Chemical Kinetics

Chapter 12

Chemical Kinetics

The area of chemistry that concerns reaction rates.

Spontaneity

• tendency for a reaction to occur.

• does not mean that the reaction will be fast!

• Diamonds will spontaneously change into graphite, but the process is so slow that it is not detectable.

• What would be 2 ways to speed up a chemical reaction?

Reaction Mechanism

• the steps by which a chemical process occurs.

• allows us to find ways to facilitate reactions.

• can be changed by the use of a catalyst.

Reaction Rate

Change in concentration (conc) of a reactant or product per unit time.

Rate = conc of A at time conc of A at time 2 1

2 1

t tt t

At

Reaction Rate

It is customary to work with positive reaction rates, so a negative sign is used in some cases to make the rate positive.

Rates determined over a period of time are called average rates.

Instantaneous rate equals the negative slope of the tangent line.

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0.000370s

O2

0.0025

0.005

0.0075

0.0100

0.0006

70s

0.0026

110 s

NO2

NO

50 100 150 200 250 300 350 400

Con

cent

ratio

ns (

mol

/L)

Time (s)

[NO2 ]

t

The concentrations of nitrogen dioxide, nitric oxide, oxygen plotted versus time.

12_1575

Time Time

(a) (b) (c)

Representation of the reaction of 2 NO2(g) ---> 2 NO(g) + O2(g).

a) t = 0 b) & c) with increased time, NO2 is changed intoNO and O2.

Rate Laws(differential)

Rate = k[NO2]n

k = rate constant

n = rate order

Types of Rate Laws

Differential Rate Law: expresses how rate depends on concentration.

Integrated Rate Law: expresses how concentration depends on time.

Rate Laws Summary• Differential rate law -- rate of a reaction depends

on concentration.• Integrated rate law -- concentration depends on

time.• Rate laws normally only involve concentrations

of reactants.• Experimental determination of either rate law is

sufficient.• Experimental convenience dictates which rate

law is determined experimentally.• Rate law for a reaction often indicates reaction

mechanism.

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.20

.40

.60

.80

1.00

400 800 1200 1600 2000

Rate = 5.4 x 10-4 mol/L.s

Rate = 2.7 x 10-4 mol/L.s[N

2O5]

(mol

/L)

Time (s)

Plot of the concentration of N2O5 as a function of time for thereaction 2N2O5(soln) → 4NO2(soln) + O2(g). Rate at 0.90 M is twice the rate at 0.45 M.

Method of Initial Rates

Initial Rate: the “instantaneous rate” just after the reaction begins.

The initial rate is determined in several experiments using different initial concentrations.

See Sample Exercise 12.1 on pages 570-571.

Overall Reaction Order

Sum of the order of each component in the rate law.

rate = k[H2SeO3][H+]2[I]3

The overall reaction order is 1 + 2 + 3 = 6.

First-Order Rate Law

Integrated first-order rate law is

ln[A] = kt + ln[A]o

y = mx + b

If a reaction is first-order, a plot of ln[A] versus time is a straight line.

Rate = A

A

t

k

For For aaA A Products in a 1st-order reaction, Products in a 1st-order reaction,

Half-Life of a First-Order Reaction

t1/2 = half-life of the reaction

k = rate constant

For a first-order reaction, the half-life does not depend on concentration.

tk1/2

0 693.

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[N2O5]0 0.1000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0.0700

0.0800

0.0900

[N2O5]02

[N2O5]04

[N2O5]08

50 150 250 350100 200 300 400

t1/2 t1/2 t1/2

Time (s)

[N2O

5](m

ol/L

)

Plot of [N2O5] versus time for the decomposition reaction of N2O5. Note that the half-life for a 1st order reaction isconstant.

Second-Order Rate Law

For aA products in a second-order reaction,

Integrated rate law is

Rate = A

A

t

k 2

1

A +

1

A o

kt

y = mx + bIf a reaction is second-order, a plot of 1/[A]

versus time is a straight line.

Half-Life of a Second-Order Reaction

t1/2 = half-life of the reaction

k = rate constant

Ao = initial concentration of A

The half-life is dependent upon the initial concentration.

tk1/2

oA

1

Zero-Order Rate LawFor aA---> products in zero-order reaction,

Rate = k[A]o = k

The integrated rate law is

[A] = -kt + [A]0

y = mx + b

If a reaction is zero-order, the plot of [A] versus time is a straight line.

Example -- surface of a solid catalyst cannot hold a greater concentration of reactant.

12_06T

Table 12.6 Summary of the Kinetics for Reactions of the Type aA Products That AreZero, First, or Second Order in [A]

Order

Zero First Second

Rate law Rate = k Rate = k[A] Rate = k[A]2

Integrated rate law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 [A]1

= kt + [A]0

1

Plot needed to give a straight line [A] versus t ln[A] versus t [A]1

versus t

Relationship of rate constantSlope = -k Slope = -k Slope = kto the slope of straight line

Half-life

2kt1/2 =

[A]0k

t1/2 =0.693

k[A]0t1/2 =

1

KNOW THIS TABLE!!!!

Rate Law Analogy

v = 8 l o Rate = k[A]0 = k Zero order

v = vertices of cube

P = 12 l 1 Rate = k[A]1 First order

P = sum of lengths of edges of cube

A = 6 l 2 Rate = k[A]2 Second order

A= total surface area of cube

V = 1 l 3 Rate = k[A]3 Third order

V = volume of cube

A Summary

1. Simplification: Conditions are set such that only forward reaction is important.

2. Two types: differential rate law integrated rate law

3. Which type? Depends on the type of data collected - differential and integrated forms can be interconverted.

A Summary (continued)

4. Most common: method of initial rates.

5. Concentration v. time: used to determine integrated rate law, often graphically.

6. For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions).

Reaction Mechanism

- The series of steps by which a chemical reaction occurs.

- A chemical equation does not tell us how reactants become products - it is a summary of the overall process.

Reaction Mechanism (continued)

The reaction

has many steps in the reaction mechanism.

6CO 6H O C H O O2 2light

6 12 6 2 6

Often Used Terms

Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product.

Molecularity: the number of species that must collide to produce the reaction indicated by that step.

Elementary Step: A reaction whose rate law can be written from its molecularity.

uni, bi and termolecular

Reaction Mechanism Requirements

1. The sum of the elementary steps must give the overall balanced equation for the reaction.

2. The mechanism must agree with the experimentally determined rate law.

Rate-Determining Step

In a multistep reaction, it is the slowest step. It therefore determines the rate of reaction.

Fast Equilibrium Reaction

When the 1st step is not the slow step, but a fast equilibrium, the rate can be determined as follows:

2A + B → C 2nd order in B, 1st order in A.

Step 1 A + B ↔ D (Fast equilibrium)

Step 2 D + B → E Slow

Step 3 E + A → C + B Fast

A + B ↔ D

D + B → E

E + A → C + B

2 A + B → C

1st requirement that elementary stepsequal overall reaction is met.

Fast Equilibrium Reaction(continued)

kf[A][B] = kr[D] (fast)

[D] = ([A][B])

Rate = k2[D][B] (slow)

Substitute fast equation in terms of [D] into slow reaction.

Rate = ([A][B][B])

Rate = k[A][B]2

2nd requirement is also met, this is, then, a possible mechanism.

f

r

k

k

2 f

r

k k

k

Collision Model

Key Idea: Molecules must collide to react.

However, only a small fraction of collisions produces a reaction. Why?

Arrhenius: An activation energy must be overcome.

a) The change in potential energy as a function of reaction progress.Ea is the activation energy and E is the net energy change --exothermic. b) Molecular representation of the reaction.

Three possible collision orientations-- a) & b) produce reactions,while c) does not.

Activation Energy, Ea

Activation energy for a given reaction is a constant and not temperature dependent.

The rate constant (k) is temperature dependent.

Arrhenius Equation

- Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy).

- Orientation of reactants must allow formation of new bonds.

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T1

T2

00

EaEnergy

T2 > T1

Plot showing the number of collisions with a particular energy at T1& T2, where T2 > T1 -- Boltzman Distribution.

Arrhenius Equation (continued)

k = rate constant

A = frequency factor

Ea = activation energy

T = temperature

R = gas constant

k Ae E RT a /

Arrhenius Equation

If the natural logarithm of each side of the Arrhenius Equation is taken, the following equation results:

ln(k) = + ln(A)y = mx + b

m = when ln(k) is plotted versus .

See Sample Exercise 12.7.

a-E

R

a-E

R

1

T

1

T

CatalysisCatalyst: A substance that speeds up a reaction without being consumed

Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.

Homogeneous catalyst: Present in the same phase as the reacting molecules.

Heterogeneous catalyst: Present in a different phase than the reacting molecules.

12_303

E

Reactants

Products

Catalyzedpathway

Uncatalyzedpathway

Reaction progress

Ene

rgy

Energy plots for a catalyzed and an uncatalyzed pathway for an endothermic reaction.

12_304

Ea (uncatalyzed )

Effectivecollisions(uncatalyzed)

Effectivecollisions(catalyzed)

Ea (catalyzed )

(a) (b)

Nu

mbe

r of

col

lisio

nsw

ith a

giv

en e

nerg

y

Nu

mbe

r of

col

lisio

nsw

ith a

giv

en e

nerg

y

Energy Energy

Effect of a catalyst on the number of reaction-producingcollisions. A greater fraction of collisions are effective for the catalyzed reaction.

Heterogeneous Catalysis

1. Adsorption and activation of the reactants.

2. Migration of the adsorbed reactants on the surface.

3. Reaction of the adsorbed substances.

4. Escape, or desorption, of the products.

Steps:Steps:

Homogeneous Catalysis

Catalyst is in the same phase as the reacting molecules.

NO(g) + O2(g) → NO2(g)

NO2(g) → NO(g) + O(g)

O2(g) + O(g) → O3(g)

O2(g) → O3(g)

What is the catalyst in this reaction?

What are the intermediates?

3

2

1

2

NO

NO2 & O

Final Notes on Kinetics

• As T increases, so does k.

• Ea & E are independent of T.

• A catalyst lowers Ea and increases the rate of both kf & kr.