22 Nov. 2010

Objective: SWBAT write rate expressions and calculate reaction rates for chemical reactions.

Do now: Describe one very slow reaction that you’ve seen, and one very fast reaction.

1

Agenda

I. Do nowII. Kinetics notesIII. Reaction Rates DemonstrationsIV. Rate constant and reaction rates

problems.Homework: p. 602 #2, 3, 5, 7, 12, 13,

15, 17, 19

3

How can we predict whether or not a reaction will take place? Thermodynamics

Once started, how fast does the reaction proceed? Chemical kinetics: this unit!

How far will the reaction go before it stops? Equilibrium: next unit

Chemical Kinetics

The area of chemistry concerned with the speeds, or rates, at which a chemical reaction occurs.

reaction rate: the change in the concentration of a reactant or product with time (M/s) Why do reactions have such very

different rates? Steps in vision: 10-12 to 10-6 seconds! Graphite to diamonds: millions of years! In chemical industry, often more

important to maximize the speed of a reaction, not necessarily yield.

Chemical Kinetics

Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

A B

rate = -[A]t

rate = [B]t

[A] = change in concentration of A over time period t

[B] = change in concentration of B over time period t

Because [A] decreases with time, [A] is negative.

6

A B

rate = -[A]t

rate = [B]t

7

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

time

393 nmlight

Detector

[Br2] Absorption

red-brown

t1< t2 < t3

8

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -[Br2]t

= -[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent slope of

tangent

instantaneous rate = rate for specific instance in time

rate [Br2]

rate = k [Br2]

k = rate[Br2]

= rate constant

= 3.50 x 10-3 s-1

2H2O2 (aq) 2H2O (l) + O2 (g)

PV = nRT

P = RT = [O2]RTnV

[O2] = PRT1

rate = [O2]t RT

1 Pt=

measure P over time

11

12

Reaction Rates and Stoichiometry

2A B

Two moles of A disappear for each mole of B that is formed.

rate = [B]t

rate = -[A]t

12

aA + bB cC + dD

rate = -[A]t

1a

= -[B]t

1b

=[C]t

1c

=[D]t

1d

Example

Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

14

Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

rate = -[CH4]

t= -

[O2]t

12

=[H2O]

t12

=[CO2]

t

Practice Problems

Write the rate expressions for the following reactions in terms of the disappearance of the reactants and appearance of products.a. I-(aq) + OCl-(aq) Cl-(aq) + OI-(aq)b. 4NH3(g) + 5O2(g) 4NO(g) +

6H2O(g)

Using Rate Expressions

Consider the reaction: 4NO2(g) + O2(g) 2N2O5(g)

Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s.

a. At what rate is N2O5 being formed?

b. At what rate is NO2 reacting?

Consider the reaction:4PH3(g) P4(g) + 6H2(g)

Suppose that, at a particular moment during the reaction, molecular hydrogen is being formed at the rate of 0.078 M/s.

a. At what rate is P4 being formed?

b. At what rate is PH3 reacting?

The Rate Law

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB cC + dD

Rate = k [A]x[B]y

Reaction is xth order in A

Reaction is yth order in B

Reaction is (x +y)th order overall

x and y are determined experimentally!

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

rate = k [F2]x[ClO2]y

Double [F2] with [ClO2] constant Rate doubles x = 1 Quadruple [ClO2] with [F2] constant Rate quadruples y = 1

20

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

Rate Laws

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

1

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Experiment [S2O82-] [I-]

Initial Rate (M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

22

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Experiment [S2O82-] [I-]

Initial Rate (M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

rate = k [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

y = 1

Double [S2O82-], rate doubles (experiment 2 & 3)

x = 1

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

Practice Problems The reaction of nitric oxide with

hydrogen at 1280oC:2NO(g) + 2H2(g) N2(g) + 2H2O(g)

From the following data collected at this temperature, determine (a) the rate law, (b) the rate constant and (c) the rate of the reaction when [NO] = 12.0x10-3 M and [H2] = 6.0x10-3 M

23

Experiment [NO] M [H2] M Initial Rate (M/s)

1 5.0x10-3 2.0x10-3 1.3x10-5

2 10.0x10-3 2.0x10-3 5.0x10-5

3 10.0x10-3 4.0x10-3 10.0x10-5

23 Nov. 2010

Objective: SWBAT write rate expressions and calculate reaction rates for chemical reactions.

Do now: Calculate the rate constant (k) from the data below:

24

Agenda

I. Do nowII. Practice ProblemsHomework: Revisit last night’s

assignment #19 hint: Graph lnP vs. time. If

linear, it is 1st order. slope = -k Excel works great.

25

Write the reaction rate expressions for the following in terms of the disappearance of the reactants and the appearance of products:a) 2H2(g) + O2(g) 2H2O(g)

b) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Consider the reactionN2(g) + 3H2(g) 2NH3(g)

Suppose that at a particular moment during the reaction molecular hydrogen is reacting at a rate of 0.074 M/s.

a) At what rate is ammonia being formed?b) At what rate is molecular nitrogen

reacting?

Calculate the rate of the reaction at the time when [F2] = 0.010 M and [ClO2] = 0.020 M.

F2(g) + 2ClO2(g) 2FClO2(g)

[F2] (M) [ClO2] (M) Initial Rate (M/s)

0.10 0.010 1.2x10-3

0.10 0.040 4.8x10-3

0.20 0.010 2.4x10-3

Consider the reaction X + Y ZFrom the following data, obtained at 360

K, a) determine the order of the reactionb) determine the initial rate of

disappearance of X when the concentration of X is 0.30 M and that of Y is 0.40 M

29

Initial Rate of Disappearance of X (M/s)

[X] (M) [Y] (M)

0.053 0.10 0.50

0.127 0.20 0.30

1.02 0.40 0.60

0.254 0.20 0.60

0.509 0.40 0.30

Consider the reaction A B.The rate of the reaction is 1.6x10-2

M/s when the concentration of A is 0.35 M. Calculate the rate constant if the reaction isa. first order in Ab. second order in A

30

24 Nov. 2010

Objective: SWBAT determine reaction order graphically and relate concentration of 1st order reactions to time.

Do now: What is the overall order of reaction for the data shown below?

What is the rate constant?Run # Initial [A] ([A]0) Initial [B] ([B]0) Initial Rate (v0)

1 1.00 M 1.00 M 1.25 x 10-2 M/s

2 1.00 M 2.00 M 2.5 x 10-2 M/s

3 2.00 M 2.00 M 2.5 x 10-2 M/s

Agenda

I. Do nowII. Homework solutionsIII. First order reactionsIV. Time calculations for first-order

reactions.Homework: p. 603 #19, 21, 22, 23,

25-29

The rate laws can be used to determine the concentrations of any reactants at any time during the course of a reaction.

29 Nov. 2010 Take Out Homework p. 603 #19, 21, 22,

23, 25-29 Objective: SWBAT compare 1st order, 2nd

order, and zero order reactions, and describe how temperature and activation energy effect the rate constant.

Do now: Calculate the half life of the reaction F2(g) + 2ClO2(g) 2FClO2(g), with rate data shown below:

34

[F2] (M) [ClO2] (M) Initial Rate (M/s)

0.10 0.010 1.2x10-3

0.10 0.040 4.8x10-3

0.20 0.010 2.4x10-3

Agenda

I. Homework solutionsII. Review 1st order reactionsIII. Second and zero order reactionsIV. Activation EnergyV. Problem SetQuiz Weds.p. 31, 32, 35, 37, 39, 42+ Problem set

35

First-Order Reactions

A product rate = -[A]t

rate = k [A]

k = rate[A]

= 1/s or s-1M/sM

=[A]t

= k [A]-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

[A] = [A]0e−kt ln[A] = ln[A]0 - kt

37

2N2O5 4NO2 (g) + O2 (g)

Graphical Determination of k

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?

39

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?

ln[A] = ln[A]0 - kt

kt = ln[A]0 – ln[A]

t =ln[A]0 – ln[A]

k= 66 s

[A]0 = 0.88 M

[A] = 0.14 M

ln[A]0

[A]

k=

ln0.88 M

0.14 M

2.8 x 10-2 s-1=

The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-4 s-1 at 500oC.

a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minutes?

b) How long, in minutes, will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M?

c) How long, in minutes, will it take to convert 74% of the starting material?

Practice Problem

The reaction 2A → B is first order in A with a rate constant of 2.8×10-2s-1 at 80oC. How long, in seconds, will it take for A to decrease from 0.88 M to 0.14 M?

The rate of decomposition of azomethane (C2H6N2) is studied by monitoring partial pressure of the reactant as a function of time:

CH3-N=N-CH3(g) → N2(g) + C2H6(g)

The data obtained at 300oC are shown here:

Are these values consistent with first-order kinetics? If so, determine the rate constant.

Time (s) Partial Pressure of Azomethane (mmHg)

0 284

100 220

150 193

200 170

250 150

300 132

The following gas-phase reaction was studied at 290oC by observing the change in pressure as a function of time in a constant-volume vessel: ClCO2CCl3(g) 2COCl2(g) Determine the order of the reaction

and the rate constant based on the following data, where P is the total pressure

43

Time (s) P (mmHg)

0 400

2,000 316

4,000 248

6,000 196

8,000 155

10,000 122

Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows:

C2H5I(g) → C2H4(g) + HI(g)

From the following data, determine the order of the reaction and the rate constant:

Time (min) [C2H5I] (M)

0 0.36

15 0.30

30 0.25

48 0.19

75 0.13

First-Order Reactions

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

ln[A]0

[A]0/2

k=t½

ln 2k

=0.693

k=

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

How do you know decomposition is first order?

46

First-Order Reactions

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

ln[A]0

[A]0/2

k=t½

ln 2k

=0.693

k=

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln 2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order?

units of k (s-1)

47

A product

First-order reaction

# of half-lives [A] = [A]0/n

1

2

3

4

2

4

8

16

The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36x10-4 s-1 at 700oC:

C2H6(g) 2CH3(g)

Calculate the half-life of the reaction in minutes.

48

Calculate the half-life of the decomposition of N2O5:

2N2O5 4NO2(g) + O2(g)

49

t (s) [N2O5] (M) ln [N2O5]

0 0.91 -0.094

300 0.75 -0.29

600 0.64 -0.45

1200 0.44 -0.82

3000 0.16 -1.83

50

Second-Order Reactions

A product rate = -[A]t

rate = k [A]2

k = rate[A]2

= 1/M•sM/sM2=

[A]t

= k [A]2-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

1[A]

=1

[A]0

+ kt

t½ = t when [A] = [A]0/2

t½ =1

k[A]0

Iodine atoms combine to form molecular iodine in the gas phase:

I(g) + I(g) I2(g)

This reaction follows second-order kinetics and has the high rate constant 7.0x109/M·s at 23oC.

a. If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 minutes.

b. Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M.

The reaction 2A → B is second order with a rate constant of 51/M·min at 24oC.

a. Starting with [A]o = 0.0092 M, how long will it take for [A]t = 3.7x10-3 M?

b. Calculate the half-life of the reaction.

53

Zero-Order Reactions

A product rate = -[A]t

rate = k [A]0 = k

k = rate[A]0

= M/s[A]t

= k-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t = 0

t½ = t when [A] = [A]0/2

t½ =[A]0

2k

[A] = [A]0 - kt

54

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A] = ln[A]0 - kt

1[A]

=1

[A]0

+ kt

[A] = [A]0 - kt

t½ln 2k

=

t½ =[A]0

2k

t½ =1

k[A]0

Activation Energy and Temperature Dependence of Rate Constants Reaction rates increase with

increasing temperature Ex: Hard boiling an egg Ex: Storing food

How do reactions get started in the first place?

55

Collision Theory

Chemical reactions occur as a result of collisions between reacting molecules.

reaction rate depends on concentration But, the relationship is more

complicated than you might expect! Not all collisions result in reaction (p.

582-583)

56

In order to react, colliding molecules must have a total KE ≥ activation energy (Ea)

Ea: minimum amount of energy required to initiate a chemical reaction

activated complex (transition state): a temporary species formed by the reactant molecules as a result of the collision before they form the product.

57

Exothermic Reaction Endothermic Reaction

The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.

=a barrier that prevents less energetic molecules from reacting

A + B AB C + D++

59

Temperature Dependence of the Rate Constant

Ea is the activation energy (J/mol)

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature

A is the frequency factor (constant across wide range of temps.)

ln k = -Ea

R1T

+ lnA

(Arrhenius equation)

)/( RTEaeAk

Alternate format (y=mx+b):

60

Alternate Form of the Arrhenius Equation

At two temperatures, T1 and T2

or

The rate constants for the decomposition of acetaldehyde:

CH3CHO(g) → CH4(g) + CO(g)

were measured at five different temperatures. The data are shown below. Plot lnk versus 1/T, and determine the activation energy (in kJ/mol) for the reaction. (Note: the reaction is order in CH3CHO, so k has the units of )

2

3

sM 2

1

/1

k T (K)

0.011 700

0.035 730

0.105 760

0.343 790

0.789 810

)/1( 2

1

sM

slope = -2.19x104 K

slope = REa

The second order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures. Determine graphically the activation energy for the reaction.

63

k T (oC)

1.87x10-3 600

0.0113 650

0.0569 700

0.244 750

)/1( sM

30 Nov. 2010 Take Out Homework p. 605# 31,

32, 35, 37, 39 Objective: SWBAT use the Arrhenius

equation to solve for rate constants and temperatures, and solve practice problems on kinetics.

Do now: Match

64

Order Rate Law Conc-Time Eq. Half Life Eq.

2 rate = k[A] [A]=[A]0-kt t1/2=1/k[A]o

1 rate = k[A]2 1/[A]=1/[A]0 + kt t1/2=ln2/k

0 rate = k ln[A]=ln[A]0 –kt t1/2=[A]0/2k

Agenda

I. Homework solutionsII. Using the Arrhenius equation part 2III. Molecular orientationIV. Problem Set work timeHomework: Complete problem set and

p. 605 #40, 42Quiz tomorrow

65

The rate constant of a first order reaction is 3.46x10-2 /s at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol?

66

The first order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32x10-10/s at 25oC. Calculate the rate constant at 40oC if the activation energy is 116 kJ/mol.

67

Homework

Quiz Weds.p. 604 #40, 42+ Problem set #2, 18, 20, 25, 29, 32,

37, 42, 55, 57: Weds.

68

Problem Set

Handout #2, 18, 20, 25, 29, 32, 37, 42, 55, 57: Weds.

Quiz Weds.

69

2 Dec. 2010

Objective: SWBAT describe a reaction in terms of elementary steps and determine the rate law.

Do now: Which graph represents an exothermic reaction? Which is endothermic? Why?

70

Agenda

I. Do nowII. Elementary steps in reactionsIII. CatalystsHomework: p. 605 #45, 47, 48, 52,

54, 55, 56, 58, 62: Mon.

71

For simple reactions (between atoms, for example) the frequency factor (A) is proportional to the frequency of collision between the reacting species.

“Orientation factor” is also important.

72

73

Importance of Molecular Orientation

effective collision

ineffective collision

A balanced chemical equation doesn’t tell us much about how the reaction actually takes place.

It may represent the sum of elementary steps

Reaction mechanism: the sequence of elementary steps that leads to product formation.

74

75

Reaction Mechanisms

The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the reaction mechanism.

2NO (g) + O2 (g) 2NO2 (g)

N2O2 is detected during the reaction!

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

76

2NO (g) + O2 (g) 2NO2 (g)

Mechanism:

77

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.

An intermediate is always formed in an early elementary step and consumed in a later elementary step.

The molecularity of a reaction is the number of molecules reacting in an elementary step.

• Unimolecular reaction – elementary step with 1 molecule

• Bimolecular reaction – elementary step with 2 molecules

• Termolecular reaction – elementary step with 3 molecules

78

Unimolecular reaction A products rate = k [A]

Bimolecular reaction A + B products rate = k [A][B]

Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

Writing plausible reaction mechanisms:

• The sum of the elementary steps must give the overall balanced equation for the reaction.

• The rate-determining step should predict the same rate law that is determined experimentally.

The rate-determining step is the slowest step in the sequence of steps leading to product formation.

79

Sequence of Steps in Studying a Reaction Mechanism

The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:

Step 1: NO2 + NO2 NO + NO3

Step 2: NO3 + CO NO2 + CO2

What is the equation for the overall reaction?

What is the intermediate?

What can you say about the relative rates of steps 1 and 2?

81

The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:

Step 1: NO2 + NO2 NO + NO3

Step 2: NO3 + CO NO2 + CO2

What is the equation for the overall reaction?

NO2+ CO NO + CO2

What is the intermediate?

NO3

What can you say about the relative rates of steps 1 and 2?

rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2

rate determining step: the slowest step in the sequence of steps leading to product formation.

82

Decomposition of Hydrogen Peroxide

2H2O2(aq) 2H2O(l) + O2(g)

Can be catalyzed using iodide ions (I-)rate = k[H2O2][I-] Why?!

Determined experimentally.Step 1: H2O2 + I- H2O + IO-

Step 2: H2O2 + IO- H2O + O2 + I-

83

7 Dec. 2010

Objective: SWBAT describe catalysis and how ammonia is synthesized through the Haber process.

Do now: The rate for the reaction 2H2O2(aq) 2H2O(l) + O2(g) is

rate = k[H2O2][I-]

Step 1: H2O2 + I- H2O + IO-

Step 2: H2O2 + IO- H2O + O2 + I-

Which is the rate determining step? Why?

84

Agenda

I. Do nowII. Homework solutionsIII. Catalysts and catalyzing

mechanismsIV. AP Kinetics problem setHomework: p. 605 #55, 56, 58Bring your book on Thursday!

85

Example The gas-phase decomposition of nitrous

oxide (N2O) is believed to occur via two elementary steps:

Step 1: N2O N2 + O

Step 2 N2O + O N2 + O2

Experimentally the rate law is found to be rate = k[N2O].

a) Write the equation for the overall reaction.b) Identify the intermediates. c) What can you say about the relative rates

of steps 1 and 2?

86

For the decomposition for H2O2, the reaction rate depends on the concentration of I- ions, even though I- doesn’t appear in the overall equation.

I- is a catalyst for the reaction.

87

88

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.

Ea k

ratecatalyzed > rateuncatalyzed

Ea < Ea′

UncatalyzedCatalyzed

)/( RTEaeAk

Catalysts

forms an alternative reaction pathway for example, it might form an

intermediate with the reactant. Ex: 2KClO3(s) 2KCl(s) + 3O2(g)

Very slow, until you add MnO2, a catalyst. The MnO2 can be recovered at the end of the reaction!

89

90

In heterogeneous catalysis, the reactants and the catalysts are in different phases (usually, catalyst is a solid, reactants are gases or liquids).

In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.

• Haber synthesis of ammonia

• Ostwald process for the production of nitric acid

• Catalytic converters

• Acid catalysis

• Base catalysis

91

N2 (g) + 3H2 (g) 2NH3 (g)Fe/Al2O3/K2O

catalyst

Haber ProcessSynthesis of Ammonia

Extremely slow at room temperature. Must be fast and high yield!Process occurs on the surface of the Fe/Al2O3/K2O catalyst, which weakens the covalent N-N and H-H bonds.

92

Ostwald Process

Pt-Rh catalysts usedin Ostwald process

4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)Pt catalyst

2NO (g) + O2 (g) 2NO2 (g)

2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)

93

Catalytic Converters

CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic

converter

2NO + 2NO2 2N2 + 3O2

catalyticconverter

94

Enzyme Catalysisbiological catalysts

95

Binding of Glucose to Hexokinase

p. 605 #45, 47, 48, 52, 54, 55, 56, 58, 62

96