Transcript Chemistry XXI

Unit 5
How do we predict chemical change?
The central goal of this unit is to help you identify and
apply the different factors that help predict the
likelihood of chemical reactions.
Chemistry XXI
M1. Analyzing Structure
M2. Comparing Free Energies
Comparing the relative stability
of different substances
Determining the directionality and
extent of a chemical reaction.
M3. Measuring Rates
Analyzing the factors that
affect reaction rate.
M4. Understanding Mechanism
Identifying the steps that
determine reaction rates.
Unit 5
How do we predict
chemical change?
Chemistry XXI
Module 3: Measuring Rates
Central goal:
To analyze the effect of
concentration and
temperature on the rate
of chemical reactions.
The Challenge
Transformation
How do I change it?
Chemistry XXI
Imagine that you were
interested in comparing the
rates at which different
substances appeared or
were decomposed on the
primitive Earth.
How could we evaluate the kinetic stability of a
substance?
How could we determine the effect of
concentration and temperature on reaction rates?
Time Issues
Determining DGorxn or K for a chemical reaction
allows us to predict the directionality and extent
of the process, but tell us nothing about how long
it will take to happen.
Chemistry XXI
Consider these two possible routes for
the synthesis of glycine, the simplest
amino acid, on the primitive Earth:
CH2O(g) + HCN(g) + H2O(l)  C2H5NO2(s) DGorxn = -154. kJ
2 CH4(g) + NH3(g) + 5/2 O2(g)  C2H5NO2(s) + 3 H2O(l)
DGorxn = -965. kJ
Activation Energy
Ea
DG
2 CH4(g) + NH3(g)
+ 5/2 O2(g)
Thermo vs.
Kinetics
Occurs readily
at 25 oC
(Low Ea)
CH2O(g) +
HCN(g) + H2O(l)
Chemistry XXI
C2H5NO2(s)
Themodynamically
favored,
but does not occur for
all practical purposes
(High Ea)
C2H5NO2(s) + 3 H2O(l)
Reaction Coordinate
Analyzing Stability
Chemistry XXI
Analyzing chemical systems from both the
thermodynamic and kinetic point of view is crucial
in making decisions about the actual
“stability” of substances.
For example, the
decomposition or
transformation of a
substance may be
favored
thermodynamically,
but can take millions
of years to occur.
How stable is it then?
C(diamond)  C(graphite)
DGotr = -2.9 kJ/mol
Ea ~ 728 kJ/mol
Kinetic Stability
Chemistry XXI
The analysis of the kinetic stability of biomolecules
has been crucial in the analysis of different
theories about the origin of life.
For example, it has been
proposed that amino acid
synthesis could have
occurred deep in the Earth's
crust and that these amino
acids were subsequently
shot up along with
hydrothermal fluids into
cooler waters.
CH4 and NH3 are abundant
in hydrothermal vent
regions (60-400 oC).
How stable are amino acids under such conditions?
Unstable?
Many aqueous solutions of amino acids are
“thermodinamically unstable.”
Let’s consider the case of alanine:
2
2
3
Chemistry XXI
Alanine (Ala)
3
+
Decarboxylation
2
DGorxn < 0
Ethyl Amine
The kinetics of this reaction has been thoroughly
explored by measuring the concentration of
alanine [Ala] as a function of time (t) in aqueous
solutions at various temperatures.
Let’s Think
How would you quantify the rate of
decomposition of alanine at any given time?
o
[Ala] (mM)
Chemistry XXI
T = 200 C (473 K)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
3
6
9 12 15 18 21 24 27 30 33 36
Time (days)
Reaction Rate
Chemistry XXI
[Ala] (mM)
T = 200 oC (473 K)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
D[ Ala ]
Rate Average  
Dt
Rate Average  
D[Ala]
Dt
0
3
6
[0.5  0.7]
mM
 0.02
[18  9]
year
RateInst
d [ Ala ]

dt
9 12 15 18 21 24 27 30 33 36
Time (days)
Let’s Think
Hint: How does the
rate change with
C and T?
(The higher the rate,
the lower the kinetic
stability)
[Ala] (mM)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10
12
14
16
18
20
Time (Years)
T = 200 oC (473 K)
[Ala] (mM)
Chemistry XXI
What does this data
tell you about the
kinetic stability of
alanine as a function
of concentration and
temperature?
T = 150 o C (423 K)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
3
6
9 12 15 18 21 24 27 30 33 36
Time (days)
Chemistry XXI
The rate of reaction
increases with
increasing
temperature T.
Kinetic stability is a
function of [R] and T.
[Ala] (mM)
In general, the rate of
reaction decreases
as the concentration
of the reactants [R]
decreases.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
The slope decreases
0
2
4
6
8
10
12
14
16
18
20
Time (Years)
T = 200 oC (473 K)
[Ala] (mM)
Reaction
Rate
T = 150 o C (423 K)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
3
6
9 12 15 18 21 24 27 30 33 36
Time (days)
Rate Laws
The effect of temperature and concentration on
reaction rates can be modeled mathematically:
xA+yB+zC wD+yE+zF
RATE LAW
Rate = k
[A]a
[B]b
[C]c
Reaction
order
Chemistry XXI
Rate Constant
k depends on the value of T, Ea, and other
relevant factors for each reaction.
k = f (T, Ea, surface area….)
Concentration Effects
How can determine reaction orders and rate constants?
Rate
Law
Rate  k[ Ala ]
a
Reaction order?
Rate constant?
What are their
values?
Chemistry XXI
We may assume values for the reaction order a
and analyze the implications:
If a = 1 (first-order):
d [ Ala ]
Rate  
 k[ Ala ]
dt
By integration of this
differential equation we get:
[ Ala ]  [ Ala ]o e  kt
ln[ Ala]  ln[ Ala]o  kt
Graphical Analysis
If the reaction is first-order:
ln[ Ala]  ln[ Ala]o  kt
y
=
b
For example, is the
decomposition of alanine
at 150 oC (423 K)
1st order?:
C3H7NO2  C2H7N + CO2
+ mx
t (years) [Ala] (mM)
ln[Ala]
Chemistry XXI
ln[Ala]o
m = -k
t
0
1.000
2
0.8705
4
0.7578
6
0.6597
8
0.5743
10
0.5000
12
0.4352
14
0.3789
16
0.3298
Reaction Order
t (Years)
C3H7NO2  C2H7N + CO2
0
2
4
6
8
10
12
14
16
18
0
-0.2
ln[Ala]
0
1.000
0
2
0.8705
-0.1387
-0.4
-0.6
ln([Ala])
t (years) [Ala] (mM)
-0.8
-1
-1.2
4
0.7578
-0.2773
ln([Ala]) = -0.0693t
-1.4
Chemistry XXI
-1.6
6
0.6597
-0.4160
8
0.5743
-0.5546
10
0.5000
-0.6931
12
0.4352
-0.7914
14
0.3789
-0.9704
16
0.3298
-1.1092
-1.8
We have a first-order
reaction Rate = k[Ala]a
with:
a=1
k = 0.0693 years-1
Rate = 0.0693[Ala]
20
Let’s Think
Given the Rate Law:
Rate = 0.0693[Ala]
C3H7NO2  C2H7N + CO2
Chemistry XXI
[Ala] = [Ala]oe-0.0693t
If [Ala]o = 1.00 mM, predict the time it will take for
[Ala] to reach the values
0.50 mM, 0.25 mM and 0.125 mM.
How long does it take to halve the concentration?
t1 = 10 y
t2 = 20 y
t3 = 30 y
It takes 10 years to decrease the
concentration by half,
independent of the concentration.
Half Life
1
A half-life is the time
it takes for the
concentration of a
reactant to be
reduced in half.
0.75
[Ala] (mM)
Chemistry XXI
t=0
t = 1 half-life
T = 150 oC (423 K)
0.875
0.625
1 half-life
0.5
0.375
2 half-lives
0.25
3
4
0.125
0
0
t = 2 half-lives t = 3 half-lives
5
10
15
20
25 30
t (Years)
35
40
Half Life
For first order reactions:
Chemistry XXI
If [C] = [C]o/2
t1/ 2
[C ]
ln
  kt
[C ]o
1
ln  o  kt1/ 2
2
ln( 1 / 2)

k
t1/2
Half Life
Independent of
Concentration
t1/2 = 10 years for alanine at 150 oC.
Concentration Effects
If the reaction is first-order
Rate = k[C]:
ln[ C ]  ln[ C ]o  kt
What if Rate = k[C]a
Chemistry XXI
d [C ]
Rate  
 k[C ]2
dt
with a = 2 (second-order)?
1/[C]
By integration we get:
1
1

 kt
[C ] [C ] o
1/[C]o
m=k
t
Let’s Think
Is half-life for second order reactions independent
of the initial concentration of reactant?
Rate = k[C]2
Chemistry XXI
If [C] = [C]o/2
t1/ 2
1
1

 kt
[C ] [C ] o
1

k[C ]
t1/2
Half Life
t1/2 is only independent of [C]o for
first order processes.
Temperature Effects
How can we predict how rate varies with temperature?
Chemistry XXI
The decomposition of alanine at different
temperatures illustrates the effect of
T on the reaction rate.
T (K)
323
373
423
473
523
573
k (y-1)
t1/2 (y)
1.17 x 10-8 5.90 x 107
8.10 x 10-5 8.56 x 103
6.93 x 10-2 1.00 x 101
1.42 x
101
4.87 x
10-2
1.06 x 103 6.55 x 10-4
3.71 x 104 1.87 x 10-5
Larger T 
Larger rate constant 
Shorter half lives.
Ho do we explain
it and make
quantitative
predictions?
Collision Rate Model
According to this model:
1. For a reaction to occur, the
reactant particles must collide.
Chemistry XXI
2. Colliding particles must be
positioned so that the reacting
groups interact effectively.
3. Colliding particles must have
enough energy to reach a
transition state that leads to
the formation of the new
products.
Ep
Transition State
Ea
R
DHrxn
P
Reaction Coordinate
Arhenius Equation
The fraction of molecules
with enough energy to
react at a given T is
proportional to:
Chemistry XXI
e
Ea

RT
The rate constant k is
then given by:
k  Ae

Ea
RT
Likelihood of collisions
Ea
ln( k )  
 ln( A)
RT
y
=
y  ln(k)
mx +
b
x  1/T
m = -Ea/R b = ln(A)
Let’s Think
ln( k )  
Ea
 ln( A)
RT
T (K)
323
373
423
473
523
573
623
k (y-1)
1.17 x 10-8
8.10 x 10-5
6.93 x
10-2
1.42 x
101
1.06 x 103
3
t1/ 2
ln( 1 / 2)

k
+
15
10
ln (k) = -21310(1/T) + 47.709
5
0
0.0016
-5
0.002
0.0024
0.0028
-10
t1/2 ~ 30 s
2
3
Ea
 21310
R
3.71 x 104
7.29 x 105
2
2
ln (k)
Chemistry XXI
Use the data to estimate the activation energy Ea
for the decomposition of alanine.
Estimate t1/2 at 623 K in seconds.
-15
Ea ~177 kJ/mol
-20
1/T (1/K)
0.0032
Chemistry XXI
Let′s apply!
Assess what you know
Let′s apply!
Analyze
Chemistry XXI
Go back and analyze the notes for the decomposition
of Alanine. Based on our overall results, analyze the
likelihood of amino acids forming in hydrothermal
vents on the primitive Earth.
The strong dependence on T of the
decomposition of amino acids makes it difficult
to decide whether the “hydrothermal vents”
theory of the origin of life is plausible.
In fact, the contact of amino acids with hydrothermal
solutions during sediment and ocean recycling is
likely to be the major geochemical destruction
pathway of amino acids on Earth.
New Data
Let′s apply!
Recent experimental results indicate that there
may be other reactions that compete with the
decomposition of amino acids at T > 100 oC:
Dimerization
2
2
2
3
Chemistry XXI
+ H2O
3
The formation of dimers and
polymers may have helped
amino acids to accumulate on
the planet.
3
Peptide
Bond
2 A  A2 + B
Let′s apply!
Analyze
Go to:
http://www.chem.arizona.edu/chemt/C21/sim
(Dimerization)
or use the simulation on the next page.
Use the simulation of the dimerization of alanine to:
Chemistry XXI
 Determine the order of the reaction;
 Compare the half-lives of the process for a 1 M
solution of alanine at 100 oC and 200 oC.
 Estimate the activation energy Ea of the reaction;
Chemistry XXI
Let′s apply!
Analyze
10.16
Let′s apply!
T = 373 K
Graphical analysis
indicates this is a
second-order reaction.
1/[Ala] (1/mM)
10.14
10.12
10.1
10.08
1
1

 kt
[ Ala ] [ Ala ]o
10.06
18
T (K) k (s-1M-1)
20
22
24
26
28
30
32
t(s)
1

[ Ala ]o k
t1/2(s)
373
0.0085
1.17x102
423
0.247
4.05x100
473
3.606
2.77x10-1
1/[Ala] (1/mM)
Chemistry XXI
t1/ 2
1/[Ala] = 0.0085t + 9.8997
50
45
40
35
30
25
20
15
10
5
0
T = 473 K
1/[Ala] = 3.6056t - 62.715
15
20
25
t(s)
30
35
Let′s apply!
2
1
ln(k)
0
0.0016
-1
0.002
0.0024
0.0028
-2
-3
-4
Chemistry XXI
-5
ln(k) = -10672/T + 23.841
-6
1/T (1/K)
Ea/R = 10672
Ea = 88.7 kJ/mol
0.0032
Chemistry XXI
Identify with a partner
two important ideas discussed
in this module.
Measuring Rates
Summary
Reaction rates allow us to
follow the kinetic evolution of
a chemical process.
xAyB
Rate Inst
d [ A]

dt
Chemistry XXI
The effect of temperature and concentration on a
process’ reaction rate is summarized in the RATE LAW:
RATE LAW
Rate = k
[A]a
Reaction
order
Rate Constant
C and T Effects
Given a rate law, we can derive information
about how the concentration of reactants or
products changes with time.
xAyB
If a = 1 (first-order):
Chemistry XXI
If a = 2 (second-order):
Rate = k [A]a
ln[ A]  ln[ A]o  kt
1
1

 kt
[ A] [ A] o
Temperature effects on reaction
rate are determined by Arhenius
Equation for the rate constant k:
k  Ae
Ea

RT
Chemistry XXI
For next class,
Investigate how the overall rate of a reaction is
related to the reaction mechanism.
How can we use the reaction mechanism to
derive the rate law or use the rate law to
evaluate the reaction mechanism?