Chapter 16: KineticsRates and Mechanisms of Chemical Reactions

16.1 Factors That Influence Reaction Rate16.1 Factors That Influence Reaction Rate

16.2 Expressing the Reaction Rate16.2 Expressing the Reaction Rate

16.3 The Rate Law and Its Components16.3 The Rate Law and Its Components

16.4 Integrated Rate Laws: Concentration Changes over Time16.4 Integrated Rate Laws: Concentration Changes over Time

16.5 The Effect of Temperature on Reaction Rate16.5 The Effect of Temperature on Reaction Rate

16.6 Explaining the Effects of Concentration and Temperature16.6 Explaining the Effects of Concentration and Temperature

16.7 Reaction Mechanisms: Steps in the Overall Reaction16.7 Reaction Mechanisms: Steps in the Overall Reaction

16.8 Catalysis: Speeding Up a Chemical Reaction16.8 Catalysis: Speeding Up a Chemical Reaction

Reaction Rate: The Central Focus ofChemical Kinetics

Fig. 16.1

There are Several Important Reasons that DifferentReactions, and Even the Same Chemical Reaction

May Go at Different Rates

1) Concentration: molecules must collide in order to react. Reaction rate is proportional to the concentration of reactants. Rate = k (collision frequency) = k (concentration)

2) Physical state: molecules must mix in order to collide. The physical state (solid, liquid, gas) will affect frequency of collisions, as well as the physical size of droplets (liquid) or particles in the case of solids.

3) Temperature: molecules must collide with enough energy to react. Raising the temperature increases the reaction rate by increasing the number of collisions, and the energy of the collisions.

Expressing Reaction Rate:

Rate of Reaction = Rate of Reaction = [A][A]

tt

Rate of Reaction = = = =change in concentration of Achange in concentration of A change in time change in time

concconc A A t t

Table 16.1 (P 669)Time(s) [OTime(s) [O33](](molmol/L)/L)

0.0 3.20x10 0.0 3.20x10-5-5

10.0 2.42x1010.0 2.42x10-5-5

20.0 1.95x1020.0 1.95x10-5-5

30.0 1.63x1030.0 1.63x10-5-5

40.0 1.40x1040.0 1.40x10-5-5

50.0 1.23x1050.0 1.23x10-5-5

60.0 1.10x1060.0 1.10x10-5-5

The reaction between ethylene and ozone:

C2H4 (g) + O3 (g) C2H4O(g) + O2 (g)

Rate = – = –[C2H4] t

[O3] t

Rate = Rate = –– = = 33.50 x 10.50 x 10--77mol/L·smol/L·s(1.10x10(1.10x10-5-5mol/L) - (3.20x10mol/L) - (3.20x10-5-5mol/L)mol/L)60.0 s - 0.0 s60.0 s - 0.0 s

[O[O33]] t t

Rate = Rate = ––

Fig. 16.5

Fig. 16.6

Plan: (a) Choose CH4 for comparison, as it has a coefficient of unity. For each molecule of methane 2 molecules of O2 are required and one molecules of CO2 is produced along with 2 moles of water. (b) because [O2] is decreasing, the change is negative. Substitute the negative value into the expression and solve for H2O/ t.

Expressing Rate in Terms of Changes inConcentration with Time - I

Problem: One “clean” fuel for automobiles is methane:CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g)

.

(a) Express the rate of this reaction in terms of changes in [CH4], [O2], and [H2O] with time.(b) When [O2] is decreasing at 0.310 mol/L S at what rate is [CO2] increasing?

Expressing Rate in Terms of Changesin Concentration with Time - II

Solution: a) Expressing the rate in terms of each component:

b) b) At whatAt what rate is CO rate is CO22 increasing if O increasing if O22 is decreasing at the rate of is decreasing at the rate of 0.310 0.310 mol mol/L S?/L S?..

1122

Rate = Rate = –– = = [[OO22]]

tt∆∆[[COCO22]] ∆∆tt

[[OO22]] tt = = –– 0.310 0.310 mol mol/L S/L S..

1122

[[COCO22]] ∆∆tt = = –– == –– = = 0.1550.155 mol mol/L S/L S11

22[[OO22]]

tt ( (-- 0.310 0.310 mol mol/L /L SS)).. ..

1122

RateRate = = –– == –– == ++ == ++[[HH22O]O]

∆∆tt[[CHCH44]]

tt

[[OO22]] tt

[[COCO22]] ∆∆tt

1122

Reaction Rate Law

For a For a chemicalchemical reactionreaction:: aaAA + + b bBB + . + . .. .. ccC + C + ddD + .D + . .. . .

The Rate Law for the The Rate Law for the forwardforward reactionreaction has the form: has the form:

Rate = k [A]a[B]b . . .

• k = the reaction rate constant• the exponents m & n are called the reaction orders

•• if the rate doubles when the concentration of A doubles, the rate depends on [A]1, so a = 1; • if the rate quadruples when the concentration of B doubles, the rate depends on [B]2, so b = 2.

Problem: For the reactions below determine the reaction order with respect to each reactant and the overall order of the given rate law:

Determining Reaction Order from the Rate Law

b) 2 NOb) 2 NO(g)(g) + 2 H + 2 H2 (g)2 (g) N N2 (g)2 (g) + 2 H + 2 H22OO(g)(g) Rate = k[NO] Rate = k[NO]22[H[H22]]

a) 5 Bra) 5 Br--(aq)(aq) + BrO + BrO33

--(aq)(aq) + 6 H + 6 H++

(aq)(aq) 3 Br 3 Br2 (aq)2 (aq) + 3 H + 3 H22OO(aq)(aq)

Rate = k[Br][BrORate = k[Br][BrO33--][H][H++]]22

Solution:Solution: a) The reaction is a) The reaction is firstfirst order with respect to Br order with respect to Br-- and BrO and BrO33

-- and and secondsecond order with respect to H order with respect to H++; ; fourthfourth order overall. order overall. b) The reaction is b) The reaction is secondsecond order in NO and order in NO and firstfirst order in H order in H22; and; and thirdthird order overall. order overall.

Plan: Plan: Inspect the exponents in the rate law, Inspect the exponents in the rate law, notnot the coefficients of the the coefficients of thebalanced equation, to find the individual orders, and then take their sumbalanced equation, to find the individual orders, and then take their sumto obtain the overall reaction order.to obtain the overall reaction order.

Finding Reaction Orders by Experiment - I

Given the reaction: O2 (g) + 2 NO(g) 2 NO2(g)The rate law for this reaction in general terms is: Rate = kRate = k [[OO22]]mm[NO][NO]nn

To find the reaction orders, we run a series of experiments, each with adifferent set of reactant concentrations, and determine the initial reaction rates.

Experiment Initial Reactant Concentrations (mol/l) Initial Rate O2 NO (mol/L·s)

1 1.10 x 10-2 1.30 x 10-2 3.21 x 10-3

2 2.20 x 10-2 1.30 x 10-2 6.40 x 10-3

3 1.10 x 10-2 2.60 x 10-2 12.8 x 10-3

4 3.30 x 10-2 1.30 x 10-2 9.60 x 10-3

5 1.10 x 10-2 3.90 x 10-2 28.8 x 10-3

By using the data from experiments 1&2 we can calculate the values ofthe coefficients m & n for the rate equation:

Determining Reaction Orders by Experiment - II

Rate 2 [ORate 2 [O22]]22mm [O [O22]]22

Rate 1 [ORate 1 [O22]]11mm [O [O22]]11

= == =mmOxygenOxygen concentration changes concentration changes

forfor experiments experiments 1 & 2 but1 & 2 but NO NOis the same is the same and and k k isis the samethe same, , so theyso they will cancel will cancel in in ratioratio..

Using the concentrations Using the concentrations and rates fromand rates from the preceding the preceding table:table:

6.40 x 106.40 x 10-3-3 mol mol/L s 2.20 x 10/L s 2.20 x 10-2-2 mol mol/L/L

3.21 x 103.21 x 10-3-3 mol mol/L s 1.10 x 10/L s 1.10 x 10-2-2 mol mol/L/L

mm==

1.99 = (2.00)1.99 = (2.00)mm m = 1m = 1

The reaction is The reaction is first orderfirst order in O in O22, when O, when O22 doubles, the rate doubles. doubles, the rate doubles.

To find the order of NO we will compare experiments 3 & 1 in which oxygen is constant, but NO doubles, and the rate between the two experiments increases substantially.

Determining Reaction Orders by Experiment - III

Rate 3 k[ORate 3 k[O22]]33mm[NO][NO]33

nn

Rate 1 k[ORate 1 k[O22]]11mm[NO][NO]11

nn==

In expt’s. 1&3, kIn expt’s. 1&3, k andand the the concentrationconcentration ofof oxygenoxygen are constant andare constant and will will cancel,cancel, so onlyso only the ratethe rate andand [NO][NO] will vary.will vary.

Rate 3 [NO]Rate 3 [NO]33nn

Rate 1 [NO]Rate 1 [NO]11nn

== Substituting in the values from the table:

12.8 x 1012.8 x 10-3-3 mol mol/L s /L s (2(2.60 x 10.60 x 10-2-2 mol mol//L)L)nn==

3.21 x 103.21 x 10-3-3 mol mol/L s /L s (1(1.30 x 10.30 x 10-2-2 mol mol//L)L)nn

3.99 = (2.00)3.99 = (2.00)nn

n = 2n = 2

The reaction is The reaction is second ordersecond order in NO: when [NO] doubles, the rate in NO: when [NO] doubles, the ratequadruples, Thus the rate law is:quadruples, Thus the rate law is:

Rate = k [O2][NO]2

IntegratedRate LawsandReactionOrder

Fig. 16.7

ln [A]o–ln [A]t = kt

1

[ A]t

−1

[A]o

= kt

Determining the Reactant Concentration at a Given Time - I

Problem:Problem: CyclopropaneCyclopropane converts toconverts to propenepropene with a with a first first orderorder rate rate constantconstant ofof 1.7 x 101.7 x 10--2 2 hrhr - -11. . (a)(a) If the initial If the initial concentration ofconcentration of cyclopropanecyclopropane is 1.200 M, what is the concentration after 20.00 hrs? is 1.200 M, what is the concentration after 20.00 hrs?(b)(b) What fraction of What fraction of cyclopropane cyclopropane has decomposed has decomposed??

lnln [C [C33HH66]]00 - - ln ln [C [C33HH66]]t t == kt kt

Fraction decomposed = Fraction decomposed = [C[C33HH66]]00 - [C - [C33HH66]]tt

[C[C33HH66]]00

(b) The fraction decomposed is the concentration that has (b) The fraction decomposed is the concentration that has decomposed divided by the initial concentration:decomposed divided by the initial concentration:

Plan: Plan: (a) To find the concentration of cyclopropane at time (a) To find the concentration of cyclopropane at time tt,,[C[C33HH66]]tt , for a first-order reaction use the , for a first-order reaction use the integrated rate lawintegrated rate law::

Fraction decomposed = = 0.03417Fraction decomposed = = 0.03417(1.200(1.200 mol mol/L)/L)

(1.200(1.200 mol mol/L) - (1.159/L) - (1.159 mol mol/L)/L)

Determining the Reactant Concentration at a Given Time - II

Solution: (a) Solving the integrated rate expression for ln[ln[CC33HH66]]tt :

lnln [C [C33HH66]]t t == ln ln[C[C33HH66]]00 –– ktkt

lnln [C [C33HH66]]tt == ln ln(1.200(1.200 mol mol/L/L)–()–(1.7 x 1.7 x 1010 -2 -2 hrhr-1-1)(20.00 hr)(20.00 hr))

lnln[C[C33HH66]]tt == 0.18232 0.18232 –– 0.03400 = 0. 0.03400 = 0.1483214832

[C[C33HH66]]tt = = 1.1591.159 mol mol/L/L

(b) Finding the fraction that has decomposed after 20.00 hrs:(b) Finding the fraction that has decomposed after 20.00 hrs:

Fraction decomposed = Fraction decomposed = [C[C33HH66]]00 - [C - [C33HH66]]tt

[C[C33HH66]]00

Fig. 16.8

Linear plotLinear plot of of ln[A]ln[A]tt vsvs tt shows that the reaction is shows that the reaction isfirst order in Nfirst order in N22OO55..

Fig. 16.9

Half life, t 1/2 For first order reactions: tt 1/2 1/2 = = 0.6930.693

k k

(p. 684)

Dependence of the Rate Constanton Temperature

Fig. 16.10

k = Ae– Ea / RTArrhenius Arrhenius Equation Equation

ln k = ln A −Ea

R

1

T

lnk2

k1

= −Ea

R

1

T2

−1

T1

Fig. 16.11

ln k = ln A −Ea

R

1

T

Fig. 16.12

Fig. 16.14

Table 16.5 (p. 687)

Fig 16.15

Fig. 16.16

Fig. 16.17

Nature of the Transition State in theReaction Between CH3Br and OH–

Fig. 16.18

Fig. 16.19

Reaction Energy Diagrams and Possible Transition States for the Three Reactions

Fig. 16.20

Elementary Reactions and Molecularity

Elementary ReactionsElementary Reactions - The individual - The individual stepssteps whichwhich make up themake up the proposedproposed reaction mechanism reaction mechanism..

DecompositionDecomposition of ozone in the stratosphere of ozone in the stratosphere byby UV light UV light toto form oxygen. form oxygen.

2 O2 O3 (g)3 (g) 33 O O22(g)(g)

•• Proposed two Proposed two-step -step mechanism formechanism for this reaction. this reaction.•• The The first step first step isis a a unimolecularunimolecular reaction reaction,, the decomposition of the decomposition of anan O O33..•• The The second step is a second step is a bimolecular bimolecular reactionreaction (two(two molecules react).molecules react).•• The The overall reactionoverall reaction is the sum of the two steps. is the sum of the two steps.

OO3 (g)3 (g) O O2 (g)2 (g) + O + O (g) (g)

OO3 (g)3 (g) + O + O(g)(g) 2 O 2 O2 (g)2 (g)

2 O2 O3 (g)3 (g) 3 O 3 O2 (g)2 (g)

Molecularity -Molecularity - The number of reactant molecules (particles) The number of reactant molecules (particles) involved in the step. involved in the step.

Rate Laws for General Elementary Steps

Elementary Step Molecularity Rate Law

Table 16.6 (p. 693)Table 16.6 (p. 693)

A product UnimolecularUnimolecular Rate = k[A]

2A product Bimolecular Bimolecular Rate Rate = k[A] = k[A]22

A + B product BimolecularBimolecular RateRate = k[A][B] = k[A][B]

2A + B product TermolecularTermolecular Rate Rate = k[A] = k[A]22[B][B]

Determining Molecularity and Rate Laws for Elementary Steps

Problem: These two reactions are for the stepwise neutralization of sulfuric acid by gaseous ammonia:

(a) Write the overall balanced equation.(a) Write the overall balanced equation.(b) Determine the(b) Determine the molecularity molecularity of each step. of each step.(c) Write the rate law for each step.(c) Write the rate law for each step.

(1) NH(1) NH3 (g)3 (g) + H + H22SOSO4 (g)4 (g) NH NH44++

(g)(g) + HSO + HSO44--(g)(g)

(2) NH(2) NH3 (g)3 (g) + HSO + HSO44--(g)(g) NH NH44

++(g)(g) + SO + SO44

-2-2(g)(g)

2 NH3 (g) + H2SO4 (g) 2 NH4+

(g) + SO4-2

(g)

NH3 (g) + H2SO4 (g) NH4+

(g) + HSO4-(g)

NH3 (g) + HSO4 -(g) NH4

+(g) + SO4

-2(g)

Solution: (a) Writing the overall equation:

Plan: Plan: Find the overall equation from the Find the overall equation from the sumsum of the simple steps. The of the simple steps. The molecularity of each step equals the total number of reactant particles. molecularity of each step equals the total number of reactant particles. Write the rate law for each step using the molecularities as rxn. orders.Write the rate law for each step using the molecularities as rxn. orders.

Determining Molecularity and Rate Laws for Elementary Steps

SolutionSolution continuedcontinued

(b) Determining the(b) Determining the molecularity molecularity of each step: of each step:

(1) Step #1 has (1) Step #1 has twotwo reactants, ammonia and sulfuric acid, and reactants, ammonia and sulfuric acid, and is is therefore therefore bimolecularbimolecular(2) (2) Step #Step #2 has 2 has twotwo reactants, ammonia and hydrogen sulfate reactants, ammonia and hydrogen sulfate,, and and is therefore is therefore bimolecularbimolecular

(c) Writing the rate laws for the elementary reactions:(c) Writing the rate laws for the elementary reactions:

(1) Rate(1) Rate11 = k = k11[NH[NH33][H][H22SOSO44]](2) Rate(2) Rate22 = k = k22[NH[NH33][HSO][HSO44

--]]

The overall reaction and rate are:

Rate = k[NHRate = k[NH33]]22[H[H22SOSO44]]

2 NH2 NH3 (g)3 (g) + H + H22SOSO4 (g)4 (g) 2 NH 2 NH44++

(g)(g) + SO + SO44-2-2

Fig. 16.21

A Reaction with a Slow Initial StepA Reaction with a Slow Initial Step

1) NO2 + F2 -> NO2F + F slow, rate determining

2NO2 + F2 -> 2NO2 F overall

2) NO2 + F -> NO2F fast

Fig. 16.22

Mechanismfor theCatalyzedHydrolysisof anOrganicEster

Fig. 16.23

Hydrolysis reaction is catalyzed by acid.

The Metal-Catalyzed Hydrogenationof Ethylene

Fig. 16.24