Download Comparing Free Energies

298
The analysis of changes in the internal potential energy and the entropy of the
substances involved in a chemical reaction is useful in qualitatively predicting the
directionality of chemical reactions. However, given the often
competing effects between changes in internal potential energy and entropy during chemical processes, we need to find
a systematic way to make reliable predictions. This can be
accomplished by carefully analyzing what happens to the
entropy of both the chemical system and its surroundings
during a chemical process. Fortunately, we can take advantage of a law of nature to derive useful mathematical relationships to evaluate the directionality and extent of chemical
formation of red haze in big
reactions based on available experimental information such as Thecities
involves competing
energetic and entropic factors.
the heat of reaction and the change in molar entropies.
Most of our discussions in this module will focus on calculating the difference
in Gibbs Free Energy between products and reactants (DGrxn). As we will see, this
quantity depends on the heat of reaction DHrxn, the entropy difference DSrxn, and
the temperature T at which the chemical reaction takes place. The sign of DGrxn
will give us information about the directionality of the chemical process, while the
the ratio DGrxn/T can be used to quantitatively evaluate the extent of the conversion of reactants into products.
By Fidel Gonzalez (Own work)
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U5: MODULE 2
Comparing
Free Energies
THE CHALLENGE
Chemical Origins
Glycine, C2H5NO2, is an important amino acid that could be potentially synthesized using CH4(g), NH3(g), and H2O(l). The synthesis reaction leads to
an increase of entropy (DSrxn > 0) but the process is endothermic (DHrxn > 0).
•
•
How would you decide whether this synthetic path is feasible and may
have led to the formation of glycine on the primitive Earth?
How would temperature affect the likelihood of the reaction?
Share and discuss your ideas with one of your classmates.
This module will help you develop the type of chemical thinking that is used
to answer questions similar to those posed in the challenge. In particular, the
central goal of Module 2 is to help you understand how to quantitatively determine the directionality and extent of chemical reactions.
Chemical Thinking
U5
How do we predict chemical change?
2nd Law of Thermodynamics
The analysis of the many types of changes that occur in our world has allowed
physical scientists to identify common underlying characteristics. In particular,
changes that occur in isolated systems (systems that cannot exchange energy or
matter with their surroundings) are characterized by an increase in the total number of configurations available to the components of the system. In isolated systems, the random motion of particles are likely to lead to a more homogeneous
distribution of matter and energy. This means that whenever a change occurs in an
isolated system, the total entropy of the system STOT increases. This experimental
fact is known as the Second Law of Thermodynamics and it can be used to derive
useful relationships to quantitatively determine the directionality and extent of
chemical reactions.
Let us consider a container with a mixture of reactants in which a chemical reaction may take place. We will call this mixture the “system”. Our system is kept at
constant temperature and pressure and can exchange energy with the environment
(i.e., energy can come in or out of the system). We will call “surroundings” that
part of the environment affected by the exchange of energy with the system. In
this manner, the combination of our system and its surroundings can be thought
as an isolated system itself (Figure 5.5) in which the 2nd Law of Thermodynamics
applies.
According to the 2nd Law, if a chemical reaction is going to take place in
our system, the total entropy of the system plus its surroundings should increase.
This means, the difference between the total entropy after and before the process
should be positive (DSTOT > 0). This does not imply that the change of entropy
in the chemical system (DSsys) should also be larger than zero for the chemical
reaction to occur. DSsyscould be negative, as long as the associated change in the
entropy of the surroundings (DSsurr) is such that:
(5.2)
DSTOT = DSsys + DSsurr > 0
This relationship implies that chemical reactions will proceed in the direction in
which DSTOT increases. Our task is then to figure out how to calculate this quantity
based on experimental information about the properties of reactants and products.
The Surroundings
LET’S THINK
During a chemical reaction, a net amount of energy is either released to or absorbed
from the surroundings. This energy exchange affects the entropy of the surroundings as its particles gain or lose energy.
•
How would you expect the entropy of the surroundings to change during an
exothermic process? How would it change during an endothermic process?
•
Would you expect the value of DSsurr to depend on the temperature of the surroundings?
Share and discuss your ideas with a classmate, and justify your reasoning.
Isolated Overall
System
Surroundings
System
Energy Exchange
Figure 5.5 The combina-
tion of the system plus
its surroundings can be
thought as an isolated system in which the 2nd Law of
Thermodynamics applies.
299
300
MODULE 2
Comparing Free Energies
Gibbs Free Energy
To take advantage of the 2nd Law of Thermodynamics in predicting reaction directionality and extent, we need to find a way to determine the value of DSsys and
DSsurr for every process of interest. Given that we are considering chemical reactions, the change of entropy of the system DSsys can be calculated by taking the difference between the total entropies of products and reactants. For a generic chemical reaction of the form: a A + b B
c C + d D, DSsys would then be given by:
(5.3)
DSsys = DSrxn = (cSC + dSD) - (aSA + bSB) = SSproducts - SSreactants
where Si represents the molar entropy of species “i” and the symbol “S” indicates
the sum over all reactants or products.
The change of entropy of the surroundings DSsurr is determined by the amount
of energy exchanged in the form of heat during the chemical reaction, DHrxn, and
the temperature of the surroundings T. In particular, if we assume that the effect
on the surroundings is rather small compared to its total entropy (i.e., the surroundings are much larger than the system), DSsurr at constant temperature and
pressure can be estimated as:
DSsurr = – DHrxn / T
(5.4)
According to this relationship, the change of entropy of the surroundings would
be positive for exothermic processes and negative for endothermic reactions. A
release of energy into the surroundings (DHrxn < 0) should result in an increase
in the number of configurations available to surrounding particles, while energy
absorption (DHrxn > 0) should limit the number of states in which these particles
may exist. Energy exchange at high T has a smaller impact on the entropy of the
surroundings as its particles already have access to many different states. On the
other hand, an energy transfer at low T should affect the entropy of the surroundings considerably as the number of available states is relatively small.
The combination of Eq. (5.2), (5.3), and (5.4) indicates that, at constant temperature and pressure, chemical reactions proceed in the direction for which:
(5.5)
DSTOT = DSsys + DSsurr = DSrxn – DHrxn / T > 0
LET’S THINK
Types of Reactions
All chemical processes may be classified into one of four types based on the signs of DSrxn and DHrxn:
DSrxn > 0, DHrxn > 0
DSrxn > 0, DHrxn < 0
DSrxn < 0, DHrxn > 0
DSrxn < 0, DHrxn < 0
• Which of these processes can be expected to occur at all temperatures? Which processes
are unlikely to occur no matter what the temperature is?
•
Which processes may occur at some temperatures but not others?
Chemical Thinking
U5
301
How do we predict chemical change?
For chemical processes that take place at constant temperature and pressure, it
is common to calculate the difference in Gibbs free energy between products and
reactants, DGrxn, defined as:
DGrxn = DHrxn - T DSrxn
(5.6)
This definition, together with the 2nd Law of Thermodynamics as expressed in
Eq. (5.5), imply that chemical reactions will occur in the direction for which DGrxn
is negative (DGrxn < 0), this is, in the direction in which the Gibbs free energy of
the system decreases. The more negative the change in Gibbs free energy, the more
favored the chemical reaction and the larger the extent to which reactants can be
expected to convert into products. The definition of DGrxn allows us predict reaction extent and directionality based solely on changes in properties of the system,
without having to consider the effects on the surroundings. The structure of Eq.
(5.6), together with the condition DGrxn < 0, makes explicit the competition between energetic (DHrxn) and entropic factors (DSrxn) in determining how thermodynamically favorable a process is at a given temperature.
Based on Eq. (5.6) we can predict that processes for which DHrxn< 0 and
DSrxn > 0 will be thermodynamically favored at all temperatures. On the contrary,
chemical reactions in which DHrxn> 0 and DSrxn < 0 will have a very small probability to occur at all T. In those cases in which DHrxn and DSrxn have the same sign,
there will be a temperature at which the process will become more or less favorable
as the temperature varies. This particular temperature is determined by the condition DGrxn = 0 and can be estimated using Eq. (5.6), assuming that the values of
DHrxn and DSrxn do not change with temperature (Figure 5.6).
DHrxn < 0, DSrxn < 0
DGrxn
m = –DSrxn
T
b = DHrxn
Figure 5.6 If we assume that
DHrxn and DSrxn do not change
with temperature (T), a graph
of DGrxn as a function of T is a
straight line with a slope m equal
to –DSrxn and a y-intercept b equal
to DHrxn. The directionality of the
reaction changes when DGrxn = 0.
LET’S THINK
CO2(g) or CH4(g)?
Hydrogen gas, H2(g), is suspected to have been an important component of the primordial atmosphere. In a hydrogen-rich environment, CO2(g) may have been unstable and may have reacted to
form CH4(g). Is that a reasonable hypothesis?
•
Using the information included in the following table, evaluate how thermodynamically
favorable the formation of CH4(g) from CO2(g) and H2(g) is at 25 oC.
Chemical Reaction
CO2(s) + 4 H2(g)
CH4(g) + 2 H2O(l)
DHrxn (kJ mol-1) DSrxn (J mol-1 K-1)
–252.7
–410.3
Given the values of DHrxn and DSrxn one can expect that the above process will be more favorable
at some temperatures than others.
•
Build a graph of DGrxn as a function of T. Use the graph to predict whether the formation of
CH4(g) is favored at low temperature or at high temperatures.
•
Estimate the temperature at which CH4(g) becomes more thermodynamically favored than
CO2(g) in the presence of H2(g).
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
302
MODULE 2
Comparing Free Energies
Standard Values
To facilitate the prediction of the thermodynamic likelihood of a wide variety
of chemical processes, chemical scientists have carried out a multitude of measurements to determine the values of experimental properties that can be used
to calculate DSrxn, DHrxn, and DGrxn. These types of measurements are commonly
done under standard conditions: 25 oC and 1 atm, and the results are commonly
expressed per mole of substance. Thus, the measured quantities are referred to as
standard molar properties. In the following sections, we will discuss how to use
this valuable experimental information to calculate the quantities needed to predict reaction directionality and extent.
a) DSorxn
As discussed in the previous module, information about the value of the heat
capacity of a substance “A” at various temperatures can be used to determine its
standard molar entropy of formation Sfo[A] (J mol-1 K-1). The entropy change as
a result of a chemical reaction of the form: a A + b B
c C + d D, carried out
under standard conditions can then be calculated by taking the difference between
the total entropy of the products minus that of the reactants:
(5.7)
DSorxn = (c Sfo[C] + d Sfo[D]) - (a Sfo[A] + b Sfo[B])
LET’S THINK
Consider the formation of NH3(g) from the elemental substances:
3 H2(g) + N2(g)
2 NH3(g)
• Based on the nature of reactants and products, qualitatively
predict the sign of DSorxn for this process.
Entropy Change
Substance Sfo (J mol-1 K-1)
H2(g)
130.7
N2(g)
191.6
NH3(g)
192.8
• Use the available data to determine the actual value of DSorxn for this process.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
By Fidel Gonzalez (Own work)
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b) DHorxn
Figure 5.7 Bomb calo-
rimeters such as the one
shown in this image are
used to measure heats of
combustion.
Measurements of the energy absorbed or released as heat during a chemical reaction can be done using an experimental technique known as calorimetry (Figure
5.7). In this approach, energy transferred as heat by a chemical system is measured
indirectly by quantifying changes in the temperature of a surrounding material.
One particularly useful calorimetric measurement is the heat released or absorbed
when one mole of a given substance is synthesized from its elementary components under standard conditions (at 25 oC and 1 atm). The corresponding quantity is known as the Standard Molar Enthalpy of Formation, DHfo (kJ/mol), of the
chemical compound. The table on the next page illustrates the formation of well
known chemical substances and the associated DHfo values determined experimentally.
Chemical Thinking
U5
Synthesis Reaction
C(s, graphite) + O2(g)
H2(g) + 1/2 O2(g)
3/2 H2(g) + 1/2 N2(g)
C(s, graphite) + 2 H2(g)
How do we predict chemical change?
Substance
DHfo (kJ/mol)
CO2(g)
–393.5
H2O(l)
–285.8
NH3(g)
–45.9
CH4(g)
-74.6
CO2(g)
H2O(l)
NH3(g)
CH4(g)
Na(s) + 1/2 Cl2(g)
NaCl(s)
NaCl(s)
–411.2
Mg(s) + 1/2 O2(g)
MgO(s)
MgO(s)
–601.6
Notice that the elementary substances used for the synthesis reactions as represented in the table above are in their stable form at standard conditions (25
o
C, 1 atm). The value of DHfo for liquid water, for example, indicates that 285.8
kJ of energy are released every time that one mole of water is formed from the
needed stoichiometric amounts of hydrogen and oxygen gases. It is important to
realize that the same amount of energy will be absorbed when one mole of water
is decomposed into one mole of H2(g) and half a mole of O2(g). Thus the energy
needed to decompose a substance into elementary substance under standard conditions is given by DHdo = –DHfo. We will take advantage of this symmetric relationship between heat of formation and heat of decomposition to calculate DHorxn
for any process.
Heat of Reaction
LET’S THINK
Consider the following hypothetical data for these pair of generic chemical reactions:
A + B
B + C
•
AB
BC
Use the above information to determine the standard heat of reaction DHorxn for the process:
AB + C
•
DHfo = –200 kJ/mol
DHfo = –100 kJ/mol
BC + A
Discuss how you could generalize the calculation of DHorxn for any reaction using DHfo
values for reactants and products.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Information about the standard molar enthalpy of formation of the substances
involved in a chemical reaction can be used to predict the heat of reaction under standard conditions, DHorxn. To make this prediction, we can model chemical reactions as
processes in which all chemical compounds acting as reactants are first decomposed
into elementary substances that are then recombined to form the desired products.
Although it is unlikely that chemical reactions would actually proceed in this way,
the approach is based on the idea that, no matter what path a reaction follows,
the total energy transfer should be the same as long as we start with the same
amount of reactants and end with the same amount of products.
303
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MODULE 2
Comparing Free Energies
To illustrate this idea, let us predict the heat of reaction for a process likely
involved in the production of CO2(g) in our planet’s primordial atmosphere:
C(s) + 2 H2(g) + O2(g)
– 2 x DHfo[H2O(l)]
C(s) + 2 H2O(l)
DHfo[CO2(g)]
DH
2 H2(g) + CO2(g)
o
rxn
Figure 5.8 Alternative reaction path based on the decomposition of reactants and
the synthesis of products.
SYNTHESIS
DECOMPOSITION
C(s, graphite) + 2 H2O(l)
2 H2(g) + CO2(g)
We can think of this reaction as occurring in two steps:
1) The decomposition of two moles of H2O(l) into
H2(g) and O2(g), and 2) The combination of elementary substances to form one mole of CO2(g) (Figure
5.8). The amount of energy involved in the first step is
-2 x DHfo[H2O(l)], and that associated with the second step
is simply DHfo[CO2(g)]. The standard heat of reaction for
the reaction of interest should then be given by:
DHorxn = DHfo[CO2(g)] – 2 x DHfo[H2O(l)] = –393.5 + 2 x 285.8 = 178.1 kJ
In general, the standard heat of reaction DHorxn for any chemical process can
be calculated by taking the difference between the total standard enthalpies of
formation of the chemical compounds that are produced and the total standard
enthalpies of the chemical compounds that participate as reactants:
(5.8) DHorxn = SDHfo[product compounds] - SDHfo[reactant compounds]
Notice that no information is needed about the standard enthalpies of formation
of any elementary substance that are involved in the reaction. By convention, the
value of DHfo for any elementary substance in its stable state under standard conditions
is then taken to be equal to zero.
c) DGorxn
Once we know the values for DSorxn and DHorxn for a chemical reaction, we can use
the definition for the change in Gibbs free energy given in Eq. (5.6) to calculate
DGorxn for the process. This quantity can be used to predict the directionality and
extent of the chemical reaction (DGorxn < 0 for processes that are thermodynamically favored under standard conditions). Eq. (5.6) can also be used to estimate
DGorxn(T) = DHorxn – T DSorxn at different temperatures if we assume that the values
of DSorxn and DHorxn remain constant at all temperatures (which is not necessarily
the case, but allows us to make reasonable predictions).
To facilitate the calculation of DGorxn at 25 oC and 1 atm for a chemical reaction, chemical scientists have determined the Standard Molar Gibbs Free Energy of
Formation, DGfo (kJ/mol), of many chemical compounds. This quantity is a measure of the change in Gibbs free energy when a mole of the compound is formed
starting from its elementary components at standard T and P. The definition of
DGfo is similar to that of DHfo, and thus we can apply the ideas derived from Hess’s
Law to calculate DGorxn using DGfo values. In particular:
(5.9)
DGorxn = SDGfo[product compounds] - SDGfo[reactant compounds],
where the DGfo for elementary substances is also taken to be equal to zero.
Chemical Thinking
U5
How do we predict chemical change?
LET’S THINK
First Amino Acids
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, and CH4. Consider these two potential
paths in the synthesis of the amino acid, glycine (C2H5NO2):
2CO(g) + NH3(g) + H2(g)
2CH4(g) + NH3(g) + 2H2O(l)
C2H5NO2(s)
C2H5NO2(s) + 5H2(g)
Substance
DHfo
Sfo
DGfo
(kJ mol-1) (J mol-1 K-1) (kJ mol-1)
H2(g)
0
130.7
0
•
Calculate DSorxn, DHorxn, and DGorxn for each
reaction.
CH4(g)
–74.6
186.3
–50.5
•
Build graphs of DGorxn(T) as a function of T
and analyze the effect of temperature on the
extent of reaction.
CO(g)
–110.5
197.7
–137.2
NH3(g)
–45.9
192.8
–16.4
H2O(l)
–285.8
70.0
–237.1
C2H5NO2(s)
–528.4
103.5
–368.8
•
Determine the temperature at which each reaction becomes product-favored.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Reaction Extent
Although the change in Gibbs free energy DGrxn allows us to predict the directionality of a
chemical reaction, the actual number tells us little about the final relative amounts of product
to reactants (reaction extent). Experimental results indicate that chemical reactions reach an
“equilibrium” state in which the concentration of products and reactants does not vary over
time. In particular, it is found that, for a generic chemical reaction of the form:
aA+bB
c C + d D,
where A, B, C, and D are gaseous substances, the following ratios of the pressures of reactants
to products remain constant when chemical equilibrium is reached:
(5.10)
Kp =
PCcPDd
PAaPBb
If A, B, C, and D are substances dissolved in a solvent, then the following ratio of their molar
concentrations (expressed using square brackets) remains constant at equilibrium:
(5.11)
KC =
[C]c[D]d
[A]a[B]b
The quantities KP and KC evaluated at the equilibrium pressures or concentrations receive
the name of Equilibrium Constants, and they can be used as a measure of reaction extent. In
general, chemical reactions for which KP >1 or KC > 1 can be thought as product-favored. The
larger the value of an equilibrium constant K, the larger the extent of the process.
305
306
MODULE 2
Comparing Free Energies
LET’S THINK Equilibrium Constant
The formation of molecular oxygen O2 in Earth’s atmosphere is affected by UV radiation that
breaks these molecules apart: O2(g)
2 O(g)
•
Express the equilibrium constant Kp for this reaction. Discuss whether the value of Kp depends on the pressure of O2(g) and O(g) in any given system.
Results from experiments carried out using different initial pressures of O2(g) at a high constant
temperature T are shown in the images below:
•
Estimate the value of Kp for these two systems.
•
Compare the KP values and build a reasonable explanation for these results.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Figure 5.9 Particulate rep-
resentation of the initial and
equilibrium states of reactants
and products for the transformation of A(s) into B(g).
A(s)
B(g)
Initial
The values of the equilibrium constants KP or KC at any given temperature are
independent of the initial pressures or concentrations of the reactants and products in a chemical system. Although the equilibrium pressures and concentrations
of every species involved in the reaction will differ depending on the initial conditions, the ratios of equilibrium pressures or concentrations as defined in Eq. (5.10)
and Eq. (5.11), respectively, will remain constant.
The values of KP or KC at any given temperature are also independent of the
amounts of pure liquids or solids consumed or produced during a chemical reaction. Consider, for example, the reaction represented in
Figure 5.9 where a generic solid substance A(s) transforms
into the gaseous substance B(g). If the initial conditions are
such that the pressure of B(g) in the system is very low, we
can expect this pressure to increase as the reaction proceeds.
Although A(s) will be consumed in this process and its total
amount in the system will decrease, the concentration of A
particles (i.e., number of A particles per unit volume) will not
change. Consequently, the value of the equilibrium constant
KP only depends on the equilibrium pressure of the product
Equilibrium
and KP = PB in this case.
Chemical Thinking
KP Expressions
U5
How do we predict chemical change?
307
LET’S THINK
Consider the two hypothetical routes for the synthesis of glycine analyzed previously:
2CO(g) + NH3(g) + H2(g)
2CH4(g) + NH3(g) + 2H2O(l)
•
C2H5NO2(s)
C2H5NO2(s) + 5H2(g)
Express the equilibrium constants for following two reactions. Discuss whether the pressure
or concentration of solid and liquid substances should be included in the expressions.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
The value of the equilibrium constant KP or KC is independent of the initial pressures
and concentrations of reactants and products, as well as of the external pressure acting on the
system. However, the value of KP or KC is affected by changes in the temperature at which the
chemical reaction takes place. The following relationship expresses the effect of temperature
on the equilibrium constant:
(5.12)K = exp [ –DGorxn(T)/ (RT) ]
where K represents either KP or KC, T is the temperature, R is the ideal gas constant (R = 8.314
J mol-1 K-1), and DGorxn(T) is the standard change of Gibbs free energy for the reaction at temperature T. Equation (5.12) indicates that for product-favored processes, where DGorxn(T) < 0,
the equilibrium constant K > 1, while K < 1 when the reaction is reactant-favored. If we take
DGorxn(T)= DHorxn – TDSorxn, the dependence of the equilibrium constant on temperature can
also be expressed as:
(5.13)
K = exp [ – DHorxn / (RT) + DSorxn / R ]
Temperature Effects LET’S THINK
Consider a simple isomerization reaction in aqueous solution of the form: A(aq)
•
• Once you made your predictions, click on the image
to the right to launch a simulation that will allow
you to verify your ideas. This simulation allows you
to see the effect of independently changing the three
main variables that determine the value of KC.
Share your ideas with a classmate and justify your reasoning.
CLICK TO PLAY
Predict the effect of changing a) DHorxn , b) DSorxn, and
T on the value of the equilibrium constant KC for the
reaction and on the reaction extent.
B(aq).
http://www.chem.arizona.edu/tpp/chemthink/resources/U5_M2/gibbs.html
This relationship suggests that the effect of changing temperature on the value of the equilibrium constant is not the same for exothermic reactions as it is for endothermic reactions.
308
MODULE 2
Comparing Free Energies
According to Eq. (5.13), the equilibrium constant of endothermic reactions
increases as the temperature increases. On the other hand, the value of KP or KC
for exothermic reactions becomes smaller at higher temperatures. These effects are
independent of whether the entropy of the system increases or decreases as a result
of the chemical reaction. However, reactions where ∆Horxn < 0 and ∆Sorxn> 0 are
product-favored at all temperatures and K > 1 at all T. Similarly, reactions where
∆Horxn > 0 and ∆Sorxn< 0 are reactant-favored at all temperatures and K < 1 under
all circumstances. It is only for processes where ∆Horxn and ∆Sorxn are either both
negative or both positive that K values become larger or smaller than one when
changing temperature.
Different chemical reactions reach equilibrium states that have different proportion of products and reactants in the system. For example, combustion reactions tend to have very large K values under standard conditions:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
DGorxn = –800.9 kJ
KP = 2.41 x 10140
The value of the equilibrium constant at 25 oC is so large that this reaction can be
assumed to go to completion. Given that ∆Horxn < 0 and ∆Sorxn> 0 in this process,
K > 1 at all temperatures. In these types of processes, we commonly use a single
arrow to represent the directionality of the reaction in the chemical equation.
Other processes have equilibrium constants that are not very large and the
value of KP or KC may vary considerably when changing the temperature. Consider, for example, the reaction:
CO(g) + H2O(l)
HCOOH(l)
DGorxn = –12.9 kJ
KP = 182
with ∆Horxn < 0 and ∆Sorxn< 0. This process is thermodynamically favored at 25 oC,
but the extent of the reaction decreases with increasing temperature. In cases like
this, it is better to represent the process using a double arrow connecting reactants
and products to indicate that the reaction achieves an equilibrium state in which
both reactants and products are present to a significant extent. The use of a single
or a double arrow in the representation of these types of chemical reactions may
change depending on reaction conditions.
LET’S THINK
Temperature Effects
Consider the reaction the reaction for the decomposition of molecular oxygen in the atmosphere:
O2(g)
2 O(g).
•
Express Kp of the reaction.
•
Calculate ∆H and ∆S and predict the effect of
changing T on the value of KP.
•
o
rxn
o
rxn
Calculate the values of KP at 298 K and at 4000 K. Discuss
how to best represent the process at each temperature.
Substance
DHfo
Sfo
-1
(kJ mol ) (J mol-1 K-1)
O2(g)
0
205.2
O(g)
249
161
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
Chemical Thinking
FACING THE CHALLENGE
U5
How do we predict chemical change?
monia to form amines and amides. The resulting
protein-like compounds would form a hot soup
which would aggregate to form colloidal systems
Chemical Origins
(coacervates,) from which the first heterotrophic
Chemical thinking has played a central role in the microorganisms evolved. Similar ideas were prodevelopment of models and theories about the posed by J. B. S. Haldane around the same time.
In 1953, Stanley Miller and Harold Urey puborigin of life in our planet. Although for many
years chemists thought that the worlds of the liv- lished the results of a classic experiment designed
ing and the nonliving followed different chemical to test Oparin’s and Haldane’s hypothesis that conrules, this idea was shattered in 1827 when Fried- ditions on the primitive Earth led to the synthesis
rich Whöler demonstrated that urea, a well known of organic compounds from simple inorganic preorganic compound produced by living organisms, cursors. In this experiment, a mixture of H2O(g),
CH4(g), NH3(g), and H2(g)
could be synthesized in the
in a sealed flask was conlaboratory by heating an innected to a container with
organic substance, ammonihot liquid water and exum cyanate. Later, in 1850,
posed to sparks to simulate
Adolph Strecker would be
lightning through a waterable to synthesize the amino
vapor-rich atmosphere. Afacid alanine from a mixture
ter two weeks of running
of ammonia, hydrogen cyathe experimental system,
nide, and acetaldehyde. This
Urey and Miller observed
was followed by the synthethe formation of various orsis of sugars by Butlerov usganic compounds, including
ing formaldehyde.
a large percentage of amino
During the first part of
By GYassineMrabetTalk (Own work) [Share Alike 3.0] via Wikimedia Commons
acids needed to make prothe twentieth century, several scientists developed diverse hypotheses about teins in living cells. One of the surprising results
the beginning of life in our planet. The dominant of the Miller-Urey experiment was that the proview during this time was that the first forms of life cess did not yield a random mixture of chemical
were photosynthetic microorganisms that could fix compounds, but rather a relatively small number
CO2(g) in the atmosphere and use it in combina- of substances, most of them of biochemical sigtion with water to synthesize organic compounds nificance.
The Miller-Urey experiment inspired many
(autotrophic organisms).
Alexander Oparin’s proposal about the origin others. For example, in 1960 Juan Oró reported
of life stood in sharp contrast with the prevalent the synthesis of adenine, a critical component of
ideas during his time. Oparin reasoned that since nuclei acids, from aqueous solutions of HCN and
heterotrophic microorganisms (those that do not NH3. However, there are numerous biochemicals
produce their own food) are metabolically simpler that do not appear to be prebiotically accessible.
than autotrophs, they should have evolved first. In Current theories about the origin of life suggest
his famous book “The Origin of Life” published in that it is unlikely that a single mechanisms was
1936, he proposed that hydrocarbons were formed responsible for the wide range of chemical comin our planet from the reaction of water vapor with pounds that may have been present on the primicarbon-metal compounds (carbides), in an atmo- tive Earth. The prebiotic soup was likely formed
spheric medium containing CH4(g), NH3(g), and by contributions from atmospheric synthesis,
H2O(g). The anaerobic fermentation of these hy- deep-sea hydrothermal vent synthesis, and exterdrocarbons would yield alcohols, aldehydes, ke- nal delivery from sources such as comets, meteortones, and so on, which would then react with am- ites, and interplanetary dust.
309
310
MODULE 2
Comparing Free Energies
Let’s Apply
ASSESS WHAT YOU KNOW
Chemical Origins
The mixture used by Miller and Urey in their famous experiment about the origin of life is one
of many that have been tested in an effort to better understand how complex organic molecules
could have been formed from simple inorganic compounds. Surprisingly, several of these mixtures
produce similar results. Why is that?
Primitive Mixtures
The following two mixtures have been tested as potential primitive sources of organic
molecules: 1) H2O(g), CH4(g), NH3(g), and H2(g); 2) CO(g), N2(g), and H2(g). Experimental results indicate that the elemental composition (presence of C, H, N, O) of the
mixture is more relevant than the kinds of molecules used. This seems to be due to the
existence of chemical processes that transform one set of substances into the other:
CH4(g) + 2 H2O(g)
CO2(g) + H2(g)
2 NH3(g)
•
•
CO2(g) + 4 H2(g)
CO(g) + H2O(g)
N2(g) + 3 H2(g)
Calculate DH , DS , and DG for each of these
three processes. Express and determine the value of
KP for each reaction under standard conditions.
o
rxn
o
rxn
o
rxn
Estimate the values of DGorxn(T) and KP for each of
these processes at a) 100 oC and b) 1000 oC. These
values define the range of temperatures that were
common in the primitive Earth. Build graphs of
DGorxn(T) as a function of T to analyze the effect of
changing temperature.
Substance
DHfo
Sfo
-1
(kJ mol ) (J mol-1 K-1)
H2(g)
0
130.7
CH4(g)
–74.6
186.3
CO(g)
–110.5
197.7
CO2(g)
–393.5
213.8
NH3(g)
–45.9
192.8
H2O(g)
–241.8
188.8
H2O(l)
–285.8
70.0
N2(g)
0
191.6
O2(g)
0
205.2
• Based on your results, discuss which compounds
of C, H, N, and O are more likely to exist at each temperature.
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
By NASA
Chemical Thinking
U5
How do we predict chemical change?
311
Amino Acids
Experiments such as those carried out by Miller and Urey to test hypotheses about the origin of
life in our planet have led to the formation of different types of amino acids. A general chemical
reaction associated with the synthesis of many of these substances can be represented as:
n CO2(g) + NH3(g) + (2n-2+(m-3)/2) H2(g)
•
•
Make graphs showing how DGorxn(T) and KP
are expected to change with temperature in
the range from 25 oC to 200 oC.
Discuss how temperature affects the thermodynamic likelihood of these reactions.
DHfo
S fo
-1
(kJ mol ) (J mol-1 K-1)
Substance
Glycine, C2H5NO2(s)
–527.5
103.5
Alanine, C3H7NO2(s)
–560.0
129.2
Proline, C5H9NO2(s)
–507.6
164.1
Valine, C5H11NO2(s)
–628.9
178.9
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
From mural at Page Museum
Other Paths
Other potential reactions for the synthesis of amino acids in the primitive Earth are illustrated
below for the case of glycine. Values of the associated equilibrium constants under standard
conditions are also provided:
2 CH4(g) + NH3(g) + 2 H2O(l)
2 CO(g) + NH3(g) + H2(g)
2 CH4(g) + NH3(g) + 5/2 O2(g)
C2H5NO2(s) + 5 H2(g)
C2H5NO2(s)
C2H5NO2(s) + 3 H2O(l)
KP = 6.14 x 10-40
KP = 3.25 x 1013
KP = 6.97 x 10164
•
Determine the DGorxn for each of these processes.
•
Discuss how you would expect DGorxn(T) and KP to change with increasing temperature.
•
Although the third reaction in this set has a large KP at 25 oC, if the reactants are mixed
the process does not take place. How would you explain this phenomenon?
Share and discuss your ideas with a classmate, and clearly justify your reasoning.
ASSESS WHAT YOU KNOW
•
Calculate DHorxn, DSorxn, and DGorxn for the
synthesis of the four amino included in the
table. Express and determine the value of KP
for each reaction under standard conditions.
CnHmNO2(s) + (2n-2) H2O(l)
312
MODULE 2
Comparing Free Energies
Let’s Apply
Biochemical Energetics
ASSESS WHAT YOU KNOW
The analysis of energy generation, transfer, and storage in cells is of central importance in the biochemical study of living organisms. Determining changes in Gibbs free energy in biochemical processes is critical for understanding important processes such as cell respiration and photosynthesis.
Endergonic Reactions
The synthesis reactions for the formation of many amino acids, proteins, carbohydrates, and
fats starting from simple molecules have ∆Gorxn > 0. In biochemistry, reactions with a positive DGorxn are known as endergonic reactions. Consider, for example, the synthesis of glucose,
C6H12O6(s), starting from CO2(g) and H2O(l):
6 CO2(g) + 6 H2O(g)
C6H12O6(s) + 6 O2(g)
•
Use the information provided in the table to calculate the
DGorxn and KP for this endergonic reaction.
•
Given that DGorxn > 0, this process is not thermodynamically favored. How is then possible that plants can perform
this synthesis as part of the photosynthetic process?
Substance
DGfo
(kJ mol-1)
CO2(g)
–394.4
H2O(l)
–237.1
C6H12O6(s)
–910.4
O2(g)
0
Exergonic Reactions
One central question in discussions about the origin of
life is how it was possible to induce and sustain the endergonic synthesis of complex biomolecules. An endergonic
reaction is not favored thermodynamically, but it can be
induced by “coupling” it with other favored reactions.
This is, reactions for which DGorxn < 0, also known as
exergonic reactions. In biological systems, one of such
processes is the hydrolysis of ATP into ADP and ortophosphate ion (Pi, HPO42-):
ATP(aq) + H2O(l)
•
ADP(aq) + Pi(aq)
DGorxn = –30.5 kJ/mol
The hydrolysis of ATP is not only exergonic but exothermic.
How can this reaction release heat when it involves breaking a
bond in ATP to form ADP + Pi?
Chemical Thinking
U5
How do we predict chemical change?
313
Coupling Reactions
In biological systems, endergonic processes are commonly coupled with exergonic reactions such as the hydrolysis of ATP (DGorxn = DGATP). This coupling is frequently achieved
through the intervention of enzymes that offer alternative reaction paths with a net DGorxn
< 0. For example, consider the generic endergonic process: A + B
AB, with DGorxn =
DG1. The coupling of this reaction with ATP hydrolysis is often accomplished by following
this type of two-step path:
•
The DGorxn for the net reaction in this two-step process is DGorxn = DGATP + DG1.
Discuss why the net DGorxn can be calculated by adding the corresponding values for
each of the two independent processes.
Glutamine is an amino acid synthesized many organisms from another amino acid, called
glutamic acid:
DGorxn = 14.2 kJ
AB
•
+
C
AC
+
B
Calculate the equilibrium constant for this reaction under standard conditions.
This reaction is coupled with the hydrolysis of ATP in our bodies leading to a net reaction
that is exergonic (thermodynamically favored).
•
Propose a mechanism for the coupling of the synthesis of glutamine with ATP hydrolysis. Use the schematic representation of the synthesis reaction (i.e., AB + C
AC + B) to express the reaction steps.
•
Write the chemical equation for the net reaction of the coupled process and calculate
the associated DGorxn.
•
Calculate the equilibrium constant for the net overall reaction. How many times
larger is this equilibrium constant than the equilibrium constant for the uncoupled
synthesis of glutamine?
Share your ideas with a classmate, and clearly justify your reasoning.
ASSESS WHAT YOU KNOW
1. A + ATP + H2O
A-Pi + ADP
Phosphorylation
2.
A-Pi + B
AB + Pi
Desphosphorylation
NET REACTION
A + B + ATP + H2O
AB + ADP