Chemistry part 23, Julia Burdge,2e (2009) potx

Using the data from experiment 1, we get Note that the units of this rate constant are different from those for the rate constant we calculated for the FT CI02 reaction and the bromine r

552 CHAPTER 14 Chemical Kinetics

Recall that when an exponent is 1, it need not

rate3 2.4 X 10- 3 Mis = 2 rate I 1.2 X 10- 3 Mis

This indicates that the rate is directly proportional to the concentration of fluorine and the value

of x is 1

rate = k [F2] [Cl02Y Similarly, we can compare experiments 1 and 2:

[CI02] (M)

[CIO , ] quadruples {0.01O

0.040 0.010

Initial Rate (Mis)

{1.2 X 10

-3

Rate quadruples 3

4.8 X 102.4 X 10- 3

-We find that the rate quadruples when the concentration of chlorine dioxide is quadrupled, but the fluorine concentration is held constant

[Cl02h = 0.040 M = 4 [Cl02]l 0.010M

4.8 X 10- 3 Mis = 4 1.2 X 10-3 Mis

This indicates that the rate is also directly proportional to the concentration of chlorine dioxide, so the value of y is also 1 Thus, we can write the rate law as follows:

aA + bB - - + cC + dD

which has the general rate law

rate = k[A Y [BY

Comparing experiments 1 and 2, we see that when [A] doubles, with [B] unchanged, the rate also doubles

Experiment

1

2

3 Thus, x = 1

[A] (M)

0.10 0.20 0.10

[B] (M)

[B] unchanged {0.o15

0.015 0.010

[B] (M)

0.015 0.015 0.030

Initial Rate (Mis)

{

1.2 X 10- 3

Rate doub l es 3

4.8 X 102.4 X 10-3

-Initial Rate (Mis)

2.1 X 10- 4 4.2 X 10-4 8.4 X 10-4

SECTION 14.2 Dependence of Reaction Rate on Reactant Concentration 553

1 rate = k[A] or rate = k[B]

2 rate = k[A]2, rate = k[Bf, or rate = k[A][B]

M S - 1

S-1

M-1 S - I

~2 S- 1

* Another possibility for a third- or der reaction is rate = k [Aj(Bj(C], although such reactions are very rare

Comparing experiments 1 and 3, when [B] doubles with [A] unchanged, the rate pies

Thus, the rate is n ot directly proportional to [B] to the first power, but rather it is directly

propor-tional to [B] to the second power (i.e., y = 2):

7

rate ex The overall rate law is

[B]-rate = k[A][Bf

This reaction is therefore first order in A, second order in B, and third order overall

Once again, knowing the rate law, we can use data from any of the experiments in the table

to calculate the rate constant Using the data from experiment 1, we get

Note that the units of this rate constant are different from those for the rate constant we calculated

for the FT CI02 reaction and the bromine reaction In fact, the units of a rate constant depend on

the overall order of the reaction Table 14.4 compares the units of the rate constant for reactions

that are zeroth-, first-, secondo, and third-order overall

The following are three important things to remember about the rate law:

1 The exponents in a rate law must be determined from a table of experimental data-in

general, they are not related to the stoichiometric coefficients in the balanced chemical equation

2 Comparing changes in individual reactant concentrations with changes in rate shows how

the rate depends on each reactant concentration

3 Reaction order is always defined in terms of reactant concentrations, never product

concentrations

Sample Problem 14.3 shows how to use initial rate data to determine a rate law

The gas-phase reaction of nitric oxide with h ydroge n at 1280 ° C i s

2NO(g) + 2Hig) - - N 2 (g) + 2 H 2 0(g)

From the following data collected at 1280 ° C, determine (a) the rat e law , ( b) the rate cons tant ,

including unit s, and (c) the rate of the reaction when [NOl = 4.8 X 10 - 3 M and [H2l = 6.2 X 10 - 3 M

(Continued)

554 CHAPTER 14 C hemical Kinetics

A quotient of numbers, each raised to the

same power, is equal to the quotient raised to

that power: x n /y n = {xly) n

Think About It T h e exponent for

the concentration of H z in the rate

law i s 1, whereas the coefficient for

H z in the balanced equation is 2 It

is a common error to try to write a

rate law using the stoichiometric

coefficients as the exponents

Remember that, in general, the

exponents in the rate l aw are not

related to the coefficients in the

balanced equation Rate laws mu s t

be determined by examining a table

of experimental data

concentration of each reactant

increases by approximately a factor of 4 whe n [NO] i s doubled but [H z ] is he l d constant Comparing experiments 2 and 3 s hows that the rate doub l es w h en [H z ] doubles but [NO ] is he l d constant

rat~ = 5 0 X 10- 5 ~ = 4 = k(1.0 X 10- 2 M)"( 2.0 X 10-3 M)Y

ratel 1.3 X 10- 5 ~ k(5 0 X 10- 3 M)X(2 0 X 10-3 M)Y

Canceling identical term s in the numerator and denominator gives

( 1.0 X 10-2 M)x

-' - = 2 " = 4 ( 5.0 X 10-3 M)x

Therefore, x = 2 The reac t ion i s seco nd order in NO

Dividing the rate from experiment 3 by the rate from experiment 2, we get

Therefore, y = 1 The reaction i s first order in H z The overall rate law i s

(b) We can use data from any of the experiments to calculate the va l ue and units of k Using the data

from experiment 1 gives

k = rate

[NO f[ H2J

1.3 X 10- 5 Mis = 2.6 X 10 2 M - 2 • S-I ( 5.0 X 10-3 M) 2(2 0 X 1O - 3.M)

(c) Using the rate constant determined in part ( b) and the concentrations of NO and H z given in the problem statement, we can determine the reaction rate as follow s :

rate = (2 6 X 102 M - 2 • s-I)(4.8 X 10- 3 M)\6.2 X 10- 3 M)

= 3 7 X 10- 5 Mis

From the following data collected at a certain temperature, determine the rate law and calculate the

r a te constant, including its units

Initial Rate (Mis)

2 20 X 10 - 6

2.20 X 10 - 6

SECTION 14 3 D epe n dence of Reactant Concentrat i o n on Ti m e 555

Dependence of Reaction Rate on Reactant Concentration

Answer question s 14.2.1 through 14.2.4 using the

tab l e of initial rate data for the rea c tion Ex~eriment [A](M) [B](M)

Initial Rate (Mis )

Dependence of Reactant Concentration on Time

We can use the rate law to determine the rate of a reaction using the rate constant and the reactant

A rate law can also be used to determine the remaining concentration of a reactant at a specific

time during a reaction We will illustrate this use of rate laws using reactions that are first order

overall and reactions that are second order overall

First-Order Reactions

Afirst-order reaction is a reaction whose rate depends on the concentration of one of the reac

-tants raised to the first power Two examples are the decomposition of ethane (C2H6) into highly

reactive fragments called methyl radicals ('CH3), and the decomposition of dinitrogen pentoxide

(N20 S) into nitrogen dioxide (N02) and molecular oxygen (02):

C2H6 - _ 2 ·CH3 2N20 S(g) - _ 4N0 2(g ) + 0 2 (g)

In a first-order reaction of the type

A • product

rate = k [ C ? H 6 J

rate = k[N 2 0 S J

the rate can be expressed as the rate of change in reactant concentration,

as well as in the form of the rate law:

Ll[A J

rate = - -

-I1t

rate = k[AJ

556 CHAPTER 14 Chemical Kinetics

It is not necessary for you to be ab l e to do the

calculus required to arrive at Equation 14.3, but

i t i s very important that you know how to use

Equation 14.3

The inverse of In x is f! [ ~ Appendix 1)

Think About It Don't forget

If you calculate a concentration

concentration at time 0 (o r if you

get a negative time required for the

concentration to drop to a s pecified

common error

Becau s e p res su re is proportional to

concentration , f o r gaseous reactions

[ H4 Section 11.4) Equat i ons 14.3 and

re spectively, where Po and P, are the pressures of

reactant A at times 0 and 1 respectively

Setting these two expressions of the rate equal to each other we get

where In is the natural logarithm, and [A]o and [A] t are the concentrations of A at times 0 and t,

respectively In general, time 0 refers to any specified time during a reaction-not necessarily the

beginning of the reaction Time t refers to any specified time after time O Equation 14.3 is

some-times called the integrated rate law

In Sample Problem 14.4 we apply Equation 14.3 to a specific reaction

Sample

The rate constant for thi s reaction at 20 0

0.75 M, determine (a) the concentration of HzO z remaining after 3 h and (b) how long it will take for

Equation 14.3 can be rearranged as follows:

Equation 14.4 In [AJt = -kt + In [AJo

Equation 14.4 has the form of the linear equation y = mx + b:

In [AJt = (-k)(t) + In [AJo

SECTION 14.3 Dependence of Reactant Concentration on Time 557

k ln [Ala

'-<- - slope = -k

t

Figure 14.8(a) shows the decrease in concentration of reactant A during the course of the reaction

As we saw in Section 14.1, the plot of reactant concentration as a function of time is not a straight

line For a first-order reaction, however, we do get a straight line if we plot the natural log of

reac-tant concentration (In [A]t) versus time (y versus x) The slope of the line is equal to -k [Figure

14.8(b)], so we can determine the rate constant from the slope of this plot

Sample Problem 14.5 shows how a rate constant can be determined from experimental data

Determine the rate constant of the reaction at this temperature

decomposition of azomethane is first order We do this by plotting In P against time If the reaction is

first order, we can use Equation 14.3 and the data at any two of the times in the table to determine the

Plotting these data gives a straight line, indicating that the reaction is indeed first order Thus, we can

use Equation 14.3 expressed in terms of pressure

PI

In- = -kt

Po

P t and Po can be pressures at any two times during the experiment Po need not be the pressure at

Os-it need only be at the earlier of the two times

(Continued)

Figure 14.8 First-order reaction characteristics: (a) Decrease of reactant concentration with time (b) A plot of

In [Al l versus t The slope of the line is equal to -k

Th i s grap h ica l determi n at i on is an alternative to using the method of i nitial rates to determ i ne the value of k

558 CHAPTER 14 Chemical Kinetics

Think About It We could equally

well have determined the rate

constant by calculating the s lop e

of the plot of In P versus t Using

the two points labeled on th e plot,

From the following data, determine the rate constant of this reaction

According to the definition of half-life, t = t1/2 when [A], = i [A]o, so

According to Equation 14.5, the half-life of a first-order reaction is independent of the initial

concentration of the reactant Thus, it takes the same time for the concentration of the reactant to decrease from 1.0 M to 0.50 M as it does for the concentration to decrease from 0.10 M to 0.050 M

(Figure 14.9) Measuring the half-life of a reaction is one way to determine the rate constant of a

first-order reaction

The half-life of a first-order reaction is inversely proportional to its rate constant, so a short half-life corresponds to a large rate constant Consider, for example, two radioactive isotopes used

in nuclear medicine: 24Na (t1/2 = 14.7 h) and 60CO (t1/2 = 5.3 yr) Sodium-24, with the shorter half-life, decays faster If we started with an equal number of moles of each isotope, most of the

sodium-24 would be gone in a week whereas most of the cobalt-60 would remain unchanged

SECTION 14.3 Dependence of Reactant Concentration on Time 559

Calculate the half-life of the reaction in minute s

Strategy U se Equation 14.5 to calculate t 1/2 in s econd s, and then con v elt to minute s

Practice Problem A Calculate the half-life o f the dec o mp os ition of azomethan e, di s cu ss ed in

Sample Problem 14.5

Practice Problem B C alcula t e the rate c on s tant for the fir s t-ord e r deca y of 2 4 N a (t 1/2 = 14 7 h)

Figure 14.9 A plot of [AJ versus time for the fir s t -order reaction

A • product s The half-life ofthe reaction i s 1 min The concentration of

where the rate can be expressed as

As before, we can obtain the expression for the half-life by setting [A], = i [A]o in Equation 14.6

Solving for t l/2, we obtain

con-distinguish between first-order and second-order reactions,

Sample Problem 14.7 shows how to use Equations 14,6 and 14,7 to calculate reactant centrations and the half-life of a second-order reaction,

con-Iodine atom s combine to form molecular iodine in the gas phase:

•

I(g) + I(g) - - + 12(g )

Thi s reaction i s s econd order and has a rate con s tant of7.0 X 109 M- 1 , S-1 at 23 ° C (a) If the initial concentration of I i s 0.086 M , calculate the concentration after 2.0 min (b) Calculate the half-life of the reaction when the initial concentration of I is 0.60 M and when the initial concentration of I is 0.42 M

w hen [1]0 = 0.60 M and when [1] 0 = 0.42 M

Setup

I

SECTION 14.3 Dependence of Reactant Concentration on Time 561

•

_ _ _ -:- _ :.1_-:-_ _ _ = 3.4 X 10 - 10 s (7.0 X 109 M - I • s - I)(0.42 M)

Practice Problem The reaction 2A • B is second order in A with a rate constant of

32 M- I • S - I at 25 ° C (a) Starting with [Alo = 0.0075 M, how long will it take for the concentration

of A to drop to 0.0018 M? (b) Calculate the half-life of the reaction for [Alo = 0.0075 M and for

[Alo = 0.0025 M

First- and second-order reactions are the most common reaction types Reactions of overall

order zero exist but are relatively rare For a zeroth-order reaction

A • product the rate law is given by

rate = k[AJ o = k

Thus, the rate of a zeroth-order reaction is a constant, independent of reactant concentration

Third-order and higher-Third-order reactions are quite rare and too complex to be covered in this book Table 14.5

o rate = k

1 rate = k[AJ

2 rate = k[AJ2

Kinetics ReaCtions

Integrated Rate Law

Think About It (a) Iodine, like the other halogens, exists as diatomic molecules at room temperature

It makes sense, therefore, that atomic iodine would react quickly, and essentially completely, to

form Iz at room temperature The

very low remaining concentration

of I after 2 min makes sense

(b) As expected, the half-life of this second-order reaction is not constant (A constant half-life

is a characteristic of first-order reaction s )

562 CHAPTER 14 Chemical Kinetics

Cookbooks sometimes give alternate directions

for cooking at high altitude s, where the lower

atmospheric pressure re su lts in wa ter boiling at

a lower tem p e ra ture [ ~ Section 12.6]

Kinetic energy is the result of motion of the

whole molecule, relative to its surroundings

Vibrational energy is the resu lt of motion of the

atoms in a molecule, relati ve to one another

The fir s t-order decomposition of dinitrogen

pentoxide (NzOs) is represented by

U s e the t able of data to answer que st ions 14 3.1 and 14.3.2:

14.3.1 Wh a t i s the rate constant for the

14.3.2 Approximately h ow long will it take

for [N20 S J to fall from 0.62 M t o

0.64

0.44 0.16

Calculate the half-life of a first-order reaction for which the rate constant is

2.5 X 10 - 2 s - I

a) 28 s

b ) 0.017 s c) 361 s

d ) 58 s

e) 1.4 X 103 s

A • 2B i s a second-order reaction forwhichk = 5.3 X 10 - 1 M - 1 , S - I,

Calculate t 1/2 when [Al o = 0 55 M

a) 1.3 s

b ) 0.29 s

c) 0.77 s

d ) 15 s e) 3.4 s

Dependence of Reaction Rate on Temperature

food depends largely on the boiling point of water The reaction involved in hard-boiling an egg happens faster at 100°C (about 10 min) than at 80°C (about 30 min), The dependence ofreaction rate on temperature is the reason we keep food in a refrigerator-and why food keeps even longer

in a freezer The lower the temperature, the slower the processes that cause food to spoiL

product molecule is formed by the direct combination of an A molecule and a B molecule, If we doubled the concentration of A, then the number of A-B collisions would also double, because there would be twice as many A molecules that could collide with B molecules in any given vol-ume Consequently, the rate would increase by a factor of 2 Similarly, doubling the concentration

of B molecules would increase the rate twofold Thus, we can express the rate law as

rate = k[A][B]

The reaction is first order in both A and B and is second order overalL

This view of collision theory is something of a simplification, though, because not every collision between molecules results in a reaction A collision that does result in a reaction is called

an effective collision A molecule in motion possesses kinetic energy; the faster it is moving, the greater its kinetic energy, When molecules collide, part of their kinetic energy is converted to

vibrational energy If the initial kinetic energies are large, then the colliding molecules will vibrate

-Is the Shroud of Turin Really the Burial Cloth

,

of Christ?

Scientists can determine the ages of such artifacts as the Shroud

of Turin using radiocarbon dating Earth's atmosphere is

con-stantly being bombarded by cosmic rays of extremely high

pene-trating power These rays, which originate in outer space, consist

of electrons, neutrons, and atomic nuclei One of the important

reactions between the atmosphere and cosmic rays is the capture

of neutrons by atmospheric nitrogen (nitrogen-14) to produce the

radioactive carbon-14 isotope and hydrogen:

The carbon-14 atoms eventually find their way into 14COz, which

mixes with the ordinary carbon dioxide (12C02) in the atmosphere

As the carbon-14 isotope decays, it emits {3 particles (electrons)

The rate of decay (as measured by the number of electrons

emit-ted per second) obeys first-order kinetics It is customary in the

study of radioactive decay to write the rate law as

rate = kN

where k is the first-order rate constant and N is the number of 14C

nuclei present The half-life of the decay, t l/2, is 5715 yr, so

rear-ranging Equation 14.5, we write

k = 0.693 = 1.21 X 10- 4 yr- I

5715 yr

Carbon-14 enters the biosphere when carbon dioxide is taken up

in plant photosynthesis The 14C lost by radioactive decay is

con-stantly replenished by the production of new 14C in the atmosphere

In this decay-replenishment process, a dynamic equilibrium is

established whereby the ratio of 14C to 12C remains constant in

living matter But when an individual plant or animal dies, the

car-bon-14 in it is no longer replenished, so the ratio decreases as 14C

decays This same change occurs when carbon atoms are trapped

in coal, petroleum, or wood preserved underground, and, of

course, in any object that was once living including the Shroud

of Turin, which is linen, a cloth made from the flax plant

In 1955, American chemist Willard F Libby suggested

that the 14C: 12C ratio could be used to estimate the length of time

the carbon-14 isotope in a particular specimen has been ing without replenishment Rearranging Equation 14.3, we can

decay-write

No

InN=kt

I

where No and Nt are the number of 14C nuclei present at t = 0 and

t = t, respectively Because the rate of decay is directly tional to the number of 14C nuclei present, the preceding equation

-1.21 X 10- 4 yr- 1 decay rate of old sample

Knowing k and the decay rates for the fresh sample and the old

sample, we can calculate t, which is the age of the old sample

This ingenious technique is based on a remarkably simple idea Its success depends on how accurately we can measure the rate of decay In a fresh sample, the ratio 14C/!?C is about 111012, so the

equipment used to monitor the radioactive decay must be very

sensitive Precision is more difficult with older samples because

they contain even fewer 14C nuclei Nevertheless, radiocarbon dating has become an extremely valuable tool for estimating the

age of archaeological artifacts, paintings, and other objects ing back 1000 to 50,000 years

dat-In 1988 three laboratories in Europe and the United States,

working on samples of less than 50 mg of the shroud, pendently showed by carbon-14 dating that the shroud dates

inde-from between A.D 1260 and A.D 1390 Thus, the shroud could not have been the burial cloth of Christ Libby received the Nobel Prize in Chemistry in 1960 for his work on radiocar-bon dating

so strongly as to break some of the chemical bonds This bond breaking is the first step toward

product formation If the initial kinetic energies are small, the molecules will merely bounce off of

each other intact There is a minimum amount of energy, the activation energy (Ea), required to

initiate a chemical reaction Without this minimum amount of energy at impact, a collision will be

ineffe ct iv e; that is, it will not result in a reaction

When molecules react (as opposed to when atoms react), having sufficient kinetic energy is not

the only requirement for a collision to be effective Molecules must also be oriented in a way that favors

reaction The reaction between chlorine atoms and nitrosyl chloride (NOCl) illustrates this point:

Cl + NOCl _ Cl2 + NO

This reaction is most favorable when a free Cl atom collides directly with the Cl atom in the NOCI

molecule [Figure 14 10 (a)] Otherwise, the reactants simply bounce off of each other and no

reac-tion occurs [Figure l4.1O(b)]

563

564 CHAPTER 14 Chemical Kinetics

Figure 14.10 (a ) In orderfor an

effective collision to take place, the free

Cl atom must collide directly with the

Cl atom in NOCI (b) Otherwi s e, the

reactants bounce off of one another and

the collision is ineffective-no reaction

takes place

Figure 14 11 Energy profile

for the reaction of Cl with NOCI In

addition to being oriented properly ,

reactant molecules must po ss ess

sufficient energy to overcome the

collision Figure 14.11 shows a potential energy profile for the reaction between Cl and NOCl

We can think of the activation energy as an energy barri e r that prevents less energetic

mol-ecules from reacting Because the number of reactant molecules in an ordinary reaction is very large, the speeds, and therefore also the kinetic energies of the molecules, vary greatly Normally, only a small fraction of the colliding molecules-the fastest-moving ones have sufficient kinetic

energy to exceed the activation energy These molecules can therefore take part in the reaction The relationship between rate and temperature should now make sense According to kinetic molecular theory, the average kinetic energy of a sample of molecules increases as the temperature increases

[ ~~ Section 11.6, Figure 11 19] Thus, a higher percentage of the molecules in the sample have

sufficient kinetic energy to exceed the activation energy, and the reaction rate increases

The Arrhenius Equation

The dependence of the rate constant of a reaction on temperature can be expressed by the nius equation,

where E a is the activation energy of the reaction (in kJ/mol) , R is the gas constant (8.314 J/K

mol), T is the absolute temperature, and e is the base of the natural logarithm [ ~~ Appendix ]]

The quantity A represents the collision frequency and is called the frequency factor It can be treated as a constant for a given reaction over a reasonably wide temperature range Equation

14.8 shows that the rate constant decreases with increasing activation energy and increases with

increasing temperature This equation can be expressed in a more useful form by taking the natural logarithm of both sides:

In k = In Ae -E"I RT