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Unformatted text preview: Chemical Kinetics
Chemical Kinetics 1 Chapter 15
Chapter 15 Chemical Kinetics
Chemical Kinetics 2 • But this gives us no info on HOW FAST
• But this gives us no info on HOW
reaction goes from reactants to products.
reaction goes from reactants to products. Br from biomass burning destroys
Br from biomass burning
stratospheric ozone. (See R.J. Cicerone,
stratospheric ozone. (See R.J. Cicerone,
Science,, volume 263, page 1243, 1994.)
Science volume 263, page 1243, 1994.)
Step 1:
Br + O3 > BrO + O2
Step 1:
Br + O3 > BrO
2 • KINETICS — the study of REACTION
— the study of REACTION
RATES and their relation to the way the
RATES and
reaction proceeds, i.e., its MECHANISM.
reaction proceeds, i.e., its MECHANISM
MECHANISM. Reaction Rates
Reaction Rates NET:
NET: 5 Determining a Reaction Rate
Determining a Reaction Rate Secttion 15..1
Sec ion 15 1 Screen 15.2 Dye Conc
Dye Conc • Reaction rate = change in
concentration of a reactant or
product with time.
• Three “types” of rates
–initial rate
–average rate
–instantaneous rate Step 2:
Step 2:
Step 3:
Step 3: • The reaction mechanism is our goal!
• The reaction mechanism is our goal! 4 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. Time
Time Page 1 3 The sequence of events at the molecular
level that control the speed and
outcome of a reaction. • We can use thermodynamics to tell if
• We can use thermodynamics to tell if
a reaction is product or reactant favored.
a reaction is product or reactant favored. An automotive catalytic muffler Reaction Mechanisms
Reaction Mechanisms Cl + O3 > ClO + O2
Cl + O3 > ClO
2
BrO + ClO + llight > Br + Cl + O2
BrO ClO ight
Cl
2
2 O3 > 3 O2
2 O3 > 3 O2 Factors Affecting Rates
Sectiion 15.2
Sect on 15.2 • Concentrations
• and physical state of
reactants and products (Screens
reactants
(Screens
15.315.4) • Temperature (Screen 15.11)
(Screen
• Catalysts (Screen 15.14)
(Screen 6 Factors Affecting Rates 7 Factors Affecting Rates
Factors Affecting Rates Sectiion 15.2
Sect on 15.2 • Concentrations 8 Rate with 0.3 M HCl 9 Factors Affecting Rates
Factors Affecting Rates • Physical state of reactants Catalysts: catalyzed decomp of H2O2
2 H2O2 > 2 H2O + O2 Rate with 6.0 M HCl 10 Factors Affecting Rates
Factors Affecting Rates
• Temperature 11 Concentrations and Rates
To postulate a reaction
mechanism, we study • reaction rate and
•reaction
and
• its concentration
its
•its concentration
dependence Concentrations and Rates
Take reaction
Take reaction
where Cl iin
where Cl n
cisplatin
cisplatin
[Pt(NH3)2Cl3]
[Pt(NH3)2Cl3]
is replaced
is replaced
by H2O
by H2O Rate of change of conc of Pt compd
= Page 2 Cisplatin Am't of cisplatin reacting (mol/L)
elapsed time (t) 12 Concentrations and Rates Cisplatin 13
Cisplatin Rate of change of conc of Pt compd
= 14 Am't of cisplatin reacting (mol/L)
elapsed time (t) m
n
p
Rate = k [A]m[B]n[C]p • If m = 2, rxn. is 2nd order in A.
• If m = 2, rxn.
Rate = k [A]2
Rate = k [A]2 The exponents m, n, and p
The exponents m, n, and p
• are the reaction order
are
reaction
•are Rate of reaction = k [Pt(NH 3)2Cl2]
Rate of reaction = k [Pt(NH 3)2Cl2] Doubling [A] increases rate by ________
Doubling [A] increases rate by ________
• If m = 0, rxn. is zero order.
• If m = 0, rxn.
Rate = k [A]0
Rate = k [A]0 • can be 0, 1, 2 or fractions
•can
fractions where k = rate constant
where
rate • must be determined by experiment!
•must
experiment! k is independent of conc. but increases with T
k is independent of conc. Deriving Rate Laws If [A] doubles, rate ________
If [A] doubles, rate ________ 16 Cisplatin Deriving Rate Laws Derive rate law and k for
CH3CHO(g) > CH4(g) + CO(g)
from experimental data for rate of disappearance of
CH3CHO Rate of rxn = k [CH3CHO]2
Rate of rxn = k [CH3CHO]2 Expt.
Expt. [CH3CHO]
[CH 3CHO]
(mol/L)
(mol/L) Now determine the value of k. Use expt. #3 data—
Now determine the value of k. Use expt. #3 data— 1
1
2
2 0.10
0.10
0.20
0.20 0.020
0.020
0.081
0.081 3
3
4
4 0.30
0.30
0.40
0.40 0.182
0.182
0.318
0.318 Disappear of CH3CHO
Disappear of CH3CHO
(mol/L•sec)
(mol/L•sec) 15 Rate = k [A]1
Rate = k [A]1
If [A] doubles, then rate goes up by factor of __
If [A] doubles, then rate goes up by factor of __ a A + b B > x X with a catalyst C
with a catalyst C We express this as a RATE LAW
We express this as a RATE Interpreting Rate Laws Cisplatin Rate = k [A]m[B]n[C]p
Rate = k [A]m[B]n[C]p
• If m = 1, rxn. is 1st order in A
• If m = 1, rxn. In general, for
In general, for Rate of reaction is proportional to [Pt(NH 3)2Cl2]
Rate of reaction is proportional to [Pt(NH 3)2Cl2] Cisplatin Concentrations, Rates, &
Rate Laws Here the rate goes up by ______ when initial conc.
Here the rate goes up by ______ when initial conc.
doubles. Therefore, we say this reaction is
doubles. Therefore, we say this reaction is
_________________ order.
_________________ order.
0.182 mol/L•s = k (0.30 mol/L)2
0.182 mol/L•s = k (0.30 mol/L)2 k = 2.0 (L / mol•s)
k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of [CH3CHO]
Using k you can calc. rate at other values of [CH3CHO]
at same T.
at same T. Page 3 17 Cisplatin Concentration/Time Relations Need to know what conc. of reactant is as
Need to know what conc.
function of time. Consider FIRST ORDER
function of time. Consider FIRST ORDER
REACTIONS
REACTIONS
The rate law is
The rate law is [A]
= k [ A]
time 18 Cisplatin Concentration/Time Relations
Integrating  ( [A] //
Integrating  ( [A] ln
natural
logarithm 19 Sucrose decomposes
Sucrose decomposes
to simpler sugars
to simpler sugars
Rate of disappearance
Rate of disappearance
of sucrose
of sucrose
= k [sucrose]
= k [sucrose]
k = 0.21 hr1
k = 0.21 hr1 time) = k [A], we get
time) = k [A], we get [A] =  k t
[A]o [A] // [A]00 =fraction remaining after time tt
[A] [A] =fraction remaining after time
has elapsed.
has elapsed. 21 Rate of disappear of sucrose = k [sucrose], k = 0.21 hr1. If
If
initial [sucrose] = 0.010 M, how long to drop 90% or to
0.0010 M? Sucrose Use the first order integrated rate law ln Initial [sucrose] =
Initial [sucrose] =
0.010 M
0.010 M
How long to drop 90%
How long to drop 90%
(to 0.0010 M)?
(to 0.0010 M)? [A] at time = 0 Concentration/Time Relations 20 Concentration/Time Relations 0.0010 M
0.010 M =  (0.21 hr1) t ln (0.100) =  2.3 =  (0.21 hr 1) • time time = 11 hours Called the integrated firstorder rate law .
Called
integrated Using the Integrated Rate Law
The integrated rate law suggests a way to tell
the order based on experiment.
2 N2O5(g) > 4 NO 2(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
ln
0
0.35
0.70
1.75 Rate = k [N 2O5] 22 23 Using the Integrated Rate Law Using the Integrated Rate Law 2 N2O5(g) > 4 NO 2(g) + O2(g) Rate = k [N 2O5] Plot of lln [[N22O55] vs. time
Plot of n N O ] vs. time
is a straight line!
is a straight line!
Eqn. for straight line:
Eqn. for straight line:
y = ax + b
y = ax + b
ax l n [N 2 O 5 ] v s . t i m e
0 [N 2 O 5 ] v s . t i m e l n [N 2 O 5 ] v s . t i m e 1 0 ln [N 2O5] =  kt + ln [N2O5]o
2
0
0 2
0 ti m e 5 Data of conc. vs.
time plot do not fit
straight line. 0 ti m e 5 Plot of ln [N2 O5] vs.
time is a straight
line! Page 4 ti m e 5 conc at
time t rate const
= slope conc at
time = 0 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) 24 25 26 HalfLife Properties of Reactions HalfLife Sectiion 15..4 & Screen 15..8
Sect on 15 4 & Screen 15 8 Screen 15.7 27 HALFLIFE is the time
it takes for 1/2 a
sample is disappear.
For 1st order reactions,
the concept of HALFLIFE is especially
useful. • Reaction is 1st order
decomposition of
H2O2. 28 29 30 HalfLife HalfLife HalfLife • Reaction after 654
min, 1 halflife.
• 1/2 of the reactant
remains. • Reaction after
1306 min, or 2
halflives.
• 1/4 of the reactant
remains. • Reaction after 3
halflives, or 1962
min.
• 1/8 of the reactant
remains. Page 5 31 HalfLife 32 Section 15.4 & Screen 15.8 HalfLife
HalfLife Sugar is fermented in a 1st order process (using
an enzyme as a catalyst).
sugar + enzyme > products
sugar
Rate of disappear of sugar = k[sugar]
k = 3.3 x 10 4 sec1
What is the halflife of this reaction? • Reaction after 4
halflives, or 2616
min.
• 1/16 of the
reactant remains. Rate = k[sugar] and k = 3.3 x 104 sec1. What is the halflife of this reaction? Solution
[A] / [A]0 = 1/2 when t = t 1/2
Therefore, ln (1/2) =  k • t1/2
 0.693 =  k • t 1/2 t1/2
So, for sugar, = 0.693 / k t1/2 = 0.693 / k = 2100 sec = HalfLife 34 HalfLife Section 15.4 & Screen 15.8 Section 15.4 & Screen 15.8 Rate = k[sugar] and k = 3.3 x 10 4 sec1. Halflife
is 35 min. Start with 5.00 g sugar. How much is
left after 2 hr and 20 min?
Solution
2 hr and 20 min = 4 halflives
hr
halflives
Halflife
Halflife Time Elapsed Mass Left
Time
1st
35 min
2.50 g
2nd
70
1.25 g
3rd
105
0.625 g
4th
140
0.313 g Radioactive decay is a first order process.
Tritium > electron + helium
Tritium
3H
0e
3He
1
If you have 1.50 mg of tritium, how much is left
after 49.2 years? t 1/2 = 12.3 years Page 6 33 HalfLife Section 15.4 & Screen 15.8 35 35 min
35 HalfLife Section 15.4 & Screen 15.8 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 mg
Need k, so we calc k from:
k = 0.693 / t 1/2
Obtain k = 0.0564 y 1
Now ln [A] / [A] 0 = kt =  (0.0564 y1) • (49.2 y)
=  2.77
Take antilog: [A] / [A]0 = e2.77 = 0.0627
0.0627 = fraction remaining
0.0627 fraction 36 HalfLife Section 15.4 & Screen 15.8 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 !
0.0627
fraction
Because [A] 0 = 1.50 mg, [A] = 0.094 mg
But notice that 49.2 y = 4.00 halflives
1.50 mg > 0.750 mg after 1 halflife
> 0.375 mg after 2
> 0.188 mg after 3
> 0.094 mg after 4 37 38 HalfLives of Radioactive Elements
Rate of decay of radioactive isotopes given in
terms of 1/2life.
238U > 234Th + He
4.5 x 10 9 y
> 234
14C > 14N + beta
5730 y
> 14
131I
> 131Xe + beta
8.05 d
> 131
Element 106  seaborgium
263Sg
Sg Page 7 0.9 s
0.9 39 • Darleane Hoffman of
UCBerkeley studies
the newest elements.
She is Director of
Seaborg Institute.
• Rec’d ACS Award in
Nuclear Chem, the
Garvan Medal, and
(in 2000) the ACS
highest award, the
Priestley Medal. ...
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 Fall '08
 McQuade
 Organic chemistry, Kinetics

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