Chemical KineticsChapter 13Dr. Ali Bumajdad

Chapter 13 Topics

• Rate of a Reaction• Reaction Rates and Stoichiometry• The Rate Law• Relationship between Reactant Concentration and Time1)First-order reactions2)Second-ordered reactions3)Zero-ordered reactions• The Effect of Activation Energy and Temperature on

Rate• Collision Theory• The Arhenius Equation • Finding Ea from graph and from equation• Catalysis

Chemical Kinetics

Dr. Ali Bumajdad

Chemical Kinetics = Chemistry and Time

•Thermodynamics – does a reaction take place?

•Kinetics – how fast does a reaction proceed?

- Why it is important:

To answer questions like:1) How quickly a medicine is able to work?2) How fast the depletion of Ozone?3) How rapidly food spoils?4) How fast steel rust?

•Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

A B

rate = -[A]t

rate = [B]t

[A] = change in concentration of A over time period t

[B] = change in concentration of B over time period t

Because [A] decreases with time, [A] is negative.

Rate of a reaction

A B

rate = -[A]t

rate = [B]t

time

•Kinetics can be studied by following a change in property as a function of time

This property could be:

1) UV absorption2) Pressure3) Conductivity4) Concentration5) Gas volume

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

time

393 nmlight

Detector

[Br2] Absorption

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

slope oftangent

slope oftangent slope of

tangent

average rate = -[Br2]t

= -[Br2]final – [Br2]initial

tfinal - tinitial

instantaneous rate = rate for specific instance in time(slope of a tangent)

rate [Br2]

rate = k [Br2]

k = rate[Br2]

= rate constant

= 3.50 x 10-3 s-1

2H2O2 (aq) 2H2O (l) + O2 (g)

PV = nRT

P = RT = [O2]RTnV

[O2] = PRT1

rate = [O2]t RT

1 Pt=

measure P over time

Reaction Rates and Stoichiometry

2A B

Two moles of A disappear and one mole of B that is formed.

= [B]t

rate = -[A]t

12

aA + bB cC + dD

•In this case, the rate will be different if we consider the disappearance of A than if we consider the formation of B

•In order to make it the same we should divide by the coefficient

• In general for:

rate = -[A]t

1a

= -[B]t

1b

=[C]t

1c

=[D]t

1d

(1)

Q) Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

rate = -[CH4]

t= -

[O2]t

12

=[H2O]

t12

=[CO2]

tSa. Ex. a) How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation:

2O3 (g) 3O2 (g)

b) If at certain time the rate of appearance of O2, [O2]/t = 6.0 10-5 M/s what is the value of the rate of disappearance of O3, - [O3]/t at the same time.

4.0 10-5 M/s

The Rate Law•Rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers that can be determined by experiment only.

•Rate Law is different than the rate of reaction

aA + bB cC + dD

reaction is xth order in A

reaction is yth order in B

reaction is (x +y)th order overall

rate = -[A]t

1a

Rate = k [A]x[B]y (2)

Rate = k [A]x[B]y

Reactant Reactant not productnot product

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2]x[ClO2]y

When Double [F2] with [ClO2] constant

x = 1

Quadruple [ClO2] with [F2] constant

y = 1

rate = k [F2][ClO2]

Rate doubles

Rate quadruples

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

Rate Laws

1

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

Q) Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Experiment [S2O82-] [I-]

Initial Rate (M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

rate = k [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

y = 1

Double [S2O82-], rate doubles (experiment 2 & 3)

x = 1

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

First-Order Reactions

• For reaction of the type A product

rate = -[A]t

and rate = k [A]

[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=0

Relationship between Reactant Concentration and Time

and [A]t= k-

[A]

[A]t

= k [A]-

Reactions with first overall order

Using Calculus

ln[A] = ln[A]0 - kt (3a) (3b)[A] = [A]0exp(-kt)

2N2O5 4NO2 (g) + O2 (g)

k = rate[A]

= 1/s or s-1M/sM

=Unit of first order rate constant

ln[A] = ln[A]0 - kt (3a)

Y = b + mX

Where m = slope b = intercept

Q) The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?

kt = ln[A]0 – ln[A]

t =ln[A]0 – ln[A]

k= 66 s

[A]0 = 0.88 M

[A] = 0.14 M

ln[A]0

[A]

k=

ln0.88 M

0.14 M

2.8 x 10-2 s-1=

ln[A] = ln[A]0 - kt(3a)

• half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

ln[A]0

[A]0/2

k=t½

ln2k

=0.693

k=

0.693k

=ln2k

t1/2 =(4) Independent of

initial concentration

First-Order Reactions

1) It take the same time for the concentration of reactant to decrease from 0.1M to 0.05M as it does from 0.05M to 0.025M

2) If you know t1/2 you know k

A product

First-order reaction

# of half-lives [A] = [A]0/n

1

2

3

4

2

4

8

16

Q) What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order?

units of k (s-1)

• For reactions of the type A product

rate = -[A]t

and rate = k [A]2

[A]t

= k [A]2and -

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

t½ = t when [A] = [A]0/2

Second-Order Reactions Reactions with second overall order

k = Unit of second order rate constantrate[A]2 = 1/M•sM/s

M2=

Using Calculuswhere1

[A]=

1[A]0

+ kt(5)

t½ =1

k[A]0

(6)

A product rate = -[A]t

rate = k [A]0 = k

k = rate[A]0

= M/s[A]t

= k-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

t½ = t when [A] = [A]0/2

Zero-Order Reactions

[A] = [A]0 - kt (7)(7)

t½ =[A]0

2k(8)(8)

order Zero First Second

Rate law Rate = k Rate = k[A]

Rate = k[A]2

Integrated rate law [A] = -kt + [A]o

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

1 1

[A] [A]o

kt

Plot needed to give

a straight line [A] versus t In[A] versus t 1

[A]versus t

Relationship of k to the slope of straight

line

Slope = -k Slope = -k Slope = k

Half-life 1

2

[ ]

2oA

tk

1

2

0.693t

k 1

2

1

[ ]o

tk A

Summary of the Kinetics for reactions of type aA products

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A] = ln[A]0 - kt

1[A]

=1

[A]0

+ kt

[A] = [A]0 - kt

t½ln2k

=

t½ =[A]0

2k

t½ =1

k[A]0

k Unit

s-1

M-1 s-1

M s-1

ln[A] = ln[A]0 - kt (3a)

ln [A] = - kt (3c)

[A]0

(a)(a)

(b,c)(b,c)

• Plot of ln [A] versus t should be linear for first order reactionPlot of ln [A] versus t should be linear for first order reaction

• Since [A] Since [A] P (PV=nRT), then lnP =-kt +ln P P (PV=nRT), then lnP =-kt +ln P00

0.693k

=ln2k

t1/2 =(4)

1[A]

=1

[A]0

+ kt(5)

t½ =1

k[A]0

(6)

- is the minimum amount of energy

required to initiate a chemical reaction

The Effect of Activation Energy (Ea) and Temperature on Rate

-Usually rate increase by increasing T

- Rate increase by decreasing Ea

- Unit: J/mol

- Unit: K

Collision Theory

1) How reaction Occurs?By collisions between molecules or atoms

2) Does all collisions results in a reactions? No, a) The collision should be strong enough (more than the

activation energy or the reaction) b) The collision should be with the right orientation

3) What factor affect the speed (rate) of the reactions?

a) Collision frequency (A)b) Temperaturec) Activation energy (Ea)

Orientation effect

Exothermic Reaction Endothermic Reaction

A + B AB C + D++

Ea is the activation energy (J/mol)

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature (K)

A is the frequency and orientation factor

(Arrhenius equation)

The Arhenius Equation k = A • exp( -Ea / RT ) (9)

lnk = -Ea

R1T

+ lnA(10)

Y = m X + b

k T

k 1/Ea

lnk = -Ea

R1T

+ lnA

Y = m X + b

1) Finding Ea from graph

lnk1 =Ea

R1T1

lnA -

lnk2 =Ea

R1T2

lnA -

2) Finding Ea from Equation

Subtracting 2 from 1

(1)

(2)

1T2

lnk1 - lnk2 =Ea

R1T1

- Ea

R+

1T2

lnk1 - lnk2 =Ea

R1T1

-

1T2

ln k1 =Ea

R1T1

- k2

(11)

Sa. Ex.Sa. Ex.

lnk = -Ea

R1T

+ lnA(10)

Y = m X + b

1T2

ln k1 =Ea

R1T1

- k2

(11)

•Catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.

k = A • exp( -Ea / RT ) Ea k

ratecatalyzed > rateuncatalyzed

Ea < Ea‘

Uncatalyzed Catalyzed

Catalysis

In heterogeneous catalysis, the reactants and the catalysts are in different phases.

In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.

• Haber synthesis of ammonia

• Ostwald process for the production of nitric acid

• Catalytic converters

• Acid catalysis

• Base catalysis

N2 (g) + 3H2 (g) 2NH3 (g)Fe/Al2O3/K2O

catalyst

Haber Process

Ostwald Process

Hot Pt wire over NH3 solutionPt-Rh catalysts used

in Ostwald process

4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)Pt catalyst

2NO (g) + O2 (g) 2NO2 (g)

2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)

Catalytic Converters

CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic

converter

2NO + 2NO2 2N2 + 3O2

catalyticconverter

Enzyme Catalysis: very selective biological catalyst

uncatalyzedenzyme

catalyzed

rate = [P]t

rate = k [ES]