Chemistry 213: Course Outline Chemical Kinetics Acid-Base Equilibria Chemical Equilibrium Chemistry of the Environment Chemical Thermodynamics Organic/Biological

Chemistry 213: Course OutlineChemistry 213: Course Outline

Chemical KineticsChemical Kinetics

Acid-Base EquilibriaAcid-Base Equilibria Chemical EquilibriumChemical Equilibrium Nuclear ChemistryNuclear Chemistry

Chemistry of the EnvironmentChemistry of

the Environment

Chemical Thermodynamics

Chemical Thermodynamics

Organic/BiologicalChemistry

Organic/BiologicalChemistry

ElectrochemistryElectrochemistryCoordination Compounds

Coordination Compounds

Spring 2013


Rates of Reaction

Zeroth-Order First-Order Second-Order

Rate Laws(IRL, DRL, t1/2)

Temperature KM-Model Arrhenius

Catalysis Mechanism Nuclear Chemistry

• Kinetics => study of how fast chemical reactions occur.• Four important factors which affect rates of reactions:

– reactant concentration,

– temperature,

– action of catalysts, and

– surface area.

• Goal: to understand chemical reactions at the molecular level (Mechanisms)

Factors that Affect Reaction RatesFactors that Affect Reaction RatesFactors that Affect Reaction RatesFactors that Affect Reaction Rates

• Speed of a reaction is measured by the change in concentration with time.

• For a reaction A B

Reaction RatesReaction RatesReaction RatesReaction Rates

t

B of moles

in time changeB of moles ofnumber in change

rate Average

• For the reaction A B there are two ways of measuring rate:– the speed at which the products appear (i.e. change in moles of B

per unit time), or

– the speed at which the reactants disappear (i.e. the change in moles of A per unit time).


t

A of molesA respect to with rate Average

t

B of molesB respect to with rate Average

Change of Rate with Time• Most useful units for rates are to look at molarity. Since

volume is constant, molarity and moles are directly proportional.

• Consider:

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)


Change of Rate with Time


– calculate average rate in terms of the disappearance of C4H9Cl.

– Units for average rate: mol/L·s or mol L-1 s-1 or M/s or M s-1 .– The average rate decreases with time.

– Plot [C4H9Cl] versus time.

– The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve.

– Instantaneous rate is different from average rate.– We usually call the instantaneous rate the rate.


Time / s [C4H9Cl] / M Avg Rate / M s-1

0.0 0.1000 -[(Mf-Mi)/(tf-ti)]

50.0 0.0905 1.9E-04100.0 0.0820 1.7E-04150.0 0.0741 1.6E-04200.0 0.0671 1.4E-04300.0 0.0549 1.22E-04400.0 0.0448 1.0E-04500.0 0.0368 8.0E-05800.0 0.0200 5.6E-05

10,000.0 0.0000

HomeWork - Friday Jan. 18, 2013.EXCEL graph and average rateRed Trend Line obtained without last point.Bonus Point towards Quiz Total 0.0000

0.0100

0.0200

0.0300

0.0400

0.0500

0.0600

0.0700

0.0800

0.0900

0.1000

0.0 100.0 200.0 300.0 400.0 500.0 600.0 700.0 800.0 900.0

[C4H

9Cl ]

(M)

Time (s)

Reaction of Chlorobutane

Reaction Rate and Stoichiometry• For the reaction


we know

• In general for

aA + bB cC + dD


tt

OHHCClHCRate 9494

tdtctbta

D1C1B1A1Rate

• In general rates increase as concentrations increase.

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

Concentration and RateConcentration and RateConcentration and RateConcentration and Rate

• For the reaction

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

we note – as [NH4

+] doubles with [NO2-] constant, the rate doubles,

– as [NO2-] doubles with [NH4

+] constant, the rate doubles,

– We conclude rate [NH4+][NO2

-].

• Rate law:

• The constant k is the rate constant.


]NO][NH[Rate 24k

Exponents in the Rate Law• For a general reaction with rate law

we say the reaction is mth order in reactant 1 and nth order in reactant 2.

• The overall order of reaction is m + n + ….• A reaction can be zeroth order if m, n, … are zero.• Note the values of the exponents (orders) have to be determined

experimentally. They are not simply related to stoichiometry.


nmk ]2reactant []1reactant [Rate

Using Initial Rates to Determine Rate LawsUsing Initial Rates to Determine Rate LawsUsing Initial Rates to Determine Rate LawsUsing Initial Rates to Determine Rate Laws

Given data: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)

Expt. # [NO] / M [H2] / M Rate / M s-1

1 0.10 0.10 1.23x10-3

2 0.10 0.20 2.46x10-3

3 0.20 0.10 4.92x10-3

Determine Rate Law for the reaction.

i.e. Rate = k [NO]x [H2]y ; Find x , y , and k .



1 0.10 0.10 1.23x10-3

2 0.10 0.20 2.46x10-3

3 0.20 0.10 4.92x10-3



1 0.10 0.10 1.23x10-3

2 0.10 0.20 2.46x10-3

3 0.20 0.10 4.92x10-3

First-Order Reactions• Goal: convert rate law into a convenient equation to give

concentrations as a function of time.• For a first-order reaction, the rate doubles as the

concentration of a reactant doubles.– Plot [C4H9Cl] versus time.

The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time

tkot

t

t

eAA

kt

kt

kt

][][

A

Aln

AlnAln

]A[A][

Rate

0

0

Simulation

tkot

t

tt

eAA

kt

ktorkt

kt

][][

A

Aln

AlnAlnAlnAln

]A[A][

Rate

0

00

tkot

t

tt

eAA

kt

ktorkt

kt

][][

A

Aln

AlnAlnAlnAln

]A[A][

Rate

0

00

First-Order Reactions


0AlnAln ktt

First-Order Reactions

• The first-order rate constant for the decomposition of a certain insecticide in water at 12oC is 1.45 yr-1 . A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10-7 g/cm3 of water. Assume that the average temperature of the lake is 12oC.

• (A) What is the concentration of the insecticide on June 1 of the following year?

• (B) How long would it take for the concentration of the insecticide to drop to 3.0x10-7 g/cm3 ?


0AlnAln ktt

The first-order rate constant for the decomposition of a certain insecticide in water at 12oC is 1.45 yr-1 . A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10-7 g/cm3 of water. Assume that the average temperature of the lake is 12oC.(A) What is the concentration of the insecticide on June 1 of the following year?(B) How long would it take for the concentration of the insecticide to drop to 3.0x10-7 g/cm3 ?

Second-Order Reactions• For a second-order reaction with just one reactant

• A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0

• For a second-order reaction, a plot of ln[A]t vs. t is not linear.


0

2

A

1

A

1

][][

kt

Akt

ARate

t

Second-Order Reactions


0A1

A1 kt

t

Second-Order Reactions


0A1

A1 kt

t

The NO2 reaction has a rate constant of 0.543 M-1 s-1 . If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining concentration after 0.500 hr ?

The NO2 reaction has a rate constant of 0.543 M-1 s-1 . If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining concentration after 0.500 hr ?

Zeroth-Order Reactions• For a zeroth-order reaction with just one reactant

• A plot of [A]t versus t is a straight line with slope -k and intercept [A]0

• Applicable to catalysis on metal surfaces.


0

0

][][

][][

AktA

kAkt

ARate

t

Zeroth-Order Reactions

• A zeroth-order reaction has a rate constant of 1.1x10-7 M s-1 . The reaction began with a reactant concentration of 0.0200 M . What is the fraction of reactant concentration remaining after 45.0 hr ?


0

0

][][

][][

AktA

kAkt

ARate

t

•A zeroth-order reaction has a rate constant of 1.1x10-7 M s-1 . The reaction began with a reactant concentration of 0.0200 M . What is the fraction of reactant concentration remaining after 45.0 hr ?

Solution Key

Half-Life• Half-life is the time taken for the concentration of a

reactant to drop to half its original value.

• For a first-order process, half life, t½ is the time taken for [A]0 to reach ½[A]0.

• Mathematically,


kk

t693.0ln

21

21

Half-Life• For a second-order reaction, half-life depends on the

initial concentration:

• For a zeroth-order reaction:


0A

12

1

kt

k

t

2

A 0

21

Summary of Rate LawsSummary of Rate Laws

First-Order Second-Order Zeroth-Order

DRL

(-Δ[A]/Δt)k[A] k[A]2 k

IRL[A]t = [A]oe-kt

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

1/[A]t = kt + 1/[A]o [A]t = -kt + [A]o

Linear Equation

ln[A]t vs. t 1/[A]t vs. t [A]t vs. t

Linear Plot

Half-Life ln(2)/k 1/k[A]o [A]o/2k

Units on k time-1 M-1 time-1 M time-1

m = -k

b = ln[A]o

m = k

b = 1/[A]o

m = -k

b = [A]o

Temperature and RateTemperature and RateTemperature and RateTemperature and Rate

The Collision Model

• As temperature increases, the rate increases.

The Collision Model• The collision model: in order for molecules to react they must

collide.• The greater the number of collisions the faster the rate.• The more molecules present, the greater the probability of

collision and the faster the rate.• The higher the temperature, the more energy available to the

molecules and the faster the rate.• Complications: not all collisions lead to products. In fact, only

a small fraction of collisions lead to product.


The Orientation Factor


Activation Energy• Arrhenius: molecules must posses a minimum amount of

energy to react. Why?– In order to form products, bonds must be broken in the

reactants.

– Bond breakage requires energy.

• Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.


Activation Energy


The Arrhenius Equation• Arrhenius discovered most reaction-rate data obeyed the

Arrhenius equation:

– k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.

– A is called the frequency factor.

– A is a measure of the probability of a favorable collision.

– Both A and Ea are specific to a given reaction.


RTEa

eAk

Determining the Activation Energy

• If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation:

• From the above equation, a plot of ln k versus 1/T will have slope of –Ea/R and intercept of ln A.


ARTE

k a lnln


Determining the Activation Energy• If we do not have a lot of data, then we recognize


122

1

2121

22

11

11ln

lnlnlnln

lnln and lnln

TTRE

kk

ARTE

ARTE

kk

ARTE

kARTE

k

a

aa

aa

Ea ~ 160 kJ/mol (previous slide of ln(k) versus 1/T plot)

Ea ~ 160 kJ/mol (previous slide of ln(k) versus 1/T plot)

• The balanced chemical equation provides information about the beginning and end of reaction.

• The reaction mechanism gives the path of the reaction.• Mechanisms provide a very detailed picture of which bonds

are broken and formed during the course of a reaction.

Elementary Steps• Elementary step: any process that occurs in a single step.

Reaction MechanismsReaction MechanismsReaction MechanismsReaction Mechanisms

Rate Laws for Elementary Steps

Reaction MechanismsReaction MechanismsReaction MechanismsReaction Mechanisms

• A catalyst changes the rate of a chemical reaction.

• There are two types of catalyst:– homogeneous, and

– heterogeneous.

• Chlorine atoms are catalysts for the destruction of ozone.

CatalysisCatalysisCatalysisCatalysis


Enzymes


Nuclear Equations• Nucleons: particles in the nucleus:

– p+: proton

– n0: neutron.

• Mass number: the number of p+ + n0.• Atomic number: the number of p+.• Isotopes: have the same number of p+ and different numbers of n0.• In nuclear equations, number of nucleons is conserved:

23892U 234

90Th + 42He

RadioactivityRadioactivityRadioactivityRadioactivity

Types of Radioactive Decay


Types of Radioactive Decay


Neutron-to-Proton Ratio• The heavier the nucleus,

the more neutrons are required for stability.

• The belt of stability deviates from a 1:1 neutron to proton ratio for high atomic mass.

Radioactive SeriesFor 238U, the first decay is to 234Th (-decay). The 234Th undergoes -emission to 234Pa and 234U. 234U undergoes -decay (several times) to 230Th, 226Ra, 222Rn, 218Po, and 214Pb. 214Pb undergoes -emission (twice) via 214Bi to 214Po which undergoes -decay to 210Pb. The 210Pb undergoes -emission to 210Bi and 210Po which decays () to the stable 206Pb.

Patterns of Patterns of Nuclear StabilityNuclear Stability

• 90Sr has a half-life of 28.8 yr. If 10 g of sample is present at t = 0, then 5.0 g is present after 28.8 years, 2.5 g after 57.6 years, etc. 90Sr decays as follows

9038Sr 90

39Y + 0-1e

• Each isotope has a characteristic half-life.• Half-lives are not affected by temperature, pressure or chemical

composition.• Natural radioisotopes tend to have longer half-lives than synthetic

radioisotopes.

Rates of Radioactive DecayRates of Radioactive DecayRates of Radioactive DecayRates of Radioactive Decay


• Half-lives can range from fractions of a second to millions of years.

• Naturally occurring radioisotopes can be used to determine how old a sample is.

• This process is radioactive dating.


Dating• Carbon-14 is used to determine the ages of organic compounds

because half-lives are constant.• We assume the ratio of 12C to 14C has been constant over time.• For us to detect 14C the object must be less than 50,000 years

old.• The half-life of 14C is 5,730 years.• It undergoes decay to 14N via -emission:

146C 14

7N + 0-1e


Calculations Based on Half Life• Radioactive decay is a first order process:

• In radioactive decay the constant, k, is the decay constant.• The rate of decay is called activity (disintegrations per

unit time).

• If N0 is the initial number of nuclei and Nt is the number of nuclei at time t, then

Rates of Radioactive DecayRates of Radioactive Decay

kNRate

ktNNt

0ln

Calculations Based on Half Life

• With the definition of half-life (the time taken for Nt = ½N0), we obtain


21

)2ln(tk kt

NNt

0ln

A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 s-1 . The half-life of 14C is 5715 yr. What is the age of the archeological sample? [Answer: 2,229 yr]

A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 s-1 . The half-life of 14C is 5715 yr. What is the age of the archeological sample? [Answer: 2,229 yr]

HW Key


Rates of Reaction

Zeroth-Order First-Order Second-Order

Rate Laws(IRL, DRL, t1/2)

Temperature KM-Model Arrhenius

Catalysis Mechanism Nuclear Chemistry

t

molesRateReaction

[A]t = [A]oe-kt1/[A]t = kt + 1/[A]t[A]t = -kt + [A]o

RTEa

eAk

ktNN ot )/ln(

Documents

Chemistry 213: Course Outline Chemical Kinetics Acid-Base Equilibria Chemical Equilibrium Chemistry of the Environment Chemical Thermodynamics Organic/Biological