Transcript Slide 1

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 4: Understanding Mechanism
Central goal:
To use reaction
mechanisms to make
predictions about reaction
rate and vice versa.
The Challenge
Transformation
How do I change it?
Chemistry XXI
Imagine that you were
interested in understanding
why certain types of
substances and processes
appeared in our planet.
How can we use reaction mechanisms to make
predictions?
How can we deduce reaction mechanisms based
on reaction outcomes?
Reaction Pathways
Chemistry XXI
Our ability to predict the most likely outcomes of a
chemical reaction improves considerably when we
understand the mechanism that leads from
reactants to possible products.
Most reaction
mechanisms involve
several steps. However,
some steps play a more
central role than others in
determining the overall
rate of reaction.
CO2(g) +
2 H2O(g)
CH4(g) +
O2(g)
Reaction Steps
In the mechanistic model, the overall reaction is
viewed as the result of multiple elementary reactions
or steps occurring simultaneously in the system.
For example, the overall reaction: 2A  2C + E
may involve two elementary steps:
Chemistry XXI
A+AB+C
Bimolecular
Unimolecular
B C+E
2A  2C + E
B
is an intermediate
Detecting intermediates is an important means of
investigating reaction mechanisms.
Overall Rate Order
In the mechanistic model, the overall rate of the
reaction is an “emergent property” of the rates of the
individual steps.
Chemistry XXI
For many reactions, one step is slow enough to
limit the rate of the overall reaction:
A+AB+C
Slow
B C+E
Fast
2A  2C + E
Overall Rate = k [A][A] = k [A]2
Mirror Images
Consider the following problem of central relevance
for our understanding of the origin of life:
Chemistry XXI
L
D
Most amino acids found
on Earth appear in only
one of two possible
mirror-image forms,
called enantiomers or
optical isomers.
Non-superimposable
These isomers have most of the same
properties, but react differently with L and D
isomers of other “chiral” substances.
Chiral Centers
Molecular chirality is commonly caused by the
presence of carbon atoms in a molecule attached
to four different groups:
Chemistry XXI
Non-Chiral
carbons
Chiral
carbon
Identify the chiral carbons in
this molecule:
Let′s think!
L-Glucose
Left or Right Handed
Many biologically active molecules are chiral,
including the naturally occurring amino acids,
which tend to be “left-handed” (L).
Chemistry XXI
Chirality is of central importance for many
biological functions.
Different enantiomers
interact differently with
the chiral molecules
(proteins, DNA)
in our body.
How this preferred chirality emerged on our planet?
Racemization
Pure samples of L or D amino acids eventually
convert into a mixture of both forms. This process
is called racemization. How do we explain it?
H+
DG
forward
NH2
Mechanism
1. Unimolecular Step
2. Bimolecular Step
backward
C
H3C COOH
Chemistry XXI
H 2N
H 3C
C
H
NH2
H
C
CH3
HOOC
L
COOH
Ea = 124 kJ/mol
D
DGorxn = 0
Reaction Coordinate
Mechanism
The racemization process can be explained by this
two-step mechanism:
L  C- + H+
Unimolecular (Slow)
Rate = k[L]
H+ + C-  D
Bimolecular (Fast)
Rate = k[C-][H+]
Chemistry XXI
where H+ and C- are intermediates.
The rate of the overall reaction is determined by the
slowest step (Rate Determining Step), thus:
LD
Rate = k [L]
(First Order Reaction)
The same ideas apply to the backward reaction
D  L.
Let’s Think
Chemistry XXI
According to this mechanism, the interconversion
between the L and D forms should occur
with the same probability.
Imagine that you start
with 1 M solution of L-Ala
and 30% of it transforms
to D-Ala every second.
The same percentage of
D-Ala in the system
transforms to L-Ala in
that time.
Follow the evolution of
the system. When does
the process “stop”?
t (s)
L-Ala (M)
D-Ala (M)
0
1
0
1
2
3
Let’s Think
L  D 40%
[Ala] (M)
D  L 40%
L-Ala (M)
D-Ala (M)
0
1
0
LD
0.3 
DL
0
1
0.7
0.3
1
LD
0.7 x 0.3 = 0.21 
0.8
DL
 0.3 x 0.3 = 0.09
0.6
L-Ala
D-Ala
0.4
0.2
Chemistry XXI
t (s)
0
0
1
2
3
4
t (s)
Equilibrium?
5
2
0.58
0.42
LD
0.58 x 0.3 = 0.174 
DL
 0.42 x 0.3 = 0.126
3
0.532
0.468
LD
0.532 x 0.3 = 0.1596 
DL
 0.468 x 0.3 = 0.1404
4
0.5128
Etc.
0.4872
Kinetics and Equilibrium
Chemical equilibrium is reached when the rate of
the forward reactions is equal to the rate of the
backward process.
LD
Rate = kf[L]
DL
Rate = kb[D]
Chemistry XXI
kf[L]eq = kb[D]eq
Kc 
kf
kb

[D]
L
D
[L]
o
In this case, kf = kb. Thus, Kc = 1.
Kc  e

D G rxn
RT
The reaction keeps going at equilibrium but
[L] and [D] remain constant.
1
Amino Acid Chirality
The origin of biologically active amino acids’
chirality is an unresolved problem in science.
Chemistry XXI
Many hypothesis have been suggested.
Imagine that once amino acids are
linked into proteins, the % of an L amino
acid converted to the D form is 20%
every second, versus 80% conversion
from D to L in the same period of time.
Let′s
think!
What would be [D]/[L] at equilibrium?
LD
DL
kf
kb
kb = 4 kf
Kc 
kf
kb

[D]
[L]
Kc = ¼
Let’s Think
Consider the following data
derived through our work in
this unit:
2
2
3
+
2
3
2
2
+ H 2O
2
3
Chemistry XXI
3
HOOC
Dimerization
Ea = 88.7 kJ/mol
3
H 2N
H 3C
Decomposition
Ea = 177 kJ/mol
NH2
C
H
H
C
CH3
COOH
Racemization
Ea = 124 kJ/mol
Build a hypothesis about what could have prevented
amino acid racemization on primitive Earth.
Changing Mechanism
Chemistry XXI
It has been suggested that one crucial step in the
origin of life was the synthesis of substances that
could speed up the rate of certain chemical reactions.
As we know,
these “catalysts”
act by either
reducing the
activation energy
Ea or changing
the reaction
mechanism.
Enzymes
The substances that catalyze biological processes
are called enzymes. One common model to explain
their behavior is the “lock and key” model.
Chemistry XXI
(S)
(P)
(E)
(ES)
E+S
Mechanism
(E)
ES
ES  E + P
Fast
Slow
Enzyme Kinetics
DGo
E+S
ES
ES  E + P
Fast
Slow
ES
E+S
Chemistry XXI
E+P
The rate law for this process
is determined by the second
step: Rate = k2 [ES]
This expression is not very useful given that we
cannot easily follow [ES] as a function of time.
[ ES ]
Kc 
   [ ES ]  K c [ E ][ S ]
If we assume that
[ E ][ S ]
equilibrium is reached
in the first step:
Rate = k2 Kc [E][S]
Let’s Think
For a given concentration
of enzyme [E]o, the
reaction rate is first order
in [S], but only at low
concentrations of the
substrate.
At high concentrations:
Chemistry XXI
Rate ~ k [S]0 = k (constant)
(Zeroth order process)
Use the lock and key model to explain
this behavior.
Let’s Think
Chemistry XXI
The "lock and key" model has proven inaccurate.
The induced fit model is the most currently accepted
E+S
ES
Fast
ES  EP
Slow
EP
Fast
E+P
Draw an energy profile for
this mechanism and analyze
whether the associated rate
law needs to be modified.
Chemistry XXI
Induced Fit Model
The same rate law:
Rate = k2 Kc [E][S]
Auto-Catalysis
Processes in which the reaction is catalyzed by its
own products are called auto-catalytic and may
have played a central role in the origin of life.
[B]
A+B2B
Chemistry XXI
Rate = k [A] [B]
Why this shape?
t
The presence of auto-catalytic steps in some
reaction mechanisms may explain the
appearance of metabolic cycles.
Let’s Think
Consider this mechanism:
Chemistry XXI
A+X2X
X+Y2Y
YB
a) Write the overall reaction;
b) Identify the auto-catalytic steps and the [X]
intermediate species;
c) Predict the structure of the plots [X] and
[Y] vs. t as the reaction proceeds.
Hint: Think of A as grain, X as ducks, Y as wolves,
and B as “dead” wolves.
t
Oscillating Reactions
[X]
Chemistry XXI
[Y]
A+X2X
X+Y2Y
YB
Chemistry XXI
Let′s apply!
Assess what you know
Analyze
Let′s apply!
Consider the dimerization of amino acids (Aa):
2
2
3
Chemistry XXI
+ H 2O
2
3
3
which we found to be a second order reaction
Rate = k [Aa]2
Several possible mechanisms have been
proposed for this type of reaction.
Let′s apply!
Analyze
2 Aa  Aa-Aa + H2O
DGo
2Aa
Is there a way to
determine which of
these mechanisms
is more plausible?
Aa-Aa
+ H 2O
Chemistry XXI
2 Aa  Aa-Aa*
Aa-Aa*  Aa-Aa + H2O
DGo
Aa-Aa*
2Aa
Aa-Aa
+ H 2O
Could you propose
a different
mechanism that
leads to the same
experimental
rate law
Rate = k[Aa]2?
Analyze
Let′s apply!
Both mechanisms lead to Rate = k[Aa]2.
We would have to experimentally confirm the
existence of the intermediate.
Another possibility:
Chemistry XXI
DGo
2 Aa
Aa-Aa*
Fast
Aa-Aa*  Aa-Aa + H2O Slow
Rate = k2 [Aa-Aa*]
Kc = [Aa-Aa*]/[Aa]2
Aa-Aa*
2Aa
Aa-Aa
+ H 2O
[Aa-Aa*] = Kc [Aa]2
Rate = k2 Kc [Aa]2
Chemistry XXI
Discuss with a partner one thing
you do not fully understand about
the content of this Module.
Understanding Mechanism
Summary
Reaction mechanisms allow us to understand
reaction kinetics and make predictions about most
likely outcomes.
Chemistry XXI
Most chemical processes can be thought as
occurring in a sequence of elementary steps:
A +B
 C +D
CB+E
Bimolecular
Rate = k1[A][B]
Unimolecular Rate = k2[C]
AD+E
The rate law is determined by the slowest step.
Catalysts
Chemistry XXI
The reaction mechanism can be altered by the
presence of substances that help create alternative
reaction paths.
Catalysts”
act by either
reducing the
activation energy
Ea or changing
the reaction
mechanism.
Chemistry XXI
Are You Ready?
The Quest for Ammonia
Ammonia, NH3, is one of the
most important industrial
chemical substances.
Chemistry XXI
It is widely used in the
production of fertilizers,
pharmaceuticals, refrigerants,
explosives, and cleaning agents.
It ranks as one of the 10 top
chemicals substances produced
annually in the world.
The Synthesis
Ammonia is mainly produced via this
simple chemical reaction:
Let′s think!
1/2 N2(g)
+ 3/2 H2(g)
NH3(g)
Chemistry XXI
Compare the energetic and entropic stability
of reactants and products.
Make a prediction of the signs of
DHorxn and DSorxn for this process.
Energy:
DHorxn < 0
A-A bonds  A-B bonds
Entropy:
DSorxn < 0
4 mol gas  3 mol gas
mixture  one compound
Directionality
Reaction
DHorxn (kJ)
DSorxn (J/K)
1/2 N2(g) + 3/2 H2(g)  NH3(g)
-45.9
-99.1
Use these data to calculate DGorxn.
Write Kp for this reaction and calculate its
value at 25 oC.
Chemistry XXI
Let′s think!
DGrxn = DHrxn–TDSrxn = -45.9 – 298.15*0.0991 = -16.4 kJ
Kp 
PNH 3
1/ 2
2
4
3/2
P N PH
 exp( 1 . 64 x10 /( 8 . 314 x 298 . 15 ))  7 . 47 x10
2
2
Let′s think!
Reaction
DHorxn (kJ)
DSorxn (J/K)
1/2 N2(g) + 3/2 H2(g)  NH3(g)
-45.9
-99.1
Chemistry XXI
Is the synthesis of ammonia thermodynamically
favored at low or high temperatures?
Estimate the temperature at which the directionality
switches (K = 1)?
The reaction is favored at low temperatures.
K = 1  DGrxn = 0
DGrxn = -45.9 – T*0.0991 = 0  T = 463 K
Energy Profile
Chemistry XXI
Use the following information to build the energy
profile for the reaction:
Reaction
DHorxn (kJ)
Ea (kJ/mol)
1/2 N2(g) + 3/2 H2(g)  NH3(g)
-45.9
325
Ep
1/2 N2
3/2 H2
325 kJ
Let′s think!
45.9 KJ
NH3
Reaction Coordinate
Reaction Conditions
Reaction
DHorxn (kJ)
Ea (kJ/mol)
1/2 N2(g) + 3/2 H2(g)  NH3(g)
-45.9
325
Chemistry XXI
Given its high Ea, the reaction is normally done at
high T (~500 oC) and P (~200 atm).
Discuss how the increased T will affect the:
Let′s think!
Rate (Calculate k500/k25)
Extent (Calculate K500/K25)
of the reaction.
Reaction Conditions
k  Ae
k T2
e
E  1
1 
 a
 
R  T 2 T1 
5
e
Chemistry XXI
o
K e
K T1
RT
 3 . 25 x 10  1
1 

 773 298 
8 . 314


k T1
K T2
Ea

e
o
D H rxn  1 1 

 

R  T 2 T1 
(
D H rxn
RT
 1 . 00 x10
35
 1 . 14 x10
5
o

D S rxn
)
R
4
e
4 . 59 x 10  1
1 

8 . 314  773 298 
Reaction Conditions
High temperature increases reaction rate, but
decreases reaction extent. That is why the
reaction is carried out at high P too.
Discuss why high P favors the product side
in this process:
Chemistry XXI
Let′s think!
N2(g)
+ 3 H2(g)
2 NH3(g)
The collision rate is higher in the side with more
particles. The forward reaction is favored.
Catalysts and Reaction Order
The synthesis of NH3 is carried out in the presence
of catalysts. The order of the reaction depends on
the composition and structure of this catalyst.
For example: T = 500 oC P = 200 atm
Fe catalyst
Chemistry XXI
The rate does only depends on the
concentration of N2(g).
t(s)
0
0.1
0.2
0.3
0.4
0.5
0.6
CN2(mol/L)
2.00
1.68
1.42
1.19
1.01
0.846
0.71
Let′s think!
What is the reaction order with respect
to N2(g)? What is the value of the rate
constant under these conditions?
Reaction Order
t(s)
0
0.1
0.2
0.3
0.4
0.5
0.6
CN2(mol/L)
2.00
1.68
1.42
1.19
1.01
0.846
0.71
ln(C)
0.693
0.521
0.349
0.177
0.005
-0.167 -0.339
0.8
Rate = k [N2(g)]
ln(C)= -kt + ln(Co)
0.6
Ln(C)
Chemistry XXI
0.4
First order with
respect to
[N2(g)]
0.2
0
0
0.2
0.4
0.6
0.8
-0.2
k = 1.72 s-1
-0.4
t(s)
y = -1.72x + 0.6931
Reaction Mechanism
In the presence of a solid catalyst, the reaction
takes place on the surface of the solid.
N2(g)  N2(ad)
N2(ad)  2 N(ad)
Chemistry XXI
N(ad) +H(ad)  NH(ad)
Let′s think!
What other mechanistic steps are involved in the
synthesis of NH3? Which step can be expected to be
the slowest given that Rate = k[N2]?
H2(g)   H2(ad)
Reaction
Mechanism
N2(g)  N2(ad)
N2(ad)  2 N(ad)
H2(ad)  2 H(ad)
N(ad) +H(ad)  NH(ad)
NH(ad) + H(ad)  NH2(ad)
NH2(ad) + H(ad)  NH3(ad)
Chemistry XXI
NH3(ad)   NH3(g)
Slowest:
or
N2(ad)  2 N(ad)
Rate = k’ [N2(ad)]
N2(g)  N2(ad)
Kc = [N2(ad)]/[N2(g)]
Rate = k [N2(g)]
Rate = k’ Kc [N2(g)]
Activation Energy
The presence of the catalyst reduces Ea considerably.
Use the following data to derive Ea for an Fe-based
catalyst.
Chemistry XXI
How many times faster is the reaction at
T = 500 oC in the presence of the catalyst?
(Ea = 325 kJ/mol without it)
Let′s think!
T(oC)
25
100
200
300
400
k(s-1) 3.24x10-9 2.32x10-6 5.81x10-4 0.0211 0.264
500
1.72
Activation Energy
T(oC)
25
100
200
300
400
k(s-1) 3.24x10-9 2.32x10-6 5.81x10-4 0.0211 0.264
ln( k )  
Ea 1
 ln( A )
500
1.72
Ea = 9743R
R T
Ea = 81 kJ/mol
5
0
ln(k)
Chemistry XXI
-5
0
0.001
0.002
0.003
0.004
k Ea
 e
2

1
RT
E a2  E a 1 
k Ea
-10
1
-15
-20
y = -9743x + 13.146
-25
1/T
k Ea
2
k Ea
1
 3 . 09 x10
16