Chemical Kinetics Chapter 15_2_ by pptfiles

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									Chemistry and Chemical Reactivity 6th Edition

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John C. Kotz Paul M. Treichel Gabriela C. Weaver

CHAPTER 15 Principles of Reactivity: Chemical Kinetics
Lecture written by John Kotz as modified by George Rhodes

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My background
• • • •
•

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B.S. Chemistry ASU, 1964 Ph.D. UUtah, Chemistry, 1973 Asst. Prof., USAF Academy 1968-1970 New laser research, USAF Weapons Lab 1970-1974 Directed AEC Laser Isotope Separation Program, 1974-1976 Directed ERDA/DOE Solar Design and Construction program 1976-1978 VP Energy Programs, The BDM Corp, 19781982 Sr. VP, EMP, Energy, BDM 1982-1989 VP, Technology, Quatro, Inc, 1989-1996

Additional background
• Pres/CEO, Dynamic Resonance Systems, 1996-1999 • Ex. VP & Chief Operating Officer, Avistar, Inc. (PNM Resources), 1999-2006 • Chief Scientist, PNM Resources, 2006 to present • Visiting Faculty, NAU 2006-Present • Guest Scientist, Los Alamos National Laboratory 1970-2000 • Named as Eminent Scholar, NMIT, 1989 • DOE Award for Excellence in Technology Transfer 1993 • 11 Patents (2 pending), over 50 publications • Commercial pilots license, swimming (WSI), squash, golf, PADI certified diver, fly fishing…
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First, let’s review algebra
• • • • • • (a) a+b=b+a (c) ab=ba (e) (ab)c=a(bc) (g) abac=a(b+c) (i) a+0=a (k) a1=a (b) a+(b+c)=(a+b)+c (d) a(b+c)=ab+ac (f) (ab)c=acbc (h) (ab)c=a(bc) (j) a . 1=a

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Forward and inverse operations
• • • • • • • • (a) addition a+b=c (b) multiplication ab=c (c) power ba=c (d) power ab=c (a') subtraction b=c-a (b') division b=c/a (c') root b=a√c (d') logarithm b=logac

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Logarithms
• y=logax or logax=y • Log 1000 = log (103) = 3 which is: the power to which you must raise 10 to get the number • Log 2.10 = 0.3222 or 10 0.3222 = 2.10 • and ln x = 2.303 log x • The natural logarithm is the logarithm having base e, where • This function can be defined as • • For x>0

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More on Logarithms
• ln x = 2.303 log x • So when logax=y to get the antilog you simply solve x=10y , and when using natural logs • When lny=x the antilog is: y=ex • Let’s take an example • The ln of 1000 is: 6.9078 • If you divide by 2.303 you get 3 • 103 = 1000, e6.9078 = 1000 • Isn’t it great when a plan comes together?

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Definitions: Chemical
• • • • • • A substance that: 1) An organic chemist turns into a foul odor; 2) an analytical chemist turns into a procedure; 3) a physical chemist turns into a straight line; 4) a biochemist turns into a helix; 5) a chemical engineer turns into a profit.

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The Exam Schedule
• Exams: – July 14th – Kinetics and Chemical Equilibria – July 21st – Acids and Bases and Principles of Reactivity – July 28th – Entropy and Free Energy and Electron Transfer Reactions – August 4th – Nuclear Chemistry and Carbon and the Environment – August 5th - Final

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Advice
• If you are taking another time consuming course in addition to CHM 152 and 152 Lab, consider dropping the other course if you wish to do well in chemistry. • Experience has shown that you have a very high probability of doing poorly in one or both

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The Facts
• There is no way we can cover the necessary material without using PowerPoint. • We will cover over 20 pages per day in the text, plus problems • This is a math course and you must be comfortable with algebraic operations • With this background, you will learn a lot and be well prepared for future science courses. • All lectures will be posted on the Chem Dept web site at http://www.nau.edu/chem/ • Go to online course materials and select the appropriate lecture

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Chemical Kinetics
Chapter 15

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Chapter Goals p-741-742
• Understand rates of reaction and the conditions affecting rates • Derive the rate equation, rate constant, and reaction order from experimental data • Use integrated rate laws • Understand collision theory of Rx rates and the role of activation energy • Relate Rx mechanisms and rate laws

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Thermodynamics
• We can use thermodynamics to tell if a reaction is product- or reactantfavored. • But this gives us no info on HOW FAST reaction goes from reactants to products. • This is the study of chemical kinetics

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Chemical Kinetics
• KINETICS — the study of REACTION
RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. • Our goal is to understand the reaction mechanisms so that we can tell how fast a reaction will take place

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Reaction Mechanisms
The sequence of events at the molecular level that control the speed and outcome of a reaction.
Br & Cl from biomass burning destroys stratospheric ozone.
(See R.J. Cicerone, Science, volume 263, page 1243, 1994.)

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Step 1:

Br + O3 ---> BrO + O2

Step 2: Step 3: NET:
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Cl + O3 ---> ClO + O2 BrO + ClO + light ---> Br + Cl + O2 2 O3 ---> 3 O2

Reaction Rates
Section 15.1

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• Reaction rate = change in concentration of a reactant or product with time.
• Three ―types‖ of rates –initial rate –average rate –instantaneous rate
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Determining a Reaction Rate
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.
Screen 15.2

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Dye Conc

Time
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The Reaction Rate
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

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A
D[A] rate = Dt

B

D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt
Because [A] decreases with time, D[A] is negative.

D[B] rate = Dt

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Determining a Reaction Rate

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Factors Affecting Rates
Section 15.2

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

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Concentrations & Rates
Section 15.2

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0.3 M HCl

6 M HCl

Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g)

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Factors Affecting Rates
• Physical state of reactants

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Factors Affecting Rates
Catalysts: catalyzed decomp of H2O2 2 H2O2 --> 2 H2O + O2

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A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
k = A • exp( -Ea/RT ) Ea k

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uncatalyzed

catalyzed

ratecatalyzed > rateuncatalyzed
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Ea < Ea ‘

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Factors Affecting Rates
• Temperature

Bleach at 54 ˚C

Bleach at 22 ˚C

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Iodine Clock Reaction
1. Iodide is oxidized to iodine H2O2 + 2 I- + 2 H+ -----> 2 H2O + I2 2. I2 reduced to I- with vitamin C I2 + C6H8O6 ----> C6H6O6 + 2 H+ + 2 IWhen all vitamin C is depleted, the I2 interacts
with starch to give a blue complex.
Page 705 in CCR
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Iodine Clock Reaction

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Concentrations and Rates
To postulate a reaction mechanism, we study

• reaction rate and • its concentration dependence

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Concentrations and Rates
Take reaction where Cl- in cisplatin [Pt(NH3)2Cl3] is replaced by H2O

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Rate of change of conc of Pt compd Am't of cisplatin reacting (mol/L) = elapsed time (t)
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Concentrations & Rates
Rate of change of conc of Pt compd Am't of cisplatin reacting (mol/L) = elapsed time (t)
Rate of reaction is proportional to [Pt(NH3)2Cl2] We express this as a RATE LAW

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Rate of reaction = k [Pt(NH3)2Cl2]
where k = rate constant

k is independent of conc. but increases with T

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To review
Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A
D[A] rate = Dt

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B

D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt
Because [A] decreases with time, D[A] is negative.

D[B] rate = Dt

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Concentrations, Rates, & Rate Laws
In general, for

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a A + b B + c C --> x X with a
catalyst C

Rate = k [A]m[B]n[C]p
The exponents m, n, and p

• are the reaction order • can be 0, 1, 2 or fractions • can only be determined by experiment!
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Reaction Rates and Stoichiometry
2A B

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Two moles of A disappear for each mole of B that is formed. 1 D[A] rate = 2 Dt aA + bB D[B] rate = Dt cC + dD

1 D[A] 1 D[B] 1 D[C] 1 D[D] rate = == = a Dt b Dt c Dt d Dt

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Interpreting Rate Laws
Rate = k [A]m[B]n[C]p

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• If m = 1, rxn. is 1st order in A Rate = k [A]1 If [A] doubles, then rate goes up by factor of __ • If m = 2, rxn. is 2nd order in A. Rate = k [A]2 Doubling [A] increases rate by ________ • If m = 0, rxn. is zero order. Rate = k [A]0 If [A] doubles, rate ________

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Deriving Rate Laws
Derive rate law and k for CH3CHO(g) --> CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO

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Expt. [CH3CHO] (mol/L) 1 2 3 4
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Disappear of CH3CHO (mol/L•sec) 0.020 0.081 0.182 0.318

0.10 0.20 0.30 0.40

Deriving Rate Laws
Rate of rxn = k [CH3CHO]

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2

Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ order. Now determine the value of k. Use expt. #3 data—
0.182 mol/L•s = k (0.30 mol/L)2

k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of [CH3CHO] at same T.

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Concentration/Time Relations
What is concentration of reactant as function of time? Consider FIRST ORDER REACTIONS The rate law is

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D[A] Rate  = k [A] Dtime
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Cisplatin

Concentration/Time Relations
Integrating - (∆ [A] / ∆ time) = k [A], we get

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[A] / [A]0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law.
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Write the rate expression for the following reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

D[CH4] D[CO2] 1 D[O2] 1 D[H2O] rate = = == Dt Dt Dt 2 Dt 2

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The Rate Law

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The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB

cC + dD

Rate = k [A]x[B]y

reaction is xth order in A reaction is yth order in B
reaction is (x +y)th order overall

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F2 (g) + 2ClO2 (g)

2FClO2 (g)

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rate = k [F2]x[ClO2]y

Double [F2] with [ClO2] constant
Rate doubles

x=1 Quadruple [ClO2] with [F2] constant Rate quadruples
y=1
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rate = k [F2][ClO2]

Concentration/Time Relations
Sucrose decomposes to simpler sugars

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Rate of disappearance of sucrose = k [sucrose] If k = 0.21 hr-1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)?

Glucose
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Concentration/Time Relations
Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?

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Use the first order integrated rate law

0.0010 M ln 0.010 M

= - (0.21 hr-1) t

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

time = 11 hours How many seconds?
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Using the Integrated Rate Law
The integrated rate law suggests a way to tell the order based on experiment. 2 N2O5(g) ---> 4 NO2(g) + O2(g) Time (min) 0 1.0 2.0 5.0 [N2O5]0 (M) 1.00 0.705 0.497 0.173 ln [N2O5]0 0 -0.35 -0.70 -1.75

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Rate = k [N2O5]
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Using the Integrated Rate Law
2 N2O5(g) ---> 4 NO2(g) + O2(g) Rate = k [N2O5]

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Data of conc. vs. time plot do not fit straight line.
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Plot of ln [N2O5] vs. time is a straight line!

Using the Integrated Rate Law
Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b
ln [N 2O5] = - kt + ln [N 2O5]o conc at time t rate const = slope conc at time = 0

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All 1st order reactions have straight line plot for ln [A] vs. time. (2nd order gives straight line for plot of 1/[A] vs. time)
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Properties of Reactions
page 719

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Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
Experiment

[S2O82-]

[I-]

Initial Rate (M/s) 2.2 x 10-4 1.1 x 10-4 2.2 x 10-4

1 2 3

0.08 0.08 0.16

0.034 0.017 0.017

rate = k [S2O82-]x[I-]y y=1 x=1 rate = k [S2O82-][I-]

Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) 2.2 x 10-4 M/s rate k= = = 0.08/M•s 2-][I-] [S2O8 (0.08 M)(0.034 M)
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Definitions: 1st Order Reaction
• First Order Reaction: The reaction that occurs first, not always the one desired. For example, the formation of brown gunk in an organic prep.

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Learn this equation & how to use

[ N ]0 ln  kt [ N ]t
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An Example
• A reaction is 25% complete in 42 seconds. Calculate the half life.
[ N ]t   kt [ N ]0 [ N ]0 ln  kt [ N ]t ln

or

Ln 1/.75=0.287, Div by 42 and k=6.84x10-3 T1/2 =0.693/k=101 s

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Half-Life
Section 15.4

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.

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Half-Life
• Reaction is 1st order decomposition of H2O2.

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Half-Life
• Reaction after 1 half-life. • 1/2 of the reactant has been consumed and 1/2 remains.

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Half-Life
• After 2 half-lives 1/4 of the reactant remains.

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Half-Life
• A 3 half-lives 1/8 of the reactant remains.

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Half-Life
• After 4 half-lives 1/16 of the reactant remains.

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Section 15.4 & Screen 15.8

Half-Life

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Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products
Rate of disappear of sugar = k[sugar]

k = 3.3 x 10-4 sec-1
What is the half-life of this reaction?

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Half-Life
Section 15.4

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Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the halflife of this reaction? Solution [A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = _________ Therefore, ln (1/2) = - k • t1/2 - 0.693 = - k • t1/2

t1/2 = 0.693 / k
So, for sugar, t1/2 = 0.693 / k = 2100 sec = 35
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min

Can you perform this operation on your calculator?

Half-Life
Section 15.4

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Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min 2.50 g 2nd 70 1.25 g 3rd 105 0.625 g 4th 140 0.313 g
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Half-Life
Section 15.4

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Radioactive decay is a first order process. Tritium ---> electron + helium 3H 0 e 3He -1 t1/2 = 12.3 years
If you have 1.50 mg of tritium, how much is left

after 49.2 years?

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Half-Life
Section 15.4

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Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution ln [A] / [A]0 = -kt [A] = ? [A]0 = 1.50 mg t = 49.2 y Need k, so we calc k from: k = 0.693 / t1/2 Obtain k = 0.0564 y-1 Now ln [A] / [A]0 = -kt = - (0.0564 y-1) • (49.2 y) = - 2.77 Take antilog: [A] / [A]0 = e-2.77 = 0.0627 0.0627 = fraction remaining

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Half-Life
Section 15.4

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Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution [A] / [A]0 = 0.0627 0.0627 is the fraction remaining! Because [A]0 = 1.50 mg, [A] = 0.094 mg But notice that 49.2 y = 4.00 half-lives 1.50 mg ---> 0.750 mg after 1 half-life ---> 0.375 mg after 2 ---> 0.188 mg after 3 ---> 0.094 mg after 4

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Half-Lives of Radioactive Elements
Rate of decay of radioactive isotopes given in terms of 1/2-life. 238U --> 234Th + He 4.5 x 109 y 14C --> 14N + beta 5730 y 131I --> 131Xe + beta 8.05 d Element 106 - seaborgium 263Sg 0.9 s

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A Microscopic View of Reactions
Sections 15.5

MECHANISMS

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Mechanism: how reactants are converted to products at the molecular level. RATE LAW ----> MECHANISM experiment ----> theory

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Activation Energy
Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, Ea.

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Reaction coordinate diagram
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Activation Energy
• The useful quantity of energy available in one cup of coffee

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MECHANISMS & Activation Energy

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Conversion of cis to trans-2-butene requires twisting around the C=C bond. Rate = k [trans-2-butene]
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MECHANISMS
Cis Transition state Trans

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Activation energy barrier

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MECHANISMS
Energy involved in conversion of trans to cis butene

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energy

Activated Complex -266 kJ 4 kJ/mol trans
See Figure 15.14

+262 kJ

cis

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Mechanisms
• Reaction passes thru a TRANSITION STATE where there is an that has sufficient energy to become a product.

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activated complex

ACTIVATION ENERGY, Ea
= energy req’d to form activated complex.

Here Ea = 262 kJ/mol

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MECHANISMS
Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, cis ---> trans is EXOTHERMIC This is the connection between thermodynamics and kinetics.

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Effect of Temperature
• Reactions generally occur slower at lower T.
Room temperature In ice at 0 oC

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Iodine clock reaction, Screen 15.11, and book page 705. H2O2 + 2 I- + 2 H+ --> 2 H2O + I2

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Activation Energy and Temperature
Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules.

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In general,

differences in activation energy cause
reactions to vary from fast to slow.

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Mechanisms
1. Why is trans-butene <--> cis-butene reaction observed to be 1st order? As [trans] doubles, number of molecules with enough E also doubles. 2. Why is the trans <--> cis reaction faster at higher temperature? Fraction of molecules with sufficient activation energy increases with T.

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More About Activation Energy Arrhenius equation —
Rate constant Temp (K)

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k  Ae

-Ea / RT

Frequency factor

Activation 8.31 x 10-3 kJ/K•mol energy

Frequency factor related to frequency of collisions with correct geometry.

Ea 1 ln k = - ( )( ) + ln A R T
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Plot ln k vs. 1/T ---> straight line. slope = -Ea/R

More on Mechanisms
A bimolecular reaction
Reaction of

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trans-butene --> cis-butene is UNIMOLECULAR - only one reactant is involved. BIMOLECULAR — two different molecules must collide --> products

Exo- or endothermic?
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Collision Theory
Reactions require (a) activation energy and (b) correct geometry. O3(g) + NO(g) ---> O2(g) + NO2(g)

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1. Activation energy

2. Activation energy and geometry

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Mechanisms
O3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps.
Adding elementary steps gives NET reaction.

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Mechanisms
Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] NOTE 1. Rate law comes from experiment 2. Order and stoichiometric coefficients not necessarily the same! 3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction.

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Mechanisms
Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2]
Proposed Mechanism Step 1 — slow Step 2 — fast Step 3 — fast HOOH + I- --> HOI + OHHOI + I- --> I2 + OH2 OH- + 2 H+ --> 2 H2O

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Rate of the reaction controlled by slow step —

RATE DETERMINING STEP, rds.
Rate can be no faster than rds!

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Mechanisms
2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2]
Step 1 — slow HOOH + I- --> HOI + OHStep 2 — fast HOI + I- --> I2 + OH-

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Step 3 — fast 2 OH- + 2 H+ --> 2 H2O

Elementary Step 1 is bimolecular and involves I- and HOOH. Therefore, this predicts the rate law should be
Rate  [I-] [H2O2] — as observed!!

The species HOI and OH- are reaction

intermediates.

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Rate Laws and Mechanisms
NO2 + CO reaction: Rate = k[NO2]2

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Two possible mechanisms

Two steps: step 1

Single step Two steps: step 2
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Ozone Decomposition over Antarctica

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2 O3 (g) ---> 3 O2 (g)

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Ozone Decomposition Mechanism
2 O3 (g) ---> 3 O2 (g) Proposed mechanism Step 1: fast, equilibrium O3 (g) → O2 (g) + O (g) Step 2: slow O3 (g) + O (g) ---> 2 O2 (g)

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[O 3 ]2 Rate = k [O 2 ]

CATALYSIS
Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier.
Dr. James Cusumano, Catalytica Inc.

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What is a catalyst?

Catalysts and the environment

Catalysts and society
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CATALYSIS
In auto exhaust systems — Pt, NiO

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2 CO + O2 ---> 2 CO2 2 NO ---> N2 + O2

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CATALYSIS

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2. Polymers: H2C=CH2 ---> polyethylene 3. Acetic acid:

CH3OH + CO --> CH3CO2H 4. Enzymes — biological catalysts

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CATALYSIS
Catalysis and activation energy

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MnO2 catalyzes decomposition of H2O2 2 H2O2 ---> 2 H2O + O2

Uncatalyzed reaction

Catalyzed reaction
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Iodine-Catalyzed Isomerization of cis-2-Butene

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Figure 15.16
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Iodine-Catalyzed Isomerization of cis-2-Butene

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Practice Exercises
• 2 O3 (g)  3 O2 (g) give the relative rates of disappearance and formation of the reactant and product.

1 D O3  1 D O2  Reaction Rate = -  =+  2 Dt 3 Dt

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Practice Exercises 2
15.4 if -D(H2)/ Dt = 4.5x10-4 mol/L.min, what is D(NH3)/ Dt? • N2(g) + 3H2(g)  2NH3 (g) •

D[NH3 ] D[H2 ] 2 mol NH3 4.5  10–4 mol/L 2 mol NH3 =– · = · = 3.0  10–4 mol/L  min Dt Dt 3 mol H2 min 3 mol H2

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Practice Exercises 3
• 15.11 The reaction of 2NO(g) + O2(g) 2NO2(g) • Experiment [NO] [O2] • i 0.010 0.010 2.5 x 10-5 • ii 0.020 0.010 1.0 x 10-4 • iii 0.010 0.020 5.0 x 10-5

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Practice Exercises 3’
• What is the order of each reaction? Ans This rate change was the result of doubling the concentration of NO. The order of dependence of NO must be second order, Looking at the first and third data sets, [NO] is constant and [O2] doubles. The rate also doubles from the first data set to the third, so the reaction is first order in O2.

Write the rate equation for the reaction, 2NO+O22NO2 • Rate = k[NO]2[O2] • Calculate the rate constant • 1  D[NO]  1
Rate =  2 Dt

–5 –5  = (2.5  10 mol/L  s) = 1.3  10 mol/L  s  2

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Practice Exercises 3‖
• Thus
Rate 1.3  10–5 mol/L  s k= = = 13 L2 /mol2  s [NO]2 [O2 ] (0.010 mol/L)2 (0.010 mol/L)

• What is the rate in mol/L.s when (NO)=0.015 mol/L and (O2)=0.0050 mol/L? • Rate = k[NO]2[O2] = (13 L2/mol2·s)(0.015 mol/L)2(0.0050 mol/L) = 1.4  10–5 mol/L·s
• When NO is reacting at 1.0x10-4mol/L.s, what is the rate that O2 is reacting and NO2 is forming?

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Practice Exercises 3‖’
• Rate at which O2 is reacting =
1.0  10–4 mol/L 1 mol O2 · = 5.0  10–5 mol/L  s s 2 mol NO

• Rate at which NO2 is forming=
1.0  10–4 mol/L 2 mol NO2 · = 1.0  10–4 mol/L  s s 2 mol NO

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