Chemistry 213: Course Outline Chemical Kinetics Acid-Base Equilibria Chemical Equilibrium Chemistry...
-
Upload
stanley-collins -
Category
Documents
-
view
231 -
download
3
Transcript of Chemistry 213: Course Outline Chemical Kinetics Acid-Base Equilibria Chemical Equilibrium Chemistry...
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
Chemical KineticsChemical Kinetics
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).
Reaction RatesReaction RatesReaction RatesReaction Rates
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)
Reaction RatesReaction RatesReaction RatesReaction Rates
Change of Rate with Time
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
– 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.
Reaction RatesReaction RatesReaction RatesReaction Rates
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
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
we know
• In general for
aA + bB cC + dD
Reaction RatesReaction RatesReaction RatesReaction Rates
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.
Concentration and RateConcentration and RateConcentration and RateConcentration and Rate
]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.
Concentration and RateConcentration and RateConcentration and RateConcentration and Rate
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 .
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
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
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
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
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 ?
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
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.
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
0
2
A
1
A
1
][][
kt
Akt
ARate
t
Second-Order Reactions
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
0A1
A1 kt
t
Second-Order Reactions
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
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.
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
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 ?
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
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,
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
kk
t693.0ln
21
21
Half-Life• For a second-order reaction, half-life depends on the
initial concentration:
• For a zeroth-order reaction:
The Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with TimeThe Change of Concentration with Time
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.
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
The Orientation Factor
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
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.
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
Activation Energy
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
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.
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
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.
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
ARTE
k a lnln
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
Determining the Activation Energy• If we do not have a lot of data, then we recognize
Temperature and RateTemperature and RateTemperature and RateTemperature and Rate
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
CatalysisCatalysisCatalysisCatalysis
Enzymes
CatalysisCatalysisCatalysisCatalysis
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
RadioactivityRadioactivityRadioactivityRadioactivity
Types of Radioactive Decay
RadioactivityRadioactivityRadioactivityRadioactivity
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
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.
Rates of Radioactive DecayRates of Radioactive DecayRates of Radioactive DecayRates of Radioactive Decay
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
Rates of Radioactive DecayRates of Radioactive DecayRates of Radioactive DecayRates of Radioactive Decay
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
Rates of Radioactive DecayRates of Radioactive DecayRates of Radioactive DecayRates of Radioactive Decay
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
Chemical KineticsChemical Kinetics
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(