1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005...

1

General Chemistry 4th edition, Hill, Petrucci, McCreary, PerryPrentice Hall © 2005 Chapter Thirteen

Today…• Turn in:

– Nothing

• Our Plan:– Test Results– Videos/Notes– Investigation 10 Pre-Lab

• Homework (Write in Planner):– Be prepared for Investigation 10 next class and

have report started.

2


Introduction to Chemical Kinetics

• What is Kinetics? I’ll let Hank explain…

http://www.youtube.com/watch?v=7qOFtL3VEBc• Stop at 3:20

http://www.youtube.com/watch?v=7qOFtL3VEBc


3


Kinetics in Action…

• Clock Reactions…• http://

www.youtube.com/watch?v=BqeWpywDuiY

http://www.youtube.com/watch?v=BqeWpywDuiY




4


Unit Preview…Some Reactions

• Are fast:– Acid/base

neutralization– Sodium + water – PPT reactions

• Are slow: – Aluminum oxidation– Iron oxidizing– Plastic bottle

decomposing

5


And some depend…

• On temperature: fireflies• On conditions: Iron oxide (humid or

desert) • On enzymes: many processes in our body• On a catalyst:

2CO + 2NO --> 2CO2 + N2 (automobile

pollution and catalytic converters)

6


• Chemical kinetics is the study of:– the rates of chemical reactions– factors that affect these rates– the mechanisms by which reactions occur

• Reaction rates vary greatly – some are very fast (burning, precipitation) and some are very slow (rusting, disintegration of a plastic bottle in sunlight).

Chemical Kinetics: A Preview

7


Variables in Reaction Rates

• Concentrations of reactants: Reaction rates generally increase as the concentrations of the reactants are increased.

• Temperature: Reaction rates generally increase rapidly as the temperature is increased.

• Surface area: For reactions that occur on a surface rather than in solution, the rate increases as the surface area is increased.

• Catalysts: Catalysts speed up reactions and inhibitors slow them down.

8


Think of it like this…

• http://ed.ted.com/lessons/how-to-speed-up-chemical-reactions-and-get-a-date

http://ed.ted.com/lessons/how-to-speed-up-chemical-reactions-and-get-a-date

http://ed.ted.com/lessons/how-to-speed-up-chemical-reactions-and-get-a-date

9


Theories of Chemical Kinetics: Collision Theory

• Before atoms, molecules, or ions can react, they must first collide.

• An effective collision between two molecules puts enough energy into key bonds to break them.

• The activation energy (Ea) is the minimum energy that must be supplied by collisions for a reaction to occur.

• A certain fraction of all molecules in a sample will have the necessary activation energy to react; that fraction increases with increasing temperature.

• The spatial orientations of the colliding species may also determine whether a collision is effective.

10


Distribution of Kinetic Energies

At higher temperature (red), more molecules

have the necessary activation energy.

11


Key Point Orientation of

molecules at the time of their collision will determine whether they react or not!

12


Analogy - Car Crash Example

Energy and orientation of cars during a car

crash can establish the change that occurs to the

cars.

13


Importance of Orientation

One hydrogen atom can approach another from any direction …

… and reaction will still occur; the spherical symmetry of the atoms means

that orientation does not matter.

Effective collision; the I atom can bond to the C atom to form CH3I

Ineffective collision; orientation is important

in this reaction.

14


Transition State Theory

• The configuration of the atoms of the colliding species at the time of the collision is called the transition state.

• The transitory species having this configuration is called the activated complex.

• A reaction profile shows potential energy plotted as a function of a parameter called the progress of the reaction.

• Reactant molecules must have enough energy to surmount the energy “hill” separating products from reactants.

15


Activated complex• Species formed as a result of collisions

between energetic molecules that is an intermediate between the reactants and the products of a reaction. Once formed the activated complex dissociates either into products or back to the reactants

16


Reaction Profile• A graphical representation of a

chemical reaction in terms of the energies of the reactants, activated complexes and products

17


Reaction Profile

18


Exothermic

19


Endothermic

20


CO(g) + NO2(g) CO2(g) + NO(g)

A Reaction Profile

21


An Analogy for Reaction Profiles and Activation Energy

22


Stop!• Investigation 10 - We’re going to test

the effect of different variables on the rate of reaction next class.

• Complete steps 1 – 6 on the lab handout and be prepared to conduct the lab next class.

23



–Nothing• Our Plan:

–Investigation 10• Homework (Write in Planner):

–Lab Report Due Friday

24



–Lab Report (rubric on top)• Our Plan:

–Notes & Practice• Homework (Write in Planner):

–Work on the Ch. 13 Homework

25


The Meaning of Rate

• The rate of a reaction is the change in concentration of a product per unit of time (rate of formation of product).

• Rate is also viewed as the negative of the change in concentration of a reactant per unit of time (rate of disappearance of reactant).

• The rate of reaction often has the units of moles per liter per unit time (mol∙L–1∙s–1 or M∙s–1)

26


If the rate of consumption of H2O2 is 4.6 M/h, then …

… the rate of formation of H2O must also be 4.6 M/h, and …

… the rate of formation of O2 is 2.3 M/h

27


2 H2O2 --> 2 H2O + O2

… 0.1850 mol H2O2 reacted in 60 s.

Rate =0.1850 mol H2O2/L

60 s

= 0.00131 M H2O2 s–1

1 L

2.960 g O2 (0.09250 mole) produced in 60 s means …

28


Example 13.1

Consider the hypothetical reaction

A + 2B --> 3C + 2D

Suppose that at one point in the reaction, [A] = 0.4658 M and 125 s later [A] = 0.4282 M. During this time period, what is the average (a) rate of reaction expressed in M∙s–1 and (b) rate of formation of C, expressed in M∙min–1.

29


Try It Out!

EX 13.1 A: Consider the hypothetical reaction

2A + B → 2C + D

Suppose that at some point during the reaction [D] = 0.2885 M and that 2.55 min (that is 2 min 33 sec) later [D] = 0.3546 M.

a) What is the average rate of reaction during this time period, expressed in M min-1?

b) What is the rate of formation of C expressed in M s-1?

30


Rate of Reaction Expressed as the negative of the Slope of a Tangent Line

• Average rate (green dotted line)

• Initial Rate ( blue solid line)• Instantaneous Rate (red line)• Question: Over what time

interval are the instantaneous rates greater than the average rate measured for the 600 sec period?

31


Average vs. Instantaneous Rate

Instantaneous rate is the slope of the tangent to the curve at a particular time.

We often are interested in the initial instantane-ous rate; for the initial concentrations of reactants and products are known at this time.

32


Example 13.2

Use data from Table 13.1 and/or Figure 13.5 to

(a) determine the initial rate of reaction and

(b) calculate [H2O2] at t = 30 s.

33


Try It Out

EX 13.2 A: From Figure 13.5,

a) Determine the instantaneous rate of reaction at t = 300 s.

b) Use the result of a) to calculate a value of [H2O2] at t = 310 s.

34


The Rate Law of a Chemical Reaction

• The rate law for a chemical reaction relates the rate of reaction to the concentrations of reactants.

• The exponents (m, n, p…) are determined by experiment.• Exponents are not derived from the coefficients in the

balanced chemical equation, though in some instances the exponents and the coefficients may be the same.

• The value of an exponent in a rate law is the order of the reaction with respect to the reactant in question.

• The proportionality constant, k, is the rate constant.

aA + bB + cC …→ products rate = k[A]n[B]m[C]p …

35


Rate Law Examples

Rate = k[A]1 = k[A] Reaction is first order in A

Rate = k[A]3 Reaction is third order in A

Rate = k[A]2 Reaction is second order in

A

If we triple the concentration of A in a second-order reaction, the rate increases by a

factor of ________.

36


Effect of Concentration on RateOrder of Rxn Concentration

ChangeEffect on Rate

0 Double/Triple… Nothing (20)

1 Double Double (21)

1 Triple Triple (31)

2 Double Quadruple (22)

2 Triple 8x (32)

2 Quadruple 16x (42)

3 Double 9x (23)

3 Triple 27x (33)

37


More About the Rate Constant k

• The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants.

• The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants.

• The rate and the rate constant have the same numerical values and units only in zero-order reactions.

• For reaction orders other than zero, the rate and rate constant are numerically equal only when the concentrations of all reactants are 1 M. Even then, their units are different.

38


To find the overall order of a reaction…

• Add the orders for each compound.• Example:

–rate = [A]2[B]1 is 3rd order overall–How about rate = [A]0[B]1[C]1?

39


Units of k (p. 536)

Overall Reaction Order

Units of k

Zero M s-1

First s-1

Second M-1 s-1

Third M-2 s-1

40


Method of Initial Rates

• The method of initial rates is a method of establishing the rate law for a reaction—finding the values of the exponents in the rate law, and the value of k.

• A series of experiments is performed in which the initial concentration of one reactant is varied. Concentrations of the other reactants are held constant.

• When we double the concentration of a reactant A, if:– there is no effect on the rate, the reaction is zero-order in A.– the rate doubles, the reaction is first-order in A.– the rate quadruples, the reaction is second-order in A.– the rate increases eight times, the reaction is third-order

in A.

41


The concentration of NO was held the

same in Experiments 1 and 2 …

… while the concentration of

Cl2 in Experiment 2 is twice that of Experiment 1.

The rate in Experiment 2 is twice that in Experiment 1, so the reaction must be first

order in Cl2.

Which two experiments are used to find the order of the

reaction in NO?

How do we find the value of k after obtaining the order of the

reaction in NO and in Cl2?

42


Example 13.3

For the reaction 2 NO(g) + Cl2(g) → 2 NOCl(g) described in the text and in Table 13.2, (a) what is the initial rate for a hypothetical Experiment 4, which has [NO] = 0.0500 M and [Cl2] = 0.0255 M? (b) What is the value of k for the reaction?

43


So, when looking at data…..

• Zero order reaction: initial rate is unaffected (20 )

• 1st order reaction: double the concentration, doubles the rate (21 =2)

• 2nd Order Reaction: Initial rate increases fourfold (22 =4)

• 3rd Order Reaction: Initial rate increases eightfold (23 =8)

44


Try It Out

• Below is some rate data for the hypothetical reaction, 2A + B --> C. What is the rate law for this reaction?

Experiment [A]0 [B]0 Rate (M/s)

1 2.0 M 1.0 M 0.100

2 2.0 M 2.0 M 0.400

3 4.0 M 1.0 M 0.100

45


Let’s Take a Break

• With your partner, Complete Part 1 of the Partner Review

46


•Sum of exponents is equal to zero

47


Rate of Reaction= k[A]0

•The concentration time graph is a straight line with a negative slope

Zero Order

48


Zero Order• Rate of the reaction:

–IS equal to k –remains constant throughout the

reaction –is the negative of the slope of the

line when graph Molarity vs. time (see pg. 544)

49


Rate is independent of initial concentration

A Zero-Order Reaction

rate = k[A]0

= k

50


Zero OrderIntegrated rate equation:

[A]t = -kt + [A]0

y = mx + b

51


Zero Order (cont.)

y = [A]t = conc of A at some time

x = t = time

b = [A]0

m = -k (m, the slope of the straight line)

52


First-Order Reactions

• In a first-order reaction, the exponent in the rate law is 1.• Rate = k[A]1 = k[A]

Look! It’s an equation for a straight line!

ln [A]t – ln [A]0 = –kt

ln [A]t = –kt + ln [A]0

• At times, it is convenient to replace molarities in an integrated rate law by quantities that are proportional to concentration.

• The integrated rate law describes the concentration of a reactant as a function of time. For a first-order process:

ln[A]t

[A]0

= –kt

53


Rate of reaction= k[A]1

Integrated rate equation

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

y = mx + b

54


1st Order

• Easy test for a first order reaction is to plot the natural log of reactant conc. vs. time and see if the graph is linear.

55


Decomposition of H2O2

56


Example 13.4

For the first-order decomposition of H2O2(aq), given k = 3.66 x 10–3 s–1 and [H2O2]0 = 0.882 M, determine (a) the time at which [H2O2] = 0.600 M and (b) [H2O2] after 225 s.

57


Another ExampleH202 initially at a conc of 2.32 M, is allowed to decompose. What will the [H202] be 1200 s later? Use k = 7.3 x 10-4 s-1 for this first order decomposition.

58


Try It Out

EX 13.4 A: The decomposition of nitramide, NH2NO2, is a first order reaction:

NH2NO2 (aq) → H2O (l) + N2O (g)

The rate law is rate = k[NH2NO2], with k = 5.62 x 10-3 min-1 at 15ᵒC. Starting with 0.105 M NH2NO2,

a) At what time will [NH2NO2] = 0.0250 M

b) What is [NH2NO2] after 6.00 hours?

59


Half-lifeTime required for ½ of the

reactant to be consumedEquation:

t1/2 = ln2 = 0.693

k k

60


Example 13.5

• Use data from Figure 13.7 (p. 542) to evaluate the

a) half-life and

b) Rate constant for the first order decomposition of N2O5 at 67 ᵒC:

N2O5 (g) → 2NO2 (g) + ½ O2 (g)

61


Example 13.5 A

Use the result of Example 13.5 to determine

a) The time required to reduce the quantity of N2O5 to 1/16 of its initial value and

b) The mass of N2O5 remaining after a 4.80 g sample of N2O5 has decomposed for 10.0 min.

62


Sum of exponents = 2

63


Second-Order Reactions

• A reaction that is second order in a reactant has a rate law in which the exponent for that reactant is 2.

Rate = k[A]2

• The integrated rate law has the form:

• The half-life of a second-order reaction depends on the initial concentration as well as on the rate constant k:

1 1–––– = kt + –––– [A]t [A]0

What do we plot vs. time to get a

straight line?

1t½ = ––––– k[A]0

64


Example 13.7The second-order decomposition of HI(g) at 700 K is represented in Figure 13.9.

HI(g) → ½ H2(g) + ½ I2(g)

Rate = k[HI]2

What are the: (a) rate constant and (b) half-life of the decomposition of 1.00 M HI(g) at 700 K?

65


Try It Out

EX 13.7 A: If, in the second order reaction A → products, it takes 55 s for the concentration of reactant A to fall to 0.40 M from an initial concentration of 0.80 M, what is the rate constant k for the reaction?

66


Summary of Kinetic Data

67


Short Cut!!• [A] vs time (straight line = Zero

order) • ln [A] vs. time ( straight line = first

order) • 1/[A] vs. time (straight line =

second order)

68


Try it in your note packet!

Time, min [A] ln[A] 1/[A]0 1.05 0.6310 0.4615 0.3625 0.25

69


Crash Course Review

• http://www.youtube.com/watch?v=7qOFtL3VEBc

• 3:10 – 5:30



70


Let’s Take a Break

• Complete Part 2 of the Partner Review.

71



–Get out a piece of notebook paper• Our Plan:

–Scavenger Hunt Review– Investigation 11 Pre-Lab

• Homework (Write in Planner):–Prepare Lab Report–Homework Problems (Due 2/12)

72


Pre Lab

• Complete steps 1 – 5 on the handout and have your formal lab report started.

• Be prepared to experiment next class.

73




–Investigation 11• Homework (Write in Planner):

–Lab Report due Next Class (2/10)

74



– Lab Report (rubric on top)

• Our Plan:– Review Activity– Elephant Toothpaste Demo– Notes – Catalysts & Rate Determining Steps– Finish Homework Problems

• Homework (Write in Planner):– Homework Problems

75


Effect of Temperature onthe Rates of Reactions

• In 1889, Svante Arrhenius proposed the following expression for the effect of temperature on the rate constant, k:

k = Ae–Ea/RT

• The constant A, called the frequency factor, is an expression of collision frequency and orientation; it represents the number of collisions per unit time that are capable of leading to reaction.

• The term e–Ea/RT represents the fraction of molecular collisions sufficiently energetic to produce a reaction.

76


I like this equation better…

• Look at Eq. 13.16 on p. 551 of the text!

77


Example 13.9

Estimate a value of k at 375 K for the decomposition of dinitrogen pentoxide illustrated in Figure 13.15, given that

k = 2.5 x 10–3 s–1 at 332 K.

78


Try it Out

EX 13.9 B: Di-tert-butyl peroxide (DTBP) is used as a catalyst in the manufacture of polymers. In the gaseous state, DTBP decomposes to acetone and ethane by a first-order reaction.

(C4H9)2O2 (g) → 2(CH3)2CO (g) + C2H6 (g)

The half-life of DTBP is 17.5 h at 125ᵒC and 1.67 h at 145ᵒC. What is the activation energy, Ea, of the decomposition reaction?

79


Reaction Mechanisms

• Analogy: a banana split is made by steps in sequence: slice banana; three scoops ice cream; chocolate sauce; strawberries; pineapple; whipped cream; end with cherry.

• A chemical reaction occurs according to a reaction mechanism—a series of collisions or dissociations—that lead from initial reactants to the final products.

• Like making a banana split, three molecules will not collide simultaneously very often, so steps of a reaction mechanism involve only one or two reactants at a time.

80


Reaction Mechanisms

• An elementary reaction represents, at the molecular level, a single step in the progress of the overall reaction.

• A proposed mechanism must:– account for the experimentally determined rate

law.– be consistent with the stoichiometry of the

overall or net reaction.

81


Molecularity

The molecularity of an elementary reaction refers to the number of free atoms, ions, or molecules that collide or dissociate in that step.

Termolecular processes are unusual, for the

same reason that three basketballs shot at the same time are unlikely to collide at the same

instant …

82


The Rate-Determining Step

• The rate-determining step is the crucial step in establishing the rate of the overall reaction. It is usually the slowest step.

• Some two-step mechanisms have a slow first step followed by a fast second step, while others have a fast reversible first step followed by a slow second step.

Slow

Fast

Mechanism for

2 NO + O2 --> 2 NO2

83


An Example

• Given the reaction:

2A + 2B → C + D rate = k[A]2[B]

Could take place by the following three-step mechanism:

I. A + A ↔ X (fast)

II. X + B → C + Y (slow)

III. Y + B → D (fast)

84


Intermediates

• X & Y are called intermediates because they appear in the mechanism, but they cancel out of the balanced equation.

• They are products from one reaction and then a reactant in the next.

85


An Example

• The steps of a reaction mechanism must add up to equal the balanced equation with all intermediates cancelling out. Let’s try it with our example.




86


Rate-determining Step

• As in any process where many steps are involved, the speed of the whole process can’t go faster than the speed of the slowest step in the process.

• The slowest step of a reaction is the rate-determining step.

• Because the slowest step is the most important step in determining the rate of a reaction, the slowest step and the steps leading up to it are used to see if the mechanism is consistent with the rate law for the overall reaction.

87


An Example

• Let’s look at our reaction again and show that it is consistent with the rate law:

2A + 2B → C + D rate = k[A]2[B]




88


An Example (p. 199 Cracking AP Chemistry Exam)

1. Write rate law for slow step.

2. X is an intermediate, we need to eliminate it from the rate law.

3. Solve for [X] in terms of [A].

4. Substitute for [X] in our second step.

5. Now we have the rate law.

89


Fast step: I2 2 I

Slow step: 2 I + H2 2 HI

k1

k–1

k2

Example 13.10 For the reaction H2(g) + I2(g) --> 2 HI(g), a proposed mechanism is below. What is the net equation for the overall reaction, and what is the order of the reaction according to this mechanism?

90


EX 13.10 A

The decomposition of nitrosyl choride,

2NOCl (g) → 2NO (g) + Cl2 (g)

Is a first-order reaction. Propose a mechanism for this reaction consisting of one fast step and one slow step.

91


Catalyst

• A catalyst provides an alternative reaction pathway of lower activation energy

• Participates in a chemical reaction w/o undergoing permanent change

• Speeds up without being consumed in the reaction. It is neither a reactant nor product in a reaction.

92


Catalysis – how does it work

• In general, a catalyst works by changing the mechanism of a chemical reaction.

• Often the catalyst is consumed in one step of the mechanism, but is regenerated in another step.

• The pathway of a catalyzed reaction has a lower activation energy than that of an uncatalyzed reaction, so more molecules at a fixed temperature have the necessary activation energy.

93


Effect of Catalyst on Reaction Profile and Activation Energy

A catalyst lowers the activation energy, making it

easier for the reactants to “climb the energy hill” and

form the products.

94


A Kinetics Pick Up Line…

95


Homogeneous Catalysis

Ozone decomposition catalyzed by chlorine atoms has a much lower activation energy and proceeds much more rapidly than the uncatalyzed reaction

96


Heterogeneous Catalysis

• Many reactions are catalyzed by the surfaces of appropriate solids.

• A good catalyst provides a higher frequency of effective collisions.

• Four steps in heterogeneous catalysis:– Reactant molecules are adsorbed.– Reactant molecules diffuse along the surface.– Reactant molecules react to form product molecules.– Product molecules are desorbed (released from the

surface).

97


Heterogeneous Catalysis

Hydrogen is adsorbed onto the surface of a

nickel catalyst. A C=C approaches …

… and is adsorbed.

Hydrogen atoms attach to the carbon

atoms, and the molecule is desorbed.

98


A Surface-Catalyzed Reaction

99


Enzyme Catalysis• Enzymes are high-molecular-mass proteins that usually

catalyze one specific reaction, or a set of similar reactions.• The reactant substance, called the substrate (S), attaches

itself to an area on the enzyme (E) called the active site, to form an enzyme-substrate complex (ES).

• The enzyme–substrate complex decomposes to form products (P), and the enzyme is regenerated.

100


Factors Influencing Enzyme Activity

The rates of enzyme-catalyzed reactions are influenced by factors such as concentration of the substrate, concentration of the enzyme, acidity of the medium, and temperature.

101


Mercury Poisoning: An Example of Enzyme Inhibition

When Hg reacts with an enzyme …

… the Hg binds to sulfur atoms …

… changing the shape of the active site, so

that it no longer “fits” the substrate.

102


Homogeneous Catalysis

Step 1: A + catalyst → intermediate + C

Step 2: B + intermediate → D + catalyst

Net equation: A + B → C + D

103


Identifying Catalysts and intermediate species

O3 + Cl∙ --> ClO∙ + O2

ClO· + O∙ --> Cl∙ + O2

Catalyst?

Intermediate?

Net Equation?

104


Crash Course Review

• http://www.youtube.com/watch?v=7qOFtL3VEBc

• 7:15 - end



105


Finish the HW All homework problems are due on Wednesday!

106


Today…• Before Class:

– Mark Homework Questions on the Board

• Our Plan:– Homework Questions/Check HW– Worksheet Race– Study Guide

• Homework (Write in Planner):– Test Next Class– Breakfast Club 6 am on Friday

107


Study Guide Changes

•7 – do 39 a only•11 – do 25 a and b only

108




–Study Guide Questions–Test

• Homework (Write in Planner):–Complete the POGIL (Day 1 ONLY)

1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005...

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Transcript of 1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005...