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

13.3

The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.A. Determine the molarity of an antiobiotic solution that sits for 1.00 month, if its original concentration was 5.00 x 10-3 M?

ln[A] = ln[A]0 - kt [A]0 = 5.00 x 10-3 M t = (1.00 month)(1 yr/12 months) = 0.0833 yr k = 1.29 yr-1

13.3

What is the order of the reaction? Units on rate constant tell us it is first order

ln[A] = ln (5.00 x 10-3 M) - (1.29 yr-1)(0.0833 yr)

ln[A] = -5.397take antilog

[A] = 4.53 x 10-3 M

The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.B. If the antiobiotic solution is considered to be “no longer effective” at concentrations lower than 5.00 x 10-4 M, how long will the solution be effective?

ln [A]0 = kt [A] t

[A]0 = 5.00 x 10-3 M[A]t = 5.00 x 10-4 M k = 1.29 yr-1

13.3

first order reaction

ln (5.00 x 10-3 M/ 5.00 x 10-4 M) = (1.29 yr-1) t

t = ln(10)/ (1.29 yr-1 ) = 1.78 yr

The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.C. A 70. % depletion of the initial concentration of the antiobiotic solution results in a significantly lower effectivness. How long before this lower effectiveness will be noticed?

ln [A]0 = kt [A] t

[A]t = 0.30 [A]0 k = 1.29 yr-1

13.3

first order reaction

t = ln(3.3)/ (1.29 yr-1 ) = 0.933 yr

ln [A]0 = (1.29 yr-1) t

0.30[A]0

The decomposition of a certain antiobiotic in water has a rate constant at 20°C of 1.29 yr-1.D. Determine the half-life of the antiobiotic.

t 1/2 = ln 2 / k k = 1.29 yr-1

13.3

first order reaction

t 1/2 = ln(2)/ (1.29 yr-1 ) = 0.537 yr

A second order decomposition reaction takes 2.54 x 103 seconds for the initial concenteration to fall to one half of its origianal value. What is the value of the rate constant if the initial concentration was 6.23 x 10-2 M?.

t 1/2 = 1 / k [A]0 t 1/2 = 2.54 x 103 seconds

[A]0 = 6.23 x 10-2 M

13.3

Second order reaction

k = 1 / (6.23 x 10-2 M)(2.54 x 103 seconds) = 6.32 x 10-3 /Ms

k = 1 / [A]0 t 1/2

A + B C + D

Exothermic Reaction Endothermic Reaction

The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.

13.4

Temperature Dependence of the Rate Constant

k = A • exp( -Ea/RT )

Ea is the activation energy (J/mol)

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature

A is the frequency factor

lnk = -Ea

R1T

+ lnA

(Arrhenius equation)

13.4

13.4

lnk = -Ea

R1T

+ lnA

13.5

Reaction Mechanisms

The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the reaction mechanism.

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

N2O2 is detected during the reaction!

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

13.5

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.

An intermediate is always formed in an early elementary step and consumed in a later elementary step.

The molecularity of a reaction is the number of molecules reacting in an elementary step.

• Unimolecular reaction – elementary step with 1 molecule

• Bimolecular reaction – elementary step with 2 molecules

• Termolecular reaction – elementary step with 3 molecules

Unimolecular reaction A products rate = k [A]

Bimolecular reaction A + B products rate = k [A][B]

Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

13.5

Writing plausible reaction mechanisms:

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

• The rate-determining step should predict the same rate law that is determined experimentally.

The rate-determining step is the slowest step in the sequence of steps leading to product formation.

The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:

Step 1: NO2 + NO2 NO + NO3

Step 2: NO3 + CO NO2 + CO2

What is the equation for the overall reaction?

NO2+ CO NO + CO2

What is the intermediate?

NO3

What can you say about the relative rates of steps 1 and 2?

rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2

13.5

But what happens if the first step is NOT the rate determining step?

1: NO + O2 NO 3 fast2: NO3 NO + O2 fast3: NO3 + NO 2 NO2 slow

rate = k3 [NO3][NO], but NO3 is an intermediateSince the first two steps are fast,

the rate for creating NO3 = rate for destroying NO3

k1 [NO] [O2] = k2 [NO3] + k3 [NO3][NO] negligibly small rate => very small number

solve for [NO3]: [NO3] = k1 [NO] [O2] k2

substitute into rate expression: rate = k1k3 [NO]2 [O2] k2

=> observed rate = kobs [NO]2 [O2]

What does the potential energy diagram look like for a reaction with multiple steps ?

Po

t en

tial E

ne

rgy

Reaction Progress

A + B

C + D

Ea

Po

t en

tial E

ne

rgy

Reaction Progress

A + B

C + D

Ea

Reaction with first step as slow step

Reaction with last step as slow step

Each hill represents a reactionAny dip or valley represents an intermediate

Consider the following mechanism for a gas phase reaction:

1: Cl2 Cl . fast2: Cl . + CHCl3 HCl + .CCl3 slow3: Cl . + .CCl3 CCl4 fast

A. What is the overall reaction?

Cl2 + CHCl3 HCl + CCl4

B. What are the intermediates in the reaction?Cl . & .CCl3

C. What is the molecularity of each step in the mechanism? Step 1 = unimolecular; Steps 2 & 3 = bimolecular

D. What is the rate determining step?Step 2 = slow step = rate determing step

E. What is the rate law predicted by this mechanism?Step 2 = slow step = rate determing step

rate = k2 [Cl .][CHCl3 ]

But Cl . is an intermediate formed in Step 1 and , thus, cannot appear in the rate law.

Cl . is in equilibrium with Cl2By definition, this means that the forward and reverse rates of

this reaction are equal.k1 [Cl2 ] = k-1[Cl .]2

Solving for [Cl .] in terms of [Cl2],

[Cl .]2 = k1/ k-1 [Cl2 ] => [Cl .] = {k1/ k-1 [Cl2 ]} 1/2

Substituting into the overall rate law:rate = k2 {k1/ k-1 [Cl2 ]} 1/2 [CHCl3 ]

rate = kobs [Cl2 ]1/2 [CHCl3 ]

overall reaction order is 3/2

Chain Reactions: a series of reactions in which one reaction initiates the next

Initiation: I2 2 I . slow - produces reactive intermediate

Propagation: I . + Br2 IBr + Br . fast Br . + I2 IBr + I . fast

Termination: Br . + Br . Br2 rare reactions since

I . + I . I2 concentrations of species Br . + I . IBr2 are small

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.

k = A • exp( -Ea/RT ) Ea k

uncatalyzed catalyzed

ratecatalyzed > rateuncatalyzed

Ea < Ea‘ 13.6

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 catalyses

• Base catalyses

13.6

A catalyst can change the mechanism of a reaction, but it will not cause a reaction to proceed that is not favorable.

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

catalyst

Haber Process

13.6

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)

13.6

Catalytic Converters

13.6

CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic

converter

NO + NO2 N2 + O2

catalyticconverter

Enzyme Catalysis

13.6

uncatalyzedenzyme

catalyzed

13.6