1 chemical kinetics chapter 15 an automotive catalytic muffler
TRANSCRIPT
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Chemical KineticsChemical KineticsChapter 15Chapter 15
Chemical KineticsChemical KineticsChapter 15Chapter 15
An automotive catalytic muffler.
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• We can use thermodynamics to tell if a We can use thermodynamics to tell if a reaction is product or reactant favored.reaction is product or reactant favored.
• But this gives us no info on HOW FAST But this gives us no info on HOW FAST reaction goes from reactants to products.reaction goes from reactants to products.
• KINETICSKINETICS — the study of — the study of REACTION REACTION RATESRATES and their relation to the way the and their relation to the way the reaction proceeds, i.e., its reaction proceeds, i.e., its MECHANISMMECHANISM..
• We can use thermodynamics to tell if a We can use thermodynamics to tell if a reaction is product or reactant favored.reaction is product or reactant favored.
• But this gives us no info on HOW FAST But this gives us no info on HOW FAST reaction goes from reactants to products.reaction goes from reactants to products.
• KINETICSKINETICS — the study of — the study of REACTION REACTION RATESRATES and their relation to the way the and their relation to the way the reaction proceeds, i.e., its reaction proceeds, i.e., its MECHANISMMECHANISM..
Chemical KineticsChemical KineticsChapter 15Chapter 15
Chemical KineticsChemical KineticsChapter 15Chapter 15
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Thermodynamics or Thermodynamics or KineticsKinetics
• Reactions can be one of the following:
1. Not thermodynamically favored (reactant favored)
2. Thermodynamically favored (product favored), but not kinetically favored (slow)
3. Thermodynamically favored (product favored) and kinetically favored (fast)
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Thermodynamics or Thermodynamics or KineticsKinetics
• Not thermodynamically favored (reactant favored)
• Sand (SiO2) will not decompose into Si and O2
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5Thermodynamics or Thermodynamics or KineticsKinetics
• Thermodynamically favored (product favored), but not kinetically favored (slow)
• Diamonds will turn into graphite, but the reaction occurs very slowly
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Thermodynamics or Thermodynamics or KineticsKinetics
• Thermodynamically favored (product favored) and kinetically favored (fast)
• Burning of paper in air will turn to ash very quickly
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Reaction MechanismsReaction MechanismsReaction MechanismsReaction MechanismsThe sequence of events at the molecular level that The sequence of events at the molecular level that
control the speed and outcome of a reaction.control the speed and outcome of a reaction.
Br from biomass burning destroys stratospheric ozone. Br from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, (See R.J. Cicerone, Science, volume 263, page 1243, 1994.)1994.)
Step 1:Step 1:Br + OBr + O33 ---> BrO + O ---> BrO + O22
Step 2:Step 2:Cl + OCl + O33 ---> ClO + O ---> ClO + O22
Step 3:Step 3:BrO + ClO + light ---> Br + Cl + OBrO + ClO + light ---> Br + Cl + O22
NET: NET: 2 O2 O33 ---> 3 O ---> 3 O22
The sequence of events at the molecular level that The sequence of events at the molecular level that control the speed and outcome of a reaction.control the speed and outcome of a reaction.
Br from biomass burning destroys stratospheric ozone. Br from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, (See R.J. Cicerone, Science, volume 263, page 1243, 1994.)1994.)
Step 1:Step 1:Br + OBr + O33 ---> BrO + O ---> BrO + O22
Step 2:Step 2:Cl + OCl + O33 ---> ClO + O ---> ClO + O22
Step 3:Step 3:BrO + ClO + light ---> Br + Cl + OBrO + ClO + light ---> Br + Cl + O22
NET: NET: 2 O2 O33 ---> 3 O ---> 3 O22
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•Reaction rate = change in Reaction rate = change in concentration of a reactant or concentration of a reactant or product with time.product with time.
•Know about Know about initial rateinitial rate, , average average raterate, and , and instantaneous rateinstantaneous rate. See . See Screen 15.2. Screen 15.2.
Reaction Rates Reaction Rates Section 15.1Section 15.1
Reaction Rates Reaction Rates Section 15.1Section 15.1
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Determining a Reaction RateDetermining a Reaction RateDetermining a Reaction RateDetermining a Reaction Rate
Blue dye is oxidized Blue dye is oxidized with bleach. with bleach.
Its concentration Its concentration decreases with time.decreases with time.
The rate — the The rate — the change in dye conc change in dye conc with time — can be with time — can be determined from the determined from the plot.plot.
Blue dye is oxidized Blue dye is oxidized with bleach. with bleach.
Its concentration Its concentration decreases with time.decreases with time.
The rate — the The rate — the change in dye conc change in dye conc with time — can be with time — can be determined from the determined from the plot.plot.
Dye
Co
nc
Dye
Co
nc
Dye
Co
nc
Dye
Co
nc
TimeTime
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• ConcentrationsConcentrations and and physical physical statestate of reactants and products of reactants and products (Screens 15.3 and 15.4)(Screens 15.3 and 15.4)
• TemperatureTemperature (Screen 15.11) (Screen 15.11)
• CatalystsCatalysts (Screen 15.14) (Screen 15.14)
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
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• Concentrations Concentrations
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
Rate with 0.3 M HClRate with 0.3 M HCl
Rate with 6.0 M HClRate with 6.0 M HCl
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• Physical state of reactantsPhysical state of reactants
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
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Catalysts: catalyzed decomp of HCatalysts: catalyzed decomp of H22OO22
2 H2 H22OO22 --> 2 H --> 2 H22O + OO + O22
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
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• Temperature Temperature
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2
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Concentrations and Rates Concentrations and Rates Concentrations and Rates Concentrations and Rates
To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study
•• reaction ratereaction rate and and
•• its its concentration dependenceconcentration dependence
To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study
•• reaction ratereaction rate and and
•• its its concentration dependenceconcentration dependence
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Concentrations and Rates Concentrations and Rates Concentrations and Rates Concentrations and Rates
Take reaction Take reaction where Clwhere Cl-- in in cisplatin cisplatin [Pt(NH[Pt(NH33))22ClCl33] ]
is replaced is replaced by Hby H22OO
Take reaction Take reaction where Clwhere Cl-- in in cisplatin cisplatin [Pt(NH[Pt(NH33))22ClCl33] ]
is replaced is replaced by Hby H22OO
CisplatinCisplatin
Rate of change of conc of Pt compd
= Am' t of cisplatin reacting (mol/L)
elapsed time (t)
Rate of change of conc of Pt compd
= Am' t of cisplatin reacting (mol/L)
elapsed time (t)
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Concentrations and Rates Concentrations and Rates Concentrations and Rates Concentrations and Rates
Rate of reaction is proportional to [Pt(NHRate of reaction is proportional to [Pt(NH33))22ClCl22]]
We express this as a We express this as a RATE LAWRATE LAW
Rate of reaction = k [Pt(NHRate of reaction = k [Pt(NH33))22ClCl22]]
where where k = rate constantk = rate constant
k is independent of conc. but increases with Tk is independent of conc. but increases with T
Rate of reaction is proportional to [Pt(NHRate of reaction is proportional to [Pt(NH33))22ClCl22]]
We express this as a We express this as a RATE LAWRATE LAW
Rate of reaction = k [Pt(NHRate of reaction = k [Pt(NH33))22ClCl22]]
where where k = rate constantk = rate constant
k is independent of conc. but increases with Tk is independent of conc. but increases with T
CisplatinCisplatin
Rate of change of conc of Pt compd
= Am' t of cisplatin reacting (mol/L)
elapsed time (t)
Rate of change of conc of Pt compd
= Am' t of cisplatin reacting (mol/L)
elapsed time (t)
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Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws
In general, forIn general, for
a A + b B ---> x Xa A + b B ---> x X with a catalyst Cwith a catalyst C
Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp
The exponents m, n, and p The exponents m, n, and p
•• are the reaction orderare the reaction order
•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions
•• must be determined by experiment!!!must be determined by experiment!!!
In general, forIn general, for
a A + b B ---> x Xa A + b B ---> x X with a catalyst Cwith a catalyst C
Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp
The exponents m, n, and p The exponents m, n, and p
•• are the reaction orderare the reaction order
•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions
•• must be determined by experiment!!!must be determined by experiment!!!
CisplatinCisplatin
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Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws Rate = 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 ?
Rate = 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 ?
CisplatinCisplatin
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Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws Deriving 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 disappearance of CHfrom experimental data for rate of disappearance of CH33CHOCHO
Expt. Expt. [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO
(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)
11 0.100.10 0.0200.020
22 0.200.20 0.0810.081
33 0.300.30 0.1820.182
44 0.400.40 0.3180.318
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 disappearance of CHfrom experimental data for rate of disappearance of CH33CHOCHO
Expt. Expt. [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO
(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)
11 0.100.10 0.0200.020
22 0.200.20 0.0810.081
33 0.300.30 0.1820.182
44 0.400.40 0.3180.318
CisplatinCisplatin
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Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws
Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22
Here the rate goes up by ______ when initial conc. doubles. Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ Therefore, we say this reaction is _________________ order.order.
Now determine the value of k. Use expt. #3 data—Now determine the value of k. Use expt. #3 data—
0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22
k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of [CHUsing k you can calc. rate at other values of [CH33CHO] at CHO] at
same T.same T.
Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22
Here the rate goes up by ______ when initial conc. doubles. Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ Therefore, we say this reaction is _________________ order.order.
Now determine the value of k. Use expt. #3 data—Now determine the value of k. Use expt. #3 data—
0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22
k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of [CHUsing k you can calc. rate at other values of [CH33CHO] at CHO] at
same T.same T.
CisplatinCisplatin
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23Concentration/Time Concentration/Time Relations Relations
Concentration/Time Concentration/Time Relations Relations
Need to know what conc. of reactant is as Need to know what conc. of reactant is as function of time. Consider FIRST ORDER function of time. Consider FIRST ORDER REACTIONSREACTIONS
For 1st order reactions, the rate law isFor 1st order reactions, the rate law is- (- ( [A] / [A] / time) = k [A] time) = k [A]
Need to know what conc. of reactant is as Need to know what conc. of reactant is as function of time. Consider FIRST ORDER function of time. Consider FIRST ORDER REACTIONSREACTIONS
For 1st order reactions, the rate law isFor 1st order reactions, the rate law is- (- ( [A] / [A] / time) = k [A] time) = k [A]
CisplatinCisplatin
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24Concentration/Time Concentration/Time Relations Relations
Concentration/Time Concentration/Time Relations Relations
Integrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we getIntegrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we get
CisplatinCisplatin
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25Concentration/Time Concentration/Time Relations Relations
Concentration/Time Concentration/Time Relations Relations
Integrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we getIntegrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we get
CisplatinCisplatin
[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 =fraction remaining after time
t has elapsed.t has elapsed.
[A] / [A][A] / [A]00 =fraction remaining after time =fraction remaining after time
t has elapsed.t has elapsed.
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26Concentration/Time Concentration/Time Relations Relations
Concentration/Time Concentration/Time Relations Relations
Integrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we getIntegrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we get
CisplatinCisplatin
[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 =fraction remaining after time
t has elapsed.t has elapsed.
[A] / [A][A] / [A]00 =fraction remaining after time =fraction remaining after time
t has elapsed.t has elapsed.
Called the Called the integrated first-order rate lawintegrated first-order rate law..
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Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations
Sucrose decomposes Sucrose decomposes to simpler sugarsto simpler sugars
Rate of Rate of disappearance of disappearance of sucrosesucrose= k [sucrose]= k [sucrose]
k = 0.21 hrk = 0.21 hr-1-1
Initial [sucrose] = Initial [sucrose] = 0.010 M0.010 M
How long to drop How long to drop 90% (to 0.0010 M)?90% (to 0.0010 M)?
Sucrose decomposes Sucrose decomposes to simpler sugarsto simpler sugars
Rate of Rate of disappearance of disappearance of sucrosesucrose= k [sucrose]= k [sucrose]
k = 0.21 hrk = 0.21 hr-1-1
Initial [sucrose] = Initial [sucrose] = 0.010 M0.010 M
How long to drop How long to drop 90% (to 0.0010 M)?90% (to 0.0010 M)?
SucroseSucrose
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28Concentration/Time RelationshipsConcentration/Time RelationshipsRate of disappear of sucrose = k [sucrose], k = 0.21 hrRate of disappear of sucrose = k [sucrose], k = 0.21 hr -1-1.. If If initial [sucrose] = 0.010 M, how long to drop 90% or to initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?0.0010 M?
Use the first order integrated rate lawUse the first order integrated rate law
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29Concentration/Time RelationshipsConcentration/Time RelationshipsRate of disappear of sucrose = k [sucrose], k = 0.21 hrRate of disappear of sucrose = k [sucrose], k = 0.21 hr -1-1.. If If initial [sucrose] = 0.010 M, how long to drop 90% or to initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?0.0010 M?
Use the first order integrated rate lawUse the first order integrated rate law
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30Concentration/Time RelationshipsConcentration/Time RelationshipsRate of disappear of sucrose = k [sucrose], k = 0.21 hrRate of disappear of sucrose = k [sucrose], k = 0.21 hr -1-1.. If If initial [sucrose] = 0.010 M, how long to drop 90% or to initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?0.0010 M?
Use the first order integrated rate lawUse the first order integrated rate law
ln (0.100) = - 2.3 = - (0.21 hrln (0.100) = - 2.3 = - (0.21 hr -1-1) • time) • time
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31Concentration/Time RelationshipsConcentration/Time RelationshipsRate of disappear of sucrose = k [sucrose], k = 0.21 hrRate of disappear of sucrose = k [sucrose], k = 0.21 hr -1-1.. If If initial [sucrose] = 0.010 M, how long to drop 90% or to initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?0.0010 M?
Use the first order integrated rate lawUse the first order integrated rate law
ln (0.100) = - 2.3 = - (0.21 hrln (0.100) = - 2.3 = - (0.21 hr -1-1) • time) • time
time = 11 hourstime = 11 hours
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Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawThe integrated rate law suggests a way to tell if a The integrated rate law suggests a way to tell if a
reaction is first order based on experiment. reaction is first order based on experiment.
2 N2 N22OO55(g) ---> 4 NO(g) ---> 4 NO22(g) + O(g) + O22(g)(g)
Rate = k [NRate = k [N22OO55]]
Time (min)Time (min) [N[N22OO55]]00 (M) (M) ln [Nln [N22OO55]]00
00 1.001.00 00
1.01.0 0.7050.705 -0.35-0.35
2.02.0 0.4970.497 -0.70-0.70
5.05.0 0.1730.173 -1.75-1.75
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Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate Law
2 N2 N22OO55(g) ---> 4 NO(g) ---> 4 NO22(g) + O(g) + O22(g) Rate = k [N(g) Rate = k [N22OO55]]
[N2O5] vs. time
time
1
0
0 5
[N2O5] vs. time
time
1
0
0 5
Data of conc. vs. Data of conc. vs. time plot do not fit time plot do not fit straight line.straight line.
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Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate Law
2 N2 N22OO55(g) ---> 4 NO(g) ---> 4 NO22(g) + O(g) + O22(g) Rate = k [N(g) Rate = k [N22OO55]]
[N2O5] vs. time
time
1
0
0 5
[N2O5] vs. time
time
1
0
0 5
l n [N2O5] vs. time
time
0
-2
0 5
l n [N2O5] vs. time
time
0
-2
0 5
Data of conc. vs. Data of conc. vs. time plot do not fit time plot do not fit straight line.straight line.
Plot of ln [NPlot of ln [N22OO55] vs. ] vs.
time is a straight time is a straight line!line!
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Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate Law
All 1st order reactions have straight line plot All 1st order reactions have straight line plot for ln [A] vs. time. for ln [A] vs. time.
(2nd order gives straight line for plot of 1/[A] (2nd order gives straight line for plot of 1/[A] vs. time)vs. time)
ln [N2O5] vs. time
time
0
-2
0 5
l n [N2O5] vs. time
time
0
-2
0 5
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 = ax + b y = ax + 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 = ax + b y = ax + 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
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Graphical Methods for Graphical Methods for Determining Reaction Order Determining Reaction Order
and Rate Constantand Rate Constant• Zero-order
• Straight line plot of [A]t vs. t• Slope is –k (mol/L*s)
mxby
ktoAtA
][][
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Graphical Methods for Graphical Methods for Determining Reaction Order Determining Reaction Order
and Rate Constantand Rate Constant• First-order
• Straight line plot of ln[A]t vs. t• Slope is – k (s-1)
mxby
ktoAtA
]ln[]ln[
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Graphical Methods for Graphical Methods for Determining Reaction Order Determining Reaction Order
and Rate Constantand Rate Constant• Second-order
• Straight line plot of 1/[A]t vs t• Slope is k (L/mol*s)
mxby
ktoAtA
][
1
][
1
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Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
HALF-LIFE is the time it HALF-LIFE is the time it takes for 1/2 a sample takes for 1/2 a sample is disappear.is disappear.
For 1st order reactions, For 1st order reactions, the concept of HALF-the concept of HALF-LIFE is especially LIFE is especially useful.useful.
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• Reaction is 1st order Reaction is 1st order decomposition of decomposition of HH22OO22..
Half-LifeHalf-LifeHalf-LifeHalf-Life
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Half-LifeHalf-LifeHalf-LifeHalf-Life
• Reaction after 654 Reaction after 654 min, 1 half-life.min, 1 half-life.
• 1/2 of the reactant 1/2 of the reactant remains.remains.
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Half-LifeHalf-LifeHalf-LifeHalf-Life
• Reaction after Reaction after 1306 min, or 2 1306 min, or 2 half-lives.half-lives.
• 1/4 of the reactant 1/4 of the reactant remains.remains.
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Half-LifeHalf-LifeHalf-LifeHalf-Life
• Reaction after 3 Reaction after 3 half-lives, or 1962 half-lives, or 1962 min.min.
• 1/8 of the reactant 1/8 of the reactant remains.remains.
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Half-LifeHalf-LifeHalf-LifeHalf-Life
• Reaction after 4 Reaction after 4 half-lives, or 2616 half-lives, or 2616 min.min.
• 1/16 of the 1/16 of the reactant remains.reactant remains.
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Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Sugar is fermented in a 1st order process (using Sugar is fermented in a 1st order process (using an enzyme as a catalyst).an enzyme as a catalyst).
sugar + enzyme --> productssugar + enzyme --> products
Rate of disappear of sugar = k[sugar]Rate of disappear of sugar = k[sugar]
k = 3.3 x 10k = 3.3 x 10-4-4 sec sec-1-1
What is the half-life of this reaction?What is the half-life of this reaction?
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Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
SolutionSolution
[A] / [A][A] / [A]00 = 1/2 when t = t = 1/2 when t = t1/21/2
Therefore, ln (1/2) = - k • tTherefore, ln (1/2) = - k • t1/21/2
- 0.693 = - k • t- 0.693 = - k • t1/21/2
tt1/21/2 = 0.693 / k = 0.693 / kSo, for sugar, So, for sugar,
tt1/21/2 = 0.693 / k = 2100 sec = = 0.693 / k = 2100 sec = 35 min35 min
Rate = k[sugar] and k = 3.3 x 10Rate = k[sugar] and k = 3.3 x 10-4-4 sec sec-1-1. What is the half-. What is the half-life of this reaction?life of this reaction?
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Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
SolutionSolution
2 hr and 20 min = 4 half-lives2 hr and 20 min = 4 half-lives
Half-life Half-life Time ElapsedTime Elapsed Mass LeftMass Left
1st1st 35 min35 min 5.00 g5.00 g
2nd2nd 7070 2.50 g2.50 g
3rd3rd 105105 1.25 g1.25 g
4th4th 140140 0.625 g0.625 g
Rate = k[sugar] and k = 3.3 x 10Rate = k[sugar] and k = 3.3 x 10-4-4 sec sec-1-1. Half-life . Half-life is 35 min. Start with 5.00 g sugar. How much is is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min?left after 2 hr and 20 min?
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Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
SolutionSolution
Radioactive decay is a first order process. Radioactive decay is a first order process.
Tritium ---> electron + heliumTritium ---> electron + helium
33HH 00-1-1ee 33HeHe
If you have 1.50 mg of tritium, how much is left If you have 1.50 mg of tritium, how much is left after 49.2 years? tafter 49.2 years? t1/21/2 = 12.3 years = 12.3 years
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49Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
SolutionSolution
ln [A] / [A]ln [A] / [A]00 = -kt = -kt
[A] = ?[A] = ? [A][A]00 = 1.50 mg = 1.50 mg t = 49.2 yearst = 49.2 years
Need k, so we calc k from: k = 0.693 / tNeed k, so we calc k from: k = 0.693 / t1/21/2
Obtain k = 0.0564 yObtain k = 0.0564 y-1-1
Now ln [A] / [A]Now ln [A] / [A]00 = -kt = - (0.0564 y = -kt = - (0.0564 y-1-1) • (49.2 y) ) • (49.2 y)
= - 2.77= - 2.77
Take antilog: [A] / [A]Take antilog: [A] / [A]00 = e = e-2.77-2.77 = 0.0627 = 0.0627
0.0627 is the 0.0627 is the fraction remainingfraction remaining!!
Start with 1.50 mg of tritium, how much is left after 49.2 Start with 1.50 mg of tritium, how much is left after 49.2 years? tyears? t1/21/2 = 12.3 years = 12.3 years
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50Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8
SolutionSolution
[A] / [A][A] / [A]00 = 0.0627 = 0.0627
0.0627 is the 0.0627 is the fraction remainingfraction remaining!!
Because [A]Because [A]00 = 1.50 mg, [A] = 0.094 mg = 1.50 mg, [A] = 0.094 mg
But notice that 49.2 y = 4.00 half-livesBut notice that 49.2 y = 4.00 half-lives
1.50 mg ---> 0.750 mg after 11.50 mg ---> 0.750 mg after 1
---> 0.375 mg after 2---> 0.375 mg after 2
---> 0.188 mg after 3---> 0.188 mg after 3
---> 0.094 mg after 4---> 0.094 mg after 4
Start with 1.50 mg of tritium, how much is left after 49.2 Start with 1.50 mg of tritium, how much is left after 49.2 years? tyears? t1/21/2 = 12.3 years = 12.3 years
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Half-Lives of Radioactive ElementsHalf-Lives of Radioactive ElementsHalf-Lives of Radioactive ElementsHalf-Lives of Radioactive Elements
Rate of decay of radioactive isotopes given in Rate of decay of radioactive isotopes given in terms of 1/2-life. terms of 1/2-life.
238238U --> U --> 234234Th + HeTh + He 4.5 x 104.5 x 1099 y y1414C --> C --> 1414N + betaN + beta 5730 y5730 y131131I --> I --> 131131Xe + betaXe + beta 8.05 d8.05 d
Element 106 - seaborgiumElement 106 - seaborgium263263Sg Sg 0.9 s0.9 s