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Learning Objectives:. Reaction Rate. Expressing the Reaction Rate. **The Rate Law and Its Components. **Integrated Rate Laws: Concentration Changes over Time. Catalysis: Speeding Up a Reaction. Theories of Chemical Kinetics. ID: 708139

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Slide1

Chemical Kinetics:

Rates and Mechanisms of Chemical Reactions

Slide2

Learning Objectives:

Reaction Rate

Expressing the Reaction Rate

**The Rate Law and Its Components

**Integrated Rate Laws: Concentration Changes over Time

Catalysis: Speeding Up a Reaction

Theories of Chemical Kinetics

*Reaction Mechanisms: The Steps from Reactant to Product

Slide3

Chemical reactions: Fast or Slow

Explosion

Ripening of fruits/vegetables

Digestion

Fossilization from ancient vegetation to fossil fuel

Slide4

Microscopic view between fast and slow reactions

Slide5

Factors That Influence

Reaction Rate

Particles must collide in order to react: Greater collision rate, faster rate.Higher the concentration of reactants, reaction rateGreater number of collisions. Household bleach (CloroxTM

Physical state

of the reactants influences reaction rate. -

Substances must mix in order for particles to collide. Example:

Surface area (powder form vs. chunky form)Higher the temperature, reaction rate

At higher temperatures particles have more energy and therefore collide more often and more effectively. Milk spoils faster at higher temperature.

Slide6

The effect of

Surface Area

on reaction rate

A hot steel nail glows feebly when placed in O

The same mass of steel wool (

more surface area

) bursts into flame, aka reacts much faster

Slide7

Sufficient collision energy is required for a reaction to

occur

Slide8

What is the

Reaction Rate

Reaction rate: the changes in concentrations of reactants or products per unit time.

For the general reaction A

→ B, we measure the concentration of A at

and at

t2:

“-” : because the concentration of A is decreasing. This gives the rate a positive value.

D[A]Dt

Rate =

change in concentration of A

change in time

conc A

- conc A

= -

“[xxx]” = concentration in moles per liter.

Slide9

Average rate

Instantaneous rate

= slope of tangent line at tInitial rate = instantaneous rate at t = 0

Slide10

Question

: point

A or B?______ has the higher [O3]______ has higher reaction rate After reaction begins, the rate (increase or decrease)

Slide11

Plots of [reactant] and [product] vs. time.

2H4 + O3 → C2

O + O

2] increases just as fast as [C2H4] decreases.

Rate = -

D[C2H4]Dt

= -

]

Quick Question

Which has higher

]

Which has higher reaction rate?

Key: b; a

Slide12

Plots of [reactant] and [product] vs. time.

2 + I2 → 2HI

[HI] increases twice as fast as [H

] decreases.

Rate = -

]Dt

= -

]

[HI]

]

Rate =

]

[IH]

= -2

The expression for the rate of a reaction and its numerical value depend on which substance serves as the reference.

Slide13

In general, for the reaction

A + bB → cC + dD

where

c, and d are the coefficients for the balanced equation, the rate in terms of each species is expressed as:

The Reaction Rates depends on the stoichiometry of Reaction

Analogy: For the same person, heart rate and breathing rate is not the same at the same moment.

_________________________________________

Slide14

Example:

The balanced equation for the reaction of bromate ion with bromide in acidic solution is given by

BrO3- + 5Br- + 6H+ → 3Br2 + 3H2OAt a particular instant in time, the value of -△[Br-]/△t is 1.0 × 10-2 mol/L ∙ s. What is the value of △[Br2

△

in the same unit

6.0 × 10-3 mol/L ∙ sPractice: What is the value of -△[H+]/△

t in the same unit?

Slide15

The Rate Law:

How Rate Depends on Concentrations

For any general reaction occurring at a fixed temperature

A +

→

cC + dD

Rate = k[A]m[B]n

k = Rate Constant

m and n : Reaction Orders, determined by experiment.k depends on temperature and catalyst, not affected by concentrations.

Order:

zero order;

first

order;

second

order

Slide16

Individual and Overall Reaction Orders

For the reaction 2NO(

g) + 2H2(g) → N2(

) + 2H

g):

The rate law is rate = k[NO]2[H2]

The above reaction is 2nd order for NO, 1st order with respect to H2; 3rd order overall.

If the concentration of a reactant does NOT affect the rate, the order of this reactant is ______ order.

Slide17

Plots of

Reactant concentration, [A], vs. T

ime

Slide18

Plots of

Rate vs. Reactant

Concentration [A]

Slide19

Determining Reaction Orders: A. Initial Rate Method

For the general reaction

aA + bB → d

, rate law as Rate = k[A]m[B]n

To determine the values of m and n, we run a series of experiments and find the order ONE by ONE

To find n, only ___ is changed, measure new initial rate3. Set first trial ([A]1 and [B]1

measure initial rate1 at time = 0).To find m, only [A] is changed while [B] unchanged, measure new initial rate2

Slide20

Initial Rate Method to Determine the order

Experiment

Initial Rate

(mol/L·s)

Initial [A]

(mol/L)

Initial [B]

(mol/L)

1.75x10-32.50x10-1

3.00x10-12

3.50x10-35.00x10-1

3.00x10

-1

3.95x10

-3

2.50x10

-1

4.50x10

-1

The mathematical way to find the order:

order = ______________

Slide21

Finding the order with respect to A

compare Trial# 1 and ___, where [B]

is kept

constant but [A]

changes:

Rate =

[A]

m[B]nTrial

Initial Rate(mol/L·s)

Initial [A]

(mol/L)

Initial [B]

(mol/L)

1.75x10

-3

2.50x10

-1

3.00x10

-1

3.50x10

-3

5.00x10

-1

3.00x10

-1

3.95x10

-3

2.50x10

-1

4.50x10

-1

Slide22

Practice: Find

the order with respect to

compare Trial# 1 and ___, where

[A] is kept

constant but

[B] changes:

Rate = k[A]m[B]nTrial

Initial Rate(mol/L·s)

Initial [A]

(mol/L)

Initial [B]

(mol/L)

1.75x10

-3

2.50x10

-1

3.00x10

-1

3.50x10

-3

5.00x10

-1

3.00x10

-1

3.95x10

-3

2.50x10

-1

4.50x10

-1

Slide23

Practice: Finding the rate constant

Choose

any trial to determine the rate

constant

__________

Trial 1: k

= 0.0778 mol-2 L2 s-1

Rate = k[A]1[B]2

Trial

Initial Rate

(

mol

/L·s)

Initial [A]

(

mol

/L)

Initial [B]

(

mol

/L)

1.75x10

-3

2.50x10

-1

3.00x10

-1

3.53x10

-3

5.00x10

-1

3.00x10

-1

3.92x10

-3

2.50x10

-1

4.50x10

-1

Your k

= __________________

Slide24

Easier Way to Find Reaction Orders?

Double the concentration(s)

For the simple reaction A → products:

If the rate

doubles

when [A]

doubles

, rate

 [A]1, reaction is _____ order with respect to A.

If the rate quadruples when [A] doubles, rate  [A]2, reaction is __________ order with respect to [A].

If the rate does not change when [A] doubles, rate  [A]0, reaction is _______ order with respect to A.

Slide25

Units of the Rate Constant

k for Common Reaction Orders

Overall Reaction OrderUnits of k

(

in seconds)

mol/L·s

(or mol L-1s-1)

11/s (or s-1)

2L/mol·s (or L mol-1s

-1)3L

/mol

·s

(or L

mol

-2

-1

)

General formula:

mol

unit of

order-1

Units of

The value of

is easily determined from experimental rate data. The

units

depend on the overall reaction order.

Slide26

Information sequence to determine the kinetic parameters of a reaction.

Series of plots of concentration vs. time

Initial rates

Reaction orders

Rate constant (

) and actual rate law

Determine slope of tangent at t

for each plot.

Compare initial rates when

[A]

changes and

[B]

is held constant (and vice versa).

Substitute initial rates, orders, and concentrations into

rate =

[A]

[B]

and solve for k.

Slide27

Integrated Rate Laws

An integrated rate law includes

time as a variable.

First-order rate equation:

rate = -



[A]



[A]

Second-order rate equation:

rate = -



[A]



[A]

Zero-order rate equation:

rate = -



[A]



[A]

Remember both rate and concentration depends on time

Integrated Rate Law:

_____________________

Slide28

Graphical

presentation of 1st

order reactionFirst-order reaction

[A]

integrated rate law

ln[A]

= _______ + _____

straight-line form

A plot of

ln[A]

vs.

time

gives a straight line for a first-order reaction.

Rate constant

= - slope

Slide29

Graphical

presentation of 2nd

order reactionSecond-order reaction

integrated rate law

[A]

straight-line form

[A]

lot

1/[A]

vs.

time: linear.

Rate

constant

= slope

y = mx + b

Slide30

Graphical

presentation of zero order reaction

Zero-order reaction

lot

[A] vs. time

gives a straight line for a first-order reaction.

Rate constant k = - slope

integrated rate law

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

straight-line form

[A]

= _____ + ____

Slide31

Example:

Determining the Reactant Concentration after a Given Time

At 1000

cyclobutane

H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4).

(a) If the initial C4H8 concentration is 2.00 M, what is the concentration after 0.010 s?

0.83 mol/L

Given: k = 87 s-1, t = 0.010 s, [A]0 = 2.00 M, ln([A]t/[A]0) = - ktFind: [A]t

Slide32

Example:

Apply Integrated rate law in Problem solving

At a certain temperature,

cyclobutane

) decomposes in a first-order reaction, with a rate constant of 1.0 s-1.

(b) What fraction of C4H8 has decomposed after 3.0 sec?

0.95

Given: k = 1.0 s-1, t = 3.0 s, ln([A]t/[A]0) = - kt

Find: 1 - [A]

/[A]

Slide33

Example:

Apply Integrated rate law in Problem solving

At a certain temperature,

cyclobutane

) decomposes in a first-order reaction. After 10.0 seconds, 20.0% of cyclobutane decomposed. Determine the rate constant at this temperature.

0.0223 s-1

Given: t = 10.0 s, 1 - [A]t/[A]0) = 0.200Find: k

Slide34

Graphical determination of

Reaction order: Decomposition of N

2O5

Three ways to graph:

[A] vs. time

ln[A] vs. time

1/[A] vs. time

Since the plot of ln[N2O5] vs. time gives a straight line, the reaction is first order.

Slide35

Reaction Half-life

The

half-life (t1/2) for a reaction is the time taken for the concentration of a reactant to drop to half its initial value.

For a

first-order

reaction,

1/2

does not depend on the starting concentration.

t1/2 =

ln 2

0.693

The half-life for a first-order reaction is a

constant

, independent on the concentration of reactant

Slide36

[

2O5] vs. time for three reaction half-lives.

1/2

for a first-order process

ln 2

0.693

Slide37

Half-life Equations

For a

first-order reaction, t1/2 does not depend on the initial concentration.

For a

second-order

reaction,

1/2

is inversely proportional to the initial concentration:

1k[A]0

t1/2 =

For a

zero-order

reaction,

1/2

directly

proportional to the initial concentration:

[A]

1/2

Slide38

Practice:

Determining the Half-Life of a First-Order Reaction

At 1000

C, the rearrangement reaction of cyclopropane has rate constant of 9.2 s

-1

(a)

What is the half-life of the reaction?

(b) How long does it take for the concentration of cyclopropane to reach one-tenth of the initial value?

a. Half life t1/2 = 0.693/k = 0.075 s

b. 0.25 s

Slide39

An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions

Zero Order

First Order

Second Order

Rate law

rate =

k[A]

rate = k[A]2

Units for kmol/L·s1/s

L/mol·sHalf-life

Integrated rate law

in straight-line form

[A]

= -

+ [A]

ln[A]

= -

+ ln[A]

Plot for straight line

[A]

vs.

ln[A]

vs.

Slope,

intercept

, [A]

-k

, ln[A]

[A]

ln 2

[

kt +

[

Slide40

Collision Theory and Concentration

The basic principle of

collision theory is that particles must collide in order to react.

An increase in the concentration of a reactant leads to

higher frequency

of collisions, hence increasing reaction rate.

The number of collisions depends on the

product

of the numbers of reactant particles, not their sum.

Concentrations are multiplied in the rate law, not added.

Slide41

Temperature Affects Rate Constant

Temperature has a dramatic effect on reaction rate.

For many reactions, an increase of 10°C will double or triple the rate.

Experimental data shows that

k increases exponentially as T increases

: the

Arrhenius equation

-Ea/RT

= rate constant

= frequency factor

= activation energy

Higher

increased rate

larger

-E

/RT

= rate constant

= frequency factor

= activation energy

Slide42

Increase of the rate constant with temperature for the hydrolysis of an ester.

Expt

[Ester]

(K)

Rate

(mol/L·s)

k(L/mol·s)

10.100

0.200

288

1.04x10

-3

0.0521

0.100

0.200

298

2.20x10

-3

0101

0.100

0.200

308

3.68x10

-3

0.184

0.100

0.200

318

6.64x10

-3

0.332

Reaction rate and

increase exponentially as

increases.

Slide43

Activation Energy

In order to be

effective, collisions between particles must exceed a certain energy threshold.

The

lower

the activation energy, the

faster

the reaction.

When particles collide effectively, they reach an

activated state. The energy difference between the reactants and the activated state is the activation energy (Ea) for the reaction.

Smaller Ea

increased rate

larger

Slide44

Energy-level diagram for a

reaction

Collisions must occur with sufficient energy to reach an activated state.

This particular reaction is reversible and is exothermic in the forward direction.

Slide45

Temperature and Collision Energy

An increase in temperature causes an increase in the kinetic energy of the particles. This leads to

more frequent collisions

Also, at a higher temperature, higher fraction of molecules has the kinetic energy above the activation energy

Slide46

The Effect of

Ea and T

on the Fraction (f) of Collisions with Sufficient Energy to Allow ReactionEa (kJ/mol)

(at

= 298 K)

1.70x10-9

757.03x10-14100

2.90x10-18

Tf (at E

= 50 kJ/

mol

)

25°C (298 K)

1.70x10

-9

35°C (308 K)

3.29x10

-9

45°C (318 K)

6.12x10

-9

Slide47

Calculating Activation Energy

= Ae

-E

/RT

Arrhenius equation: Exponential form vs. Logarithm form

= ln

A -

Ea 1R T

Use rates at two different temperatures:

At T

, ln

= ln

At T

, ln

= ln

= ___________________

Slide48

Graphical determination of the Activation energy:

k (as y) vs. 1/T (as x)

= ln

Ea 1R T

= (-8.314 J/mol) x slope

Unit: Joule

= exp(

intercept)

Unit: same unit as rate constant

Slide49

Find Activation Energy and A factor from graph

Expt

[Ester]

(K)

Rate

(mol/L·s)k

(L/mol·s)

10.100

0.200

288

1.04x10

-3

0.0521

0.100

0.200

298

2.20x10

-3

0101

0.100

0.200

308

3.68x10

-3

0.184

0.100

0.200

318

6.64x10

-3

0.332

= (-8.314 J/mol) x slope

exp

(

intercept

)

46.9 kJ/

mol

1.6 x 10

mol

Slide50

Practice:

The decomposition of hydrogen iodide, 2HI(

g) → H2(g) + I2(

), has

rate constants of 9.51x10

-9

L/mol·s at 500. K and 1.10x10-5 L/mol·s at 600. K. Find Ea.

Ea = 1.76x105 J/mol = 1.76x102 kJ/mol

Slide51

Molecular Structure and Reaction Rate

For a collision between particles to be effective, it must have both sufficient

energy and the appropriate relative orientation between the reacting particles.

-E

/RT

in the Arrhenius equation is the

frequency factor for the reaction.

= orientation probability factor

Z = collision frequency

depends on the specific reaction and is related to the structural complexity of the reactants.

Slide52

Molecular orientation for an Effective Collision

NO(

) + NO

(

) → 2NO2(g)

Only one relative orientation of these two molecules, Effective Orientation, that leads to an effective collision.

Think: Which bonds are broken? Which bonds are formed?

Slide53

Transition State Theory

ffective collision between particles leads to the formation of a transition state or activated complex.

The

transition state

is an

unstable

species that contains

partial bonds. It is a transitional stage between reactants and products. Transition states cannot be isolated.

The transition state exists at the point of maximum potential energy. The energy required to form the transition state is the activation energy.

Slide54

Transition state: Reaction

between BrCH3

and OH-

The transition state contains partial bonds (

dashed

) between C and Br and between C and O. It has a trigonal bypyramidal shape.

BrCH

+ OH- → Br- + CH3OH

Slide55

Quantum mechanical Study of the reaction between BrCH

and OH-.

Slide56

Reaction Energy Diagram: Endothermic vs. Exothermic Reaction

Slide57

Example

Drawing Reaction Energy Diagrams and Transition States

For reaction: O

(

) + O(

) → 2O2(g)The Ea(fwd) is 19 kJ, and the DHrxn for the reaction as written is -392 kJ. Draw a reaction energy diagram, label the reactant, product, activation energy, reaction enthalpy change, and transition state. Predict a structure for the transition state, and calculate Ea(rev).

Slide58

Reaction

Mechanism: Elementary Step

Two Step Mechanism: Decomposition of Ozone

Unlike a balanced equation, many chemical reactions do not occur with

all

the reactant molecules together turning into final products.

The

mechanism

of a reaction is the sequence of single reaction steps (elementary steps) that make up the overall equation. Driving direction from OC to LAX.

Slide59

Reaction intermediates

Reaction

intermediate

is the species (molecules, atoms, ions, etc.) produced in the initial elementary

steps

typically

unstable and reactive, later consumed in the subsequent elementary steps.

Slide60

Molecularity of Elementary Steps

Each elementary step is characterized by its

molecularity, the number of particles involved in the reaction: Unimolecular; Bimolecular; Termolecular

(three molecules)

Slide61

Elementary Step

Molecularity

Rate Law

A product

2A product

A + B product

2A + B product

Unimolecular

Bimolecular

Termolecular

Rate = [A]

Rate =

[A]

Rate =

[A][B]

Rate =

[A]

[B]

The rate law for an

elementary step

can

be deduced from the reaction stoichiometry –

reaction order equals molecularity

for an elementary step only

Slide62

Example:

Reaction mechanism, Rate law for Elementary Steps, Reaction intermediate

The following elementary steps are proposed for a reaction mechanism:

Cl(

)

→ NO2(g) + Cl(g) NO2Cl(g) + Cl(

g) → NO2(g) + Cl2(g)

Write the overall balanced equation.Write the rate law for each step.

Identify the reaction intermediate.

Slide63

The Rate-Determining Step of a Reaction

The

slowest elementary step in a reaction is the rate-determining or rate-limiting step. Analogy: Traffic between Orange County and Riverside

The rate law for the rate-determining step becomes the rate law for the overall reaction.

The reaction NO

(

) + CO(

g) → NO(g) + CO2(g) has been proposed to occur by a two-step mechanism:

NO2(g) + NO2(g) → NO3(

g) + NO(g) [slow; rate-determining](2) NO3(g) + CO(g) → NO2(g) + CO2(g) [fast]

Observed rate law: rate =

[NO

]

Slide64

Correlating Mechanism with

Rate Law

A valid mechanism must meet three criteria:

The elementary steps must add up to the overall balanced equation.

The elementary steps must be reasonable.

The mechanism must correlate with the

observed

rate law.

If first elementary step is the rate-determining, its rate law should match the experimental rate law

A mechanism is a

hypothesis

–we cannot prove it is correct, but if it is consistent with the data, and can be used to predict results accurately, it is a useful model for the reaction.

Slide65

Example: Propose a Mechanism with a Slow Initial Step

The overall reaction 2NO

2(g) + F2(g) →2NO2F(

) has an

experimental rate law Rate =

[NO2][F2].

In the first elementary step is rate-determining step, the mechanism would be:

For step 1 (the slow step), rate1 = k1 x ______ which matches the experimental rate law

Step 1:

Step 2: The second NO2 molecule needs to be reactant so that the overall reaction has two NO2 molecules

Slide66

Reaction energy

diagrams for

Two-step reaction(A) NO2 and F2 (B) NO and O2.

Each step

in multi-step mechanism

has its own transition

state

Slide67

Catalysis: Speeding up a Reaction

catalyst is a substance that increases the reaction rate without itself being consumed in the reaction.

In general, a catalyst provides an alternative reaction pathway that has a

lower total activation energy

than the

uncatalyzed

reaction.

A catalyst will speed up

BOTH the forward and the reverse reactions.

A catalyst does not affect either DH or the overall yield for a reaction.

Slide68

nergy diagram:

Catalyzed vs. Uncatalyzed

Slide69

Decomposition

of H

2O2 catalyzed by bromide ionA small amount of NaBr is added to a solution of H

Oxygen gas forms quickly as Br

-(aq) catalyzes the H2O2 decomposition; the intermediate Br2 turns the solution orange.

Slide70

Metal-catalyzed

hydrogenation of ethene

A heterogeneous catalyst is in a different phase than the reaction mixture.

Slide71

Enzyme: Protein as Biological Catalyst

Enzyme active site (“Lock”) has

fixed shape, which matches the shape of its substrate(“Key”) and be specific to its substrate.

Slide72

Induced-fit model for Enzyme

The enzyme

active site changes shape in order to bind its substrate(s) more effectively. The active site is

selective towards certain substrates.

Slide73

Chlorofluorocarbons (CFC) destroy ozone layer over the Antarctic: Satellite images of the increasing size of the Antarctic ozone hole (

purple

Slide74

Reaction energy diagram for breakdown of O

by Cl atoms. (Not drawn to scale.)

Slide75

Concentration of O

at Various Times in its Reaction with C2H4 at 303 K

Time (s)

0.0

20.0

30.0

40.0

50.0

60.0

10.0

Concentration of O

(mol/L)

3.20x10

-5

2.42x10

-5

1.95x10

-5

1.63x10

-5

1.40x10

-5

1.23x10

-5

1.10x10

-5



]



rate = -



]



= -

Practice:

Estimate the rate between 10.0s and 20.0s.

Key:

4.7x10

-6

mol/L s

Slide76

In general, for the reaction

A + bB → cC + dD

[A]

[B]

rate = -

= -

[C]

[D]

where

, and

are the coefficients for the balanced equation, the rate is expressed as:

The Reaction Rates depends on the stoichiometry of Reaction

Analogy: For the same person, heart rate and breathing rate is not the same at the same moment.

Slide77

Rate 2

Rate 1

[A] [B]

Finding

, the order with respect to A:

We compare experiments 1 and 2, where [B] is kept constant but [A] doubles:

[A]

3.50x10

-3

mol/L

·s

1.75x10

-3

mol/L

·s

5.00x10

-2

mol/L

2.50x10

-2

mol/L

Dividing, we get 2.00 = (2.00)

= 1

Rate =

[A]

[B]

Slide78

Rate 3

Rate 1

[A] [B]

Finding

, the order with respect to B:

We compare experiments 3 and 1, where [A] is kept constant but [B] doubles:

[B]

3.50x10

-3

mol/L

·s

1.75x10

-3

mol/L

·s

6.00x10

-2

mol/L

3.00x10

-2

mol/L

Dividing, we get 2.00 = (2.00)

= 1

Slide79

Practice:

Determining Reaction Orders from Rate Data

PROBLEM:

For the following reaction:

(

) + CO(

g) → NO(g) + CO2(g) rate = k[NO2]m[CO]nUse the following data to determine the individual and overall reaction orders:

ExperimentInitial Rate (mol

/L·s)Initial [NO2](mol/L)

Initial [CO]

(mol/L)

0.0050

0.10

0.080

0.40

0.10

0.0050

0.10

0.20

rate =

[NO

]

[CO]

or rate =

[NO

]

Slide80

Group Practice: What is the rate law for the following reaction?

Initial Rate

Initial Reactant Concentrations (

mol

/L)

Experiment

(

mol

/L·s)[O2

][NO]1

3.21x10-31.10x10

-21.30x10-2

6.40x10

-3

2.20x10

-2

1.30x10

-2

28.8x10

-3

1.10x10

-2

3.90x10

-2

(

) + 2NO(

)

→ 2NO

(

)

Rate law: rate =

]

[NO]

rate =

][NO]

Slide81

Integrated Rate Laws

An integrated rate law includes

time as a variable.

First-order rate equation:

rate = -



[A]



[A]

Second-order rate equation:

rate = -



[A]



[A]

Zero-order rate equation:

rate = -



[A]



[A]

- [A]

= -

Slide82

Practice:

The reaction A → B + C

is second order in A. When [A]0 = 0.100 M, the reaction is 20.0% complete in 40.0 minutes. Calculate the value of the rate constant (in L/min ∙ mol).

[A]

= 0.080 M

k = 0.61 L/min mol

Slide83

Practice:

The half-life of a certain first-order reaction is 25 minutes. What fraction of the original reactant concentration will be consumed after 2.0 hours?

= 0.02

7 min

-1

[A]

t/[A]0 = exp(-kt) = 0.036, 1 - [A]t

/[A]0 = 0.964

Slide84

Calculating Activation Energy

= Ae

-E

/RT

can be calculated from the Arrhenius equation:

so ln

k = ln A -

Ea 1R T

straight-line form

= -

−

If data is available at two different temperatures:

Slide85

Group Practice:

The decomposition of hydrogen iodide, 2HI(

g) → H2(g) + I2

(

), has

rate constants of 9.51x10

-9 L/mol·s at 500. K and 1.10x10-5 L/mol·s at 600. K. Predict the rate constant at 700C.

Ea = 1.76x102 kJ/mol

First, find the A factor in Arrhenius equation; Then use Arrhenius equation to find k

Slide86

SOLUTION:

rate

1 = k1[NO2

Cl]

rate

k2[NO2Cl][Cl]

(b)

(a) Writing the overall balanced equation:

(1) NO2Cl(g) → NO2(g) + Cl(g)

(2) NO2Cl(g) + Cl(g) → NO2(g) + Cl2(g)

2NO

Cl(

)

→ 2NO

(

) + Cl

(

)

The species does NOT appear in the overall reaction is the reaction intermediate.

(c)

PROBLEM:

The following elementary steps are proposed for a reaction mechanism:

(1) NO

Cl(

)

→ NO

(

) + Cl(

)

(2) NO

Cl(

) + Cl(

) → NO

(

) + Cl

(

)

(a)

Write the overall balanced equation.

(b)

Write the rate law for each step.

Identify the reaction intermediate

Slide87

Mechanisms with a Slow Initial Step

The overall reaction 2NO

2(g) + F2(g) →2NO2

) has an experimental rate law Rate =

[NO2][F2].

The accepted mechanism isNO2(g) + F2(g) → NO2F(g) + F(g

) [slow; rate determining]NO2(g) + F(g) → NO2F(g) [fast]

The elementary steps sum to the overall balanced equation:

2NO2(g) + F2(g)

→2NO

)

Both steps are bimolecular and are therefore reasonable.

rate

[NO

]

rate

[NO

][F]

Step 1 is the slow step, and rate

correlates with the

observed

rate law.

The mechanism is therefore reasonable.

Slide88