Chapter 15 - Chapter 15 - Chemical Chemical Kinetics Kinetics Objectives: 1. Determine rates of reactions from graphs of concentration vs. time. 2. Recall the conditions which affect the rates. 3. Recognize the order of reaction, give the rate equation, calculate the rate constant. 4. Use integrated rate laws and half-life equations in calculations. 5. Draw energy diagrams, find activation energy. 6. Identify catalysts and their properties. 7. Identify reaction intermediates. 8. Recognize the rate equation given a mechanism, and given the rate equation determine the mechanism.
Chapter 15 - Chemical Kinetics. Objectives: Determine rates of reactions from graphs of concentration vs. time. Recall the conditions which affect the rates. Recognize the order of reaction, give the rate equation, calculate the rate constant. - PowerPoint PPT Presentation
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Chapter 15 - Chapter 15 - Chemical Chemical KineticsKinetics
Objectives:1. Determine rates of reactions from graphs of concentration
vs. time.2. Recall the conditions which affect the rates.3. Recognize the order of reaction, give the rate equation,
calculate the rate constant.4. Use integrated rate laws and half-life equations in
calculations.5. Draw energy diagrams, find activation energy.6. Identify catalysts and their properties.7. Identify reaction intermediates.8. Recognize the rate equation given a mechanism, and
given the rate equation determine the mechanism.
KINETICS KINETICS — the study of — the study of REACTION RATESREACTION RATES
H2O2 decomposition in an insect
H2O2 decomposition catalyzed by MnO2
Chemical engineering, enzymology, environmental engineering, etc.
Kinetics and MechanismsKinetics and Mechanisms
• KINETICS KINETICS — the study of REACTION — the study of REACTION RATES and their relation to the way the RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM.reaction proceeds, i.e., its MECHANISM.
• The reaction mechanism is our goal!The reaction mechanism is our goal!• The sequence of events at the molecular The sequence of events at the molecular
level that control the speed and outcome level that control the speed and outcome of a reaction.of a reaction.
Br from biomass burning destroys Br from biomass burning destroys stratospheric ozone. stratospheric ozone.
(See R.J. Cicerone, (See R.J. Cicerone, ScienceScience, volume 263, page 1243, 1994.), volume 263, page 1243, 1994.)
• Identify the intermediate• Identify the catalyst(s)
Reaction RatesReaction Rates
• Reaction rate = change in concentration of Reaction rate = change in concentration of a reactant or product with time.a reactant or product with time.
Change in distance over a period of time:
distance time
concentration timeRate =
• Three “types” of Three “types” of rates: rates:
• initial rateinitial rate• average rateaverage rate• instantaneous instantaneous raterate
Reaction Rates can be Reaction Rates can be determined from a Plotdetermined from a Plot
• Blue dye is oxidized with bleach. • Its concentration decreases with time.• The rate — the change in dye conc with time — can be determined from the plot.
• The initial rate (in this case over the first minute) is calculated from a tangent line crossing the initial concentration. Then the slope of the line is determined:
Reaction Rates can be Reaction Rates can be determined from a Plotdetermined from a Plot
N2O5 NO2 + O22 4• The average rate is
calculated from a time interval.
• The instantaneous rate is calculated at a single point in time (or given concentration) by drawing a tangent line crossing the point. Then the slope of the line is determined.
• Compare average rates at the beginning and end of reaction.
Factors affecting the RatesFactors affecting the Rates
• ConcentrationsConcentrations• Physical State of Reactants and Physical State of Reactants and
• To postulate a mechanism we study: - The reaction rate
and its - Concentration dependence.
• Generate a Rate Law Equation.
Rate LawsRate Laws
• In general for:
a A + b B --> x Xa A + b B --> x X
Rate = k [A]Rate = k [A]mm[B][B]nn
The exponents m, n The exponents m, n
•• are the are the ______________________________
•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions
•• must be determined by must be determined by ________________________!!
With a catalyst C
[C][C]pp
Interpreting Rate LawsInterpreting Rate LawsRate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp
• If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A
Rate = k [A]Rate = k [A]11
If [A] doubles, then rate goes up by factor of __________ If [A] doubles, then rate goes up by factor of __________ • If m = 2, rxn. is 2nd order in A.If m = 2, rxn. is 2nd order in A.
Rate = k [A]Rate = k [A]22
Doubling [A] increases rate by _____________Doubling [A] increases rate by _____________• If m = 0, rxn. is zero order.If m = 0, rxn. is zero order.
Rate = k [A]Rate = k [A]00
If [A] doubles, rate ____________________If [A] doubles, rate ____________________
Deriving Rate LawsDeriving Rate Laws
Derive rate law and k for Derive rate law and k for CHCH33CHO(g) --> CHCHO(g) --> CH44(g) + CO(g)(g) + CO(g)
from experimental data for rate of from experimental data for rate of disappearance of CHdisappearance of CH33CHOCHO
Expt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec)
1 0.10 0.020
2 0.20 0.081
3 0.30 0.182
4 0.40 0.318
Look at exp 1 and 2: concentration doubles; rate cuadruples
Rate = k [CH3CHO]n
4 Rate = k [2 CH3CHO]n
n = 2
Deriving Rate LawsDeriving Rate LawsRate of rxn = Rate of rxn =
Here the rate goes up by ________ when initial Here the rate goes up by ________ when initial concentration doubles. Therefore, we say this reaction is concentration doubles. Therefore, we say this reaction is __________ order.__________ order.
Now determine the value of k. Use any exp. Data:Now determine the value of k. Use any exp. Data:
Using k you can calculate rate at other values of [CH3CHO] at same T.
Concentration and TimeConcentration and Time• What is the concentration of reactant
as a function of time?• Consider First Order reactions:
Rate -[A]
time = k [A]
Integrating we get:Integrating we get:
[A] = - k tln[A]o
naturallogarithm [A] at time = 0
[A] = - k tln[A]o
naturallogarithm [A] at time = 0
Integrated First Order LawIntegrated First Order Law
[A] = - k tln[A]o
naturallogarithm [A] at time = 0
[A] = - k tln[A]o
naturallogarithm [A] at time = 0
[A] / [A][A] / [A]00 =fraction remaining after time t =fraction remaining after time t
has elapsed.has elapsed.
[A] / [A][A] / [A]00 =fraction remaining after time t =fraction remaining after time t
has elapsed.has elapsed.
The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 year-1 at 12oC. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm3. 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 will it take for the concentration of the insecticide to drop to 3.0 x 10-7 g/cm3?
Using Integrated Rate LawsUsing Integrated Rate Laws
All 1st order reactions have straight line plot All 1st order reactions have straight line plot for for ln [A] vs. timeln [A] vs. time. .
And 2nd order gives straight line for plot of And 2nd order gives straight line for plot of 1/[A] vs. time1/[A] vs. time..
Using Integrated Rate LawsUsing Integrated Rate Laws
• In an experiment for:
2 N2O5(g) ---> 4 NO2(g) + O2(g)
Time (min) [N2O5]0 (M) 0 1.001.0 0.7052.0 0.4975.0 0.173
[N2O5] vs. time
time
1
0
0 5
[N2O5] vs. time
time
1
0
0 5
Data of conc. vs. time plot do not fit straight line.
If it were zero order:
Using Integrated Rate LawsUsing Integrated Rate Laws
ln [N2O5] vs. time
time
0
-2
0 5
ln [N2O5] vs. time
time
0
-2
0 5
Plot of ln [N2O5] vs. time is a straight line!
ln [Nln [N22OO55]]00
00-0.35-0.35-0.70-0.70-1.75-1.75
• In an experiment for:
2 N2O5(g) ---> 4 NO2(g) + O2(g)
Time (min) [N2O5]0 (M)
0 1.001.0 0.7052.0 0.4975.0 0.173
If it were first order:
Calculate the ln :
Using Integrated Rate LawsUsing Integrated Rate Laws
Plot of ln [NPlot of ln [N22OO55] vs. time ] vs. time
is a straight line! is a straight line! Eqn. for straight line: Eqn. for straight line: y = mx + b y = mx + b
Plot of ln [NPlot of ln [N22OO55] vs. time ] vs. time
is a straight line! is a straight line! Eqn. for straight line: Eqn. for straight line: y = mx + b y = mx + b
ln [N2O5] = - kt + ln [N 2O5]o
conc at time t
rate const = slope
conc at time = 0
ln [N2O5] = - kt + ln [N 2O5]o
conc at time t
rate const = slope
conc at time = 0
ln [N2O5] vs. time
time
0
-2
0 5
ln [N2O5] vs. time
time
0
-2
0 5
The gas phase decomposition of hydrogen peroxide at 400 oC is second order in H2O2. In one experiment, when the initial concentration of H2O2 was 0.246 M, the concentration of H2O2 dropped to 3.39 x 10-2 M after 25.9 seconds had passed. What is the rate constant for the reaction?
2 H2O2 2 H2O + O2
Half-LifeHalf-Life
HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful.
Half-LifeHalf-Life
• Reaction is 1st order Reaction is 1st order decomposition of decomposition of HH22OO22..
• Reaction after 1 Reaction after 1 half-life.half-life.
• 1/2 of the reactant 1/2 of the reactant has been consumed has been consumed and 1/2 remains.and 1/2 remains.
Half-LifeHalf-Life
• After 2 half-After 2 half-lives _____ of lives _____ of the reactant the reactant remains.remains.
• After 3 half-After 3 half-lives ____ of lives ____ of the reactant the reactant remains.remains.
Half-LifeHalf-Life
[A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = _________
ln (____) = - k • t1/2
ln [R] / [R]0 = k t
In an experiment, it is determined that 75% of a sample of HCO2H (formic acid) has decomposed in 72 seconds following first-order
kinetics. Determine t1/2 for this reaction. HCO2H CO2 + H2
ln [R] / [R]0 = k t t1/2 = 0.693 / k
MechanismsMechanisms
• Mechanism: how reactants are Mechanism: how reactants are converted to products at the converted to products at the molecular level.molecular level.RATE LAW ----> RATE LAW ---->
• Conversion of cis to trans-2-butene Conversion of cis to trans-2-butene requires twisting around the C=C bond.requires twisting around the C=C bond.
• Rate = k [trans-2-butene]Rate = k [trans-2-butene]
Transition StateTransition State
Activation energy barrier
CisCis TransTransTransition stateTransition state
MechanismsMechanisms
• Reaction passes thru a Reaction passes thru a TRANSITION STATE where there is an where there is an activated complex that that has sufficient energy to has sufficient energy to become a product. become a product.
ACTIVATION ENERGY, ACTIVATION ENERGY,
EEaa = = energy req’d to form energy req’d to form
activated complex.activated complex.
Here EHere Eaa = ___________ = ___________
MechanismsMechanisms
Also note that trans-butene is MORE STABLE Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol.than cis-butene by about 4 kJ/mol.
Therefore, cis ---> trans is Therefore, cis ---> trans is __________________________________
This is the connection between thermo-This is the connection between thermo-dynamics and kinetics.dynamics and kinetics.
Effect of TemperatureEffect of Temperature
• Reactions generally occur slower at lower T.
Iodine clock reaction, book page 705.
H2O2 + 2 I- + 2 H+ --> 2 H2O + I2
Room temperature
In ice at 0 oC
Activation Energy and Activation Energy and TemperatureTemperature
Reactions are Reactions are __________ at a higher T__________ at a higher T because a because a
larger fraction of reactant molecules have larger fraction of reactant molecules have
enough energy to convert to product molecules.enough energy to convert to product molecules.
In general, In general, differences in differences in activation energyactivation energy cause reactions to cause reactions to vary from fast to vary from fast to slow.slow.
Collision TheoryCollision Theory
• Molecules must collide with one another
• Molecules must collide with sufficient ________ to break bonds
• Molecules must collide in an orientation that can lead to rearrangement of the atoms.
Arrhenius EquationArrhenius Equation
k Ae -Ea/RTk Ae -E
a/RT
ln k = - (EaR
)(1T
) + ln Aln k = - (EaR
)(1T
) + ln A
Rate Rate constantconstant
Temp (K)Temp (K)
8.31 x 108.31 x 10-3-3 kJ/K•mol kJ/K•molActivation Activation energyenergy
Frequency factorFrequency factor
Frequency factor related to frequency of collisions with correct geometry.
Plot ln k vs. 1/T Plot ln k vs. 1/T ---> straight ---> straight
2. Activation energy and geometry1. Activation energy
MechanismsMechanisms
Most reactions involve a sequence of elementary Most reactions involve a sequence of elementary steps.steps.
2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO
Rate = k [IRate = k [I--] [H] [H22OO22]]
NOTENOTE1.1. Rate law comes from experiment.Rate law comes from experiment.2.2. Order and stoichiometric coefficients not Order and stoichiometric coefficients not
necessarily the same!necessarily the same!3.3. Rate law reflects all chemistry down to and Rate law reflects all chemistry down to and
including the slowest step in multistep reaction.including the slowest step in multistep reaction.
Rate of the reaction controlled by slow step — RATE DETERMINING STEP,
Rate can be no faster than RDS!
Most reactions involve a sequence of elementary steps.Most reactions involve a sequence of elementary steps.
2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO
Rate = k [IRate = k [I--] [H] [H22OO22]]
MechanismsMechanisms
Elementary Step 1Elementary Step 1 is is bimolecularbimolecular and involves I and involves I-- and HOOH. and HOOH. Therefore, this predicts the rate law should beTherefore, this predicts the rate law should be
Rate Rate [I [I--] [H] [H22OO22] — as observed!!] — as observed!!
The species HOI and OHThe species HOI and OH-- are are ________________.________________.