Ch13. Kinetics

1

Chapter 13Kinetics: The Study of Rates of

Reaction

Brady and Senese5th Edition

Index13.1 Five factors affect reaction rates13.2 Rates of reaction are measured by monitoring change in concentration over time13.3 Rate laws give reaction rate as a function of reactant concentrations13.4 Integrated rate laws give concentration as a function of time13.5 Reaction rate theories explain experimental rate laws in terms of molecular collisions13.6 Activation energies are measured by fitting experimental data to the Arrhenius equation13.7 Experimental rate laws can be used to support or reject proposed mechanisms for a reaction13.8 Catalysts change reaction rates by providing alternative paths between reactants and products

13.1 Five factors affect reaction rates 3

IntroductionRate of reaction - the relationship between changes

in concentration and time: the “speed” of a reaction


Factors Affecting Reaction Rate

1. Chemical nature• Bond strengths• General reactivity

2. Ability to establish contact with one another• Physical state• Surface area for liquids, solids, and heterogeneous

mixtures • Amount of mixing• Particle shape/size


Factors Affecting Reaction Rate (Cont.)

3. Concentration of reactants Molarity for solutions Pressure effects for gases Volume effects for gases

4. Temperature5. Catalysts


Your Turn!

Which of the following would speed a reaction?A. Stirring itB. Dissolving the reactants in water, if ionicC. Adding a catalystD. Grinding any solidsE. All of these

13.2 Rates of reaction are measured by monitoring change in concentration over time 7

Measuring Rates• Can be measured using any substance in the reaction• Units: M s-1

• Measured in three ways: Instantaneous rate (text uses this unless specified) Average rate Initial rate


Instantaneous Reaction Rates

• Instantaneous rate - the slope of the tangent to the curve at any specific time

• Initial rate is determined at the initial timeNO2 Appearance

-0.005

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035

0 200 400 600 800Time (s)

[NO

2]


Slope = -0.01 M/100 s

rate = 1 × 10-4 M s-1

Instantaneous Rates Changes with Time

• As reactants are consumed, reaction slows

• The process of determining rates is important for reproducibility


Average Reaction Rates rateΔtime

]Δ[Reactant

Average rate of reaction - the slope of the line connecting the starting and ending coordinates for a specified time frame

NO2 Appearance

-0.005

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035

0 100 200 300 400 500 600 700 800Time (s)

[NO

2]


Your Turn!

What is the average rate of disappearance of B between 10 and 40 s?

A. -0.006 M s-1

B. +0.006 M s-1 C. -0.002 M s-1

D. +0.002 M s-1

E. Can’t tell form the information

10 20 30 40

Time Elapsed in Reaction Progress (s)

Con

cent

ratio

n of

B (M

)

0.0

0.1

0.2

0.3

0.4


Rates and Stoichiometry

• Rates based on each substance are related to one another by the stoichiometric coefficients of the reaction

• Examine the reaction: aA + bB → dD The stoichiometric relationship between substances A

and B is given as a mole A : b mole B

RateA × (b/a) = RateB

mol A b mol B mol BL s a mol A L s


Learning Check

• In the reaction: 2A + 3B → 5D We measured the rate of disappearance of substance A to be 3.5 × 10-5 M s-1. What is the rate of appearance of D?

• In the reaction 3A + 2B → C, we measured the rate of B. How does the rate of C relate?

8.75 × 10-5 M s-1

rateC = 1/2 rateB


Your Turn!

In the reaction 2CO(g) + O2(g) → 2CO2(g), the rate of the reaction of CO is measured to be 2.0 M s-1. What would be the rate of the reaction of O2?

A. The sameB. Twice as greatC. Half as largeD. You cannot tell from the given information

13.3 Rate laws give reaction rate as a function of reactant concentrations 15

The Rate Law Depends on the Concentrations Used

• rate = k[reactant]order

• k is a reaction rate constant, a measure of time efficiency

• High values of k mean high efficiency• k must be determined experimentally• Each experiment has its own rate law• Rate law must be determined experimentally


Learning Check

The rate law for the reaction 2A + B → 3C is rate = 0.045 M-1 s-1 [A][B]

if the concentration of A is 0.2 M and that of B is 0.3 M, what will be the reaction rate?

rate = 0.0027 M s-1

rate = 0.045 M-1 s-1 (0.2 M)(0.3 M)


Determining the Rate Law

• Run reaction under the same conditions, varying only the concentrations of reactants

• A ratio of rate laws for each experiment allows us to determine the orders of each reactant

• The rate law is unique to temperature and concentration conditions

• The rate law cannot be predicted from the chemical equation, but must be determined experimentally


Orders • Indicate the degree of resistance to reaction - a high order

suggests a reactant that is slow to react• Are indicated for each reactant• The overall reaction order is the sum of individual

reactant orders• May be negative, fractional or integers, but in this course

we will usually encounter positive integers• Must be determined from experimental data• The units for k depends on the order of the reaction

zero order mol L-1 s-1 M s-1

1st order s-1 s-1

2nd order L mol-1 s-1 M-1 s-1


Use Rate Laws to Determine Orders

• Select 2 rate laws that vary in concentration for only one of the substances

• Next choose 2 rate laws where the unknown changes

• Since we know the exponent on the other, at this stage it doesn’t matter if it changes

-1 x y

-1 x y0.048 s [0.015 ] [0.015 ]0.192 s [0.030 ] [0.015 ]

M k M MM k M M

exp [NO] [O2] RNO2 M s-1

1 0.015 0.015 0.0482 0.030 0.015 0.1923 0.015 0.030 0.096

-1 x y

-1 x y0.096 s [0.015 ] [0.030 ]0.048 s [0.015 ] [0.015 ]

M k M MM k M M

y = 1x = 2 rate = k[NO]2[O2]

2NO(g) + O2(g) → 2NO2(g)


Determining the Value of k

At this stage, we can solve for k. Use any rate law and substitute the now known orders.

exp [NO] [O2] RNO2 M s-1

1 0.015 0.015 0.0482 0.030 0.015 0.1923 0.015 0.030 0.096

rate = k[NO]2[O2]

0.048 M s-1 = k[0.015 M]2[0.015 M]

1.4 × 104 M-2 s-1 = k


Determine the Rate Law from Given Data

[A] [B] [C] Rate M s-1

0.01 0.02 0.15 0.00020.02 0.02 0.15 0.0004

0.01 0.01 0.15 0.0001

0.01 0.02 0.30 0.0002

-1 a b c

-1 a b c0.0002 s [0.01] [0.02] [0.15]0.0004 s [0.02] [0.02] [0.15]

M kM k

-1 a b c

-1 a b c0.0002 s [0.01] [0.02] [0.15]0.0002 s [0.01] [0.02] [0.3]

M kM k

-1 a b c

-1 a b c0.0002 s [0.01] [0.02] [0.15]0.0001 s [0.01] [0.01] [0.15]

M kM k

c = 0a = 1 b = 1

rate = k[A][B]


Your TurnFor the following data, determine the order of NO2 in the reaction at 25 °C

2NO2(g) + F2(g) → 2NO2F(g):

Exp. [NO2] [F2] Rate NO2 disappearance (M s-1)

1 0.001 0.005 2 × 10-4

2 0.002 0.005 4 × 10-4

3 0.006 0.002 4.8 × 10-4

A. 0 B. 1 C. 2 D. 3 E. Not enough information given


Your TurnChlorine dioxide, ClO2, is a reddish-yellow gas that is soluble in water. In basic solution it gives ClO3

- and ClO2- ions.

2ClO2(aq) + 2OH-(aq) → 6ClO3- (aq) + ClO2

- (aq) + H2O(l)

The rate law is Rate = k[ClO2]2[OH-], what is the value of the rate constant given that when [ClO2] = 0.060 M, [OH-] = 0.030 M, the reaction rate is 0.0248 M s-1.

A. 0.02 M-1 s-1

B. 0.02 M s-1

C. 0.02 s-1

D. None of these 2.3 × 102 M-2 s-1


Rate Law Overview

• Rate laws relate the rate of the reaction to the reactant concentrations used for a given set of conditions

• Because the value of k varies with temperature, so will the rate law

13.4 Integrated rate laws give concentration as a function of time 25

Integrated Rate Laws

• Integrated rate laws tell us how the reactant concentration varies with time during the course of a reaction.

• Derived from one-reactant systems and a plot of the relationships of [reactant] and time zero order: linear plot [reactant ] vs. time first order: linear plot ln [reactant] vs. time second order: linear plot 1/[reactant] vs. time


2N2O5(g ) → 4NO2(g ) + O2(g )Time (s) [N2O5] [NO2] [O2]

0 0.0200 0.0000 0.0000100 0.0169 0.0063 0.0016200 0.0142 0.0115 0.0029300 0.0120 0.0160 0.0040400 0.0101 0.0197 0.0049500 0.0086 0.0229 0.0057600 0.0072 0.0256 0.0064700 0.0061 0.0278 0.0070

Learning Check Determine the order of the reactant graphically

0 order plot

1st order plot

2nd order plot


Zero-Order Reactions

• Rate=k• Plot of [reactant ] vs. time will be linear• The equation of the line will be [A]=[A0]-kt

A= amount remaining after elapsed time, t. Ao=original amount

• Diffusion controlled - usually are fast reactions in viscous media

• Rate is independent of concentrations of reactants, but the reaction still requires reactants


Learning Check

The rate law for the reaction of A→B is zero order in A and has a rate constant of 0.02 M/s. If the reaction starts with 1.50 M A, how much is present 15 seconds after the reaction begins?

•[A]=[A0]-kt

•[A]=1.2M


Learning Check

The rate law for the reaction of A→2B is zero order in A and has a rate constant of 0.12 M/s. If the reaction starts with 1.50 M A, after what time will the concentration of A be 0.90M?

•[A]=[A0]-kt

•t=5 s


Your Turn!

Which of the following is the correct set of units for the rate constant for a zero order reaction?

A. M/sB. M-1/sC. M-2/sD. Can’t tell from the given data


Your Turn!

What was the starting quantity if 2.5 M remain after 5 hr. k=2.3×10-5 M/s

A. Ao=2.9 M

B. Ao=2.5 M

C. Ao=2.1 M

D. None of these


First Order Reactions

• Rate = k[A]1

• Typically these reactions are decomposition type, or radioactive decay

• If the rate law is specified as d[A]/dt = k[A] or Integrating the equation gives us:

0ln t

Akt

A


Learning Check

The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s-1?

100ln (21 day)33

k

0ln t

Akt

A

k = 6.11×10-7 s-1

k = 0.0528 day-1


Half-Life of First Order Reaction

• In monitoring decay processes, a half-life (t1/2) is often recorded, in lieu of k

• t1/2 is the time needed for exactly half of the substance to decay. At this time,

[A]t = [A]0/2


How Does t1/2 Relate to k for a First Order Reaction?

0ln t

Akt

AAt t1/2, [A]0/2 = [A]

012

0

ln12

A

ktA

1/2ln 2 kt 1/ 2ln 2

tk


Learning Check

0ln t

Akt

A

12

ln 2t

k

The half-life of I-132 is 2.295 h. What percentage remains after 24 hours?

0.302 h-1 = k

A = 0.0711%


Your Turn!

What is the half-life of a new element, Barclium-146, if, after 2.2 h, 1.3% remains?

A. 2.0 hB. 0.35 hC. 0.51 hD. None of these


Second Order Reaction

• Are of several types: Rate = k[A]2

Rate = k[A]1[B]1

Rate = k[A]2[B]0

etc…• The integrated equation is of the form

0

1 1[ ] [ ]

t

ktA A


Learning Check

The rate constant for the second order reaction 2A → B is 5.3×10-5 M-1s-1. What is the original amount present if, after 2 hours, there is 0.35 M available?

0

1 1[ ] [ ]

t

ktA A

5

0

1 1 5.3 10 7200 s0.35 s

A M

[A]0 = 0.40 M


Second Order Half-Life

1/20 0

1 10.5 A

ktA

• Depends on the amount present at the start of the time period

• What is the relationship between k and t1/2 for this reaction type?

1/20

1t

k A


Learning Check

1/20

1t

k AThe rate constant for a second order reaction is 4.5 × 10-4 M-1 s-1. What is the half-life if we start with a reactant concentration of 5.0 M?

1/2-4 -1 -1

1

4.5 10 s 5

t

M M

t1/2 = 440 s

= 7.4 min


Your Turn!

Which order has a half-life that is independent of the original amount?

A. ZeroB. FirstC. SecondD. None depend on the original quantity


Your Turn!

A 0.10 M solution of moxium, a new antidepressant is bottled. The drug decays to fortium, a toxic chemical as a second order process. The rate constant is 2.3 × 10-3 M-1 h-1. What quantity of moxium is present after 90. days?

A. 0.098 MB. 5.5 × 10-5 MC. 0.067 MD. None of the above

13.5 Reaction rate theories explain experimental rate laws in terms of molecular collisions

44

Collision Theory of ReactionsFor a reaction to occur, three conditions must be met:1. Reactant particles must collide 2. Collision energy must be enough to break bonds/initiate3. Particles must be oriented so that the new bonds can form


45

Potential Energy Diagrams

• Demonstrate the energy needs and products as a reaction proceeds

• Tell us whether a reaction is exothermic or endothermic

• Tell us if a reaction occurs in one step or several steps

• Show us which step is the slowest


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Potential Energy Diagrams


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Features of Potential Energy Diagrams

Activation Energies

Pote

ntia

l Ene

rgy

Reaction Coordinate(progress of reaction)

Activated Complexes

Product Energy

Enthalpy of Reaction

Reactant Energy


48

Your Turn!Examine the potential energy diagram. Which is the slowest (rate determining) step?A. Step 1B. Step 2C. Can’t tell from the given information

Has greatest Ea

Pote

ntia

l Ene

rgy

Reaction Coordinate (progress of reaction)

Step 1

Step 2

13.6 Activation energies are measured by fitting experimental data to the Arrhenius equation

49

Temperature Effects

Changes in temperature affect the rate constant, k, according to the Arrhenius equation:

p is the steric factor Z is the frequency of collisions. Ea is the activation energy R is the Ideal Gas Constant (8.314 J/mol K) T is the temperature (K) A is the frequency factor

a- / E RTk pZe a- / E RTk Ae


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Working with the Arrhenius Equation

Linear Form: To determine the Ea and A value

a 1ln ln

Ek A

R T

Ratio form: Can be used when A isn’t known.

a2

1 2 1

1 1ln

Ekk R T T


51

Learning CheckGiven that k at 25 °C is 4.61 × 10-1 M s-1 and that at 50 °C it is 4.64 × 10-1 M s-1, what is the activation energy for the reaction?

1 -1

1 -1 1 14.64 10 s 1 1ln

323 K 298 K4.61 10 s 8.314 J mol K

aEM M

208 J mol-1 = Ea

a2

1 2 1

1 1ln

Ekk R T T


52

Working with the Arrhenius Equation

Given the following data, predict k at 75 °C using the graphical approach

k (M s-1) T°C

0.000886 250.000894 500.000918 1500.000908 100graph

ln k = -0.0278/T - 0.1917

k = 8.25 × 10-1

a 1ln ln E

k AR T


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Your Turn!In the reaction 2N2O5(g) 4NO2(g) + O2(g) the following temperature and rate constant information is obtained. What is the activation energy of the reaction?

A. 108 kJB. -108 kJC. 1004 kJD. -1004 kJE. None of these

T (K) k (s-1)

338328318

4.87 × 10-3

1.50 × 10-3

4.98 × 10-4

13.7 Experimental rate laws can be used to support or reject proposed mechanisms for a reaction

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Reaction Mechanisms

• Tell what happens on the molecular level, and in what order

• Tell us which steps in a reaction are fast and slow

• The rate determining step is the slowest step of the reaction that accounts for most of the reaction time

• Elementary steps sum to the overall reaction


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Elementary Steps

Elementary Reactions Molecularity Rate Law

A → Products Unimolecular Rate = k[A]A + A → ProductsA + B → Products

Bimolecular Rate = k[A]2

Rate = k[A][B]A + A + B → Products3A → ProductsA + B + C → Products

Termolecular Rate = k[A]2[B]Rate = k[A]3

Rate = k[A][B][C]


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Intermediates and Catalysts

• Because catalysts interact with the reactant, they will appear in the mechanism

• Catalysts are consumed in an early step and are regenerated in a subsequent step

• Intermediates are temporary products• Intermediates are formed in an early step and

consumed in a later step


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Learning Check

The reaction mechanism that has been proposed for the decomposition of H2O2 is

1. H2O2 + I- → H2O + IO- (slow)

2. H2O2 + IO- → H2O + O2 + I- (fast)

• Which is the rate determining step?• Are there any intermediates?

step 1

IO-


58

Rate Laws and Mechanisms

• The majority of the reaction time is taken by the rate determining step

• Substances that react in this step have the greatest effect on the reaction rate

• The observed rate law usually matches the rate law based on the rate determining step where the order of each reactant is its stoichiometric coefficient


59

Learning Check

The reaction mechanism that has been proposed for the decomposition of H2O2 is

1. H2O2 + I- → H2O + IO- (slow)

2. H2O2 + IO- → H2O + O2 + I- (fast)

What is the expected rate law?

rate = k[H2O2][I-]


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Learning Check

The reaction: A + 3B → D + F was studied and the following mechanism was finally determined

1. A + B → C (fast)2. C + B → D + E (slow)3. E + B → F (very fast)

What is the expected rate law?

rate = k[C][B]

13.8 Catalysts change reaction rates by providing alternative paths between reactants and products

61

Catalysts• Speed a reaction, but are not

consumed by the reaction• May appear in the rate law• Lower the Ea for the reaction.• May be heterogeneous or

homogeneous


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Catalytic Actions

• May serve to weaken bonds through induction• May serve to change polarity through

amphipathic/surfactant effects• May reduce geometric orientation effects• Heterogeneous catalyst - reactant and product

exist in different states.• Homogeneous catalyst - reactants and catalyst

exist in the same physical state


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Heterogeneous Catalysts

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Ch13. Kinetics