6 Adjusting G to different temperatures
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7.6 Adjusting
G to different temperatures
153
the change in free energy during hydrolysis is
r Gm
=
◦
r G310
+ RT ln Q
= −32.55 kJ mol−1
+ (8.314 472 × 10−3 kJ K−1 mol−1 )(310 K) ln
(0.3 × 10−3 )(2 × 10−3 )
4 × 10−3
= −55.25 kJ mol−1
This is the maximum amount of non-pressure–volume work available. Of course, muscles
mainly do non-pV work, since their volume is relatively constant, being made primarily
of incompressible components. Therefore, this is the maximum work that can be done by
muscle, per mole of ATP hydrolyzed.
Note that in the last calculation of the example, we must use a standard free energy of
reaction adjusted to the temperature of the calculation. We can’t use r G◦ (or r G◦ ) from
25 ◦ C and a different T in this equation.
Exercise group 7.6
(1) Calculate the maximum work available from the oxidation of glucose at 37 ◦ C and
solute concentrations of 0.005 mol L−1 of glucose, 0.001 mol L−1 of dissolved oxygen
and 0.03 mol L−1 of dissolved carbon dioxide. Compare your answer to the calculation
of Example 7.3.
(2) (a) Based on thermodynamic considerations alone, can liquid boron trichloride be
made by reacting diborane (B2 H6(g) ) with chlorine gas if the pressures of the two
reactants are both 0.5 bar and the pressure of hydrogen gas is 0.03 bar at 25 ◦ C?
(b) What if the reaction was run at −10 ◦ C instead of 25 ◦ C? Assume that r Hm◦ and
◦
r Sm are independent of temperature.
(3) Creatine phosphate is an alternative energy storage compound found in many cells.
Just as in the case of ATP, considerable free energy can be released by the hydrolysis
of creatine phosphate:
creatine phosphate + H2 O → creatine + Pi
with a r G◦ of −37.7 kJ mol−1 at 310 K. Suppose that, in a certain muscle with a
volume of 1 L, the concentrations of creatine and creatine phosphate are both approximately 1 mmol L−1 and that the concentration of phosphate is 2 mmol L−1 . What is
the maximum work that can be done by such a muscle if all the concentrations are
held constant by homeostatic mechanisms when the entire store of creatine phosphate
is turned over once? How far would this amount of work lift a 1 kg mass?
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Free Energy
Key ideas and equations
r Reversible processes are maximally efficient.
r A = U − T S is the Helmholtz free energy.
◦ A < 0 for a thermodynamically allowed process at constant T and V .
◦ − A is the maximum work that can be done by a machine operating isothermally.
r G = H − T S is the Gibbs free energy.
◦ G < 0 for a thermodynamically allowed process at constant T and p.
◦ − G is the maximum non-pV work that can be done by a machine operating isothermally and at constant pressure.
r Definition of standard state (1 bar, 25 ◦ C, solutes at 1 mol L−1 , everything behaving
ideally)
◦ Biochemists’ standard state: lump together species that differ by protons, pH = 7
r r Gm = r G◦m + RT ln Q
r Activities of ideal substances:
Phase
a
Solid
Liquid
Gas
Solute
1
X
p/p ◦
c/c◦
r To adjust G to different temperatures, it is often adequate to treat
independent of temperature.
H and
S as being
Suggested reading
For detailed explanations of the biochemists’ standard state, see
Robert A. Alberty, Biochim. Biophys. Acta 1207, 1 (1994).
Robert A. Alberty, Biochem. Ed. 28, 12 (2000).
This paper includes a large table of thermodynamic data in the biochemists’ standard
state.
If you ever need to convert data from the old standard pressure of 1 atm to the new
standard pressure of 1 atm, see the following paper:
Richard S. Treptow, J. Chem. Ed. 76, 212 (1999).
Because the formation reactions for most compounds involve some gaseous elements, this
conversion can affect free energies of formation for compounds in any state of matter.
Review exercise group 7.7
(1) Many reactions involve water as a reactant or product. When treating such reactions,
we generally ignore the activity of the water, i.e. set its value to unity. Why is it often
reasonable to do this?
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7.6 Adjusting
October 29, 2011
G to different temperatures
155
(2) What is wrong with the following statement?
G gives the maximum work that a system can perform at constant temperature and pressure.
(3) For processes involving only solids or liquids, what is the relationship between G
and A?
(4) A common dry cell (“battery”), also known as a Leclanch´e cell, has the following
overall reaction:
2+
Zn(s) + 2NH+
4(aq) + 2MnO2(s) → Zn(aq) + 2NH3(aq) + H2 O(l) + Mn2 O3(s)
Suppose that a dry cell operating at 25 ◦ C starts off with the following concentra−1
−1
2+
and [NH3(aq) ] = 0.12 mol L−1 .
tions: [NH+
4 ] = 1.5 mol L , [Zn ] = 0.03 mol L
The solid reactants zinc and manganese (IV) oxide are present in excess. The water
exists in a concentrated paste rather than as a free solution (hence the name “dry
cell”). For the sake of argument, assume that aH2 O is constant and has the value 0.5.
Calculate the maximum work that can be performed by this cell per mole of zinc
consumed:
(a) initially, and
(b) when half the ammonium ion available has reacted.
(5) While ATP is the compound used to directly power most metabolic processes, it is
not a very good long-term energy storage compound because of its low stability. ATP
is generally made from ADP by harnessing the energy obtained when sugars are
oxidized. A muscle can therefore be considered to be powered, indirectly at least, by
the oxidation of sugars. A teaspoon of sucrose (ordinary table sugar) weighs about
4 g. To what maximum height would the oxidation of a teaspoon of sugar allow an
organism working at 298 K to raise a 1 kg mass under standard conditions? The molar
mass of sucrose is 342.299 g mol−1 . Note that when this process occurs in a cell,
aqueous oxygen and carbon dioxide are involved instead of the gaseous molecules.
(6) CODATA, the Committee on Data for Science and Technology, produces a table that
gives generally accepted values of thermodynamic properties for a broad range of
substances. Their table is a little different from the one in most textbooks. Here is an
excerpt from that table, with notations adjusted to match the ones used in this book:
Substance State
Al
Al3+
H+
H2
s
aq
aq
g
◦
(298.15 K) Sm◦ (298.15 K) Hm◦ (298.15 K) − Hm◦ (0 K)
kJ mol−1
J K−1 mol−1
kJ mol−1
fH
0
−538.4
0
0
28.30
−325
0
130.680
4.540
8.468
(a) Calculate the standard free energy of formation of Al3+
(aq) .
(b) To which thermodynamic property of a substance is the last column of the table
related? Give the precise mathematical relationship.
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Free Energy
(7) Industrially, methanol (CH3 OH) is made by reaction of carbon monoxide with hydrogen. The reaction is carried out at high temperatures in the presence of a catalyst.
Suppose, however, that you wanted to carry out the reaction at 25 ◦ C, the pressures of
CO and of H2 both being held constant at 0.5 bar. Is there any thermodynamic reason
why this could not be done?
Note: Methanol boils at 65 ◦ C.
(8) The equation
r Gm
=
◦
r Gm
+ RT ln Q
is used to compute the Gibbs free energy of reaction under particular conditions. The
enthalpy change depends extremely weakly on the reaction conditions. Use this fact
to derive an equation relating the entropy change to the standard entropy change and
to ln Q.
(9) While ATP is the principal short-term energy storage compound in living organisms,
many other compounds are used for this purpose. Glucose-6-phosphate (G6P) is an
intermediate in glucose metabolism, but it can be and sometimes is used as a source
of work. Its hydrolysis reaction is
G6P + H2 O → glucose + phosphate
For this reaction, r G◦m = −16.7 kJ mol−1 and r Hm◦ = −35.1 kJ mol−1 at 298 K.
Suppose that the concentration of G6P in an organelle at 37 ◦ C is 50 μmol L−1 , the
concentration of glucose is 20 μmol L−1 and the total concentration of phosphates
is 2 mmol L−1 . What is the maximum work available from this reaction per mole of
G6P?
(10) The following reactions have been used in voltaic cells (batteries):
2Na(s) + 5S(s) → Na2 S5(s)
VH(s) + NiOOH(s) → V(s) + Ni(OH)2(s)
◦
r Gm
◦
r Gm
= −401 kJ mol−1
= −130 kJ mol−1
(a) What is the standard free energy of formation of Na2 S5 ?
(b) Battery designers don’t generally care about the amount of electrical work per
mole. What they really care about is the work per unit mass, heavy batteries
having a number of obvious disadvantages. Assuming a stoichiometric mixture
of the reactants in each case, batteries based on which of the above reactions will
store the most energy for a given mass?
(c) The reactions shown above involve solid reactants and products only. This makes
the calculations required for the last question much more straightforward than
they would otherwise be. Why? What additional complication(s) would arise if
some of the reactants or products were solutes?
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8
Chemical equilibrium and coupled reactions
Life is a far-from-equilibrium phenomenon, so one might reasonably question the utility
of equilibrium theory to biochemists. There are a few reasons why you should care about
equilibrium. One is that we still do a lot of experiments in test tubes, and under those
conditions, systems eventually go to equilibrium. Another reason is that some reactions,
like acid–base reactions, equilibrate much faster than others, so they can be treated as
being in equilibrium. There is another, more subtle reason: We saw earlier that reversible
processes, those in which the system is constantly in equilibrium, are the most efficient
ones possible. Equilibrium therefore sets a limit to the efficiency of chemical processes, an
idea we will use when we discuss how one reaction can drive another forward.
8.1 What does
r Gm
mean?
The molar free energy of reaction, r Gm , (as well as other quantities in thermodynamics)
is a funny quantity. We typically think of it as the difference in free energy between reactants
and products, since that’s how we calculate it. However, its meaning is a bit more subtle
than that. To make things specific, imagine that we have a reaction
reactants → P + some other products
in a system held at constant temperature and pressure. r Gm represents the change in free
energy when we convert one molar equivalent of reactants to one mole of P under specified
conditions (concentrations, pressures, etc.). In other words, the composition of the system
should not change as our mole of P is made. However, if we carry out this conversion, we will
change the quantities of reactants and products, and thus the composition. There is a way
out of this apparent contradiction: Instead of making one mole of P, imagine that we convert
some very small fraction of the reactants to products. Let n be the number of moles of P at
some particular point in time. If we know the initial composition and the stoichiometry of
the reaction, then we can calculate the number of moles of any component of the reaction
if we know n. We can therefore think of G as a function of the one composition variable
n. (This assumes there is only one reaction going on in our system. In general, we need
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Equilibrium and coupled reactions
G
n2
n1
n
Figure 8.1 Schematic showing how the free energy of a system, G, varies with the number of moles
of a product, n. r Gm is the slope of the graph of G vs. n. This sketch shows two tangent lines at two
different values of n. At n = n1 , the slope of the tangent is negative, so r Gm < 0, indicating that
the reaction is thermodynamically allowed, i.e. that it will proceed to make more product (increase
n) if kinetic factors don’t prevent it from proceeding. At n = n2 , r Gm is positive, so the reverse
reaction, consuming products and decreasing n, is thermodynamically allowed.
as many variables as there are reactions.) Now suppose that, over some period of time, the
reaction causes a change in n of n, where n is sufficiently small that it makes only a
negligible difference to the overall composition of the system. We divide the change in the
free energy during this period of time, r G, by the number of moles of P made, n, to get
r Gm :
r Gm
rG
=
n
(8.1)
Since we converted only a very small fraction of the reactants to products, the reaction
conditions didn’t change significantly during this operation. The quantities on the right of
the equality are ordinary differences: r G is the change in the total free energy of the
system when n moles of P were produced or, if you prefer, G(n + n) − G(n). Now if
you think back to your calculus course, you may recognize that if n is small, then the
right-hand side of Equation (8.1) is in fact a derivative of G with respect to n. In other
words, the quantity we refer to as r Gm is actually
r Gm
=
dG(n)
dn
Figure 8.1 shows the relationship of G(n) and
tangent line to G(n) at a particular n.
r Gm ,
which is actually the slope of the
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8.2 Free energy and equilibrium
159
8.2 Free energy and equilibrium
For thermodynamically allowed processes at constant pressure and temperature, r Gm <
0. This implies that the Gibbs free energy of a system held under these conditions will
decrease until it hits a minimum. Referring to Figure 8.1, we could start to the left of the
equilibrium, where r Gm < 0, and move to increasing values of n (more product), until
we hit the minimum of G, where r Gm = 0. If, on the other hand, the initial composition
was such that we were to the right of the equilibrium, where r Gm > 0, then the reverse
reaction, consuming product, would be thermodynamically allowed, and the system would
again move toward the minimum in G. At the minimum, r Gm = 0, meaning that neither
the forward nor the reverse process is thermodynamically favored. The system is then at
equilibrium.
Equation (7.6) gives the dependence of r Gm on the reaction quotient Q. At equilibrium,
G
r m = 0 so that
◦
r Gm
= −RT ln K
(8.2)
or
K = exp(−
◦
r Gm /RT )
(8.3)
where K is the value of Q at equilibrium. Given that Q is a ratio of activities of products
and reactants, you will probably recognize K as the equilibrium constant which you have
encountered in previous chemistry courses.
There are a few things to note:
(1) The equilibrium constant is a ratio of activities. Activities are dimensionless, hence the
equilibrium constant is dimensionless.
(2) r G◦m depends on the standard state, so the value of the equilibrium constant also
depends on the standard state. This may seem a little strange because the standard state
is arbitrary. However, the equilibrium constant is constructed from activities which
themselves depend on the standard state. The dependence of r G◦m and of the activities
on the standard state offset each other in such a way that any calculation of observable
quantities from equilibrium relationships will yield identical results, regardless of the
standard state chosen, provided it is consistently used.
Example 8.1 Gas solubility For the process
O2(g)
◦
r Gm
=
◦
f G (aq)
−
O2(aq)
◦
f G (g)
−1
= 16.35 − 0 kJ mol
= 16.35 kJ mol−1
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Equilibrium and coupled reactions
We can use this to compute the equilibrium constant for this process at 298 K:
K = exp(−
◦
r Gm /RT )
= exp
−16.35 × 103 J mol−1
(8.314 472 J K−1 mol−1 )(298.15 K)
= 1.37 × 10−3
The equilibrium constant for this process corresponds to the ratio
K = aaq /ag = [O2 ]/PO2
(give or take the standard concentration and standard pressure, both of which are 1 in
appropriate units). This can be rearranged to the form
[O2 ] = KpO2
which is Henry’s law of gas solubility. Henry’s law is valid for dilute solutions. Note that
the choice for the activity of a solute made on page 150 is required to make ideal solution
theory consistent with Henry’s law. This is in fact why this choice was made.
The experimental value1 of the Henry’s law constant for oxygen is 1.26 × 10−3 . Because
of the exponential function that appears in Equation (8.3), small errors in the thermodynamic
data are amplified. The difference between the calculated and experimental equilibrium
constants is therefore not terribly disturbing. In fact, if we work backwards and calculate
◦
−1
which,
f G (O2 , aq) from the experimental Henry’s law constant, we get 16.55 kJ mol
considering the typical spread of values for measurements of this kind, is probably within
the statistical uncertainty of this quantity.
The partial pressure of oxygen in air at sea level is about 21 kPa. The activity of oxygen
would be pO2 /p◦ = 21 kPa/100 kPa = 0.21. Using our calculated Henry’s law constant,
we would calculate a dissolved oxygen concentration of
[O2 ] = (1.37 × 10−3 )(0.21) = 2.9 × 10−4 mol L−1
at sea level.
Example 8.2 Solvent vapor pressure Suppose that we want to know the vapor pressure
of a solution made by dissolving 0.2 mol of sodium chloride in 100 mL of water at 25 ◦ C.
Since sodium chloride is involatile, the vapor pressure of the solution is entirely due to the
liquid–vapor equilibrium of water:
H2 O(l)
H2 O(g)
To determine the equilibrium constant for this process, we calculate
◦
r Gm :
◦
r Gm
∴
1
= f G◦m (g) − f G◦m (l)
= −228.582 − (−237.140) kJ mol−1 = 8.558 kJ mol−1
.
−8.558 × 103 J mol−1
=
0.03167
K = exp
(8.314 472 J K−1 mol−1 )(298.15 K)
J.C. Kotz and P. Treichel, Chemistry & Chemical Reactivity, 3rd edn; Harcourt Brace: Fort Worth, 1996; p. 664.
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8.2 Free energy and equilibrium
161
The equilibrium constant for this process is K = ag /al = pH2 O /(p ◦ XH2 O ), so we have
pH2 O = Kp◦ XH2 O , which is a form of Raoult’s law: for the pure solvent, XH2 O = 1, so if
we call the vapor pressure of the pure solvent pH• 2 O , we get pH• 2 O = Kp ◦ . We can now use
this equation to eliminate K from our equation for the vapor pressure of the solution, and
we get
pH2 O = pH• 2 O XH2 O
which now looks like Raoult’s law as you may have seen it in your introductory chemistry
course. Raoult’s law is known to be valid for dilute solutions. Again, it was necessary to
choose a = X for solvents on page 149 in order to make the dilute solution theory agree
with Raoult’s law.
We now need to calculate the mole fraction of water in this problem: the mole density
of water at 25 ◦ C is 55.33 mol L−1 so the number of moles of water is
nH2 O = (55.33 mol L−1 )(0.100 L) = 5.53 mol
Since NaCl dissociates into its ions in solution,
XH2 O =
nH2 O
5.53
=
= 0.93
nH2 O + nNa+ + nCl−
5.53 + 2(0.2)
Thus
pH2 O = (0.03167)(0.93) = 0.030 bar
Example 8.3 The solubility product Suppose that we want to calculate the solubility
product of silver sulfide at 25 ◦ C. This is the equilibrium constant for the reaction
Ag2 S(s)
All we have to do is calculate
◦
r Gm
∴
◦
r Gm
2−
2Ag+
(aq) + S(aq)
and then Ksp :
= 2 f G◦ (Ag+ , aq) + f G◦ (S2− , aq) − f G◦ (Ag2 S)
= 2(77.11) + 79 − (−40.7) kJ mol−1 = 274 kJ mol−1
Ksp = exp
−274 × 103 J mol−1
(8.314 472 J K−1 mol−1 )(298.15 K)
= 1.0 × 10−48
Example 8.4 Standard free energies of formation from solubility measurements The
solubility of iron (III) hydroxide in water is 1.1 × 10−15 mol L−1 at 25 ◦ C. If the standard
−1
and the free energy of formation of the
free energy of formation of Fe3+
(aq) is −4.6 kJ mol
−1
hydroxide ion is −157.220 kJ mol , what is the standard free energy of formation of solid
Fe(OH)3 ?
First note that the solubility product is the equilibrium constant for the reaction
Fe(OH)3(s)
−
Fe3+
(aq) + 3OH(aq)
We can use the solubility to compute the solubility product: at equilibrium, aFe3+ = 1.1 ×
10−15 . While it is tempting to say that aOH− = 3aFe3+ , we must consider the autoionization
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October 29, 2011
Equilibrium and coupled reactions
of water which produces aOH− = 10−7 by itself. Since the amount of hydroxide generated
by the latter process is much, much greater than that generated by the dissociation of iron
(III) hydroxide, we need not consider the simultaneous equilibria, the autoionization of
water being by far the more important process. Therefore aOH− = 10−7 ,
Ksp = (aFe3+ )(aOH− )3 = 1.1 × 10−36
The free energy change for the reaction is
◦
r Gm
= −RT ln Ksp = −(8.314 472 J K−1 mol−1 )(298.15 K) ln(1.1 × 10−36 )
= 205 kJ mol−1
However
◦
r Gm
∴
fG
◦
=
fG
(Fe(OH)3 ) =
fG
◦
(Fe3+ , aq) + 3
fG
◦
(Fe3+ , aq) + 3
fG
◦
(OH− , aq) −
◦
(OH− , aq) −
◦
f G (Fe(OH)3 )
◦
r Gm
−1
= −4.6 + 3(−157.220) − 205 kJ mol−1 = −682 kJ mol .
Equilibrium problems often involve acids and bases. You will already be familiar with
the pH scale, but you were likely told that pH = − log10 [H+ ]. This is wrong. First of
all, taking the logarithm of a dimensioned quantity ([H+ ] has units of mol L−1 ) is not
mathematically sensible. [What are the units of log(mol L−1 )?] Secondly, pH is usually
measured electrochemically (Chapter 10). Electrochemical measurements, as we will see,
are equivalent to measuring the change in free energy, and the latter quantity is related to
activities, and not directly to concentrations. The correct definition of pH is
pH = − log10 aH+
Example 8.5 Acid dissociation constants are also equilibrium constants. For instance,
for acetic acid, Ka is the equilibrium constant for the reaction
CH3 COOH(aq)
◦
r Gm
∴
+
CH3 COO−
(aq) + H(aq)
= f G◦ (CH3 COO− , aq) − f G◦ (CH3 COOH, aq)
= −369.31 − (−396.39) kJ mol−1 = 27.08 kJ mol−1
Ka = exp
−27.08 × 103 J/mol
(8.314 472 J K−1 mol−1 )(298.15 K)
= 1.80 × 10−5 .
Given the dissociation constant, we can compute the pH of a solution of known formal
concentration. If for instance we have a 0.20 mol L−1 solution of acetic acid in water, we
are saying that
aCH3 COOH + aCH3 COO− = 0.20
It is often the case in these problems that the autoionization of water is a negligible source
of protons. If we make this assumption, which we can check later, the number of moles of
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8.2 Free energy and equilibrium
163
protons in solution will be equal to the number of moles of CH3 COO− so
Ka = 1.80 × 10−5 =
aH2 +
0.20 − aH+
After some rearrangement, this gives the quadratic equation
aH2 + + 1.80 × 10−5 aH+ − 3.60 × 10−6 = 0
The physically reasonable solution of this quadratic equation is
aH+ = 1.89 × 10−3
Note that aH+
10−7 , which confirms that we were justified in ignoring the autoionization
of water. The pH is therefore
pH = − log10 aH+ = 2.72
Exercise group 8.1
(1) Calculate the equilibrium constant (Kw ) at 298.15 K for the autoionization of water
H2 O → H+ + OH− .
(2) Predict the solubility of nickel (II) chloride in water at 25 ◦ C. Comment on the magnitude of the value computed.
(3) Calculate the pH of a solution with a formal concentration of 2 × 10−7 mol L−1 of the
strong acid HI at 25 ◦ C.
(4) A sample of 25 g of sodium acetate (CH3 COONa) is dissolved in 1 L of water at 25 ◦ C.
What is the pH of the solution? The base ionization constant (Kb ) of acetate at this
temperature is 5.6 × 10−10 .
(5) The ozone layer is approximately 20 km above sea level. At this altitude, the mean
temperature is 210 K and the partial pressure of oxygen is 1.2 kPa.
(a) Accurately calculate r G◦m for the reaction
2O3(g) → 3O2(g)
at 210 K, taking into account the temperature dependence of r Hm◦ and r Sm◦ .
(b) Calculate r G◦m for the above reaction at 210 K under the assumption that neither
◦
◦
r H nor
r S vary significantly with temperature. Is the result significantly
different?
(c) Calculate the equilibrium pressure of ozone under the conditions prevailing at an
altitude of 20 km.
(d) The above calculation gives a value that is several orders of magnitude smaller
than the actual partial pressure of ozone at 20 km. Why?
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