CHEMICAL KINETICS CHAPTER 17, Kinetics Fall 2009, CHEM 1310 1.

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CHEMICAL KINETICS CHAPTER 17, Kinetics Fall 2009, CHEM 1310 1

Transcript of CHEMICAL KINETICS CHAPTER 17, Kinetics Fall 2009, CHEM 1310 1.

Page 1: CHEMICAL KINETICS CHAPTER 17, Kinetics Fall 2009, CHEM 1310 1.

CHEMICAL KINETICSCHAPTER 17, KineticsFall 2009, CHEM 1310

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KeKin

Kinetics vs

Thermodynamics

A: ReactantsB: Transition state

C: productsE: Forward Activation

Free EnergyF: Reverse Activation

Free Energy

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• Most reactions occur through several steps and are not single step reactions.

• Each step in a multi-step reaction is called an elementary reaction.

Types of elementary reactions1. Unimolecular (a single reactant)

2. Bimolecular

3. Termolecular (very unlikely)

3

Reaction Mechanisms

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• Each step of this reaction is an “elementary step”. Each elementary step has reactant(s), a transition state, and product(s). Products that are consumed in subsequent elementary reaction are called intermediates.

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Reaction Rates: To measure a reaction rate we could monitor the disappearance of reactants or appearance of products.

e.g., 2NO2 + F2 → 2NO2F

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Gen. Rxn: aA + bB → cC + dD

tD

dtC

c

tB

btA

arate

][1][1

][1][1

tX

tt

XX

if

if

][][][

rate average

NO2 + CO → NO + CO2

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Order of a ReactionThe power (n) to which the concentration of A is raised in the rate expression describes the order of the reaction with respect to A.

k[A]rate

Products k

aA n

e.g.,

Do not confuse the order (n) with the stoichiometric coefficient (a).

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mth order in [A]

nth order in [B]

aA + bB k→ products

rate = k[A]m[B]n

as before

rate = k[A] m[B] n

= −1

a ⎛ ⎝

⎞ ⎠d[A]

dt= −

1

b ⎛ ⎝

⎞ ⎠d[B]

dt

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A + B → C

rate = k[A]m[B]n (this is the "rate law"

for this reaction)

If you know rates at various concentrations,

take the ratio of rates, solve for n or m.

rate2

rate1

=k [A]2( )

m[B]2( )

n

k [A]1( )m

[B]2( )n

[A] (mol L-1) [B] (mol L-1) Rate (mol L-1 s-1)

1 1.0x10-4 1.0x10 -4

2.8x10 -6

2 1.0x10-4 3.0x10 -4

8.4x10 -6

3 2.0x10-4 3.0x10 -4

3.4x10 -5

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[A] (mol L-1) [B] (mol L-1) Rate (mol L-1 s-1)

1 1.0x10-4 1.0x10 -4

2.8x10 -6

2 1.0x10-4 3.0x10 -4

8.4x10 -6

3 2.0x10-4 3.0x10 -4

3.4x10 -5

rate2

rate3

=k [A]2( )

m[B]2( )

n

k [A]3( )m

[B]3( )n

8.4 x 10-6( )

3.4 x 10-5( )=

1

4=

1m

2m

m = 2

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Example: At elevated temperatures, HI reacts according to the chemical equation

2HI → H2 + I2The rate of reaction increases with concentration of HI, as shown in this table.Data [HI] RatePoint (mol L-1) (mol L-1 s-1)

1 0.005 7.5 x 10-4

2 0.010 3.0 x 10-3

3 0.020 1.2 x 10-2

a) Determine the order of the reaction with respect to HI and write the rate expressionb) Calculate the rate constant and give its unitsc) Calculate the instantaneous rate of reaction for a [HI] = 0.0020M 11

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INTEGRATED RATE LAWS

• Single Reactant (three cases)– Zero-Order Rate Law (n = 0)– First-Order Rate Law (n = 1)– Second-Order Rate Law (n = 2)

• More than one Reactant– Must state the order of the reaction with respect

to each reactant (rate = k[A]n[B]m[C]p)

A → B

rate = k[A]n

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A k ⏐ → ⏐ B

rate = −d[A]

dt= k[A]n

rate =d[A]

[A]nAo

A

∫ = −k dtt=0

t

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INTEGRATED RATE LAWSn=0,1,2

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In the real world, if we do not know the order of the reaction we can use experimental plots

to estimate the order.If a plot of [A] vs t is a straight line, then the reaction is zero order.

If a plot of ln[A] vs t is a straight line, then the reaction is 1st order.

If a plot of 1/ [A] vs t is a straight line, then the reaction is 2nd order.

First Order Reaction

-6

-5

-4

-3

-2

-1

0

0.0E+00 2.0E+04 4.0E+04 6.0E+04

Time (in seconds)

ln [

A}

(in

mo

l /

L

Second Order Reaction

0

50

100

150

200

250

0 500 1000

Time (in seconds)1/

[A] (

L /m

ol)

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INTEGRATED RATE LAWS

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for zero order (n = 0)

A k ⏐ → ⏐ products

rate = −d[A]

dt= k[A]0

Zero Order Reactions

[A] - [Ao] = -kt

Graph [A] vs t

Slope = -k, intercept = [Ao]

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INTEGRATED RATE LAWS

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for first order (n =1)

A k ⏐ → ⏐ products

rate = −d[A]

dt= k[A]1

First Order Reactions

ln[A]-ln[Ao] = -kt

Graph ln[A] vs t

Slope = -k, intercept = [Ao]

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ktAA 0]ln[]ln[

In[N205] versus time.

Slope = - k

2 N2O5 (g) → 4 NO2 (g) + O2 (g)

This graph gives a straight line, and so is

First order with respect to the decomposition of N2O5

If a plot of ln[A] vs t is not a straight line, the

reaction is not first order!

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Dimerization

Data set provided [C4H6] vs time

2 C4H6 (g) → C8H12 (g) [C4H6]˚ = 0.01M

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• Most reactions proceed not through a single step but through a series of steps

• Each Step is called an elementary reaction

Types of elementary reactions1. Unimolecular (a single reactant)

E.g., A → B + C (a decomposition)

2. Bimolecular (most common type)E.g., A + B → products

3. Termolecular (less likely event)E.g., A + B + C → products

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Reaction Mechanisms

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(fast) CO NO CO NO 2.)

(slow) NO NO NO NO 1.)

223

322

Notice that NO3 is formed and consumed. This is called a __________________________________.Notice also that Step 1 is bimolecular and Step 2 is bimolecular

223

32

CO NO NO NO

CO NO 2NO

sum

CO NO CO NO overall 22

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CHEMICAL EQUILIBRIUM

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A direct connection exists between the equilibrium constant of a reaction and the rate constants.

a) at equilibrium: forward reaction rate = reverse reaction rate.

b) Keq = kf / kr (same as K = k1/k-1)

A ⇌ B

kf

kr

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REACTION MECHANISM & RATE LAWS

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(fast) F2NO k

F 2NO 2.)

(slow) FF2NOk

2F 2NO 1.)

2

1

F2

2NO 2

F 2

2NO

Typically with a reaction one of several elementary step reaction is the slowest step. This is called the Rate Determining Step (RDS)

Case #1: When the RDS occurs first, the first step is slow and determines the rate of the overall reaction.

Example 15.6

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Reaction Progress

Energy

F + NO2F

NO2+ F2

slow fast

NO2F

RDS theis 1 Step

(fast) F2NO k2

F 2NO 2.)

(slow) FF2NOk1

2F 2NO 1.)

F2

2NO 2

F 2

2NO

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Chem 1310 Spring 2009 stop here

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Reaction Progress

Energy

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

22

1-k

1k

22 2NO O 2NO

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Need to express [intermediates] in terms of other reactants

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

22

1-k

1 k

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]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

22

1-k

1k

Substituting for [N2O2] in the rate expression above

22 2NO O 2NO

reaction Overall

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Reaction Progress

Energy

N2O2 + O2

slow

fast

2NO2NO2

]][OO[Nk rate

(slow) 2NO O ON 2.)

(fast) ON NO NO 1.)

2222

2k2

222

22

1-k

1 k

][O[NO]Kk rate 22

12

22 2NO O 2NO

1

11 k

kK

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Reaction Progress

Energy

Reaction Mechanism

• Intermediates

• Transition states

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A MODEL FOR CHEMICAL KINETICS

RT

E a

eA k

Equation Arrenhius

k of units the has andconstant a is

factor" lexponentia-pre" the is A

and

moleper energy are units

energy n Activatiothe is a

E

where

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ave

ave

or

PV 2 = RT = (KE)3n

3(KE) = RT2

Chapter 5: The Kinetic Molecular Theory of Gases

M

3RTu2

The Meaning of Temperature: temperature is a measure of the average kinetic energy of the gas particles.

The Kelvin temperature of a gas is a measure of the random motions of the particles of gas. With higher temperature, greater motion.

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Chapter 5: Speed Distribution Curves

Maxwell-Boltzmann speed distribution

Temperature is a measure of the average kinetic energy of molecules when their speeds have Maxwell Boltzmann distribution. i.e., the molecules come to thermal equilibrium.

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Transition State, also called Activated Complex

Two requirements must be satisfied for reactants to collide successfully to rearrange to form products

1.

2.

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Consider two different temperatures.

1.) Collisions must have enough energy to produce a reaction. Not all collisions have enough energy to make product

Ecollision > Eact

RT

E a

eA k

Num

ber o

f col

lisio

ns

Distribution of velocities

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2 BrNO (g) → 2 NO (g) + Br2 (g)2.) Molecular Orientation

Relative orientations of the reactants must allow formation of any new bonds to produce products.

Orientation a or b lead to product,

c does not.

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Find the rate constant k at several temperatures. Plot of In(k) versus 1/T for the reaction

Aek RTEa

Equation Arrenhius

RTE

lnA lnk a

y = mx + b

Slope =

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Reaction Progress

Energy

Ear

Eaf

Transition State

The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products

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Reaction Progress

Energy

ΔE = Eaf - Ea

r

Ear

Eaf

Transition State

The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products

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CHEMICAL KINETICS

• Catalyst

• Inhibitor

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KINETICS OF CATALYSIS• A catalyst has no effect on the

thermodynamics of the overall reaction• It only provides a lower energy path• Examples

– Pt and Pd are typical catalysts for hydrogenation reactions (e.g., ethylene to ethane conversion)

– Enzymes act as catalysts• Phases

– Homogenous catalysis – the reactants and catalyst are in the same catalyst (gas or liquid phase)

– Heterogeneous catalysis – reaction occurs at the boundary of two different phases (a gas or liquid at the surface of a solid)

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Effect of a catalyst on the number of reaction-producing collisions.

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