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. When splashed with cold water, a lizard's body simply gets colder. The drop in body temperature immobilizes the lizard because its movement depends on chemical reactions that occur within its muscles, and the rates of those reactions—how fast they occur—are highly sensitive to temperature. In other words, when the temperature drops, the reactions that produce movement occur more slowly; therefore the movement itself slows down. Cold reptiles are lethargic, unable to move very quickly. For this reason, reptiles try to maintain their body temperature in a narrow range by moving between sun and shade. The rates of chemical reactions, and especially the ability to control those rates, are important phenomena in our everyday lives. For example, the human body's ability to switch a specific reaction on or off at a specific time is achieved largely by controlling the rate of that reaction through the use of enzymes. Chemical kinetics is an important subject to chemists and engineers. The launching of a rocket depends on controlling the rate at which fuel bums—too quickly and the rocket can explode, too slowly and it will not leave the ground. The rate of nuclear decay in a nuclear power plant must be carefully controlled in order to provide electricity safely and efficiently. Chemists must always consider reaction rates when synthesizing compounds. No matter how stable a compound might be, its synthesis is impossible if the rate at 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 products (usually in terms of concentration) divided by the change in time. Typical units are molarity 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 reactant divided by the change in time, because reactant concentrations decrease as a reaction proceeds; therefore the change in the concentration of a reactant is negative. The negative sign in the definition thus makes the overall rate positive. In other words, the negative sign is the result of the convention that reaction rates are usually reported as positive quantities. Since the product concentrations are increasing, the concentration of a product divided by the change in time is positive.

13.5

The average rate of the reaction can be calculated for any time interval as l[A], 2 -[A] ( l l[B], 2 -[B] (l l[C] f 2 -[C] f l l[D] ( 2 -[D] ( l Rate = - = - - —- = 1 = ~— for the a 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 at any one point in time, represented by the instantaneous slope of the plot of concentration versus time at that point. We can obtain the instantaneous rate from the slope of the tangent to this curve at the point of interest.

501

502 13.6

For a zero order reaction, Rate = k [A]° = k, so doubling the concentration of A does nothing to the reaction rate. For a first order reaction, Rate = k [A]1 = k [A], so doubling the concentration of A doubles the reaction rate. 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 by running controlled experiments where the concentrations of the reactants are varied and the reaction rates are 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 and n 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. The integrated 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\

Chapter 13 Chemical Kinetics

\

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 second 1 order reaction, tin = , r , . *[A]0

13.12

The rates of chemical reactions are, in general, highly sensitive to temperature. Reaction rates increase as the temperature 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 in kelvin (T), is as follows: k = A e~E*/RT, where R is the gas constant (8.314 J/mol • K), A is a constant called the frequency factor (or the pre-exponential factor), and Ea is called the activation energy (or activation barrier). The frequency factor is the number of times that the reactants approach the activation barrier per unit time. The exponential factor (-Ea/RT) is the fraction of approaches that are successful in surmounting the activation barrier and forming products. The exponential factor increases with increasing temperature, but decreases with an increasing value for the activation energy. As the temperature increases, the number of collisions increases and the number of molecules having enough thermal energy to surmount the activation 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 have enough 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 having enough 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 temperature in 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 two reactant molecules. In collision theory, therefore, each approach to the activation barrier is a collision between the reactant molecules. The value of the frequency factor should simply be the number of collisions that occur per second. In the collision model k = pze~E°/RT, where the frequency factor (A) has been separated into two separate parts—p is called the orientation factor and z is the collision frequency. The collision frequency is simply the number of collisions that occur per unit time. The orientation factor says that if two molecules are to react with each other, they must collide in such a way that allows the necessary bonds to break and form. The small orientation factor indicates that the orientational requirements for this reaction are fairly stringent—the molecules must be aligned in a very specific way for the reaction to occur. When two molecules with sufficient energy and the correct orientation collide, something unique happens: The electrons on one of the atoms or molecules are attracted to the nuclei of the other; some bonds begin to weaken while other bonds begin to form and; if all goes well, the reactants go through the transition state and 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 reaction, not the series of individual steps by which the reaction occurs. The overall equation simply shows the substances present at the beginning of the reaction and the substances formed by the reaction—it does not show the intermediate steps that may be involved. A reaction mechanism is a series of individual chemical steps 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 simpler steps—they occur as they are written. Elementary steps are characterized by their molecularity, the number of reactant particles involved in the step. The molecularity of the three most common types of elementary steps are as follows: unimolecular - A —> products and Rate = k [A]- bimolecular - A + A —> products and Rate = k [A]2; and bimolecular - A + B -> products and Rate = k [A][B]. Elementary steps in which three reactant particles collide, called termolecular steps, are very rare because the probability of three particles simultaneously colliding is small.

13.18

For a proposed reaction mechanism to be valid—mechanisms can only be validated, not proven—two conditions must be met: (1) the elementary steps in the mechanism must sum to the overall reaction; and (2) the rate 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 another step. An intermediate is not found in the balanced equation for the overall reaction, but plays a key role in the mechanism.

13.20

A catalyst is a substance that increases the rate of a chemical reaction but is not consumed by the reaction. A catalyst works by providing an alternative mechanism for the reaction—one in which the rate-determining step 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 catalyst is 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 they approach each other; (3) reaction—the reactants react to form the products; and (4) desorption—the products desorb from the surface into the gas phase.

13.23

Enzymes are biological catalysts that increase the rates of biochemical reactions. Enzymes are large protein molecules with complex three-dimensional structures. Within that structure is a specific area called the 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 key fitting 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 Rates 1 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 rate Conceptual Plan: rx, t^ [HBrJj, [HBr]2 —> average rate lA[HBr] Rate = - —

Rate = -

2

'

2

*2 - h

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

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

504_ (c)

Chapter 13 Chemical Kinetics

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

molBr M = -^

_. Solution: Rate = 1 .76 x 10 J M • s

A[Br2] A[Br2] = —-- = -—-——- . Rearrange to solve for A[Br2l. Af 15.0s - 0.0s ,M rnolBr A [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 sense because the time is shorter than in part (b) and we need to divide by 2 and multiply by 1.5 because of the stoichiometric coefficient difference and the volume of the vessel, respectively. 13.26

(a) Rate = (b)

1 A[N20] 1 A[N2] A[02] --= -- = 2 Af 2 Af Af

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

A[O2J Rate = —— A(

molo, 0.015 mol O2 Solution: At 1 1 = 0 s we have no O2. M = —-— = — T T -- = °-030 M then I-*™ - [Qzltj 0.030M - 0.000M , Rate = - = 2.0xlO~ 3 M-s~ 1 . t-i - fi 15.0s-0.0s Check: The units (M • s"1) are correct. The magnitude of the answer (10~3 M • s"1) makes physical sense because rates are always positive and we are not changing the concentration much in 15 s. (c)

A[N2O] —

Given: part (b) data Find:

A[N2O] Conceptual Plan: average rate —» — 1 A[N20] _ A[QJ

1 A[N20] A[02] A[N20] = . Rearrange to solve for . 2 Af Af Af A[N2O] » „ „ ,,,-3», -1 ™,,^, -i n «i^zi = - 2 x 2 . 0 x l O ~ 3 M - s ~ 1 = -rt 0.0040M-s" Af Af Check: The units (M • s"1) are correct. The magnitude of the answer (- 0.004 M • s"1) makes physical sense because we multiply by 2 because of the different stoichiometric coefficients. The change in concentration with time is negative since this is a reactant. Solution: Rate =

1 A[A]

A[B]

1 A[C]

A [A] A[B] A[C] Given: - — = - 0.100 M/s Find: —— , and Af Af Af A [A] A[B] A[C] Conceptual Plan: -- * —-—, and —-— Af Af Af _ 1 A[A^_ _ A[B] _ i A[q 2 At ° " Af "3 At

Solution: Rate = values. so— Af

1 A[A] --=

1 -0.100M --= s = 0.150 M-s -1 .

A[B] 1 A[C] -- = - —-— . Substitute in value and solve for the two desired A[B] -so Af

A[B] —— = - 0.0500 M - s Af

1 - 0.100 M and - — s

1 A[C] 3 At

Chapter 13 Chemical Kinetics

505

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

Af (b)

Af

2 Af

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

A[C] Conceptual Plan: —

A[B] A[A] —, and ~^A[A]

Rate=__ =

A[B] 1 A[C] _2_ = __

A[A] A[B] 1 A[C] Solution: Rate = -- -— = - 2 —-— = - —-— . Substitute in value and solve for the two desired Af Af 2 Af A[B] 10.025M A[B] A[A] 1 0.025 M values. -2 - = - - so - =-0.0063M-s l and -- = -- so A* 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 physical sense because fewer moles of B are reacting for every mole of C being formed and the change in concentration 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 the change in concentration with time is negative since this is a reactant. B has the smallest stoichiometric 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 Rate Conceptual Plan: write the expression for the rate with respect to each species then Rate =

rate expression,

A[C12] -= At

i A[F2] i A[C1F3] --= -3 At 2 Af

A[C12] A[F2] A[C1F3] -- -> —-—; —--; Rate AICIJ _ At

i A[F2]

! A[C1F3] At

3 M '2

A[C12] 1 A[F2] 1 A[C1F3] -- = --- :— = — —:- so Af 3 Af 2 Af A[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 Af A[C12] 1 A[C1F3] A[C1F3] and -= - —-- . Rearrange to solve for —-- . Solution: Rate =

Check: The units (M • s"1) are correct. The magnitude of the answers (- 0.036 M • s"1 and 0.024 M • s"1) makes physical sense because F2 is being used at three times the rate of C12, C1F3 is being formed at two times the rate 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/s Find: A[O2]/Af; A[H 2 O]/Af; A[S8]/Af; and Rate Conceptual Plan: write the expression for the rate with respect to each species then Rate

! A[H2S] = 8 At

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

506

Chapter 13 Chemical Kinetics rate expression,

A[H2S] Af

Rate - - 8

A[02] A[H20] A[S8] Af ; Af '' Af '

A[H2S]

A[S8]

A/

4

8

At

A(

A[02] A[H2S] A[H20] Solution: Rate = • so 8 Af 4 Af 8 Af Af 1 A[H2S] A[02] A[02] . Rearrange to solve for —-—. 8 Af 4 Af A[02] = 4 A[H2S] = -(- 0.080M-s'1) = -0.040M-S"1 Af 8 Af A[H2S] _ I A[H2O] A[H20] and-. Rearrange to solve for ~~ 8 Af 8 Af Af A[H20] 8 A[H2S] = -(-0.080 M-s' 1 ) = 0.080 M - s,-1 8 Af Af A[H2S] A[S8] 1 A[H2S] 1 and - = --(-0.080 M-s" 1 ) = 0.010 M-s" 1 . Af Af 8 Af 8 Check: The units (M• s'1) are correct. The magnitude of the answers (- 0.040 M-s" 1 - 0.080 M• s"1, and 0.010 M • s"1) makes physical sense because O2 is being used at one-half times the rate of H2S, H2O is being formed at two times the rate of H2S disappearance, S8 is being formed at one-eighth times the rate of H2S disappearance, 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 s Conceptual Plan: t\, t^ [CtH8]i, [C4H8]2 —» average rate A[C4H8] Rate = - -

[QH8]f2 - [C4H8]f] o .913 M - 1.000 M Solution: For 0 to 10 s Rate = = 8.7xlO~ 3 M-s -1 and f2-fa 10. s - O . s [C4H8](2 - [C4H8](l 0.637M - 0.697M for 40 to 50 s: Rate = = 6.0xlO~ 3 M-s -l f 2 - fi 50.S-40.S Check: The units (M • s"1) are correct. The magnitude of the answer (10"3 M • s"1) makes physical sense because rates are always positive and we are not changing the concentration much in 10 s. Also reactions slow as they proceed because the concentration of the reactants is decreasing.

(b)

Given: [C4H8] versus time data Find:

—— between 20 and 30 s Af A[C2H4] [C4H8]2 Af

Conceptual Plan: tir

Rate = -

Solution: Rate = -

[C4H8](2 - [C4H8]t]

=

1 A[C2H4]

~ 2

Af

0.763M - 0.835M 30.s-20.s

-3

Af ,-i

Rearrange to solve for

—. So — = 2(7.2 x 10" Af Af Check: The units (M • s" ) are correct. The magnitude of the answer (10~2 M • s"1) makes physical sense because rate of product formation is always positive and we are not changing the concentration much in 10 s. The rate of change of the product is faster than the decline of the reactant because of the stoichiometric coefficients. 1

13.32

(a)

Given: [NO2] versus time data Find: average rate between 10 and 20 s, and between 50 and 60 s Conceptual Plan: tlr f^ [NO2]lr [NO2]2 —> average rate A[N02] Rate = - Af

Chapter 13 Chemical Kinetics

507

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

0.904M - 0.951 M = 4.7xlO~ 3 M-s~ 1 and 20.s - 10.s [N02](2 - [N02](] 0.740M - 0.778M 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 sense because rates are always positive and we are not changing the concentration much in 10 s. Also reactions slow as they proceed because the concentration of the reactants is decreasing. (b)

A[O2] Given: [NO2] versus time data Find: ——— between 50 and 60 s At A[02] Conceptual Plan: average rate from part a) —• At A[N0 ]

2 ___

A[0 ]

2 ___

A[NO2] A[O2] Solution: Rate = -- -- = 2 - . Substitute in value and solve for the desired value. A* At A[O2] A[02] = 2-^80 Af Af Check: The units (M • s"1) are correct. The magnitude of the answer (10~3 M • s"1) makes physical sense because rate of product formation is always positive and we are not changing the concentration much in 10 s. The rate of change of the product is slower than the decline of the reactant because of the stoichiometric coefficients. (a)

Given: [Br2] versus time plot Find: (i) average rate between 0 and 25 s; (ii) instantaneous rate at 25 s; and (iii) instantaneous rate of HBr formation at 50 s Conceptual Plan: (i) fj, t^ [B^, [Br2]2 —* average rate then Rate =

A[Br2] —

(ii) draw tangent at 25 s and determine slope —* instantaneous rate then A[Br2] Rate = - At

(iii) draw tangent at 50 s and determine slope —* instantaneous rate

A [HBr]

At A[HBr]

A[Br2]

Solution: [Br2]t2 - [Br2](j 0.75 M - 1.00 M t2-t, 25s - O . s 2 1 - 1.0x10 M - s !-2~ u and (ii) at 25 s: ' Ay 0.68 M - 0.85 M a pe ° Ax" 35s -15s °'87_ 1 - S.SxlO^M-s" — 07 A[Br2] A[Br2] 5 un f i _ since the slope = A f and Rate , ^ -° Ai then Rate = - (- 8.5 x 10'3 M • s'1) = 8.5 x 10'3 M • s'1

(iii) at 50 s: Ay Ax

0.4 -

i >X

S

i-"x

__ ^ks v.

\,*"'~-.

•—"•••» ,

0.53 M -0.66M 60 s -40.s

_ 6 5 x 10-3M.S-1 A[Br2]

no 0

10 20 30 40 50 60 70 80 90 1 0 0 1 1 0 1 2 Time (s)

508

Chapter 13 Chemical Kinetics 1 A[HBr] then Af 2 Af A [HBr] A[Br2] = -2 = -2(-6.5xlO" 3 M-s~ 1 ) = LSxlO^M-s" 1 Af Af Check: The units (M • s"1) are correct. The magnitude of the first answer is larger than the second answer 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 positive and we are not changing the concentration much. Rate = -

(b)

A[Br2]

Given: [Br2] versus time data; and [HBr]0 = 0 M Find: plot [HBr] with time A[Br2] 1 A [HBr] Conceptual Plan: Since Rate = — . The rate of change of [HBr] will be twice Af 2 Af that of [Br2]. The plot will start at the origin. Solution: 1.6 1.4

1.2

0

20

40

60 Time (s)

80

100

120

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

13.34

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

(b)

A[H202] 2

A(

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

(d)

[H202]0 „ [H202]5o s -* A [H202] -» A [02] then A [O2], V I A[H202] A[H202] = [H202]50s - [H202]0

A[02]

M =

A[02] Af A(

mol o2

509

Chapter 13 Chemical Kinetics Solution: (a)

1.2

Rate -

v\

[H2O2]f2 - [H2O2]tl *2 ~ ' 1

n Q -i 08

0.55M-0.75M 20. s - 10. s 2

1

— 2 0 x 10~ M-s~ and

^nf.-

sk.

\s

iv

(D)

at M s. jiope

.2x10 Rate = - -

M-s

Ay Ax

0.28 M - 0.52 M 40. s - 20. s

since the slope -

At

and

,then

\%

-t

0.3

' —1

o.i

"*>*- '

nn0

10

20

30

40 50 Time (s)

60

, •—„ 70

80

90

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 = and Ax 60. s - 40.s At A[H202] A[02] A[02] M • s'1) = 3.3 x 10"3 M • s then —TT— = (-0.5)(-6.5 x Rate = 2 At At At (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 = —~ so 2 *-> 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 the concentrations and times involved (1M / 100 s). (b) 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 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 is decreasing as the reaction proceeds, (d) The units (mol) are correct. We expect an answer less than 1 mol since the drop in reactant concentration is less than 1 M, we have 1.5 L, and only half as much O2 is generated as hydrogen peroxide is consumed.

he Rate Law and Reaction Orders (a)

Given: Rate versus [A] plot Find: reaction order Conceptual 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 time Conceptual Plan: Using the result from part (a), shape plot of [A] versus time should be curved with [A] decreasing. Use 1.0 M as initial concentration.

510

Chapter 13 Chemical Kinetics Solution:

100 200 300 Time (s) Check: The plot has a shape that matches the one in the text for first order plots. (c)

Given: part (a) Find: write a rate law and estimate k Conceptual Plan: Using result from part (a), the slope of the plot is the rate constant. 0.010-- 0.00s s Solution: Slope = — = = 0.010 s"1 so Rate = k [A]1 or Rate = k [A] or 1.0M- 0.0M Ax 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 because of the rate and concentration data. Remember that concentration is in units of M, so plugging the rate constant into the equation has the units of the rate as M • s"1, which is correct.

(a)

Given: Rate versus [A] plot Find: reaction order Conceptual 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 is zero order with respect to A. Check: The order of the reaction is a common reaction order.

(b)

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

1.0q 0.8 o

| 0.6I 0.4o

0.2 0.0-

80 100 40 60 Time (s) Check: The plot has a shape that matches the one in the text for zero order plots. 0

(c)

20

Given: part (a) Find: write a rate law and estimate k Conceptual 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 sense because of the rate and concentration data. Plugging the rate constant into the equation, the rate has the units of M • s"1, which is correct.

Chapter 13 Chemical Kinetics 13.37 J

511

Given: reaction order: (a) first-order; (b) second-order; and (c) zero-order Find: units of k Conceptual 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

M (a)

(b)

Rate Rate = k [A]J = k [A] so k = "[AT

Rate = k [A]2 so k =

Rate

[A]2

M

=s

M 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 that the 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: rate Conceptual 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 to choose the appropriate units for k.

(a)

Rate = k [N2O5]1 = k [N2O5] = ^ x 0.055 M = 2.9X10-3-; s s Rate = k [N2O5]2 = k [N2O5] =

x (0.055 M)2 = 1.6 x 10~4 —; and Ms s 0 2 Rate = Jt [N2O5] = k = 5.3 x 10~ — 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 the order the lower the rate since the concentration is less than 1 M.

(b)

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

Solution: (a)

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

(b)

overall reaction order = m +

tt+p

]0 or Rate = k [A][B]2. 0 = 3soitisa third order reaction overall.

512

Chapter 13 Chemical Kinetics (0

-

and [Ah . 2 |All, IB,2 , (B,,, [Ck . ,C],, so

Ratel /c[AK[B]f reaction rate doubles (factor of 2).

-

_2

Rate 2 fcfAB]. Rate 2 [A]2 [A] [B]2 2 [B] [C]2 [C] S ialeT - j i i «* ' '< = >= '> ° RltTT ' the reaction rate quadruples (factor of 4).

= * =4 so

reaction rate is unchanged (factor of 1). = ~ f §and [Ak = 2 Mi, Pk = 2 [B]i, [C]2 = 2 [Ch, so

(0

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 is larger than any of the individual orders. The factors are consistent with the reaction orders. The larger the order, the larger the factor. When all concentrations are changed the rate changes the most. If a reactant is not 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 second order in C. Find: (a) rate law; (b) overall order of reaction; (c) factor change in rate if [A] doubled; (d) factor change in rate if [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]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 reaction overall.

= Q + 1/2 + 2 = 5/2 = 2.5 so it is a two and a half order

^[B]1/2[C]2

Rate 2

Rate 2

reaction rate is unchanged (factor of 1). k [B]l/2 [C]2

Rate 2 =

Rate 2 [

= IAh [

=

[ ]l [ L = ICh S

SaleT ^BiFtcI ^ ' ^ ' ° the reaction rate increases by a factor of 21/2 or V2 or 1.414.

k (2f%)1/2 T6]2.

"

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

Rate 2 = EfBfi^ Vm^ _ ^Z = = 2 = 4 Ratel

so the reaction rate quadruples (factor of 4).

=

'

„

Chapter 13 Chemical Kinetics

(f)

Rate 2 Ratel

*[B]1/2[C]2 k[B]\/2[C]i

513 and [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 is larger that any of the individual orders. The factors are consistent with the reaction orders. The larger the order, the larger the factor. When all concentrations are changed the rate changes the most. If a reactant is not in the rate law, then changing its concentration has no effect on the reaction rate.

13.41

Given: table of [A] versus initial rate Find: rate law and k Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order. Rate 2 _ M A 2

Ratel

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

Rate 2 Solution: Rate 1

k [A]? 0. fc(0.200M)" a Comparing the first two sets of data and 3.9623 = 2" k [A]" 0.053 MA ~~ fc (0.100 M)" Ht (0300 M)8 0.473 so n = 2. If we compare the first and the last data sets 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. So Rate = k [A]2. Selecting the second data set and rearranging the rate equation

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 more rapidly than the concentration, so second order is consistent. The rate constant is consistent with the units necessary to get rate as M/s and the magnitude is reasonable since we have a second order reaction.

k =

13.42

Given: table of [A] versus initial rate Find: rate law and k Conceptual 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]. Selecting the second data set and rearranging the rate equation k =

Rate [A]

°-°167 = 5.3xlO~ 2 s~ 1 0.30 M

so 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 rapidly as the concentration is consistent with first order. The rate constant is consistent with the units necessary to get 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 order Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order of each reactant. 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

514

Chapter 13 Chemical Kinetics 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 rapidly as each concentration is changing, which is consistent with first order in each reactant. The rate constant is consistent with the units necessary to get rate as M/s and the magnitude is reasonable since we have a second order reaction. Given: table of [CH3C1] and [C12] versus initial rate Find: rate law, k, and overall order Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order of each reactant. Be sure to choose data that changes only one concentration at a time. Ratel

Then 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 ma 0 .029 MA fc (0^00 M)m (0 .050 M)" but are not necessary. They should reinforce these values of the reaction orders. So Rate = 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 reasonable. The rate is changing as rapidly as the CH3C1 concentration is changing, which is consistent with first order in this reactant. The rate is changing a bit more slowly than the C12 concentration, which is consistent with half order in this reactant. The rate constant is consistent with the units necessary to get rate as M/s and the magnitude is reasonable since we have a one and a half order reaction.

The Integrated Rate Law and Half-Life ( B.45 \J

(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)

The reaction is first order. The expression for the half-life of a first order reaction is 11/2 = ' is independent of the reactant concentration.

(c)

The reaction is second order. The integrated rate expression for a second order reaction is

[A], 13.46

(a)

, which

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

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.

,

Chapter 13 Chemical Kinetics

517

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, the rate constant is the negative of the slope.

(b)

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

(c)

0.693 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. So ln[A], = - (0.0045/*)(225 s) + In 0.250 M = - 2.39879 and [AJjsos = *™ + U»)

then A[AB] = [AB]0s - [AB]75s = 0.250 M - 0.12308 M = 0.12692 MAB so 0.12692

L

x

J^IA = 0.13 M A and 0.12692 IffrofAB

x

= 0.13 M B.

-

518

Chapter 13 Chemical Kinetics Check: The units (M"1 • s"1, none, s, and M) are correct. The rate law is a common form. The rate constant is consistent with value of the slope. The half-life is consistent with a small value of k. The concentration at 75 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 M Conceptual Plan: (a)

k -* tyz 0.693 '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:

(a)

0.693 tm = ~ - =

0.693 - = 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 solve 1 [SO2C12], 1 0.78M forf f = ln 1 7xlo3s - -^[s^2^ = -T^^^ Oo^ = -

(d)

[SO2C12]0 = 0.150 M and 2.00 x 102 s in ln[SO2Cl2]( = - fl.42xHT 4 s [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~ 4 s [SO2C12], = e-1-9*812 = 0.140M Check: The units (s, s, s, and M) are correct. The rate law is a common form. The half-life is consistent with a small value of k. The time to 25% is consistent with two half-lives. The time to 0.78 M is consistent with being less 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 to 0.062 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 M Conceptual Plan: (a)

[XY]0,k -» r1/2 i

(b)

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

[A]0

Chapter 13 Chemical Kinetics (c)

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

1

[A], "

(d)

+

519

1

[A]0

[XYJo, t, k -» [XY], l i [Al = + [Ak

Solu tion: 1J./±/. AI „ -in3 „ A 1U >

(a)

A

(b)

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

fc[XY]ci

t

( l k V[XY],

l

-

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

L_| [XYlo^

1 * f 3 1 (7.02 x 10- M^-s' ) V0.0125M

[XY], = 0.125 [AY Jo — u.izj x u.zuu M — u.u/au M. since

(c)

1 \

1

[XY]o/

3

k V[XYJ,

M/

[XYJo/

— Kt i

(7.02x lO" !^^^- ) V0.0250M ft i

[XYJj

r-v>-v/i

| q q 7 v i n 3= 0.100 M/

1

rwi n^o- [A i Jo — u.ic>u m. Dince rvvl [Xrjj —n U.U6z,

1/ 1

(d)

/

[XY], I

1

a (7.02 x 10

i

1

M^i • s

rvv.i

[XY JQ

(

[XY]0 1 \

rearrange to solve ror c.

0.200M/

rearrange to soive >r r.

l

l

iv I n n/-^ v*

) \0-062 M

}

n i en T^A I

0.150 M/

3 nxio '* * s°* i

J

±u

\A and ^ c n ^ 1 nl s c. i^* 5.0x10 m ^ -l'*-t W I [xy]o [XYJo = 0-050 M r

1

.. -in— 3 n x — I . >^rl\/c nU ., A i1unl v.\ i

(/.UZAIU

m

-s^v^-

1

k K = "C + 273.15

1000 J ——

56.8 H Solution: T = 25°C + 273.15 = 298 K and mol

. k = A e E«'Ja

1000J - x1

J - = 5.68 x 104 - - then fej mol

Chapter 13 Chemical Kinetics

523

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/mol E./I 1000 J

(b) E» klf TJ, T2 -> fc2 /^2\ Ez / i In I — 1 = — I —

l\ I

Solution: (a)

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

-

Ti (b)

.

T2/

V400.K

5 J

.w

molX1000j"

,« mol"

450. K

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-mol fc2 = «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 constant 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 °C Conceptual 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 7 2 (a) Rearrange to solve for £a.

_ VTj (b)

_ T2

f_J

!_1

V300.K \300.K

mol"1000J

~mol'

350. K/

In [ — J = — ( jr] withki, = 0.000122/s, Tj = 300. K, T2 = 17°C + 273.15 = 290K. Ui/ R \T] T2/ Rearrange to solve for k2. "

.

^

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

Chapter 13 Chemical Kinetics

525

(b)

The intermediates are the species that are generated by one step and consumed by other steps. These areaCl(£)andCCl 3 (£).

(c)

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

1/2

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

13.76

(a)

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)

(b)

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

(c)

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

Catalysis 13.77

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

13.78

The initial and final energies (reactants and products) remain the same. The activation energy drops, from 75 kj/mol to a smaller value, for example 30 kj/mol. There are usually more steps in the reaction progress diagram.

without catalyst

, cc tt

= 75 kj/mol

with catalyst

op

OJ

s

products __ Reaction progress

13.79

Reaction progress

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!

M

| _ mei j298K 2

22.199

= 1012-1.25X10

5

fnel 298 K