion frequency is impo~ant because moleculesmust collide in order to react.

.~ ene_&Qgzg2~t~ted to th~_e formatio~a~__ctivated co__ _m~p_l~ex.~Thea tivated c le i t ~ er intermediate species that d~~.Figure 13.3 shows that t_he ~n ener.qy is the differe e in e et ~

..Le, a~ants and the activate_~d._c~:aj:zL&x. The~ ener~kihet c energ~of ra idl movin molecules during_O_ollisiq.l:z~ ~s must "get over the ~they become products. The factor e-EdR~- that.~_t_g_Epears in the Arr " . i ifraction of molecules with ener,q es equal to or~[eater than the activation ene~o_.rchart ’ in ficantl ’ . As t.~perature increases a qreater fractio_l~qfmolecules have an energy equa~r than, the activation energy.

Activatedcomplex

A+B

C+D

Reaction Progress

Figure 13.3 Molecules of A and B with average energies must firstacquire an energy Ea before they can react.


From the Arrhenius equation we can point out that reactions for which Ea is_~much slower than those for which Ea is small. A ’ ,SEa Inc, reases, the n~tive exponent

~ and so kA convenient equation that can be used to calculate the activation energy is

where kl is the rate constant at temperature T1, and k2 is the rate constanLat.~Use of this equation retiuires t~hat rate constants k~ and k2 be

/~<red at two temperatures T1 and T2, respectively.

t~-R.~ ~_c~h_an’LSl~ (Section 13.5) ~ ,~\-r~,~-~p,~L~_~_~~,~

~ Elementa~ Steps. Th~ur_ose of a. r bn mechanism ~thereactants are ~onve~ed into products. A reaction ~hani~m consists of a set ofelementaH step~. A mechanism indicates which molec~~

~. Each elementa~ step marks the progress of the conveyor.sof reactants into products. A m~di~~ediates that are for~~~~ as, which steps are ~st an~~. Thisinformation is n~t suppliedJ lied bb thethe balabala ch~che ’ ~1 e uation because ........ it is intoned only

~icate the number of moles of one reactant that are consumed per mole of the otherr£8c~ ~.

Elementa~ steps have a pro~e~ularit~ which pe~ains to the number ofmolecules that must collide in a si~. An e emeutar~ that involves three~olecules is called termol~cular. A ~ that involves two molecules is bimolecular, and~step in which one molecule de~ses or rearranges is called unimolecular. Thus the

represents a unimolecular reaction in which a sin~q~e molecule of A reacts to for~e~B. The following equation represents a bimolecular reaction.

A+B->C

That is one in which a molecule of A collides with a molecule of B, and as a moult amolecule of C is formed.~--~

Reaction Order for Elementary Steps. T~e rate of a unimolecular elementary step

A-eB

will__depend on the number of molecules of A per unit volume of t~’ e~r. Therefore,unimolecutar r~w first-order rate laws.

rate = k[A]

Th.e, zatb~q~_ bimolecular ste A +~ill d..~_pend on the number of molecules of A per ounit volume times the number of molecules of B per unit volume.

~ rate =-~[A][a]

Therefore, bimolecular steps follow second-order rate laws.


This also can be understood by realizing that A and B must collide wi h each other if they.Lare to react, and so the rate of a bimolecular process de[aends on the r_ate of collisions of A

"-a~d B. Do~ub~the concentration of A will double the roEg...~_bility of collisi~~B.~ilarly, doubli the concentration o ’1 also double th r t

See Figure 13.14 in the textbook. Reasoning further along these lines, we conclude thattermolecular reactions are third order,

~-~ Rate-Determining Step. It often turns out that one of the elementa~mechanism ~than all the rest. ::~~Y~es the overall rate of reaction

others~-v~ thro-u~h the line. The s_~_low_.~est step in a sequence of elemen[ary~the rat~determ~his allows us to say that the overali-law~d_b~mecha...___~nism will be the one corresponding to the rate--ling step.

Te~anism~ The.sequence of events in a kinetic study of a reaction thatleads to a proposed mechanism is as follows:

1. Determine the rate la~.2, Pr._~9~ose a mechanism for the reaction.3. Test the mechanism.

~osed mechanism, assume ones~ to be the rate-determining Examplestep, Then establish the rate law~fo_r that step_~ves the ra~redicte~’ b~L13.!0the mechanism. If the mechanism is ade UCLM~ then when the.oredicted rate law is~ the e~jaerimental rate law, the two will m~at~. On the other hand, if~. I Exercisesth~predlcted and expenmantal rate laws do not match, the mechanism has been 13-16-13-18proved inadequate.~iy those mechanisms consistent with all the data can be

~ered adequ-~e.

t~C~t~lysis (Section t3.6)

"-’~l~’echanism era Catalyzed Reaction. A c~i~talyst is a substance that increases the ra~te~--In th~rea~or this r~ason a-~a-at.alystdoes not appear I~ the stoichiometric equatio~n. Chromium III ox~e, Cr~Q , is t~-e~s_t

.~sent ir~ ~atal t~ converters in Ame.rican automobiles. This c~ound c eler ~ereaction of~onoxide.with oxygen. By converting CO to CO~n the exhaust stream~,CO emissions are reduce~d,

Cr203

CO(g) + ~O2(g) ~ CO2(g)

The lifetime of catalytic converters is quite long; the Cr203 and other catalysts do not needto be replaced.

~ rea&tion rates by providing a new reaction pathwa_~ [m_echanisro)which has a lower activation energy.~Se~ ’ igure 1 ~.. 17 (textboo~),, In the catalytic converte,~__~..~ ~_"~ ~.both CO and O2 are chemically adsorbed on the catalyst’s surface. Adsorption of 02 byCr2Q3 weakens the Q--Q bond enough so that oxygen atoms can react with adsorbed CO.The reaction path involving a weakened Q--Q bond has a significantly lower activationenergy than that of the reaction occurring purely in the gas phase. Recall that as theactivat~n energy is lowered the rate constant for a reaction i~After CO2 is formed,it desorbs from the surface, leaving the Cr203 catalyst chemically unaltered. The catalyst is

~ ~O Chemistry, Ch. 13: Chemical Kinetics

not consumed in the reactio_j~Catalysts are regenerated in one of the last steps of themechanism. The textbook describes t_~hre~e t_y~_es of catalysts, which we will review here.

Heterogeneous Catalysis. In heterogeneous catalysis the reactants are in one phas~eand the catalyst is in another phase. The catal~d_~and the reactants a_~

_~. Ordinarily the s~te" ’-’-~~-~’ of the reaction is the surface of the solid catalyst. Manyindustrially important reactions involve gases and are catalyzed by solid surfaces. For

ex..~le~, the Haber£rocess for the synthe~i~,ef ammonia is catalyzed by iron plus a fewpercent ofthe oxides of potassium~h-d a u~qQ~qRQTn~ Th-~f~s-Zi~ - a alyst or

~~ polye~ polymer contains triethylaluminum andtitanium or vanadium salts.

~omoq~eneous Catalysis. In homocjAneous catalysis the reactants ,oroducts,_and_catalysts are all in the same phase, which is usually the ga~s or the li uq~id ~yreactions ar~catal~d~he decomposition of formic acid (HCO2H) is an example.

HCO2H -~ H20 + CO

Formic acid is normally stable and lasts a long time on the shelf. However, when sulfuricacid is added, bubbles of carbon monoxide gas can be observed immediately. Thehydrogen ions from the sulfuric acid initiate a new reaction path. The mechanism is

HCO2H + H÷ --> HCO2H~HCO2H~ ,~ H20 + HCO+

HCO÷ -4. CO + H+

HCO2H -+ CO + H20

\Note that the H+ ion is consumed in the first step, but, as for all catalysts, H+ is regenerated.The net reaction is the sum of the three steps. The intermediate HCO2H~ cancels out asdoes the catalyst.

Enzyme Cata~s.is. Nea__&&~ all chemical reactions that occur in living organisms requirecat~~~’e biolo~a! c~s s. enzymes are~rotein m61~ul~

mo’"~ar masses~er 10,000 amu~_.~AII enzyme molecules are highly specific with~:the reactions they catalyze. That is, they can only affect the reaction rates of afew specific reactant molecules¯ T~ypically, an enz ta! zes ’ action, or a~g~roup

~ ose ions¯ The_~_molecule on which an enzyme acts is called a substrate.The simplest mechanism which explains enzyme activity, and is consistent

with these trends is one in which the substrate S and the enzyme E form anenzyme-substrate complex, ES. This complex has a lower energy than theactivated complex without the enzyme (Figure 13¯25 of the text). The enzyme-substrate complex can either dissociate back into E and S, or break apart into theproducts P and the regenerated enzyme E. The simplest mechanism has the twosteps shown below.

Example13.11

Exercises13-19-13-20

E+S ~ ESES->E+P

WORKED EXAMPLES

EXAMPLE 13,1 Expressing the Rate of Reaction

Write expressions for the rate of the following reaction in terms of each of the reactants andproducts.

2 N2Os(g) -~ 4 NO2(g) + O2(g)

¯ Solution

Recall that the rate is defined as the change in concentration of a reactant or product withtime. Each "change-in-concentration" term is divided by the corresponding stoichiomelriccoefficient. Terms involving reactants are preceded by a minus sign. Therefore, the rate isexpressed as

rate = A[N2Os] - A[NO2] - z~[O2]2 At 4 At At


EXAMPLE 13.2 Rate of Reaction

Oxygen gas is formed by the decomposition of nitric oxide:

2 NO(g) --> O2(g) + N2(g)

If the rate of formation of 02 is 0.054 M/s, what !s~t~e rate of change of NO concentration?

¯ Solution

From the stoichiometry of the reaction, 2 mol NO react for each mole of 02 that forms.

rate = - A[NO] ~ A[O2]2 At At

A[NO]_ 2A[O2] _ 2(0.054 M/s)At At

=-0.11 M/s

EXAMPLE 13.3 Calculation of the Average Rate

Experimental data for the hypothetical reaction:

A-->2B

are listed in the following table.

Time (s) [AI (mol/L)0.00 1.00010.0 0.89120.0 0.79430.0 0.70740.0 0.630

a. Calculate the average rates of change of [A], and the average reaction rates for the twotime intervals from 0 to 10 s and from 30 to 40 s.

b. Why does the rate decrease from one time interval to the next?

¯ Solution

The average rate of change of [A] is given by:

A[A] [A]2 [A]IAt t2 - t1

For the time interval 0 - 10 s, we get:

A[A] _ 0.891 mol/L 1.00 mol/L -0.109 mol/L= = - 0.0109 mol/UsAt 10.0 s-0 s 10.0 s


Since the average reaction rate - A[A], then:At

average reaction rate = - (- 0.0109 mol/L.s) = 0.0109 mol/L.s

For the time interval 30 - 40 s, the rate of change of [A] is:

A[A] = (0.630 tool/L-0.707 tool/L) -0,77 mol/L 0.0077 mol/L-sAt 40.0 s-30.0s 10.0s

And the average reaction rate is:

average rate = - A[A] = 0.0077 mol/L.s

b. The reaction rate decreases as the total reaction time increases because the rate is ¯proportional to the concentration of the reactant A, and the concentration of A decreasesas the time of reaction increases.

EXAMPLE 13.4 Concentration Effect on the Rate

The reaction A + 2B ---> products was found to have the rate law: rate = k[A][B]3. By whatfactor will the rate of reaction increase if the concentration of B is increased from c to 3c,while the concentration of A is held constant?

¯ Solution

Nrite a ratio of the rate law expressions for the two different concentrations of B.

rate (expt 2) k[A][3c]3 33c3

rate (expt 1) k[A][c]3 c~

The rate will increase 27-fold when [B] is increased three-fold.

EXAMPLE 13.5 Finding the Rate LawThe following rate data were collected for the reaction:

2NO+2H2~N2+2H20

Experiment [NO]0 [H2]0 A[N2]/At(M) (M) (M/h)

1234

0.60 0.15 0.0760.60 0.30 0.150.60 0.60 0.301.20 0.60 1.21

a. Determine the rate law,b. Calculate the rate constant.


o Solution

a. We want to determine the exponents in the equation

rate = k[NO]x[H2]Y

To determine the order with respect to H2, first find two experiments in which [NO] isl~eld constant. This can be done by comparing the data of experiments 1 and 2. When theconcentration of H2 is doubled, the reaction rate d~’~bles. Thus, the reaction is first order inH2. When the NO concentration is doubled (experiments 3 and 4) the reaction ratequadruples. Therefore, the reaction is second order in NO.

The rate law is

rate = k[NO]2[H2]

b. Rearrange the rate law from part (a).

ratek=

[NO]2[H2]

Then substitute data from any one of the experiments. Using experiment 1:

0.076 M/hk = = 1.4/M2 h

(0.60 M12(0.15 M)

¯ Comment

You should get the same value of k from all four experiments. Note that the units of k arethose of a third-order rate constant. Take a closer look at proving that x = 2. Write the ratioof rate laws for experiments 4 and 3.

rate4 k[NO]X[H2]rate3 k[NO]×[H2]

1,21 k(1.20)x(0.60) (1.20tx0.30 k(0.60)×(0.60) \0.60J

4.0 = 2.0xTherefore x = 2

EXAMPLE 13.6 First-Order Reaction

Methyl isocyanide undergoes a rearrangement to form methyl cyanide that follows first-orderkinetics.

CH3NC(g) --> CH3CN(g)


The reaction was studied at 199 °C, The initial concentration of CH~NC was 0,0258 mol/L,and after 11.4 min analysis showed the concentration of product was 1.30 x 10-3 mol/L.a, What is the first-order rate constant?b. What is the half-life of methyl isocyanide?c. How long will it take for 90 percent of the CH3NC to react?

¯ Solution for (a)

This problem illustrates the use of the integrated first-order rate equation

In [CH3NC] --kt[CH~NC]~

Where k is the rate constant and [CH3NC] is the reactant concentration at time t. The initialconcentration is [CH3NC]o = 0.0258 M. After 11.4 rain the product concentration is 1.30 x10-3 M. This implies that the concentration of CH3NC remaining unreacted is 0.0258 -0.0013 = 0.0245 M.

Substitution gives

0.0245MIn = -k (11.4 min)

0.0258 M

In 0.950 = -k (11.4 rain)

k = 4.54 x 10-3/min

, Solution for (b)

For a first-order reaction the half-life equation is

0,693 0,693 - 153 mint~z2 = ~-4.54 x 10-~/min

¯ Solution for (c)

If 90 percent of the initial CH3NC is consumed, then 10 percent remains. Therefore:

[CH3NC] = 0.10 [CH3NC]0

Substitution into the rate equation gives

In 0’10[CH3NC]° -[CH3NC]0

Solving for t, we get:

In 0,10t-

4.54 x 10-3/min =

(4.54 x 10-3/min)t

2.303- 507 min

4.54 x 10 3/min


EXAMPLE 13.7 First-Order Reaction

At 230 °C the rate constant for methyl isocyanide isomerization is 9,25 × 10~/s,

CH3NC -+ CH3CN

a. What fraction of the original isocyanide will remain after 60.0 min?b. What is the half-life of methyl isocyanide at this temperature?

¯ Solution -~

a, Again applying the first-order equation:

In [CH3NC]- =-kt[CH3NC]0

The fraction of methyl isocyanide remaining is given by the fraction [CH3NC][CH3NC]0

Plug in the rate constant and the time, and solve for the fraction remaiming:

In [CH3NC] _ 9.25 × 10-~/s (60 rain) × 60~s =_3.33[CH~NC]~ 1 min

Taking the antilog of both sides gives:

[CH3NC~]] = e_3.33 = 0.0358

[CH~NC]o

b. The half-life is

0.6939.25 ×~O~/s

= 749 s or 12.4 min

¯ Comment

Note the much shorter half-life at 230 °C than at 199 °C (Example 13.6).

EXAMPLE 13.8 Calculating Activation E~nergy

For the reaction

NO + 03 ~ NO2 + 02

the following rate constants have been obtained:

Temperature, °C k(1/M.s)

10.0 9.30 x 10-~-30.0 1.25 × 107

Calculate the activation energy for this reaction.

Chemistry, Ch. 13: Chemical Kinetics 2"71

o Solution

Here we are given two rate constants c~)rresponding to two temperatures. The activationenergy (Ea) can be calculated by substituting into the equation given earlier that shows thetemperature dependence of the rate constant.

Let kl = 9.3 x 10~/M.s at T1 = 273 + 10.0 : 283 Kk2 = 1.25 x 107/M,s at T2 = 273 + 30.0 = 303 K

Recall R = 8.31 J/moI.K. Substitute into the preceding equation to solve for Ea:

1.25x107/M.s - 8,314J/moI.K~,(283 K)(303 ~-)

Solving for Ea yields:

(in 0.744)(8,314 J/moi,K) = Ea 857~-)

(-0.296)(8.314 J/mol K) = Ea(-2.33 × lob/K)

Ea = 1.06 x 104 J/tool (or 10.6 kJ/mol)

EXAMPLE 13.9 Reaction Energy Profile

Draw a reaction energy profile for the following endothermic reaction:

2 HI(g) --> H2(g) + 12(g) z~H%n = 12.5 kJ

Given the activation energy Ea = 185 kJ/mol. Label the activation energy and the activatedcomplex. What is the activation energy for the reverse reaction?

¯ Solution

The reaction is endothermic so the products are at a higher level than the reactants.

Activatedcomplex

2HI

185kJ

/H2+I2

AH = 12.5 kJI

Reaction Progress


Using the diagram we can see that the activation energy for the reverse reaction is 185 kJ -~12.5 kJ = 173 kJ.

EXAMPLE 13,10 Reaction Mechanism

The rate law for the following reaction has been experimentally determined to be third order:

2 NO(g) + O2(g) --,’, 2 NO2(g) rate = k[N~312[O2]

Which of the two proposed mechanisms that follow is more satisfactory?

kl8. NO + NO --~ N202 slow b. NO + NO ~ N202 fast

k2 k2N202 + 02 -+ 2 NO2 fast N202 + 02 ---> 2 NO2 slow

Solution

Testing mechanism (a) first, we note that the first step is the rate-determining step. The ratetaw for this bimolecular step is:

rate = k~[NO]2 I

This mechanism predicts a rate law that is second order in NO concentration and zero orderin 02. By comparison, the experimental rate law is second order in NQ and first order in 02.The predicted rate law and the experimental rate law do not match, and so mechanism (a) isnot satisfactory.

Testing mechanism (b), we find that the rate-determining step is the second step.Since it is bimolecular, its rate law should be:

rate = k2[N202][O~] II

It is not possible to compare this predicted rate law directly with the experimental rate lawbecause of the unique term, which is the concentration of N202. ~n this mechanism, N202 iSan intermediate. Being formed in step 1 and consumed in step 2, its concentration is alwayssmall and usually not measurable.

The way around this situation is to find a mathematical substitution for [N202]. Notethat step 1 is reversible and equilibrium can be established. This means that the rates of theforward and reverse reactions are equal:

kl[NO]2 = k-1[N202]

where k~ is the late constant for the forward reaction, and k-1 for the reverse.Rearranging terms gives us:

[N202] = ~ [NO]2


We now have an expression for [N202], which we can substitute into equation II. This yields:

rate- k2kl [NO]2[O2]kl

Whenever a reaction step is reversible, we can use tl]e equality of the forward and reverserates to form an expression to substitute for the coh’~entration of an intermediate.

Now we compare this rate law with the experimental rate law, which is

rate = k[NO]2[O2]

Ne see that this mechanism predicts that the reaction will be second order in NQ and firstorder in 02, just as observed. Also, the collection of constants k2kl/k_l will equal the rateconstant.

Therefore, the second mechanism predicts a rate law that matches the experimental orderof reaction.

EXAMPLE 13.11 Intermediates and Catalysts

3iven the following mechanism for the decomposition of ozone in the stratosphere, identifythe intermediate and the catalyst:

03 + CI -~ 02 + C}OCIO + 0-~ CI + 02

¯ Solution

Adding the two steps gives the overall reaction

03+0-->.202

An intermediate is formed in an early step and removed in a later step and does not appearin the overall equation; therefore ClO is the intermediate. A catalyst is consumed in an earlystep and is later regenerated. Atomic chlorine, Cl, is the catalyst.

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Chapter 13. Chemical Kinetics · 2013-02-05 · Chapter 13. Chemical Kinetics The Rate of Reaction (Section 13.1) Rate Laws (Section 13.2) The Relation Between Reactant Concentration