Determining the Components of the Rate Equation

aA + bB yY + zZ

Rate k[A] [B]

The coefficients and components of the rate equationMust be found by experimentCannot be deduced from stoichiometryDo not necessarily tell us the physical mechanismAlso, the reaction orders can be fractional, negative, zero, or even not-definable

Rate LawsWe have seen how to obtain the differential form of rate lawsbased upon experimental observation.  As they involve derivatives,we must integrate the rate equations to obtain the timedependence of concentrations.We will do this for a few cases, all involving these empirical rate laws. Here  the 

rate law need not bear any relationship to the stoichiometry of the reaction. Our next step will be to understand the origin of the empirical laws.  This leads to the concept of elementary reactions where the reaction is direct 

and occurs (or nor) in a single encounter.  We will see how the more complex rate laws arise from multiple elementary processes.Understanding the nature of the elementary reactions and the role of 

elementary reactions in complex processes will occupythe remainder of the semester.

First Order Reactions, n=1A  X

Differential form: [A] k[A]t

dd

[A] [A]k[A] or t [A]

d d kdtd

Integrated form:

Integrating,

0

[ ]

[ ] 0 0

0

[A] [A] or ln[A] [A]

which says that

A t

Akt

d k dt kt

A A e

= ‐k[A]1

[A]

X X X X X X XX X X X X X XX X X X X X XX X X X X X XX X X X X X X

X X X X X X XX X X X X X XX X X X X X XX X X X X X XX X X X X X XX X X X X X X

X X X X X XX X X X X XX X X X X XX X X X X XX X X X X X

X X X X X X XX X X X X X XX X X X X X XX X X X X X XX X X X X X X

ConcepTest 1

What is the order of this reaction?

A. Zero orderB. First orderC. Second order

Different Types ofComposite Reactions

Simultaneous:A Y and A Z

Opposing:A B Z

Consecutive: A X Y

Consecutive with opposition:A X Y

In order for the system to be at complete equilibrium, thetime derivative of all species concentrations must be zero!

Kinetics for a Simple Consecutive Reaction

1

1

1 2

2 1

2 1 0

2 1 0

20

2 1

(2)

becomes

a "standard form" o.d.e. with solution

kt

kt

kt k t

d Xk X k A

dt

d Xk X k A e

dtd X

k X k A edt

kX e e Ak k

1 2 and Xk kA X Z 1

0k tA A e with

2 11 2

0 02 1

1k t ktk e k eZ A A X Ak k

and

Quasi Steady State Approximation

Consider the differential rate equations for [A], [X] and [Z].

11 10

2

1

2

2 1

2

(1)

0

or (

(2)

(3)

2.5)ktk kX

d Ak A

dtd X

k X k Adt

d Zk X

A

dt

A ek k

Let [A]t=0 = [A]0and [X]t=0 = [Z]t=0 = 0

1

0ktA A e

1 2 and Xk kA X Z

For (2) and (2.5) to be compatible , we must have (k1/k2) << 1

Make QSS approx

1

1

1 00

01

tkt

kt

Z k A e dt

e A

Compare Exact and Steady State Solutions for k2 = 20 k1

k1t

Sample ProblemGiven reaction with stoichiometry:

A + B Y + ZThe reaction follows the mechanism:

A X

X + B Y + ZObtain an expression for the rate using the steady state treatmenta. What is the rate determining step if the second reaction is slow

relative to the initial equilibrium?b. What is the rate determining step if the second reaction is rapid

relative to the initial equilibrium?

k1

k–1

k2

ConcepTest  1

A.  Rate = k [O3]2

B. Rate = k [O3] [O2]C.  Rate = k [O3]

2 [O2]–1

B. Rate = k [O3 ]–1 [O2]

Reaction:       2 O3 3 O2

Mechanism: O3 O + O2 (fast)

O + O3 2 O2        (slow)What is the rate equation for O3 loss?

k1k–1

k2

SolutionReaction:       2 O3 3 O2

Mechanism: O3 O + O2 (fast)

O + O3 2 O2        (slow)What is the rate equation for O3 loss?

k1k–1

k2

1 3 1 2

1 3

1 2

2

3 31 32 3

1 22 3

2

1 equilibrium :

with a slow second step, the equilibrium is maintained

Rate

Rxn k O k Ok Ok O

d O O

O

k

O

k OO k O k

dtO

k O O

Lindemann – Hinshelwood Mechanism

A + B → AB*Complex with excess energy

AB* → A + BGiven enough time, it will fall apart

AB*+ M → AB + M*

Unless a collisionremoves the excess energy

2

'

[A*] A + A A* + A [A]

[A*]A* + A A + A [A*][A]

[A*] A* P [A*]

a

a

b

d kdt

d kdt

d kdt

Lindemann – Hinshelwood MechanismConsider only one species A.  This is easy to generalize.

The rate law for formation of A* is (with s.s. approximation)

2 '[A*] A A A* A* 0a a bd k k k

dt

Lindemann – Hinshelwood Mechanism

2

'

AA*

Aa

b a

kk k

Given

2

'

[ ][P] A*A

a bb

b a

k k Ad kdt k k

With the result

This is not a first order reaction. However, if the rate for stabilization of A* is much greater than the unimolecular decay, A*A,ka’[A*][A] >> kb[A*]  or just ka’[A] >> kb , then

'

[ ][P] where k a beff eff

b a

k k Ad k Adt k k A

Such a reaction goes from 2nd order to 1st order as the pressure increases. This is a very common gas phase mechanism, as collisions supply (take away) energy.

'1 1a

eff a b a

kk k k k A

Lindemann – Hinshelwood Mechanism

O + NO NO2*NO2* + M NO2 + M*

Normally, bath gas dominates

Example3. The effective rate constant for a gaseous reaction which has a Lindemann–

Hinshelwood mechanism is 2.50 ´ 10‐4 s‐1 at 1.30 kPa and 2.10 ´ 10‐5 s‐1 at 12 Pa. Calculate the rate constant for the activation step in the mechanism.

First, we check to see if  the reaction is in the Lindemann mechanism high pressure limit.In that limit, the “rate constant would be simply kB and would be pressure independent.Is it? It is not, so we use the full Lindemann expression:

Now 

Subject matter:KineticsRates, rate laws, mechanisms, method of initial ratesArrhenius rate law – physical interpretationReaction paths, potential energy surfacesMolecular basis of reactionsElementary reactions, composite reactionsFast and slow rate approximations; steady state

rate determining steps; catalyzed reactions

Exam 3

Exam 3

~ 5 problems (weights given) – budget your time Closed book Don’t memorize formulas/constants

You will be given things you need Specifically, you will be given the integrated rate laws. However, you may well be asked to conjure a mechanism that is

with a given set of observations. You might well be asked to use the steady state approximation,and to explain why/when it is useful.

This exam will have some numeric problems, similar to the homeworkin complexity.If a problem seems lengthy, do another problem & come back later

Understanding homework will be quite useful

EXAM held in JILA Auditorium, Dec 1, 10 AM