1 principles of reactivity: chemical kinetics chapter 15 1-17 + all bold numbered problems

11

PRINCIPLES OF REACTIVITY: PRINCIPLES OF REACTIVITY:

CHEMICAL KINETICSCHEMICAL KINETICS

PRINCIPLES OF REACTIVITY: PRINCIPLES OF REACTIVITY:

CHEMICAL KINETICSCHEMICAL KINETICS

CHAPTER 15

1-17 + all bold numbered problems

22

Chapter 15 OverviewChapter 15 Overview

• Calculate each part of the rate law:Calculate each part of the rate law:

rate (law) = k [A]rate (law) = k [A]mm[B][B]nn[C][C]pp

• Memorize the rate law/decay formula for:Memorize the rate law/decay formula for:

–0 order0 order

–11stst order order

–22ndnd order order

Purpose of study rate laws is to find the Purpose of study rate laws is to find the mechanism of a reactionmechanism of a reaction

33

Chemical KineticsChemical KineticsChapter 15Chapter 15

Chemical KineticsChemical KineticsChapter 15Chapter 15

An automotive catalytic muffler.

44

CHAPTER OVERVIEWCHAPTER OVERVIEWCHAPTER OVERVIEWCHAPTER OVERVIEW

• This chapter examines factors that This chapter examines factors that determine the speed of reactions, determine the speed of reactions, reaction rates, and the steps that reaction rates, and the steps that occur during the reaction process, occur during the reaction process, reaction mechanisms. reaction mechanisms.

• Reaction rates are generally affected Reaction rates are generally affected by concentration of reactants, by concentration of reactants, temperature, and the presence of a temperature, and the presence of a catalyst.catalyst.

55

• 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 But this gives us nono info on HOW FAST info on HOW FAST reaction goes from reactants to products.reaction goes from reactants to products.

• KINETICS KINETICS — 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 KineticsChemical KineticsChemical Kinetics

66

Reaction MechanismsReaction MechanismsReaction MechanismsReaction Mechanisms

The sequence of events at the molecular level The sequence of events at the molecular level that control the speed and outcome of a that control the speed and outcome of a reaction.reaction.

Br from biomass burning destroys stratospheric Br from biomass burning destroys stratospheric ozone. ozone.

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

77

• 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 initial rate, average rate, Know about initial rate, average rate, and instantaneous rate. and instantaneous rate.

Reaction Rates Reaction Rates Section 15.1Section 15.1

Reaction Rates Reaction Rates Section 15.1Section 15.1

88Determining a Reaction RateDetermining a Reaction RateDetermining a Reaction RateDetermining a Reaction Rate

Blue dye is oxidized with Blue dye is oxidized with bleach. bleach.

Its concentration Its concentration decreases with time.decreases with time.

The rate — the change in The rate — the change in dye concentration with dye concentration with time — can be time — can be determined from the determined from the plot.plot.

Dye

Co

nc

Time

99

RATES OF CHEMICAL REACTIONSRATES OF CHEMICAL REACTIONSRATES OF CHEMICAL REACTIONSRATES OF CHEMICAL REACTIONS• In chemical reactions, the rate is usually In chemical reactions, the rate is usually

measured as change in concentration per unit measured as change in concentration per unit time, -time, -ΔΔ(A)/ (A)/ ΔΔt, for reactant A. t, for reactant A.

• The rate can also be measured as the rate of The rate can also be measured as the rate of appearance of the product. appearance of the product.

• Rates are related stoichiometrically. Rates are related stoichiometrically.

• Reaction rates are always positive numbers.Reaction rates are always positive numbers.

• In the text, the rate for the decomposition of In the text, the rate for the decomposition of

NN22OO55 is calculated at several different is calculated at several different

concentrations.concentrations.

1111

RATES OF REACTIONSRATES OF REACTIONSRATES OF REACTIONSRATES OF REACTIONS

• The average rate can be calculated between The average rate can be calculated between any two points. any two points.

• The instantaneous rate at any specific time is The instantaneous rate at any specific time is the slope of the tangent line at that point for the slope of the tangent line at that point for the graph of concentration vs. time. the graph of concentration vs. time.

1212

• Concentrations and physical Concentrations and physical state of reactants and products state of reactants and products

• TemperatureTemperature

• CatalystsCatalysts

Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2

Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2

1313

• Concentrations Concentrations

Factors Affecting RatesFactors Affecting RatesFactors Affecting RatesFactors Affecting RatesRate with 0.3 M HCl

Rate with 6.0 M HCl

1414

• Physical state of reactantsPhysical state of reactants

Factors Affecting RatesFactors Affecting RatesFactors Affecting RatesFactors Affecting Rates

1515

Catalysts: catalyzed decomp of HCatalysts: catalyzed decomp of H22OO22

2 H2 H22OO22 --> 2 H --> 2 H22O + OO + O22


1616

• Temperature Temperature


1717

• In general, the rate of reaction is a function of reactant concentration.

• This function is determined experimentally, frequently using initial rates.

• The outcome of the experiment is expressed The outcome of the experiment is expressed in a form called the in a form called the rate equationrate equation or or rate lawrate law. .

15.3 EFFECTS OF CONCENTRATION ON 15.3 EFFECTS OF CONCENTRATION ON REACTION RATEREACTION RATE

15.3 EFFECTS OF CONCENTRATION ON 15.3 EFFECTS OF CONCENTRATION ON REACTION RATEREACTION RATE

1818

CONCENTRATION AND RATECONCENTRATION AND RATE

• For the general equation,For the general equation,

a A + b B ====> x X with catalyst Ca A + b B ====> x X with catalyst C

• rate = k [A]rate = k [A]mm[B][B]nn[C][C]pp

• m, n, and p are frequently positive whole m, n, and p are frequently positive whole numbers, but they can be negative, fractions, numbers, but they can be negative, fractions, or even zero. or even zero.

• They are determinedThey are determined EXPERIMENTALLY EXPERIMENTALLY

1919

The Rate Constant, kThe Rate Constant, k

• The rate constant, k, is determined The rate constant, k, is determined experimentallyexperimentally from the rate data, and has from the rate data, and has units consistent with the rate equation. units consistent with the rate equation.

• These units include These units include timetime-1-1 and and concentration concentration to some power.

• The magnitude of k is a function of the The magnitude of k is a function of the temperature for the same reaction. temperature for the same reaction.

• Larger k's mean faster reactions.Larger k's mean faster reactions.

2020

The Order of ReactionThe Order of Reaction

• The total order of the reaction is the sum of The total order of the reaction is the sum of the orders for each of the species in the the orders for each of the species in the rate law. rate law.

• In the previous example the total order is: In the previous example the total order is: m + n + pm + n + p

• As the total order varies, the units on k will As the total order varies, the units on k will vary. vary.

• For zero order, k has units of M/time. For zero order, k has units of M/time. • For first order, k has units of 1/time or timeFor first order, k has units of 1/time or time --

11, etc., etc.

2121

Rate Laws: What UseRate Laws: What Use

• There are two uses for the Rate Law:There are two uses for the Rate Law:

–1. The ability to find a “future” 1. The ability to find a “future” concentrationconcentration

• The we “Integrate” the Rate Law:The we “Integrate” the Rate Law:

–2. This gives a time dependent 2. This gives a time dependent function that tells us “how long” it function that tells us “how long” it takes to get to the “future” takes to get to the “future” concentrationconcentration

2222

Determination of the Determination of the Rate EquationRate Equation

• The rate equation is frequently determined The rate equation is frequently determined using using initial ratesinitial rates. .

• These experiments measure the rate of the These experiments measure the rate of the reaction keeping all but one reactant reaction keeping all but one reactant concentration essentially constant and concentration essentially constant and allowing the other to change by only about 1% allowing the other to change by only about 1% or 2%.or 2%.

• The order for this reactant is then determined. The order for this reactant is then determined.

2323

Determination of the Determination of the Rate EquationRate Equation

• This process is continued until all the This process is continued until all the reactant orders have been determined. reactant orders have been determined.

• The order for each reactant is determined The order for each reactant is determined algebraically by choosing an appropriate pair algebraically by choosing an appropriate pair of experiments and taking a ratio. of experiments and taking a ratio.

• The rate constant, k, is then determined by The rate constant, k, is then determined by substitution once all the reaction orders have substitution once all the reaction orders have been calculated. been calculated.

• Examples 15.3,4 and Exercises 15.3, 4.Examples 15.3,4 and Exercises 15.3, 4.

2424

Concentrations and RatesConcentrations and Rates Concentrations and RatesConcentrations and Rates

Take reaction Take reaction where Clwhere Cl-- in in cisplatin cisplatin [Pt(NH[Pt(NH33))22ClCl22] ] is replaced is replaced by Hby H22OO

CisplatinCisplatin

Rate of change of conc of Pt compound

= Am' t of cisplatin reacting (mol/L)

elapsed time (t)

2525

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

CisplatinCisplatin

Rate of change of conc of Pt compound

= Am' t of cisplatin reacting (mol/L)

elapsed time (t)

2626

Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws

In general, for In general, for

a A + b B ---> x X a 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

2727

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, reaction is 1st order in AIf m = 1, reaction 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, reaction is 2nd order in A.If m = 2, reaction is 2nd order in A.


Doubling [A] increases rate by ?Doubling [A] increases rate by ?

• If m = 0, reaction is zero order.If m = 0, reaction is zero order.


If [A] doubles, rate ?If [A] doubles, rate ?

CisplatinCisplatin

2828

Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws

Derive rate law and k forDerive rate law and k forCHCH33CHO(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 CH33CHO.CHO.

Expt. [CHExpt. [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

2929

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 Here the rate goes up by ______ when initial concentration doubles. concentration doubles.

Therefore this reaction is _________________ Therefore this reaction is _________________ order.order.

Now determine the value of k. Now determine the value of k.

Use expt. #3 data 0.182 mol/L•s = k (0.30 mol/L)Use expt. #3 data 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 calculate the rate at other values Using k you can calculate the rate at other values of [CHof [CH33CHO] at same T…”1. Future Conc”CHO] at same T…”1. Future Conc”

CisplatinCisplatin

3030

Concentration/Time Concentration/Time Relations…”2. how long”Relations…”2. how long”

Concentration/Time Concentration/Time Relations…”2. how long”Relations…”2. how long”

Need to know that concentration of reactant Need to know that concentration of reactant is as function of time. is as function of time.

Consider FIRST ORDER REACTIONSConsider FIRST ORDER REACTIONS

For For 1st order1st order reactions, the rate law is reactions, the rate law is

rate = - (rate = - (ΔΔ[A] / [A] / ΔΔtime) = k [A]time) = k [A]

CisplatinCisplatin

3131

Concentration/Time Concentration/Time RelationsRelations

Concentration/Time Concentration/Time RelationsRelations

Integrating - (Integrating - ( [A] / [A] / time) = k [A], we get time) = k [A], we get

CisplatinCisplatin

[A] / [A]0 =fraction remaining after time t has elapsed.

Called the integrated first-order rate lawintegrated first-order rate law..

[A] = - k tln[A]o

naturallogarithm [A] at time = 0

3232

Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations

Sucrose decomposes to Sucrose decomposes to simpler sugarssimpler sugars

Rate of disappearance of Rate of disappearance of sucrosesucrose= k [sucrose]= k [sucrose]

k = 0.21 hrk = 0.21 hr-1-1

Initial [sucrose] = 0.010 MInitial [sucrose] = 0.010 M

How long to drop 90% How long to drop 90%

( to 0.0010 M )?( to 0.0010 M )?

SucroseSucrose

3333Concentration/Time RelationshipsConcentration/Time RelationshipsRate of disappearance of sucrose = k Rate of disappearance of sucrose = k [sucrose], [sucrose], k = 0.21 hrk = 0.21 hr-1-1.. If initial [sucrose] = 0.010 M, If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?how long to drop 90% or to 0.0010 M?

ln (0.100) = - 2.3 = - (0.21 hrln (0.100) = - 2.3 = - (0.21 hr -1-1) • time) • time

time = 11 hourstime = 11 hours

Use the first order integrated rate lawUse the first order integrated rate law

3434

Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate Law

The integrated rate law suggests a way to tell if The integrated rate law suggests a way to tell if a reaction is first order based on experiment. a reaction is first order based on experiment.

2 N2 N22OO55(g) (g) 4 NO 4 NO22(g) + O(g) + O22(g)(g)

Rate = k [NRate = k [N22OO55]]

Time (min)Time (min) [N[N22OO55]]oo (M) (M) ln [N ln [N22OO55]]oo

00 1.00 1.00 0 01.01.0 0.705 0.705 -0.35-0.352.02.0 0.497 0.497 -0.70-0.705.05.0 0.173 0.173 -1.75-1.75

3535


2 N2 N22OO55(g) (g) 4 NO 4 NO22(g) + O(g) + O22(g) (g)

Rate = k [NRate = 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.

l n [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. ] vs.

time is a straight time is a straight line!line!

3636


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 is ] vs. time is

a straight line! a straight line! Eqn. for straight line: Eqn. for straight line:

y = ax + by = ax + b ln [N2O5] = - kt + ln [N2O 5]o

conc at time t

rate const = slope

conc at time = 0

3737

15.4 Relationships Between 15.4 Relationships Between Concentration and TimeConcentration and Time15.4 Relationships Between 15.4 Relationships Between Concentration and TimeConcentration and Time

• The order for a specific reactant can be The order for a specific reactant can be determined graphically if the function that determined graphically if the function that produces a linear plot for each order is produces a linear plot for each order is known. known.

• We will derive the algebraic relationship for We will derive the algebraic relationship for three orders, zero, first, and second. three orders, zero, first, and second.

• For the general equation: For the general equation: A ===> Products A ===> Products

• rate = -d[A]/dt = k [A]rate = -d[A]/dt = k [A]xx , x is the order. , x is the order.

3838

Relationships Between Concentration and TimeRelationships Between Concentration and Time

• The linear plot always has time on the The linear plot always has time on the

horizontal axis and a function of concentration horizontal axis and a function of concentration

on the vertical axis. on the vertical axis.

• The half-life, tThe half-life, t1/21/2 , is defined as the time , is defined as the time

required for one-half of the reactant to be required for one-half of the reactant to be

consumed:consumed:

[A] = [A][A] = [A]oo/2/2

3939

Zero-OrderZero-OrderZero-OrderZero-Order

• rate = - d[A]/dt = k [A]rate = - d[A]/dt = k [A]00 = k = k • This is the differential form of the rate This is the differential form of the rate

expression, since it contains a differential. expression, since it contains a differential. • It is also the definition of zero-order. It is also the definition of zero-order. • The integrated form for zero-order can be The integrated form for zero-order can be

found by separation of variables and found by separation of variables and integration. integration.

• The result is [A] = - kt + [A]The result is [A] = - kt + [A]oo. . • The linear plot is [A] vs. timeThe linear plot is [A] vs. time• The half-life is, tThe half-life is, t1/21/2 = [A] = [A]oo/2k. /2k. • From the graph, m = - k .From the graph, m = - k .

4040

First-OrderFirst-OrderFirst-OrderFirst-Order• rate = -d[A]/dt = k [A]rate = -d[A]/dt = k [A]11

• This is the differential form of the rate This is the differential form of the rate expression, since it contains a differential. expression, since it contains a differential.

• It is also the definition of first-order. It is also the definition of first-order.

• The integrated form for first-order can be The integrated form for first-order can be found by separation of variables and found by separation of variables and integration. integration.

• The result is The result is lnln[A] = - kt + [A] = - kt + lnln[A][A]oo..

• The linear plot is The linear plot is lnln[A] vs. time [A] vs. time

• The half-life is, tThe half-life is, t1/21/2 = ( = (lnln2)/k. 2)/k.

• From the graph, m = - k .From the graph, m = - k .

4141

Second-OrderSecond-OrderSecond-OrderSecond-Order• rate = - d[A]/dt = k [A]rate = - d[A]/dt = k [A]22

• This is the differential form of the rate This is the differential form of the rate expression, since it contains a differential. expression, since it contains a differential.

• It is also the definition of second-order. It is also the definition of second-order.

• The integrated form for second-order can The integrated form for second-order can be found by separation of variables and be found by separation of variables and integration. integration.

• The result is 1/[A] = + kt + 1/[A]The result is 1/[A] = + kt + 1/[A]oo..

• The linear plot is 1/[A] vs. timeThe linear plot is 1/[A] vs. time

• The half-life is, tThe half-life is, t1/21/2 = 1/k[A] = 1/k[A]oo. .

• From the graph, m = + k .From the graph, m = + k .

4242

SummarySummary

• 0 order: [A] = - kt + [A]0 order: [A] = - kt + [A]oo

tt1/21/2 = [A] = [A]oo/2k/2k

• 11stst order: order: lnln[A] = - kt + [A] = - kt + lnln[A][A]oo

tt1/21/2 = ( = (ln1/2ln1/2)/-k)/-k

• 22ndnd order: 1/[A] = + kt + 1/[A] order: 1/[A] = + kt + 1/[A]o, o,

tt1/21/2 = 1/k[A] = 1/k[A]oo

4343

Half-LifeHalf-LifeHalf-LifeHalf-Life

HALF-LIFE is the HALF-LIFE is the time it takes for time it takes for 1/2 a sample is 1/2 a sample is disappear.disappear.

For 1st order For 1st order reactions, the reactions, the concept of HALF-concept of HALF-LIFE is especially LIFE is especially useful.useful.

4444

• Reaction is 1st Reaction is 1st order order decomposition decomposition of Hof H22OO22..


4545


• Reaction after Reaction after 654 min, 1 half-654 min, 1 half-life.life.

• 1/2 of the 1/2 of the reactant reactant remains.remains.

4646


• Reaction after Reaction after 3 half-lives, or 3 half-lives, or 1962 min.1962 min.

• 1/8 of the 1/8 of the reactant reactant remains.remains.

4747


Sugar is fermented in a 1st order process Sugar is fermented in a 1st order process (using an enzyme as a catalyst).(using 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?

4848


Solution: Solution: lnln[A] = - kt + [A] = - kt + lnln[A][A]oo

[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-life of this reaction?What is the half-life of this reaction?

4949


SolutionSolution2 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 2.50 g2.50 g

2nd2nd 7070 1.25 g1.25 g

3rd3rd 105105 0.625 g0.625 g

4th4th 140140 0.313 g0.313 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 is 35 min. Start with 5.00 g sugar. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min?How much is left after 2 hr and 20 min?

5050

Or using the rate lawOr using the rate law

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 is . Half-life is 35 min. Start with 5.00 g sugar. How much is left 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (8400 sec)?after 2 hr and 20 min (8400 sec)?

• lnln[A] = - kt + [A] = - kt + lnln[A][A]oo (first order) (first order)

43.3 10 8400 ln5.00.313e g

5151


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 If you have 1.50 mg of tritium, how much is left after 49.2 years? tis left after 49.2 years? t1/21/2 = 12.3 years = 12.3 years

•This has been experimentally determined This has been experimentally determined to be a 1to be a 1stst order reaction. order reaction.

5252

SolutionSolution

ln [A] / [A]ln [A] / [A]00 = -kt, however 2 unknown variables = -kt, however 2 unknown variables

[A] = ?[A] = ? [A][A]00 = 1.50 mg = 1.50 mg t = 49.2 yearst = 49.2 years

Need k, so we calc k:Need k, so we calc k: k = 0.693 / t 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)

ln [A] / [A]ln [A] / [A]00 = - 2.77 = - 2.77

Take antilog: [A] / [A]Take antilog: [A] / [A]00 = e = e-2.77-2.77 = 0.0627 = 0.0627

0.06270.0627×1.50mg = .0940mg×1.50mg = .0940mg is the fraction is the fraction remaining !remaining !

Start with 1.50 mg of tritium, how much is left Start with 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


5353

SolutionSolution

[A] / [A][A] / [A]00 = 0.0627 = 0.0627

0.0627 is the 0.0627 is the fraction remaining fraction 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 ---> 0.0940 mg after 4---> 0.0940 mg after 4


Start with 1.50 mg of tritium, how much is left Start with 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

5454

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

Try: start with 10g of Sg, see how much is left Try: start with 10g of Sg, see how much is left after an 8 hour work day, assume 1after an 8 hour work day, assume 1stst order. order.

1 principles of reactivity: chemical kinetics chapter 15 1-17 + all bold numbered problems

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