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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 2: Comparing Free Energies
Central goal:
To quantitatively
determine the
directionality and extent
of chemical reactions.
The Challenge
Transformation
How do I change it?
Chemistry XXI
Imagine that you were interested
in QUANTITATIVELY determining
which of two or more chemical
processes could have occurred
on the primitive Earth.
What measurable properties of the system can be
used to make the prediction?
How could we actually quantify the directionality
and extent of a chemical reaction?
Relevant Factors
Chemistry XXI
The directionality and extent of a chemical
reaction depend on three main factors:
ENERGETIC FACTORS
DHrxn
ENTROPIC FACTORS
DSrxn
TEMPERATURE T
But
how?
What Changes?
Chemistry XXI
Consider this chemical process:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Is this a favored process? Why?
The Facts
It may be favored if the process:
a) leads to more energetically stable compounds.
 energy is released (Exothermic: DHrxn < 0).
q
Chemistry XXI
DHrxn < 0 (-)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
The Facts
It may be favored if the process:
b) Leads to the formation of substances with
more distinguishable configurations:
DSrxn> 0 (+)
Chemistry XXI
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
DSrxn > 0 (+)
Certainly favored:
DHrxn < 0 (-) (exothermic)
DSrxn > 0 (+)
Let’s Think
Consider the possible decomposition of CO2 in
the atmosphere:
Chemistry XXI
CO2(g)  C(s) + O2(g)
Analyze the sign of DHrxn and DSrxn
and decide whether the process is favored or not.
The Facts
If, as in the previous case,
DHrxn > 0 (+) (endothermic)
DSrxn < 0 (-)
the process is certainly unfavored.
Chemistry XXI
But what happens if:
or
DHrxn > 0 (+) (exothermic)
DSrxn < 0 (-)
?
DHrxn < 0 (-) (endothermic)
DSrxn > 0 (+)
?
The outcome depends on the TEMPERATURE.
The Law
In general, energetic effects become less relevant at
high temperature, where entropic effects dominate.
The sign of this quantity (Gibbs Free Energy, G):
DGrxn = DHrxn – TDSrxn
Chemistry XXI
can be used to determine process directionality.
2nd Law of Thermodynamics
At constant T and P:
DGrxn < 0
for thermodynamically favored processes.
Reaction Extent
Chemistry XXI
Experimental results indicate that processes at
constant T and P occur in the direction in which
the Gibbs free energy G of the system decreases.
AB
DGrxn = GB - GA < 0
Exergonic  Favored
CD
DGrxn = GD – GC > 0
Endergonic  Unfavored
The more negative DGrxn,
the larger the extent of the reaction.
The more negative DGrxn,
the more thermodynamically stable the
products than the reactants.
Evaluating Likelihood
The quantitative analysis of the extent of reactions is
crucial to determine, for example, what were the
likely components of our primitive atmosphere.
Chemistry XXI
H2 is suspected to be an
important component of the
very early atmosphere.
Was CO2 or CH4 more stable
in this environment?
How does temperature affect
the answer?
Let’s explore it!
Hydrogen constitutes ~75% of
the Universe’s elemental mass
Let’s Think
Consider this data:
Reaction
CO2(g) + 4 H2(g)  CH4(g) + 2 H2O(l)
DHorxn (kJ) DSorxn (J/K)
-252.7
-410.3
Chemistry XXI
Which compound of C (CO2 vs CH4) is most stable in
the presence of H2 at 25 oC?
Hint: Are reactants or products favored in the
reaction?
Let’s Think
Reaction
DHorxn
(kJ)
DSorxn
(J/K)
CO2(g) + 4 H2(g)  CH4(g) + 2 H2O(l)
-252.7
-410.3
DGrxn = DHrxn–TDSrxn = -252.7 – 298.15*0.4103 = -130.4 kJ
Chemistry XXI
CH4 is the stable forms at 25 oC in the
presence of H2.
Are there any temperatures at which CO2 is
favored?
Let’s Think
Reaction
DHorxn
(kJ)
DSorxn
(J/K)
CO2(g) + 4 H2(g)  CH4(g) + 2 H2O(l)
-252.7
-410.3
As T increases, the energetic contribution becomes
less important. There is a temperature at which:
Chemistry XXI
DGrxn = DHrxn–TDSrxn = 0.
Above this temperature, the reverse reaction is
favored.
DGrxn = -252.7 – T*0.4103 = 0  T = 615.9 K
Quantitative Data
Chemistry XXI
Our discussion highlights the importance of
measuring or calculating DHrxn, DSrxn, and DGrxn if
we want to make predictions about the
thermodynamic likelihood of a reaction.
We have already
A+BC+D
discussed how to
o
o
o
DS
=
S
–
S
rxn
products
reactants
calculate DSrxn given
available data for the
Basic Assumption:
standard molar entropies
So (perfect crystal) = 0
at 0 K.
of formation Sof.
How do we get DHrxn and DGrxn for any process?
Heat of Reaction
We have seen that the heat of a reaction DHrxn can
be measured by calorimetry.
However, we can also calculate this quantity using
measured standard enthalpies of reaction DHof for
the substances involved in the process.
Chemistry XXI
Let’s consider a reaction involved in the production of
CO2 in the primitive atmosphere:
C(graphite) + 2 H2O(l)  CO2(g) + 2 H2(g)
How could we evaluate DHorxn given this information:
C(graphite) + O2(g)  CO2(g)
H2(g) + ½ O2(g)  H2O(l)
DHof = -393.5 kJ/mol
DHof = -285.8 kJ/mol
Alternative Routes
C(graphite) + 2 H2O(l)  CO2(g) + 2 H2(g)
We could think of the reaction as combination of a
decomposition and a combination steps:
1st: 2 H2O(l)  2 H2(g) + O2(g)
Chemistry XXI
2nd:
C(graphite) + O2(g)  CO2(g)
DHorxn = 2 (+285.8) kJ
DHorxn =
-393.5 kJ
+
C(graphite) + 2 H2O(l)  CO2(g) + 2 H2(g)
DHorxn = 2 (285.8) - 393.5 = +178.1 kJ
This approach is based on the central idea
that, no matter what path we follow,
the total energy transfer should be the same.
Hess’s
Law
Generalizing
All reactions can be conceived as a sequence of
decompositions and combinations steps:
Take the general reaction:
aA+bBcC+dD
Chemistry XXI
Imagine it happens in the following steps:
Reaction
Decomposition of a moles of A
Energy Transfer
a x (-DHof,A)
Decomposition of b moles of B
b x (-DHof,B)
Formation of c moles of C
c x DHof,C
Formation of d moles of D
d x DHof,D
DHorxn = c DHof,D + d DHof,D – a DHof,A – b DHof,B
DHorxn = S DHof,products - S DHof,reactants
Reference State
DHorxn = S DHof,products - S DHof,reactants
It is important to notice that in this relationship,
we only need the DHof of the
chemical COMPOUNDS involved, not of any
elements.
Chemistry XXI
HOW SO?
DH
0
Elements
Reactants
Products
We are taking
DHof = 0 for
the elements
in their
standard
state.
Gibbs Free Energy
An identical procedure can be used to calculate the
DGorxn given the DGof for the different
chemical COMPOUNDS involved in the reaction:
DGorxn = S DGof,products - S DGof,reactants
DGof = 0 for elements in their standard state.
Chemistry XXI
Alternatively, we can also calculate DGorxn as:
DGorxn = DHorxn–TDSorxn
This route is useful
in estimating the
effect of T on the
value of DG.
Let’s Think
One central question in the theories about the
origin of life is how complex organic compounds
were synthesized from simpler molecules such as
H2, N2, CH4, NH3, and H2O.
Consider these possibilities in the synthesis of the
simplest amino acid, glycine (C2H5NO2):
Chemistry XXI
2 CO(g) + NH3(g) + H2(g)  C2H5NO2(s)
2 CH4(g) + NH3(g) + 2 H2O(l)  C2H5NO2(s) + 5 H2(g)
1. Calculate DHorxn, Sorxn, and DGorxn for these two
processes.
Let’s Think
2 CO(g) + NH3(g) + H2(g)  C2H5NO2(s)
Chemistry XXI
2 CH4(g) + NH3(g) + 2 H2O(l)  C2H5NO2(s) + 5 H2(g)
Substance
DHof (kJ/mol)
DGof (kJ/mol)
Sof (J/(mol K)
H2(g)
0
0
130.7
CH4(g)
-74.6
-50.5
186.3
CO(g)
-110.5
-137.2
197.7
NH3(g)
-45.9
-16.4
192.8
H2O(l)
-285.8
-237.1
70.0
C2H5NO2(s)
-528.4
-368.8
103.5
Let’s Think
DHorxn
DSorxn
DGorxn
(kJ)
(J/K)
(kJ)
2 CO(g) + NH3(g) + H2(g) 
C2H5NO2(s)
-261.5
-615.4
-78.0
2 CH4(g) + NH3(g) + 2 H2O(l) 
C2H5NO2(s) + 5 H2(g)
238.3
51.6
223.
Chemistry XXI
Reaction
2. Identify which reactions are product-favored
and discuss the effect of temperature on the
extent of each process.
Let’s Think
DHorxn
DSorxn
DGorxn
(kJ)
(J/K)
(kJ)
2 CO(g) + NH3(g) + H2(g) 
C2H5NO2(s)
-261.5
-615.4
-78.0
2 CH4(g) + NH3(g) + 2 H2O(l) 
C2H5NO2(s) + 5 H2(g)
238.3
51.6
223.
Chemistry XXI
Reaction
The first reaction is product-favored at 25 oC; it
becomes reactant-favored at high T.
The second reaction is reactant-favored at 25 oC;
it becomes more product-favored at high T.
Reaction Extent
Although the change in Gibbs free energy DGorxn
allows us to make predictions about reaction
directionality, the actual number tells us little about
the final relative amounts of product to reactants.
Chemistry XXI
How can we better quantify “reaction extent”?
Experimental results
indicate that chemical
reactions tend to reach an
“equilibrium” state in which
the concentration of
products and reactants
does not vary over time.
C
o
n
c
e
n
t
r
a
t
i
o
n
Time
Equilibrium Constant
At chemical equilibrium, the following ratio of the
concentrations, or gas pressures,
of reactants to products remains constant:
Equilibrium Constant
Chemistry XXI
aA+bBcC+dD
For reaction involving
gases, the equilibrium
constant may be
expressed in terms of
the partial pressures
exerted by each gas:
c
d
[C ] [ D]
Kc 
a
b
[ A] [ B]
c d
C D
a b
A B
P P
Kp 
P P
DGorxn vs. K
The value of the equilibrium constant K is directly
related to the DGorxn for the process:
o
DGrxn
(
)
o
RT

DGrxn   RT ln K
K e
Chemistry XXI
Then:
K e
o
o
DH rxn
DS rxn
(

)
RT
R
In general:
R = 8.314 J/(K mol)
Product-Favored Process:
DGorxn < 0 
K>1
Reactant-Favored Process:
DGorxn > 0 
K<1
Chemistry XXI
Let’s Think
Predict the
effect of
changing
DHorxn, DSorxn,
and T on the
value of K and
the reaction
extent.
K e
o
o
DH rxn
DS rxn
(

)
RT
R
Reaction Extent
As we have seen, different reaction reach
equilibrium states with different proportion of
products and reactants in the system.
For example, combustion reactions tend to have
very large K values:
CH4(g) + O2(g)  CO2(g) + 2 H2O(g)
DGorxn = -800.9 kJ
Chemistry XXI
K = 2.07 x 10140 at 25 oC
For all practical purposes, this reaction fully goes to
completion at all temperatures (DHorxn < 0, DSorxn> 0):
CH4(g) + O2(g)
CO2(g) + 2 H2O(l)
Reaction Extent
But what about a reaction such as this:
CO(g) + H2O(l)  CH2O2(l)
DGorxn = -12.9 kJ
K = 182.0 at 25 oC (DHorxn < 0, DSorxn < 0)
Chemistry XXI
which may have played a central role in the
formation of complex molecules in our planet?
Whether this reaction is product- or reactant-favored
depends on the actual conditions of the process.
Thus, it is better to represent it as:
CO(g) + H2O(l)
CH2O2(l)
to highlight the importance of chemical equilibrium.
Chemical Evolution
In 1956, Urey and Miller conducted a classic
experiment on the origin of life.
Chemistry XXI
The experiment
showed that biological
molecules, such as
amino acids, can form
from simple reactants.
Since then, many
experiments have been
performed with different
reactant mixtures and
sources of energy
(heat, UV, X-rays, etc.)
Primitive Mixtures
These are two of the reactant
mixtures that have been tested:
H2O, CH4, NH3, H2
CO, N2, H2
Results indicate that the elemental composition
(presence of C, H, N, O) of the mixture is more
relevant than the kinds of molecules used. Why?
Chemistry XXI
No matter the mixture, the concentration of species
seems to be controlled by these equilibriums:
CH4(g) + 2 H2O(g)
CO2(g) + 4 H2(g)
CO2(g) + H2(g
CO(g) + 3 H2O(g)
N2(g) + 3 H2(g)
2 NH3(g)
Likely
occurring
between
500-1000 oC.
Let’s Think
CH4(g) + 2 H2O(g)
CO2(g) + H2(g)
N2(g) + 3 H2(g)
CO2(g) + 4 H2(g)
CO(g) + H2O(g)
2 NH3(g)
Chemistry XXI
Substance DHof (kJ/mol) Sof (J/(mol K)
Estimate K for
these processes
at 500 oC
and 800 oC.
H2(g)
0
130.7
CH4(g)
-74.6
186.3
CO2(g)
-393.5
213.8
CO(g)
-110.5
197.7
NH3(g)
-45.9
192.8
H2O(g)
-241.8
188.8
N2(g)
-108.6
191.6
Let’s Think
o
o
DH rxn
DS rxn
K  exp( 

)
RT
R
DHor
(kJ)
DSor K500
(J/K)
K800
CO2(g) + 4 H2(g) 164.7
172.7 0.0078
10.1
42.0
1.54
Reactions
CH4(g) + 2 H2O(g)
Chemistry XXI
CO2(g) + H2(g)
N2(g) + 3 H2(g)
R = 8.314 J/(K mol)
CO(g) + H2O(g)
2 NH3(g)
41.2
-91.8
0.257
-198.1 7.16 x 1.32x
10-5
10-6
Discuss which compounds 500 oC  N2, H2, H2O, CH4
of C, H, N, and O are more
oC  N , H , CO , CO
800
2
2
2
likely to exist at each T.
Chemistry XXI
Let′s apply!
Assess what you know
Let′s apply!
Going Up-Hill
The synthesis reactions for the formation of many
amino acids, proteins, carbohydrates and fats
starting from simple molecules have DGorxn > 0.
Chemistry XXI
One central question in the origin of life is how it
was possible to induce and sustain the
synthesis of these types of substances.
6 CO2(g) + 6 H2O(l)  C6H12O6(s) + 6 O2(g)
DGorxn = 2880 kJ
Glucose
Let′s apply!
Forced Reactions
A chemical reaction with DGorxn > 0 is not favored
thermodynamically, but can be forced to happen
by coupling it with other favored reactions.
This commonly happens in biological systems.
For example, consider an hypothetical reaction:
Chemistry XXI
A + B  AB
DGo1 > 0.
Reactions like this in living organisms are often
coupled with the favored hydrolysis of ATP:
ATP(aq) + H2O(l)  ADP(aq) + Pi(aq)
DGo2 = -30.5 kJ
Let′s apply!
Forced Reactions
Chemistry XXI
The coupling is often accomplished by the
following mechanism:
The formation of
A-Pi facilitates
the addition of B.
A + ATP + H2O  A-Pi + ADP
A-Pi + B  AB + Pi
A + B + ATP + H2O  AB + ADP + Pi
DGorxn = DGo1 + DGo2 < 0
Predict
Let′s apply!
For example, glutamine is an amino acid
synthesized many organisms from another amino
acid, called glutamic acid.
B
C
Chemistry XXI
AB
DGorxn = 14.2 kJ
Glutamic Acid
AB + C  AC + B
AC
Glutamine
Express and calculate K for this reaction at 25 oC.
K = [AB][C]/[AC][B] = 3.25 x 10-3
Let′s apply!
Predict
Chemistry XXI
Write the overall reaction when the synthesis of
glutamine is coupled to the hydrolysis of ATP and
calculate its K at 25 oC.
AB + C  AC + B
DGo1 = 14.2 kJ
ATP(aq) + H2O(l)  ADP(aq) + Pi(aq)
DGo2 = -30.5 kJ
AB + C + ATP + H2O  AC + B + ADP + Pi
DGorxn = DGo1 + DGo2 = -16.3 kJ
K = 717
Propose a mechanism for the coupled process.
Predict
Let′s apply!
Glutamic Acid (AB)
Glutamine (AC)
AB + C  AC + B
DGo1 = 14.2 kJ
ATP(aq) + H2O(l)  ADP(aq) + Pi(aq)
DGo2 = -30.5 kJ
Chemistry XXI
AB + ATP + H2O  AB-Pi + ADP
AB-Pi + C  AC + Pi + B
AB + C + ATP + H2O  AC + B + ADP + Pi
Chemistry XXI
Discuss with a partner how the
values of DHorxn, DSorxn, and DGorxn
affect the directionality and extent
of a chemical process.
Comparing Free Energies
Summary
At constant T and P:
DGrxn = DHrxn – TDSrxn < 0
for thermodynamically favored processes.
Chemistry XXI
a A + b B  c C + d D
DHorxn DSorxn
+
+
+
+
DGorxn
-
DIRECTIONALITY?
Product-favored
+
Reactant favored
Depends on T P fav at low T
Depends on T P fav at high T
Equilibrium Constant
a A + b B  c C + d D
c
EXTENT?
d
[C ] [ D]
Kc 
a
b
[ A] [ B]
DG
o
rxn
Chemistry XXI
K e
  RT ln K
o
o
DH rxn
DS rxn
(

)
RT
R
In general:
Product-Favored Process:
DGorxn < 0 
K>1
Reactant-Favored Process:
DGorxn > 0 
K<1
Chemistry XXI
For next class,
Investigate how the rate of a chemical reaction
can be measured or calculated.
How can we use experimental data to track the
rate of a chemical reaction
as a function of time?