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Chemical Kinetics

Review Questions

13.1 Unlike mammals, which actively regulate their body temperature through metabolic activity,lizards are ectotherms—their body temperature depends on their surroundings. Whensplashed with cold water, a lizard's body simply gets colder. The drop in body temperatureimmobilizes the lizard because its movement depends on chemical reactions that occurwithin its muscles, and the rates of those reactions—how fast they occur—are highly sensitiveto temperature. In other words, when the temperature drops, the reactions that producemovement occur more slowly; therefore the movement itself slows down. Cold reptiles arelethargic, unable to move very quickly. For this reason, reptiles try to maintain their bodytemperature in a narrow range by moving between sun and shade.

The rates of chemical reactions, and especially the ability to control those rates, are importantphenomena in our everyday lives. For example, the human body's ability to switch a specificreaction on or off at a specific time is achieved largely by controlling the rate of that reactionthrough the use of enzymes. Chemical kinetics is an important subject to chemists and engi-neers. The launching of a rocket depends on controlling the rate at which fuel bums—tooquickly and the rocket can explode, too slowly and it will not leave the ground. The rate ofnuclear decay in a nuclear power plant must be carefully controlled in order to provide elec-tricity safely and efficiently. Chemists must always consider reaction rates when synthesizingcompounds. No matter how stable a compound might be, its synthesis is impossible if the rateat which it forms is too slow. As we have seen with reptiles, reaction rates are important to life.

13.3 The rate of a chemical reaction is measured as a change in the amounts of reactants or prod-ucts (usually in terms of concentration) divided by the change in time. Typical units aremolarity per second (M/s), molarity per minute (M/min), and molarity per year (M/yr),depending on how fast the reaction proceeds.

13.4 The reaction rate is defined as the negative of the change in concentration of a reactantdivided by the change in time, because reactant concentrations decrease as a reaction pro-ceeds; therefore the change in the concentration of a reactant is negative. The negative sign inthe definition thus makes the overall rate positive. In other words, the negative sign is theresult of the convention that reaction rates are usually reported as positive quantities. Sincethe product concentrations are increasing, the concentration of a product divided by thechange in time is positive.

13.5 The average rate of the reaction can be calculated for any time interval asl[A],2-[A]( l l[B],2-[B](l l[C] f2-[C] f l l[D](2-[D](l

Rate = - = - - —- - = 1 - = ~ — for thea t2 ~ fi b t2 ~ t\ c t2 ~ t\ d t2 - ti

chemical reaction aA + bB —> cC + dD. The instantaneous rate of the reaction is the rate atany one point in time, represented by the instantaneous slope of the plot of concentrationversus time at that point. We can obtain the instantaneous rate from the slope of the tangentto this curve at the point of interest.

501

502 Chapter 13 Chemical Kinetics

13.6 For a zero order reaction, Rate = k [A]° = k, so doubling the concentration of A does nothing to the reactionrate. For a first order reaction, Rate = k [A]1 = k [A], so doubling the concentration of A doubles the reactionrate. For a second order reaction, Rate = k [A]2, so doubling the concentration of A quadruples the reaction rate.

13.7 The reaction order cannot be determined by the stoichiometry of the reaction. It can only be determined byrunning controlled experiments where the concentrations of the reactants are varied and the reaction ratesare measured and analyzed.

13.8 When multiple reactants are present, Rate = k [A]m[B]", where m is the reaction order with respect to A andn is the reaction order with respect to B. The overall order is simply the sum of the exponents (m + n).

13.9 The rate law shows the relationship between the rate of a reaction and the concentrations of the reactants. Theintegrated rate law for a chemical reaction is a relationship between the concentration of a reactant and time.

13.10 / For a zero order reaction, [A]( = —kt + [A]o- For a first order reaction, ln[A]( = - k t + ln[A]0. For a second

J

r \\ 13.11

order reaction, - - = kt + - —.[A], [A]0

The half-life (^1/2) of a reaction is the time required for the concentration of a reactant to fall to one-half of

\^__y its initial value. For a zero order reaction, tj/2 = ~^~- For a first order reaction, f j/2 = -^—. For a second1

order reaction, tin = , r , .*[A]0

13.12 The rates of chemical reactions are, in general, highly sensitive to temperature. Reaction rates increase as thetemperature increases. The temperature dependence of the reaction rate is contained in the rate constant (k)which is actually a constant only when the temperature remains constant.

13.13 The modern form of the Arrhenius equation, which relates the rate constant (k) and the temperature inkelvin (T), is as follows: k = A e~E*/RT, where R is the gas constant (8.314 J/mol • K), A is a constant calledthe frequency factor (or the pre-exponential factor), and Ea is called the activation energy (or activation bar-rier). The frequency factor is the number of times that the reactants approach the activation barrier per unittime. The exponential factor (-Ea/RT) is the fraction of approaches that are successful in surmounting theactivation barrier and forming products. The exponential factor increases with increasing temperature, butdecreases with an increasing value for the activation energy. As the temperature increases, the number ofcollisions increases and the number of molecules having enough thermal energy to surmount the activa-tion barrier increases. At any given temperature, a sample of molecules will have a distribution of energies,as shown in Figure 13.14. Under common circumstances, only a small fraction of the molecules haveenough energy to make it over the activation barrier. Because of the shape of the energy distribution curve,however, a small change in temperature results in a large difference in the number of molecules havingenough energy to surmount the activation barrier.

13.14 An Arrhenius Plot is a plot of the natural log of the rate constant (In k) versus the inverse of the temperaturein Kelvin (l/T). It yields a straight line with a slope of -£a/R and a y-intercept of In A.

13.15 In the collision model, a chemical reaction occurs after a sufficiently energetic collision between the tworeactant molecules. In collision theory, therefore, each approach to the activation barrier is a collisionbetween the reactant molecules. The value of the frequency factor should simply be the number of collisionsthat occur per second. In the collision model k = pze~E°/RT, where the frequency factor (A) has been sepa-rated into two separate parts—p is called the orientation factor and z is the collision frequency. The collisionfrequency is simply the number of collisions that occur per unit time. The orientation factor says that if twomolecules are to react with each other, they must collide in such a way that allows the necessary bonds tobreak and form. The small orientation factor indicates that the orientational requirements for this reactionare fairly stringent—the molecules must be aligned in a very specific way for the reaction to occur. Whentwo molecules with sufficient energy and the correct orientation collide, something unique happens: Theelectrons on one of the atoms or molecules are attracted to the nuclei of the other; some bonds begin toweaken while other bonds begin to form and; if all goes well, the reactants go through the transition stateand are transformed into the products and a chemical reaction occurs.

Chapter 13 Chemical Kinetics 503

13.16 When we write a chemical equation to represent a chemical reaction, we usually represent the overall reac-tion, not the series of individual steps by which the reaction occurs. The overall equation simply shows thesubstances present at the beginning of the reaction and the substances formed by the reaction—it does notshow the intermediate steps that may be involved. A reaction mechanism is a series of individual chemicalsteps by which an overall chemical reaction occurs.

13.17 An elementary step is a single step in a reaction mechanism. Elementary steps cannot be broken down into sim-pler steps—they occur as they are written. Elementary steps are characterized by their molecularity, the numberof reactant particles involved in the step. The molecularity of the three most common types of elementary stepsare as follows: unimolecular - A —> products and Rate = k [A]- bimolecular - A + A —> products andRate = k [A]2; and bimolecular - A + B -> products and Rate = k [A][B]. Elementary steps in which threereactant particles collide, called termolecular steps, are very rare because the probability of three particlessimultaneously colliding is small.

13.18 For a proposed reaction mechanism to be valid—mechanisms can only be validated, not proven—two con-ditions must be met: (1) the elementary steps in the mechanism must sum to the overall reaction; and (2) therate law predicted by the mechanism must be consistent with the experimentally observed rate law.

Reaction intermediates are species that are formed in one step of a mechanism and consumed in anotherstep. An intermediate is not found in the balanced equation for the overall reaction, but plays a key role inthe mechanism.

13.20 A catalyst is a substance that increases the rate of a chemical reaction but is not consumed by the reaction. Acatalyst works by providing an alternative mechanism for the reaction—one in which the rate-determiningstep has a lower activation energy.

13.21 In homogeneous catalysis, the catalyst exists in the same phase as the reactants. In heterogeneous catalysis,the catalyst exists in a phase different from the reactants. The most common type of heterogenous catalystis a solid catalyst.

13.22 Heterogeneous catalysis (involving solid catalysts) occurs by the following four-step process: (1) adsorption—the reactants are adsorbed onto the solid surface; (2) diffusion—the reactants diffuse on the surface until theyapproach each other; (3) reaction—the reactants react to form the products; and (4) desorption—the productsdesorb from the surface into the gas phase.

13.23 Enzymes are biological catalysts that increase the rates of biochemical reactions. Enzymes are large pro-tein molecules with complex three-dimensional structures. Within that structure is a specific area calledthe active site. The properties and shape of the active site are just right to bind the reactant molecule,usually called the substrate. The substrate fits into the active site in a manner that is analogous to a keyfitting into a lock.

13.24 The general mechanism by which an enzyme (E) binds a substrate (S) and then reacts to form the products(P) is as follows: (1) E + S ;==i ES (fast); and (2) ES -» E + P (slow, rate-limiting).

Reaction Rates1 A[HBr] A[H2] A[Br2]13.25) (a) Rate = -__L-£l = _LJi = -LM

(b) Given: first 25.0 s; 0.600 M to 0.512 M Find: average rateConceptual Plan: rx, t^ [HBrJj, [HBr]2 —> average rate

lA[HBr]Rate = - —

Rate = - ' 2 " = -1°-512M - °-600M - l.76xlO-*M..-» - 1-8 x IQ^M-s"2 *2 - h 2 25.0s-0.0s

Check: The units (M • s"1) are correct. The magnitude of the answer (10"3M • s"1) makes physical sensebecause rates are always positive and we are not changing the concentration much in 25 s.

504 _ Chapter 13 Chemical Kinetics

(c) Given: 1.50 L vessel, first 15.0 s of reaction, and part (b) data Find: molBr2 formedConceptual Plan: average rate, t\, t & — » A [Br2] then A [Br2], L — > molBr2 formed

A[Br2] molBrRate = ~ M = -^

_. A[Br2] A[Br2]Solution: Rate = 1 .76 x 10 J M • s = — - - = -— - — — - . Rearrange to solve for A[Br2l.Af 15.0s - 0.0s

, M rnolBrA [Br2] = 1 .76 x HT3 — x 15 .0 s = 0 .0264 M then M = . Rearrange to solve for molBr2.

0.264 x 1.50 L = 0.040 mol Br2.L

Check: The units (mol) are correct. The magnitude of the answer (0.04 mol) makes physical sensebecause the time is shorter than in part (b) and we need to divide by 2 and multiply by 1.5 becauseof the stoichiometric coefficient difference and the volume of the vessel, respectively.

1 A[N20] 1 A[N2] A[02]13.26 (a) Rate = --- = -- = -

2 Af 2 Af Af

(b) Given: first 15.0 s; 0.015 mol O2 in 0.500 L Find: average rateConceptual Plan: mo!O2 , L — » M then tlf t-^ [O2]i, [O2]2 — * average rate

molo A[O2JM = - - Rate = — —

L A(

molo, 0.015 mol O2Solution: At 1 1 = 0 s we have no O2. M = — - — = — T T -- = °-030 M then

I-*™ - [Qzltj 0.030M - 0.000M ,Rate = - - = 2.0xlO~3M-s~1.

t-i - fi 15.0s-0.0sCheck: The units (M • s"1) are correct. The magnitude of the answer (10~3 M • s"1) makes physical sensebecause rates are always positive and we are not changing the concentration much in 15 s.

A[N2O](c) Given: part (b) data Find: —

A[N2O]Conceptual Plan: average rate —» —

1 A[N20] _ A[QJ

1 A[N20] A[02] A[N20]Solution: Rate = = . Rearrange to solve for .

2 Af Af Af

= -2x2 .0x lO~ 3 M-s~ 1 = - 0.0040M-s"A[N2O] n «i^zi » „ „ ,,,-3», -1 rt™,,^, -iAf Af

Check: The units (M • s"1) are correct. The magnitude of the answer (- 0.004 M • s"1) makes physicalsense because we multiply by 2 because of the different stoichiometric coefficients. The change in con-centration with time is negative since this is a reactant.

1 A[A] A[B] 1 A[C]

A [A] A[B] A[C]Given: - — = - 0.100 M/s Find: — — , and -

Af Af AfA [A] A[B] A[C]

Conceptual Plan: -- * — - — , and — - —Af Af Af

_ 1 A[A^_ _ A[B] _ i A[q

2 At ° " Af "3 At

1 A[A] A[B] 1 A[C]Solution: Rate = --- = -- = - — - — . Substitute in value and solve for the two desired

1 -0.100M A[B] A[B] 1 - 0.100 M 1 A[C]values. -- -= -- so — — = - 0.0500 M-s and -— -

s Af Af s 3 At

so— = 0.150 M-s-1.Af


Check: The units (M • s'1) are correct. The magnitude of the answer (- 0.05 M • s'1) makes physicalsense because fewer moles of B are reacting for every mole of A and the change in concentration withtime is negative since this is a reactant. The magnitude of the answer (0.15 M • s"1) makes physicalsense because more moles of C are being formed for every mole of A reacting and the change in con-centration with time is positive since this is a product.

Af Af 2 Af

(b) Given: -±- = 0.025 M/s Find: -—, and — —

A[C] A[B] A[A]Conceptual Plan: — — , and ~^-

A[A] A[B] 1 A[C]R a t e =__ = _2_ = __

A[A] A[B] 1 A[C]Solution: Rate = -- - — = - 2 — - — = - — - — . Substitute in value and solve for the two desired

Af Af 2 AfA[B] 10.025M A[B] A[A] 1 0.025 M

values. -2 - = - - so - =-0.0063M-s l and -- = -- soA* 2 s Af Af 2 s

-~" - 0.013 M-s-1.At

Check: The units (M • s"1) are correct. The magnitude of the answer (- 0.006 M • s"1) makes physicalsense because fewer moles of B are reacting for every mole of C being formed and the change in con-centration with time is negative since this is a reactant. The magnitude of the answer (- 0.01 M • s"1)makes physical sense because fewer moles of A are reacting for every mole of C being formed and thechange in concentration with time is negative since this is a reactant. B has the smallest stoichiomet-ric coefficient and so its rate of change has the smallest magnitude.

13.29 Given: Cl2(g) + 3 F2fe) -» 2 OF3(g); A[Cl2]/Af = - 0.012 M/s Find: A[F2]/Af; A[OF3]/Af; and RateConceptual Plan: write the expression for the rate with respect to each species then

A[C12] i A[F2] i A[C1F3]Rate = -- = --- = --

At 3 At 2 AfA[C12] A[F2] A[C1F3]rate expression, -- -> — - — ; — - - ; Rate

AICIJ _ i A[F2] ! A[C1F3]

At 3 M '2 At

A[C12] 1 A[F2] 1 A[C1F3]Solution: Rate = -- = --- : — = — — : - so

Af 3 Af 2 AfA[C12] 1 A[F2] A[F2] A[F2] A[C12]— . Rearrange to solve for - — . -7^ = 3 - - = 3(- 0.012 M • s'1) = -0.036 M • s"1

Af 3 Af A/ Af AfA[C12] 1 A[C1F3] A[C1F3]and -- = - — - - . Rearrange to solve for — - - .

Check: The units (M • s"1) are correct. The magnitude of the answers (- 0.036 M • s"1 and 0.024 M • s"1) makesphysical sense because F2 is being used at three times the rate of C12, C1F3 is being formed at two times therate of C12 disappearance, and C12 has a stoichiometric coefficient of 1.

13.30 Given: 8 H2S(g) + 4 O2(g) -» 8 H2O(g) + S8(g); A[H2S]/Af = - 0.080 M/sFind: A[O2]/Af; A[H2O]/Af; A[S8]/Af; and RateConceptual Plan: write the expression for the rate with respect to each species then

! A[H2S] = i AlOj = ! A[H20] = A^]Rate 8 At 4 At ~ 8 At A/


13.32

rate expression,

Rate - - -

A[H2S]AfA[H2S]

A[02] A[H20] A[S8]

Af ; Af '' Af 'A[S8]

Solution: Rate = -

1 A[H2S]

8 A/ 4A[H2S]

8 AtA[02]

A(

A[H20]

8 Af 4 AfA[02] = 4 A[H2S]

Af 8 AfA[H2S] _ I A[H2O]

~ ~ 8

8 Af 4 Af 8 AfA[02] A[02]

. Rearrange to solve for —-—.

Af• so

= -(- 0.080M-s'1) = -0.040M-S"1

and--

A[H20]

Af

and - -

Af 8 Af8 A[H2S]8

. Rearrange to solve forA[H20]

Af

AfA[H2S] A[S8]

= -(-0.080 M-s'1) = 0.080 M-s,-1

1 A[H2S] 1= --(-0.080 M-s"1) = 0.010 M-s"1.

8 Af 8Af AfCheck: The units (M• s'1) are correct. The magnitude of the answers (- 0.040 M-s"1 - 0.080 M• s"1, and0.010 M • s"1) makes physical sense because O2 is being used at one-half times the rate of H2S, H2O is beingformed at two times the rate of H2S disappearance, S8 is being formed at one-eighth times the rate of H2Sdisappearance, and Sg has a stoichiometric coefficient of 1.

(a) Given: [C4H8] versus time data Find: average rate between 0 and 10 s, and between 40 and 50 sConceptual Plan: t\, t^ [CtH8]i, [C4H8]2 —» average rate

A[C4H8]Rate = - -

Solution: For 0 to 10 s Rate = -[QH8]f2 - [C4H8]f] o .913 M - 1.000 M

for 40 to 50 s: Rate = -

f 2 - f a

[C4H8](2 - [C4H8](l

10. s - O . s0.637M - 0.697M

= 8.7xlO~3M-s-1and

= 6.0xlO~3M-s-l

(b)

f2 - fi 50.S-40.SCheck: The units (M • s"1) are correct. The magnitude of the answer (10"3 M • s"1) makes physical sensebecause rates are always positive and we are not changing the concentration much in 10 s. Also reac-tions slow as they proceed because the concentration of the reactants is decreasing.

Given: [C4H8] versus time data Find: —— between 20 and 30 sAf

Conceptual Plan: tir [C4H8]2

Rate = -

Solution: Rate = -[C4H8](2 - [C4H8]t]

A[C2H4]

Af= 1 A[C2H4]

~ 2 Af

0.763M - 0.835M30.s-20.s

-3Af

,-iRearrange to solve for —. So — = 2(7.2 x 10"Af Af

Check: The units (M • s"1) are correct. The magnitude of the answer (10~2 M • s"1) makes physical sensebecause rate of product formation is always positive and we are not changing the concentration muchin 10 s. The rate of change of the product is faster than the decline of the reactant because of the sto-ichiometric coefficients.

(a) Given: [NO2] versus time data Find: average rate between 10 and 20 s, and between 50 and 60 sConceptual Plan: tlr f^ [NO2]lr [NO2]2 —> average rate

Rate = - -A[N02]

Af


(b)

Solution: For 10 to 20 s Rate = -[NO2](2 - [NO2](] 0.904M - 0.951 M

[N02](2 - [N02](]

20.s - 10.s0.740M - 0.778M

= 4.7xlO~3M-s~1and

for 50 to 60 s Rate = -t2 - h 60.S-50.S

Check: The units (M • s"1) are correct. The magnitude of the answer (10~3 M • s"1) makes physical sensebecause rates are always positive and we are not changing the concentration much in 10 s. Also reac-tions slow as they proceed because the concentration of the reactants is decreasing.

A[O2]Given: [NO2] versus time data Find: ——— between 50 and 60 s

At

Conceptual Plan: average rate from part a) —•

A[N02]

A[02]AtA[02]___ ___

A[NO2] A[O2]Solution: Rate = -- - - = 2 - . Substitute in value and solve for the desired value.

A*A[O2]

= 2-^80

AtA[02]

Af AfCheck: The units (M • s"1) are correct. The magnitude of the answer (10~3 M • s"1) makes physical sensebecause rate of product formation is always positive and we are not changing the concentrationmuch in 10 s. The rate of change of the product is slower than the decline of the reactant because ofthe stoichiometric coefficients.

(a) Given: [Br2] versus time plotFind: (i) average rate between 0 and 25 s; (ii) instantaneous rate at 25 s; and (iii) instantaneous rate ofHBr formation at 50 sConceptual Plan: (i) fj, t^ [B ,̂ [Br2]2 —* average rate then

A[Br2]Rate = —

(ii) draw tangent at 25 s and determine slope —* instantaneous rate thenA[Br2]

Rate = - -At

(iii) draw tangent at 50 s and determine slope —* instantaneous rateA[Br2]

A [HBr]

AtA[HBr]

Solution:[Br2]t2 - [Br2](j 0.75 M - 1.00 M

t2-t, 25s -O.s- 1.0x10 2M-s 1 !-2~

and (ii) at 25 s: u 'Ay 0.68 M - 0.85 M

a°pe A x " 35s -15s °'87_

- S.SxlO^M-s"1 — 07A[Br2] A[Br2] 5 n f i _

since the slope = and Rate , ^ u-°A f A i

then Rate = - (- 8.5 x 10'3 M • s'1) = 8.5 x 10'3 M • s'1 0.4 -

(iii) at 50 s:Ay 0.53 M -0.66MAx 60 s -40.s no

>

i

XSi-"x

^k__

.s v\,

*"'~-.

• — ,"•••»

_ 65x 10-3M.S-1 0 10 20 30 40 50 60 70 80 90 1 0 0 1 1 0 1 2Time (s)

A[Br2]


13.34

Rate = -

A [HBr]

A[Br2] 1 A[HBr]

= -2

Af 2A[Br2]

Afthen

= -2(-6.5xlO"3M-s~1) = LSxlO^M-s"1

Af AfCheck: The units (M • s"1) are correct. The magnitude of the first answer is larger than the secondanswer because the rate is slowing down and the first answer includes the initial portion of the data.The magnitudes of the answers (10~3 M • s"1) make physical sense because rates are always positiveand we are not changing the concentration much.

(b) Given: [Br2] versus time data; and [HBr]0 = 0 M Find: plot [HBr] with time

Conceptual Plan: Since Rate = —A[Br2] 1 A [HBr]

Af 2that of [Br2]. The plot will start at the origin.

Solution:1.61.4

1.2

Af. The rate of change of [HBr] will be twice

0 20 40 60 80 100 120Time (s)

Check: The units (M versus s) are correct. The plot makes sense because the plot has the same gen-eral shape of the original plot, only we are increasing instead of decreasing and our concentration axisby a factor of two (to account for the difference in stoichiometric coefficients).

Given: [H2O2] versus time plot; and 1.5 L H2O2 initiallyFind: (a) average rate between 10 and 20 s; (b) instantaneous rate at 30 s; (c) instantaneous rate of O2 forma-tion at 50 s; and (d) molo-, formed in first 50 sConceptual Plan: (a) ty t^ [H2O2]j, [H2O2]2 —* average rate then

A[H202]Rate =

2 A((b) draw tangent at 30 s and determine slope —> instantaneous rate then

1 A[H202]

(c) draw tangent at 50 s and determine slope —» instantaneous rate •

Rate

A[02]

Af

A(

(d) [H202]0 „ [H202]5o s -*

A[H202] = [H202]50s - [H202]0

A [H202] -» A [02] then A [O2], VI A[H202] A[02]

molo2

M =


Solution: 1.2

[H2O2]f2 - [H2O2]tl(a) Rate - -

*2 ~ ' 1 n Q -i0.55M-0.75M 08

20. s - 10. s— 2 0 x 10~2M-s~1 and ^nf.-

Ay 0.28 M - 0.52 M(D) at M s. jiope 0.3

Ax 40. s - 20. s

.2x10 M-s since the slope - and o.iAt n n -

v\sk.\

siv\' — 1

%-t "*>*- ' ,

• — „

Rate = - - ,then 0 10 20 30 40 50 60 70 80 90Time (s)

Rate = (-0.5X-l.2xNT2M-s"1) = 6.0x lO^M-s'1

(c) at 50 s:Ay 0.15M-0.28M A[H2O2]Slope = — = = - 6.5 x 10 M • s since the slope = andAx 60. s - 40.s At

Rate =2

A[H202]

At

A[02] A[02]then —TT— = (-0.5)(-6.5 x

At AtM • s'1) = 3.3 x 10"3 M • s

(d) A[H2O2] = [H2O2]50s - [H2O2]0s = 0.23 M - LOOM = 0. 77 M since

A[O2] = - -A[H2O2] = (-0.5)(0.77M) = 0.385 M then M = — ~ so2 *->

mol O?molo2 = M - L = 0.385 — ~xl.5^ = 0.58 mol O2

Check: (a) The units (M • s"1) are correct. The magnitude of the first answer is reasonable considering theconcentrations and times involved (1M / 100 s). (b) The units (M • s"1) are correct. We expect the answer inthis part to be less than in the first part because the rate is decreasing as the reaction proceeds, (c) The units(M • s"1) are correct. We expect the answer in this part to be less than in the first part because the rate isdecreasing as the reaction proceeds, (d) The units (mol) are correct. We expect an answer less than 1 molsince the drop in reactant concentration is less than 1 M, we have 1.5 L, and only half as much O2 is gener-ated as hydrogen peroxide is consumed.

he Rate Law and Reaction Orders

(a) Given: Rate versus [A] plot Find: reaction orderConceptual Plan: Look at shape of plot and match to possibilities.Solution: The plot is a linear plot, so Rate a [A] or the reaction is first order.Check: The order of the reaction is a common reaction order.

(b) Given: part (a) Find: sketch plot of [A] versus timeConceptual Plan: Using the result from part (a), shape plot of [A] versus time should be curvedwith [A] decreasing. Use 1.0 M as initial concentration.


Solution:

(c)

(a)

(b)

(c)

100 200 300Time (s)

Check: The plot has a shape that matches the one in the text for first order plots.

Given: part (a) Find: write a rate law and estimate kConceptual Plan: Using result from part (a), the slope of the plot is the rate constant.

Solution: Slope = — =Ax

0.010-- 0.00-s s

= 0.010 s"1 so Rate = k [A]1 or Rate = k [A] or1.0M- 0.0M

Rate = 0.010 s'1 [A]Check: The units (s"1) are correct. The magnitude of the answer (10~2 s"1) makes physical sense becauseof the rate and concentration data. Remember that concentration is in units of M, so plugging the rateconstant into the equation has the units of the rate as M • s"1, which is correct.

Given: Rate versus [A] plot Find: reaction orderConceptual Plan: Look at shape of plot and match to possibilities.Solution: The plot is a linear plot that is horizontal, so rate is independent of [A] or the reaction iszero order with respect to A.Check: The order of the reaction is a common reaction order.

Given: part (a) Find: sketch plot of [A] versus timeConceptual Plan: Using the result from part (a), shape plot of [A] versus time should be a straightline with [A] decreasing. Use 1.0 M as initial concentration.Solution:

1.0-

q 0.8o

| 0.6-

I 0.4-o

0.2

0.0-0 20 80 10040 60

Time (s)

Check: The plot has a shape that matches the one in the text for zero order plots.

Given: part (a) Find: write a rate law and estimate kConceptual Plan: Using result from part (a), the rate is equal to the rate constant.Solution: Rate = k [A]° or Rate = k or Rate = 0.011 M • s'1

Check: The units (M • s"1) are correct. The magnitude of the answer (10"2 M • s"1) makes physical sensebecause of the rate and concentration data. Plugging the rate constant into the equation, the rate hasthe units of M • s"1, which is correct.


13.37 J Given: reaction order: (a) first-order; (b) second-order; and (c) zero-order Find: units of kConceptual Plan: Using rate law, rearrange to solve for k.

Rate = k [A]", where n = reaction order

Solution: For all cases rate has units of M • s"1 and [A] has units of M

(a) Rate = k [A]J = k [A] so k =

(b) Rate = k [A]2 so k =Rate

[A]2

MRate"[AT M

Ms

= s

M-M

(c) Rate = * [A]° = * = M-s"1.Check: The units (s"1, M"1 • s"1 and M • s"1) are correct. The units for k change with the reaction order so thatthe units on the rate remain as M-s"1.

Given: k = 0.053/s and [N2Os] = 0.055 M; reaction order: (a) first-order, (b) second-order, and zero-order(change units on k as necessary) Find: rateConceptual Plan: Using rate law, substitute in values to solve for Rate.

Rate = k [NjCy, where n = reaction order

Solution: For all cases Rate has units of M • s"1 and [A] has units of M. Use the results from Problem 35 tochoose the appropriate units for k.

(a)

(b)

Rate = k [N2O5]1 = k [N2O5] =

Rate = k [N2O5]2 = k [N2O5] =

^ x 0.055 M = 2.9X10-3-;s s

Rate = Jt [N2O5]0 = k = 5.3 x 10~2 —

s

x (0.055 M)2 = 1.6 x 10~4 — ; andMs s

Check: The units (M • s"1) are correct. The magnitude of the rate changes as the order of the reaction changes,because we are multiplying by the concentration a different number of times in each case. The higher theorder the lower the rate since the concentration is less than 1 M.

Given: A, B, and C react to form products. Reaction is first order in A, second order in B, and zero order in CFind: (a) rate law; (b) overall order of reaction; (c) factor change in rate if [A] doubled; (d) factor change in rateif [B] doubled; (e) factor change in rate if [C] doubled; and (f) factor change in rate if [A], [B], and [C] doubled.Conceptual Plan:

(a) Using general rate law form, substitute in values for orders.Rate = k [A]"1 [B]" [C]p, where m, n, and p = reaction orders

(b) Using rate law in part (a) add up all reaction orders.overall reaction order = m + n + p

(c) through (f) Using rate law from part (a) substitute in concentration changes.

]0 or Rate = k [A][B]2.

0 = 3soitisa third order reaction overall.

Solution:

(a) m = 1, n = 2, and p = 0 so Rate = k [

(b) overall reaction order = m + tt+p


(0 - and [Ah . 2 |All, IB,2 , (B,,, [Ck . ,C],, so - _ 2Ratel /c[AK[B]freaction rate doubles (factor of 2).

Rate 2 fcfAB]. Rate 2ialeT - j i i «* [A]2 ' [A]'< [B]2 = 2 [B]>- [C]2 = [C]'> S° RltTT ' = * = 4 so

the reaction rate quadruples (factor of 4).

reaction rate is unchanged (factor of 1).

(0 = ~f§ and [Ak = 2 Mi, Pk = 2 [B]i, [C]2 = 2 [Ch, so

Rate 2 * (2 r _2— - ~ = - i - s - = 2 x 2 = 8 so the reaction rate goes up by a factor of 8.Ratel fclfcgtBg

Check: The units (none) are correct. The rate law is consistent with the orders given and the overall order islarger than any of the individual orders. The factors are consistent with the reaction orders. The larger theorder, the larger the factor. When all concentrations are changed the rate changes the most. If a reactant isnot in the rate law, then changing its concentration has no effect on the reaction rate.

Given: A, B, and C react to form products. Reaction is zero order in A, one-half order in B, and secondorder in C.Find: (a) rate law; (b) overall order of reaction; (c) factor change in rate if [A] doubled; (d) factor change in rateif [B] doubled; (e) factor change in rate if [C] doubled; and (f) factor change in rate if [A], [B], and [C] doubledConceptual Plan:

(a) Using general rate law form, substitute in values for orders.Rate = k [A]m [B]" [Cf, where m, n, and p = reaction orders

(b) Using rate law in part (a) add up all reaction orders.overall reaction order = m + n + p

(c) through (f) Using rate law from part (a) substitute in concentration changes.Rate2

Solution:

(a) m = 0,n = 1/2, and p = 2 so Rate = k [A]°[B]1/2[C]2 or Rate = k [B]1/2[C]2.

(b) overall reaction order = m + n+p = Q + 1/2 + 2 = 5/2 = 2.5 so it is a two and a half orderreaction overall.

Rate 2 ^[B]1/2[C]2 Rate 2

reaction rate is unchanged (factor of 1).

Rate 2 k [B]l/2 [C]2 Rate 2 k (2f%)1/2 T6]2. „

SaleT = ^BiFtcI [ ^ = IAh' [ ^ = [ ]l' [ L = ICh S° " = 'the reaction rate increases by a factor of 21/2 or V2 or 1.414.

Rate 2 = EfBfi^ Vm^ _ ̂ =

Ratel

, [Bfc ~ [B]i, [C]2 - 2 [C]!, so

= 2Z = 4 so the reaction rate quadruples (factor of 4).


13.41

13.42

(f) Rate 2Ratel

*[B]1/2[C]2

k[B]\/2[C]iand [A]2 = 2 [A],. [B]2 = 2 [Bh, [C]2 = 2 [Ch, so

or 5.66.Check: The units (none) are correct. The rate law is consistent with the orders given and the overall order islarger that any of the individual orders. The factors are consistent with the reaction orders. The larger theorder, the larger the factor. When all concentrations are changed the rate changes the most. If a reactant isnot in the rate law, then changing its concentration has no effect on the reaction rate.

Given: table of [A] versus initial rate Find: rate law and kConceptual Plan: Using general rate law form, compare rate ratios to determine reaction order.

Rate 2 _ MA2

and 3.9623 = 2"

RatelThen use one of the concentration/initial rate pairs to determine k.

Rate = k[A]"

Rate 2 k [A]? 0.Solution:

Rate 1 k [A]"a Comparing the first two sets of data fc(0.200M)"

0.053 MA ~~ fc (0.100 M)"

so n = 2. If we compare the first and the last data sets0.473 Ht (0300 M)8

and 8.9245 = 3" son = 2.0.053 MA It (0.100 M)"

This second comparison is not necessary, but it increases our confidence in the reaction order. SoRate = k [A]2. Selecting the second data set and rearranging the rate equation

k =

0.210

= 5.25 M'1 • s'1 so Rate = 5.25[A]2 (0.200 M)2

Check: The units (none and M"1 • s"1) are correct. The rate law is a common form. The rate is changing morerapidly than the concentration, so second order is consistent. The rate constant is consistent with the unitsnecessary to get rate as M/s and the magnitude is reasonable since we have a second order reaction.

Given: table of [A] versus initial rate Find: rate law and kConceptual Plan: Using general rate law form, compare rate ratios to determine reaction order.

Rate 2

Rate 1 k [AJJ

Then use one of the concentration/initial rate pairs to determine k.Rate - it [A]"

Rate 2 k[A\2 0.016 MA fc(0.30M)"Solution: „ . , = TTT-̂ T Comparing the first two sets of data _______ = , . ...... „ and 2 = 2" so

Rate 1 k [A]" 0.008 MA fc(0.15M)"0.032 MA fc(0.032M)"

n = 1. If we compare the first and the last data sets ^^ = and 4 = 4 so n = 1. This

second comparison is not necessary, but it increases our confidence in the reaction order. So Rate = k[A].

Rate °-°167Selecting the second data set and rearranging the rate equation k = = 5.3xlO~2s~1

[A] 0.30 Mso Rate = 5.3 x lO~2s~l[A].Check: The units (none and s"1) are correct. The rate law is a common form. The rate is changing as rapidlyas the concentration is consistent with first order. The rate constant is consistent with the units necessary toget rate as M/s and the magnitude is reasonable since we have a first order reaction.

Given: table of [NO2] and [F2] versus initial rate Find: rate law, k, and overall orderConceptual Plan: Using general rate law form, compare rate ratios to determine reaction order of eachreactant. Be sure to choose data that changes only one concentration at a time.

Then use one of the concentration/initial rate pairs to determine k.Rate = J


Rate 2 k [NO2]2m

Solution: - - - = — — m „ Comparing the first two sets of data:

0 .051 MA fc (0 .200 M)m "(0.100 NDI

0.103 MA fc(Ch29QM)m(0.200M)"SetS: = JkCOSOQMT (0.100 M>" and 2 = 2 S° " = L ̂ comparisons can be made, but

are not necessary. They should reinforce these values of the reaction orders. So Rate = k [NO2][F2]. Selecting

the last data set and rearranging Ihe, ate equation t , ^ - (Q 400 M)(0 M) = 2.57M-'.»-'so

Rate = 2.57 M"1 • s~! [NO2][F2] and the reaction is second order overall.Check: The units (none and M"1 • s"1) are correct. The rate law is a common form. The rate is changing as rap-idly as each concentration is changing, which is consistent with first order in each reactant. The rate con-stant is consistent with the units necessary to get rate as M/s and the magnitude is reasonable since we havea second order reaction.

Given: table of [CH3C1] and [C12] versus initial rate Find: rate law, k, and overall orderConceptual Plan: Using general rate law form, compare rate ratios to determine reaction order of eachreactant. Be sure to choose data that changes only one concentration at a time.

RatelThen use one of the concentration/initial rate pairs to determine k.

Rate = *r[CH3CI]m[Cl2]"

Solution: = Comarin ** first two sets of data:

2m SQ n = L If we

0.014 MA fc (0.050 M)m (Cr^SQ M)"0.041 MA fc(07MQM)m(0.100M)"

data sets: - = - - - - and 1.414 = 2 so n = 1/2. Other comparisons can be made,0 .029 MA fc (0^00 M)m (0 .050 M)"

but are not necessary. They should reinforce these values of the reaction orders. SoRate = k [CH3C1] [C12]

1/2. Selecting the last data set and rearranging the rate equation

01113*= 1.29 M-^-s'1 so Rate = l.29M^/2-s^[CH3a] [C12]

1/2

[CH3C1] [C12]1/2 (0.200 M)(0.200 M)1/2

and the reaction is one and a half order overall.Check: The units (none and M"1/2 • s"1) are correct. The rate law is not as common as others, but is reason-able. The rate is changing as rapidly as the CH3C1 concentration is changing, which is consistent with firstorder in this reactant. The rate is changing a bit more slowly than the C12 concentration, which is consistentwith half order in this reactant. The rate constant is consistent with the units necessary to get rate as M/sand the magnitude is reasonable since we have a one and a half order reaction.

The Integrated Rate Law and Half-Life(a) The reaction is zero order. Since the slope of the plot is independent of the concentration, there is no

dependence of the concentration of the reactant in the rate law.( B.45

\J(b) The reaction is first order. The expression for the half-life of a first order reaction is 11/2 = ' , which

is independent of the reactant concentration.

(c) The reaction is second order. The integrated rate expression for a second order reaction is

= fc t + - -, which is linear when the inverse of the concentration is plotted versus time.[A], [A]0

13.46 (a) The reaction is second order. The expression for the half-life of a second order reaction, fj/2 = ,which shows that the half-life decreases as concentration increases.


o^ T\ 13.51 I Given: plot of In [A] versus time has slope = - 0.0045/s; [A]0 = 0.250 M•^ -- / Find: (a) k; (b) rate law; (c) f1/2; and (d) [A] after 225 s

Conceptual Plan:

(a) A plot of In [A] versus time is linear for a first order reaction. Using ln[A]t = — k t + ln[AJo, therate constant is the negative of the slope.

(b) Rate law is first order. Add rate constant from part (a).

0.693(c) For a first order reaction, t\/2 = — - — . Substitute in k from part (a).

K

(d) Use the integrated rate law, ln[A]( = - k t + ln[A]o, and substitute in k and the initial concentration.Solution:

(a) Since the rate constant is the negative of the slope, k = 4.5 x 10"3 s"1.

(b) Since the reaction is first order, Rate = 4.5 x 10'3 s"1 [A].

0.693 0.693 -

(d) ln[A]f = -kt + ln[A]0, and substitute in k and the initial concentration. Soln[A], = - (0.0045/*)(225 s) + In 0.250 M = - 2.39879 and [AJjsos = <T2-3?879 = 0.0908 M.

Check: The units (s"1, none, s, and M) are correct. The rate law is a common form. The rate constant is con-sistent with value of the slope. The half-life is consistent with a small value of k. The concentration at 225 sis consistent with being between one and two half-lives.

Given: plot of 1/[AB] versus time has slope = 0.055/M s; [A]0 = 0.250 MFind: (a) k; (b) rate law; (c) tl/2 when [AB]0 = 0.55 M; and (d) [A] and [B] after 75 sConceptual Plan:

(a) A plot of I/ [AB] versus time is linear for a second order reaction. Using - -- = k t + - — , the[ABJf [AB]0

rate constant is the slope.

(b) Rate law is second order. Add rate constant from part (a).

(c) For a second order reaction, f]/2 = . Substitute in k from part (a).k

(d) Use the integrated rate law, - - — = k t + - - — , and substitute in k, t, and the initial concentration[AB]t [ABJ0

to get the [AB]. Then [AB]0, [AB] -» [A], [B].

A[AB] = [ABh. - [AB)75s with -̂ !A and .1 mol AB

Solution:

(a) Since the rate constant is the slope, A: = 5.5 x 10"2 M"1 • s"1.

(b) Since the reaction is second order, Rate = 5.5 x 10'2 M'1 • s"1 [AB]2.

(c) fi /•> = so fi n = = 33 s.v ' */• I. r A Ul_ V* ._ - , ~_9 »_. —1 _—1\ //-> r-irr> »*\

(d) -^— = kt + —"— so[AB]t = ^—= 7 ^ = 0.12308X11ABl1 (AB" "^ ***»-w->*™ + U») -then A[AB] = [AB]0s - [AB]75s = 0.250 M - 0.12308 M = 0.12692 MAB

so 0.12692 x J^IA = 0.13 M A and 0.12692 x = 0.13 M B.L IffrofAB


Check: The units (M"1 • s"1, none, s, and M) are correct. The rate law is a common form. The rate constantis consistent with value of the slope. The half-life is consistent with a small value of k. The concentration at75 s is consistent with being about one half-life.

Given: decomposition of SO2C12/ first order; k = 1.42 x 1CT4 s"1

Find: (a) f1/2; (b) t to decrease to 25% of [SO2C12]0; (c) t to 0.78 M when tSO2Cl2]0 = 1.00 M; and (d) [SO2C12]after 2.00 x 102 s and 5.00 x 102 s when [SO2C12]0 = 0.150 MConceptual Plan:

(a) k -* tyz0.693

'1/2 = —

(b) [SO2Cl2]o, 25% of [SO2C12]0, k -» t, = -kt + ln[A]0

(c) [S02Cl2]o, [S02Cl2]t, k -* t, = -kt + ln[A]0

(d) [S02C12]0, t, k -» [S02C12],ln[A], = — k t + ln[A] Q

Solution:

0.693 0.693(a) tm = ~ - = - = 4.88 x 103 s

I.42xlO-*s~l

(b) [SO2C12]( = 0.25 [SO2C12]0. Since ln[SO2C!2], = - k t + ln[SO2C!2]0 rearrange to solve for t.i , [so2ci2], i o.25tse>2eye -

- = 9.8x 10 s

(c) [SO2Cl2]t = 0.78 M; [SO2C12]0 = 1.00 M. Since ln[SO2C!2]( = -kt + ln[SO2Cl2]0 rearrange to solve1 [SO2C12], 1 0.78M -

forf- f = -^[s^2^ = -T^^^lnOo^ = 1-7xlo3s-

(d) [SO2C12]0 = 0.150 M and 2.00 x 102 s in

ln[SO2Cl2]( = - fl.42xHT4s<Y2.00xl02&] + In0.150M = -1.92552 ->

[S02C12]( = (T1-92552 = 0.146M

[SO2C12]0 = 0.150 M and 5.00 x 102 s in/ A/ , \

ln[SO2C!2]j = - ( 1.42 xlO~4s<J[ 5.00 xlO2!*) + InO.lSOM = - 1.96812->

[SO2C12], = e-1-9*812 = 0.140MCheck: The units (s, s, s, and M) are correct. The rate law is a common form. The half-life is consistent with asmall value of k. The time to 25% is consistent with two half-lives. The time to 0.78 M is consistent with beingless than one half-life. The final concentrations are consistent with the time being less than one half-life.

Given: decomposition of XY, second order in XY; k = 7.02 x 10"3 M'1 • s"1

Find: (a) J1/2 when [XY] 0 = 0.100 M; (b) t to decrease to 12.5% of [XY]0 = 0.100 M and 0.200 M; (c) t to0.062 M when [XY]0 = 0.150 M; and (d) [XY] after 5.0 x 101 s and 5.50 x 102 s when [XY]0 = 0.050 MConceptual Plan:

(a) [XY]0,k -» r1/2i

(b) [XY]0, 12.5 % of [XY]0, k -» t-i- = kt + —[A], [A]0


(c)

(d)

Solu

(a)

(b)

(c)

(d)

TYV1 TYYl If — > fI-** * JO/ I'* * Jf/ **1 1

[A], " + [A]0

[XYJo, t, k -» [XY],l i

[Al= +[Aktion:

A 1 AI „ -in3 „fc[XY]c

[XY], = 0.125

t l( l

k V[XY],

[XY], = 0.125

rwi n nô-[Xr j j — U.U6z,

1 / 1k V[XYJ,

[XYJo = 0-050

- J./±/. A 1U >

i (7.02xlO~3M::is~1)(0.100M)

[XYJo - 0.125 x U.1UU M - O.Ulzi) M. bmce ..-_,, — Kt + rearrange to solve tor t.[XYJ, [XYJo

L_| * f 1 1 | q q 7 v i n 3 =[XYlo^ (7.02 x 10-3 M^-s'1) V0.0125M 0.100 M/

[AY Jo — u.izj x u.zuu M — u.u/au M. since — Kt i rearrange to solve ror c.[XY], [XY]0

1 \ 1 / I 1 \

[XY]o/ (7.02x lO"3!^^^-1) V0.0250M 0.200M/

[A i Jo — u.ic>u m. Dince rvvl ft i rvv.i rearrange to soive >r r.[XYJj [XY JQ

M a ( l l } nxio3sr-v>-v/i / i 1 iv I n n/-^ v* n i en TÂ I i'*J * ±u °*[XYJo/ (7.02 x 10 Mî • s ) \0-062 M 0.150 M/

\A ^ c n ^ 1 nl c. i^* - l '*- tM and 5.0x10 sm^ W I [xy]o

1 23.51.. -in— 3 n x — I . >^rl\/c n ., 1 nl v.\ i «. TVA/1 n nxo \>f ™

r 1 ( / .UZAIU m - s^v^- U A i u <V ' nni50M M s" 1/V1J "•"•*-• ivi aiiu

1 1 23.861[XY]0 = 0-050 M and 5.50 x 102 s in — — = (7.02 x 10~3 M'1 • §<)(5.50 x 102 S) +

U-UC)U JVl JVl

so [XY] = 0.042 M.Check: The units (s, s, s, s, M, and M) are correct. The rate law is a common form. The half-life is consistentwith a small value of k. The time to 12.5% is consistent with three half-lives, where the half-life time isincreasing. The next time (5000 s) is shorter because the initial concentration is higher. The last time is theshortest because it is less than a half-life with an intermediate concentration. The final concentrations areconsistent with the time being much less than one half-life.

13.55 \ Given: t\/i for radioactive decay of U-238 = 4.5 billion years and independent of [U-238]0

j Find: t to decrease by 10%; number U-238 atoms today, when 1.5 x 1018 atoms formed 13.8 billion years agoX.^X Conceptual Plan: ty2 independent of concentration implies first order kinetics, ti/2 — * k then

0.693'V2- —

90 % of [U-238]0, k -» t and [U-238]0, t, k -» [U-238]t

, = - kt + ln[A]0 ln[A], = - kt + ln(A]0

Solution: t-tn = — - rearrange to solve for k. k = - - = - - = 1.54 x 10~10 yr"1 thenk t\fi 4.5xl09yr

[U-238]t = 0.10 [U-238]0. Since ln[U-238], = -kt + ln[U-238]0 rearrange to solve for t.

1 [U-238], 1 0.90[tm8fe 8t = -- In * - — = -- In - = 6.8 x 10 yr

k [U-238]0 1.54 x 10-10 yr-1 ttMSgfe

and [U-238]0 = 1.5 x 1018 atoms; t = 13.8 x 109 yrIn [U-238]( = - kt + In [U-238J0 = - (1.54x 10-10ylî)(13.8x 109ys)

+ ln(1.5 x 1018atoms) = 39.726797 -+

[U-238], = e39™797 = 1.8x 1017atoms.


Check: The units (yr and atoms) are correct. The time to 10% decay is consistent with less than one half-life.The final concentration is consistent with the time being about three half-lives.

Given: f1/2 for radioactive decay of C-14 = 5730 yearsFind: t to decrease by 25%; mmol C-14 atoms left, after 2255 yr in sample initially contains 1.5 mmol C-14Conceptual Plan: radioactive decay implies first order kinetics, ti/2 —* k then

0.693

75% of [C-14]0, k -» t and [C-14]0, t, k -> [C-14],ln[A], = - kt + ln[A]0 ln[A], = - kt + ln[A]0

0.693 0.693Solution: 1i/2 = —-— rearrange to solve for k. k = —

K M

0.693rj/2 5730 yr

[C-14], = 0.75 [C-14]0. Since ln[C-14], = - kt + ln[C-14]0 rearrange to solve for f.1, [C-14], 1

= 1.20942 xlO~ 4y r then

In = 2.4 x 10V and [C-14]0 = 1.5 mmol;1.20942 xlO^yr'1

t = 2255 yr

ln[C-14], = -kt + ln[C-14]0 = - (1.20942 x 10~4yr^)(2255y*) + ln(1.5mmol) = 0.132741 -»

[C-14], = eai3?741 = 1.1 mmol.Check: The units (yr and mmol) are correct. The time to 25% decay is consistent with less than one half-life.The final concentration is consistent with the time being less than one half-life.

The Effect of Temperature and the Collision Model

13.57

I

activation energy

d - enthalpy of reaction

b - products

Reaction progress

13.58

c ̂ activation energy

d - enthalpy of reaction

a - reactants

b - products

13.59

Reaction progress

Given: activation energy = 56.8 kj/mol, frequency factor = 1.5 x 1011 /s, 25 °C Find: rate constantConceptual Plan: °C -» K and kj/mol -» J/mol then Ev T, A -> k

1000 J .K = "C + 273.15 —— k = A e E«'Ja

56.8 H 1000J JSolution: T = 25°C + 273.15 = 298 K and - - x - - = 5.68 x 104 - - then

mol 1 fej mol


iiven: rate constant = 0.0117/s at 400. K, and 0.689/s at 450. K Find: (a) Ea and (b) rate constant at 425 K/Conceptual Plan:

(a) klr TI kfr T2 -» £a then J/mol -» kj/molE./I

1000 J

(b) E» klf TJ, T2 -> fc2/^2\ Ez / i l \

In I — 1 = — I — I

Solution:

(a) In I — I = -^ ( - - — ). Rearrange to solve for £a.V fci / R \ Ti Tj/

.w ,«r - . 5J mol X 1000 j " mol"

Ti T2/ V400.K 450. K

(b) In ( 77 ) = -f [ ̂ - - =- ] with ̂ , = 0.0117/s, Tj = 400. K, T2 = 425 K. Rearrange to solve for kz.\M/ R \TI TI/1 22 x 105 — —

K-molfc2 = «T2-2922 = 0.101 s-1.

Check: The units (kj/mol and s"1) are correct. The activation energy is typical for a reaction. The rate con-stant at 425 K is in between the values given at 400 K and 450 K.

13.68 Given: rate constant = 0.000122/s at 27 °C, and 0.228/s at 77 °C Find: (a) £a and (b) rate constant at 17 °CConceptual Plan:

(a) °C — > K then fcj, 7\ fc^ T2 -> £a then J/mol -^ kj/mol;'

(b) °C -> K then £„ fcj, T! T2 -* fc2

Solution: T1= 27°C + 273.15 = 300. K and T2 = 77°C + 273.15 = 350. K then In ( -- } = -^ ( J- - ~\KI/ K \Ji 72

(a) Rearrange to solve for £a.

f_J !_1V300.K 350. K/

In [ — J = — ( jr] withki, = 0.000122/s, Tj = 300. K, T2 = 17°C + 273.15 = 290K.Ui/ R \T] T2/

_ _ mol"1000J ~mol'VTj T2 \300.K 350. K/

(b)

Rearrange to solve for k2.

" ^

.K-mol

-^ jt2 = e-iO-8298 = 0.0000198s"1 = l.QSxlO^s"1.Check: The units (kj/mol and s"1) are correct. The activation energy is typical for a reaction. The rate con-stant at 17 °C is smaller than the values given at 27 °C.


(b)

(c)

13.76 (a)

The intermediates are the species that are generated by one step and consumed by other steps. TheseareaCl(£)andCCl3(£).

Since the second step is the rate determining step, Rate = k3 [Cl] [CHCy. Since Cl is an intermediate,its concentration cannot appear in the rate law. Using the fast equilibrium in the first step, we see that

= k2 [Cl]2 or [Cl] = J-^- [C12]. Substituting this into the first rate expression we get that Rate =V ^2

--t [C12]1/2[CHC13]. Simplifying this expression we see Rate = k [C12]

1/2 [CHC13].

(b)

(c)

The overall reaction is the sum of the steps in the mechanism:

N02(g) + Cl2(g) -+ C1N02(£) + CHg)

NC»2 (g) + CH#) -£» C1NO2 (g)

2 N02 (g) + C12 (g) -» 2 C1N02 (g)

The intermediates are the species that are generated by one step and consumed by other steps. Thisis Cl (g)..

Since the first step is the rate determining step, Rate = ki [NO2] [C12]. Since both of these speciesreactants, this is the predicted rate law.

are

Catalysis13.77

13.78

Heterogeneous catalysts require a large surface area because catalysis can only happen at the active sites onthe surface. A greater surface area means greater opportunity for the substrate to react, which results in aspeedier reaction.

The initial and final energies (reactantsand products) remain the same. The acti-vation energy drops, from 75 kj/mol to asmaller value, for example 30 kj/mol.There are usually more steps in the reac-tion progress diagram.

,ccttOJs

without catalyst with catalyst

= 75 kj/mol op

products __

Reaction progress Reaction progress

13.79 Assume rate ratio oc k ratio (since concentration terms will cancel each other) and k = A e E*/RT.T=25°C + 273.15 = 298 K, Ea, = 1.25 x 105 J/mol, and £32 = 5.5 x 104 J/mol. Ratio of rates will be

-5.5x10* —are!

M2|_meij298K 22.199

-1.25X105

fnel

= 1012-

298 K

chemical kinetics - gordon state collegeptfaculty.gordonstate.edu/lgoodroad/summer 2011/che… ·...

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