Chapter Thirteen

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Transcript Chapter Thirteen

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Investigation 10 Pre-Lab

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Be prepared for Investigation 10 next class and have report started.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry 1

Introduction to Chemical Kinetics

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Kinetics in Action…

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Unit Preview… Some Reactions

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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 --> 2CO 2 + N 2 (automobile pollution and catalytic converters) Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 5

Chemical Kinetics: A Preview

•

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).

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 6

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 7

Think of it like this…

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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 (E

)

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 9

Distribution of Kinetic Energies

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry

At higher temperature (red), more molecules have the necessary activation energy.

Chapter Thirteen

Key Point

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 11

Analogy - Car Crash Example

Energy and orientation of cars during a car crash can establish the change that occurs to the cars.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Importance of Orientation One hydrogen atom can approach another from any direction … Effective collision; the I atom can bond to the C atom to form CH 3 I

… and reaction will still occur; the spherical symmetry of the atoms means that orientation does not matter.

Ineffective collision; orientation is important in this reaction.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 14

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 15 Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Reaction Profile

• A graphical representation of a chemical reaction in terms of the energies of the reactants, activated complexes and products

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Reaction Profile

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Exothermic

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Endothermic

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

A Reaction Profile

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

An Analogy for Reaction Profiles and Activation Energy

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Today…

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Turn in:

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Our Plan:

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Investigation 10

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Homework (Write in Planner):

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Lab Report Due Friday

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 23

Today…

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Turn in:

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Lab Report (rubric on top)

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Our Plan:

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Notes & Practice

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Work on the Ch. 13 Homework

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 24

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).

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 25

If the rate of consumption of H 2 O 2 is 4.6 M/h, then … … the rate of formation of H 2 O must also be 4.6 M/h, and … … the rate of formation of O 2 is 2.3 M/h

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

2 H 2 O 2 --> 2 H 2 O + O 2

1 L

2.960 g O 2 (0.09250 mole) produced in 60 s means …

… 0.1850 mol H 2 O 2 reacted in 60 s.

0.00131 M H 2 O 2 s –1

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 27

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 .

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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 ?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 29

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

30 • 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? Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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 [H 2 O 2 ] at

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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 [H 2 O 2 ] at t = 310 s.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 33

The Rate Law of a Chemical Reaction

• The

rate law

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

aA + bB + cC …→ products rate = k[A]

[B]

[C]

…

• 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,

, is the

rate constant

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 34

Rate Law Examples

Rate =

[A] 1 =

[A] Rate =

[A] 2 Rate =

first

order in A

second

order in A Reaction is

third

order in A

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 35

Effect of Concentration on Rate Order of Rxn

Concentration Change

Double/Triple…

Effect on Rate

Nothing (2 0 ) Prentice Hall © 2005 1 1 2 2 2 3 3 Double Triple Double Triple Quadruple Double Triple Double (2 1 ) Triple (3 1 ) Quadruple (2 2 ) 8x (3 2 ) 16x (4 2 ) 9x (2 3 ) 27x (3 3 ) Chapter Thirteen

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry 36

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 37

To find the overall order of a reaction…

• Add the orders for each compound.

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Units of k (p. 536) Overall Reaction Order

Zero First Second Third

Units of k

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 39

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

• 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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 40

The concentration of NO was held the same in Experiments 1 and 2 … … while the concentration of Cl 2 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 Cl 2 .

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 Cl 2 ?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.3

For the reaction 2 NO(g) + Cl 2 (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 [Cl 2 ] = 0.0255 M?

(b)

What is the value of

for the reaction?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

So, when looking at data…..

• Zero order reaction: initial rate is unaffected (2 0 ) • 1 st order reaction: double the concentration, doubles the rate (2 1 • 2 nd Order Reaction: Initial rate increases fourfold (2 2 =4) • 3 rd Order Reaction: Initial rate increases eightfold (2 3 =8) =2) Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 43

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry 0.400

0.100

Chapter Thirteen 44

Let’s Take a Break

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

•Sum of exponents is equal to zero

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 46

Zero Order

Rate of Reaction= k[A]

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 47

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 48

A Zero-Order Reaction rate = k[A] 0 = k Rate is independent of initial concentration

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Zero Order

Integrated rate equation:

[A]

= -kt + [A]

y = mx + b

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Zero Order (cont.)

y = [A]

= conc of A at some time x = t = time b = [A]

m = -k (m, the slope of the

straight line)

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 51

First-Order Reactions

• In a

first-order reaction

, the exponent in the rate law is 1.

• Rate =

[A] 1 =

[A] • The

integrated rate law

describes the concentration of a reactant as a function of time. For a first-order process: [A]

ln = –

ln [A]

[A] 0 – ln [A] 0 = –

= –

+ ln [A] 0

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

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

proportional

to concentration.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 52

Rate of reaction= k[A]

Integrated rate equation 53

ln[A]

= -kt + ln[A]

y = mx + b

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

1 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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Decomposition of H 2 O 2

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.4

For the first-order decomposition of H 2 O 2 (aq), given

3.66 x 10 –3 s –1 and [H 2 O 2 ] 0 = 0.882 M, determine

(a)

= the time at which [H 2 O 2 ] = 0.600 M and

(b)

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Another Example

H 2 0 2 initially at a conc of 2.32 M, is allowed to decompose. What will the [H 2 0 2 ] be 1200 s later? Use k = 7.3 x 10 -4 s -1 for this first order decomposition.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Try It Out

EX 13.4 A: The decomposition of nitramide, NH 2 NO 2 , is a first order reaction: NH 2 NO 2 (aq) → H 2 O (l) + N 2 O (g) The rate law is rate = k[NH 2 NO 2 ], with k = 5.62 x 10-3 min-1 at 15ᵒC. Starting with 0.105 M NH 2 NO 2 , a) At what time will [NH 2 NO 2 ] = 0.0250 M b) What is [NH 2 NO 2 ] after 6.00 hours?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 58

Half-life

Time required for ½ of the reactant to be consumed Equation: t

1/2

= ln2 = 0.693

k k

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 59

Example 13.5

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 60

Example 13.5 A

Use the result of Example 13.5 to determine a) The time required to reduce the quantity of N 2 O 5 to 1/16 of its initial value and b) The mass of N 2 O 5 sample of N 2 O 5 remaining after a 4.80 g has decomposed for 10.0 min.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Sum of exponents = 2

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 62

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 =

[A] 2 • The integrated rate law has the form: 1 –––– =

[A]

1 + –––– [A] 0

What do we plot vs. time to get a straight line?

• The

half-life

of a second-order reaction depends on the initial concentration as well as on the rate constant

½ 1 = –––––

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 63

Example 13.7

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

HI(g) → ½ H 2 (g) + ½ I 2 (g) Rate =

[HI] 2 What are the:

(a)

rate constant and

(b)

half-life of the decomposition of 1.00 M HI(g) at 700 K?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Summary of Kinetic Data

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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)

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Try it in your note packet!

Time, min [A] 0 1.0

5 10 0.63

0.46

15 25 0.36

0.25

ln[A]

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry

1/[A]

Chapter Thirteen 68

Crash Course Review

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Let’s Take a Break

• Complete Part 2 of the Partner Review.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Today…

•

Turn in:

–

Get out a piece of notebook paper

•

Our Plan:

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Scavenger Hunt Review

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Investigation 11 Pre-Lab

•

Homework (Write in Planner):

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Prepare Lab Report

–

Homework Problems (Due 2/12)

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 71

Pre Lab

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

• Be prepared to experiment next class.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Today…

•

Turn in:

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Nothing

•

Our Plan:

–

Investigation 11

•

Homework (Write in Planner):

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Lab Report due Next Class (2/10)

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 73

Today…

•

Turn in:

–

Lab Report (rubric on top)

•

Our Plan:

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Review Activity

–

Elephant Toothpaste Demo

–

Notes – Catalysts & Rate Determining Steps

–

Finish Homework Problems

•

Homework (Write in Planner):

–

Homework Problems

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 74

Effect of Temperature on the Rates of Reactions

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

e –

a /

• The constant

, 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 –

a /

represents the fraction of molecular collisions

sufficiently energetic

to produce a reaction.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 75

I like this equation better…

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.9

Estimate a value of

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

= 2.5 x 10 –3 s –1 at 332 K.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

(C 4 H 9 ) 2 O 2 (g) → 2(CH 3 ) 2 CO (g) + C 2 H 6 (g) The half-life of DTBP is 17.5 h at 125ᵒC and 1.67 h at 145ᵒC. What is the activation energy,

E a

, of the decomposition reaction?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 78

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 79

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 80

Molecularity

The

molecularity

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry

Termolecular processes are unusual, for the same reason that three basketballs shot at the same time are unlikely to collide at the same instant …

Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 82

An Example

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 83

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 85

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 86

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] I. A + A ↔ X (fast) II. X + B → C + Y (slow) III. Y + B → D (fast) Prentice Hall © 2005

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 87

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Example 13.10

For the reaction H 2 (g) + I 2 (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? 89

Fast step:

I 2

–1

Slow step:

2 I + H 2 2 I

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

EX 13.10 A

The decomposition of nitrosyl choride, 2NOCl (g) → 2NO (g) + Cl 2 (g) Is a first-order reaction. Propose a mechanism for this reaction consisting of one fast step and one slow step.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 91

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 92

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

A Kinetics Pick Up Line…

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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).

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 96

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

A Surface-Catalyzed Reaction

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Mercury Poisoning: An Example of Enzyme Inhibition When Hg reacts with an enzyme …

101

… the Hg binds to sulfur atoms … … changing the shape of the active site, so that it no longer “fits” the substrate.

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Homogeneous Catalysis

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

O 3

Identifying Catalysts and intermediate species

+ Cl∙ --> ClO∙ + O 2 ClO· + O∙ --> Cl∙ + O 2

Catalyst? Intermediate?

Net Equation?

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 103

Crash Course Review

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Finish the HW

All homework problems are due on Wednesday!

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 106

Study Guide Changes

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

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen

Today…

• • •

Turn in:

–

Nothing Our Plan:

–

Study Guide Questions

–

Test Homework (Write in Planner):

–

Complete the POGIL (Day 1 ONLY)

General Chemistry

4 th edition, Hill, Petrucci, McCreary, Perry Chapter Thirteen 108

Chapter Thirteen

Transcript Chapter Thirteen

Today…

Key Point

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

Energy and orientation of cars during a car crash can establish the change that occurs to the cars.

Activated complex

Reaction Profile

• A graphical representation of a chemical reaction in terms of the energies of the reactants, activated complexes and products

Reaction Profile

Exothermic

Endothermic

Stop!

Today…

Turn in:

Our Plan:

Homework (Write in Planner):

Today…

Turn in:

Our Plan:

Homework (Write in Planner):

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

•Sum of exponents is equal to zero

Zero Order

Rate of Reaction= k[A]

•

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

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

Zero Order

Integrated rate equation:

[A]

= -kt + [A]

y = mx + b

y = [A]

= conc of A at some time x = t = time b = [A]

m = -k (m, the slope of the

straight line)

Rate of reaction= k[A]

ln[A]

= -kt + ln[A]

y = mx + b

1

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.

Another Example

Half-life

Time required for ½ of the reactant to be consumed Equation: t

= ln2 = 0.693

k k

Sum of exponents = 2

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)

• Complete Part 2 of the Partner Review.

Today…

Today…

Turn in:

Our Plan:

Homework (Write in Planner):

Today…

Identifying Catalysts and intermediate species

Catalyst? Intermediate?

Net Equation?

Finish the HW

All homework problems are due on Wednesday!

Today…

Study Guide Changes

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

Today…

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