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Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 18
Entropy, Free Energy,
and Equilibrium
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
18
Entropy, Free Energy, and Equilibrium
18.1 Spontaneous Processes
18.2 Entropy
A Qualitative Description of Entropy
A Quantitative Description of Entropy
18.3 Entropy Changes in a System
Calculating ΔSsys
Standard S°
Qualitatively Predicting ΔS°sys
18.4 Entropy Changes in the Universe
Calculating ΔSsurr
The Second Law of Thermodynamics
The Third Law of Thermodynamics
18.5 Predicting Spontaneity
Gibbs Free-Energy Change, ΔG
Standard Free-Energy Changes, ΔG°
Using ΔG and ΔG° to Solve Problems
18.6 Free Energy and Chemical Equilibrium
Relationship Between ΔG and ΔG°
Relationship Between ΔG° and K
18.7 Thermodynamics in Living Systems
18.1 Spontaneous Processes
A process that does occur under a specific set of conditions is
called a spontaneous process.
A process that does not occur under a specific set of conditions is
called nonspontaneous.
Spontaneous Processes
A process that results in a decrease in the energy of a system often
is spontaneous:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH° = -890.4 kJ/mol
The sign of ΔH alone is insufficient to predict spontaneity in every
circumstance:
H2O(l)
H2O(s)
T > 0°C; ΔH° = -6.01 kJ/mol
18.2
Entropy
To predict spontaneity, both the enthalpy and entropy must be
known.
Entropy (S) of a system is a measure of how spread out or how
dispersed the system’s energy is.
Entropy
Spontaneity is favored by an increase in entropy.
S = k ln W
k is the Boltzmann constant (1.38 x 10–23 J/K)
W is the number of different arrangements
The number of arrangements possible is given by:
W = XN
X is the number of cells in a volume
N is the number of molecules
Entropy
Entropy
There are three possible states for this system:
1) One molecule on each side (eight possible arrangements)
2) Both molecules on the left (four possible arrangements)
3) Both molecules on the right (four possible arrangements)
The most probable state has the largest number of arrangements.
18.3
Entropy Changes in a System
The change in entropy for a system is the difference in entropy of the
final state and the entropy of the initial state.
ΔSsys = Sfinal – Sinitial
Alternatively:
Ssys
Vfinal
 nR ln
Vinitial
Worked Example 18.1
Determine the change in entropy for 1.0 mole of an ideal gas originally confined
to one-half of a 5.0-L container when the gas is allowed to expand to fill the entire
container at constant temperature.
Strategy This is the isothermal expansion of an ideal gas. Because the
molecules spread out to occupy a greater volume, we expect there to be an
increase in the entropy of the system. Use ΔSsys = Sfinal – Sinitial to solve for ΔSsys.
R = 8.314 J/K∙mol, n = 1.0 mole, Vfinal = 5.0 L, and Vinitial = 2.5 L.
Solution
ΔSsys = nR ln
Vfinal
8.314 J
5.0 L
= 5.8 J/K
= 1.0 mol ×
×ln
Vinitial
K ∙ mol
2.5 L
Think About It Remember that for a process to be spontaneous, something
must favor spontaneity. If the process is spontaneous but not exothermic (in this
case, there is no enthalpy change), then we should expect ΔSsys to be positive.
Worked Example 18.1 (cont.)
Solution These equilibrium concentrations are then substituted into the
equilibrium expression to give
(x)(0.050 + x)
-5
1.8×10 = 0.010 – x
Because we expect x to be very small (even smaller than 1.34×10-3 M–see
above), because the ionization of CH3COOH is suppressed by the presence of
CH3COO-, we assume
(0.10 – x) M ≈ 0.10 M
and
(0.050 + x) M ≈ 0.050 M
Therefore, the equilibrium expression simplifies to
(x)(0.050)
-5
1.8×10 = 0.010
and x = 3.6×10-5 M. According to the equilibrium table, [H+] = x, so
pH = –log(3.6×10-5) = 4.44.
Entropy Changes in a System
The standard entropy is the absolute entropy of a substance at 1 atm.
Temperature is not part of the standard state definition and must be
specified.
Entropy Changes in a System
There are several important trends in entropy:
 S°liquid > S°solid
 S°gas > S°liquid
 S° increases with molar mass
 S° increases with molecular complexity
 S° increases with the mobility of a phase (for an element with two
or more allotropes)
Entropy Changes in a System
In addition to translational motion, molecules exhibit vibrations and
rotations.
Entropy Changes in a System
For a chemical reaction
aA + bB → cC + dD
ΔS°rxn = [cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
Alternatively,
ΔS°rxn = ΣnS°(products) – ΣmS°(reactants)
Worked Example 18.2
From the standard enthalpy values in Appendix 2, calculate the standard entropy
changes for the following reactions at 25°C.
(a) CaCO3(s) → CaO(s) + CO2(g)
(b) N2(g) + 3H2(g) → 2NH3(g)
(c) H2(g) + Cl2(g) → 2HCl(g)
Strategy Look up standard enthalpy values and calculate ΔS°rxn. Just as we did
when we calculated standard enthalpies of reaction, we consider stoichiometric
coefficients to be dimensionless–giving ΔS°rxn units of J/K∙mol.
From Appendix 2, S°[CaCO3(s)] = 92.9 J/K∙mol, S°[CaO(s)] = 39.8 J/K∙mol,
S°[CO2(g)] = 213.6 J/K∙mol, S°[N2(g)] = 191.5 J/K∙mol, S°[H2(g)] = 131.0
J/K∙mol, S°[NH3(g)] = 193.0 J/K∙mol, S°[Cl2(g)] = 223.0 J/K∙mol, and S°[HCl(g)]
= 187.0 J/K∙mol.
Worked Example 18.2 (cont.)
Solution
(a) S°rxn = [S°(CaO) + S°(CO2)] – [S°(CaCO3)]
= [(39.8 J/K∙mol) + (213.6 J/K∙mol)] – (92.9 J/K∙mol)
= 160.5 J/K∙mol
(b) S°rxn = [2S°(NH3)] – [S°(N2) + 3S°(H2)]
=(2)(193.0 J/K∙mol) – [(191.5 J/K∙mol) + (3)(131.0 J/K∙mol)]
= –198.5 J/K∙mol
(c) S°rxn = [2S°(HCl)] – [S°(H2) + S°(Cl2)]
= (2)(187.0 J/K∙mol) – [(131.0 J/K∙mol) + (223.0 J/K∙mol)]
= 20.0 J/K∙mol
Think About It Remember to multiply each standard entropy value by the
correct stoichiometric coefficient. Like Equation 10.18, Equation 18.5 can only be
used with a balanced chemical equation.
Entropy Changes in a System
Several processes that lead to an increase in entropy are:
 Melting
 Vaporization or sublimation
 Temperature increase
 Reaction resulting in a greater number of gas molecules
Entropy Changes in a System
The process of dissolving a substance can lead to either an increase or
a decrease in entropy, depending on the nature of the solute.
Molecular solutes (i.e. sugar): entropy increases
Ionic compounds: entropy could decrease or increase
Worked Example 18.3
For each process, determine the sign of ΔS for the system: (a) decomposition of
CaCO3(s) to give CaO(s) and CO2(g), (b) heating bromine vapor from 45°C to
80°C, (c) condensation of water vapor on a cold surface, (d) reaction of NH3(g)
and HCl(g) to give NH4Cl(s), and (e) dissolution of sugar in water.
Strategy Consider the change in energy/mobility of atoms and the resulting
changeThink
in number
of possible
positionsinvolving
that eachonly
particle
can and
occupy
in each case.
About
It For reactions
liquids
solids,
An increase
in thethe
number
arrangements
corresponds
to in
anmany
increase
in entropy
predicting
sign ofofΔS°
can be more
difficult, but
such
and therefore
positive ΔS.
cases ana increase
in the total number of molecules and/or ions is
accompanied
increase
of entropy.
Solution
Increasesby
in an
entropy
generally
accompany solid-to-liquid, liquid-togas, and solid-to-gas transitions; the dissolving of one substance in another; a
temperature increase; and reactions that increase the net number of moles of gas.
ΔS is (a) positive
(b) positive
(d) negative
(e) positive
(c) negative
18.4
Entropy Changes in the Universe
Correctly predicting the spontaneity of a process requires us to
consider entropy changes in both the system and the surroundings.
An ice cube spontaneously melts in a room at 25°C.
Perspective
Components
ΔS
System
ice
positive
Surroundings
everything else
negative
A cup of hot water spontaneously cools to room temperature.
Perspective
Components
ΔS
System
hot water
negative
Surroundings
everything else
positive
The entropy of both the system AND surroundings are important!
Entropy Changes in the Universe
The change in entropy of the surroundings is directly proportional to
the enthalpy of the system.
Ssurr 
Hsys
T
The second law of thermodynamics states that for a process to be
spontaneous, ΔSuniverse must be positive.
ΔSuniverse = ΔSsys + ΔSsurr
Entropy Changes in the Universe
The second law of thermodynamics states that for a process to be
spontaneous, ΔSuniverse must be positive.
ΔSuniverse = ΔSsys + ΔSsurr
ΔSuniverse > 0 for a spontaneous process
ΔSuniverse < 0 for a nonspontaneous process
ΔSuniverse = 0 for an equilibrium process
Worked Example 18.4
Determine if each of the following is a spontaneous process, a nonspontaneous
process, or an equilibrium process at the specified temperature: (a) H2(g) + I2(g)
→ 2HI(g) at 0°C, (b) CaCO3(s) → CaO(s) + CO2(g) at 200°C, (c) CaCO3(s) →
CaO(s) + CO2(g) at 1000°C, (d) Na(s) → Na(l) at 98°C. (Assume that the
thermodynamic data in Appendix 2 do not vary with temperature.)
Strategy For each process, use ΔS°rxn = ΣnS°(products) – ΣmS°(reactants) to
determine ΔS°sys; ΔSsurr = (–ΔHsys/T)and ΔH°sys = ΣnΔH °(products)
–
f
ΣmΔH °(reactants)
to determine ΔH°sys and ΔS°surr. At the specified temperature, the
f
process is spontaneous if ΔS°sys and ΔS°surr sum to a positive number,
nonspontaneous is they sum to a negative number, and an equilibrium process if
they sum to zero. Note that because the reaction is the system, ΔSrxn and ΔSsys are
used interchangeably.
Worked Example 18.4 (cont.)
Solution (a) S°[H2(g)] = 131.0 J/K∙mol, S°[I2(g)] = 260.57 J/K∙mol, S°[HI(g)] =
206.3 J/K∙mol; ΔHf°[H2(g)] = 0 kJ/mol, ΔHf°[I2(g)] = 62.25 kJ/mol, ΔHf°[HI(g)]
= 25.9 kJ/mol.
ΔS°rxn = [2S°(HI)] – [S°(H2) + S°(I2)]
= (2)(206.3 J/K∙mol) – [131.0 J/K∙mol + 160.57 J/K∙mol] = 21.03 J/K∙mol
ΔH°rxn = [2 ΔH°f (HI)] – [ΔH°f (H2) + ΔH°f (I2)]
= (2)(25.9 kJ/mol) – [0 kJ/mol + 62.26 kJ/mol] = −10.5 kJ/mol
ΔSsurr =
−ΔHrxn
−(−10.5 kJ/mol)
=
= 0.0385 kJ/K∙mol = 38.5 J/K∙mol
T
273 K
ΔSuniverse = ΔSsys + ΔSsurr = 21.03 J/K∙mol + 38.5 J/K∙mol = 59.5 J/K∙mol
ΔSuniverse is positive; therefore the reaction is spontaneous at 0°C.
Worked Example 18.4 (cont.)
Solution (b) In Worked Example 18.2(a), we determined that for this reaction,
ΔS°rxn = 160.5 J/K∙mol; ΔH°f [CaCO3(s)] = −1206.9 kJ/mol, ΔH°f [CaO(s)] =
−635.6 kJ/mol, ΔH°f [CO2(g)] = −393.5 kJ/mol.
(b), (c)
ΔS°rxn = 160.5 J/K∙mol
ΔH°rxn = [ΔH°f (CaO) + ΔH°f (CO2)] – [ΔH°f (CaCO3)]
= [-635.6 kJ/mol + (–393.5 kJ/mol)] – (–1206.9 kJ/mol) = 177.8 kJ/mol
(b) T = 200°C and
ΔSsurr =
−ΔHsys
−(177.8 kJ/mol)
=
= −0.376 kJ/K∙mol = −376 J/K∙mol
T
473 K
ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K∙mol + (−376 J/K∙mol) = − 216 J/K∙mol
Δsuniverse is negative, therefore the reaction is nonspontaneous at 200°C.
Worked Example 18.4 (cont.)
Solution (c) T = 1000°C and
ΔSsurr =
−ΔHsys
−(177.8 kJ/mol)
=
= −0.1397 kJ/K∙mol = −139.7 J/K∙mol
T
473 K
ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K∙mol + (−139.7 J/K∙mol) = 20.8 J/K∙mol
In this case, ΔSuniverse is positive; therefore, the reaction is spontaneous at 1000°C.
Worked Example 18.4 (cont.)
Solution (d) S°[Na(s)] = 51.05 J/K∙mol, S°[Na(l)] = 57.56 J/K∙mol; ΔHf°[Na(s)]
= 0 kJ/mol, ΔHf°[Na(l)] = 2.41 kJ/mol.
ΔS°rxn = S°[Na(l)] – S°[Na(s)]
= 57.56 J/K∙mol – 51.05 J/K∙mol = 6.51 J/K∙mol
ΔH°rxn = ΔHf°[Na(l)] – ΔHf°[Na(s)]
Think About It Remember that standard enthalpies of formation
= 2.41 kJ/mol – 0 kJ/mol = 2.41 kJ/mol
have units of kJ/mol, whereas standard absolute entropies have units
−(2.41
kJ/mol)
of −ΔH
J/K∙mol.
sure
that you convert kilojoules to joules, or vice
rxn Make
ΔSsurr =
=
= −0.0650 kJ/K∙mol = −6.50 J/K∙mol
371 K the terms.
versa,Tbefore combining
ΔSuniverse = ΔSsys + ΔSsurr = 6.51 J/K∙mol − 6.50 J/K∙mol = 0.01 J/K∙mol ≈ 0
ΔSuniverse is zero; therefore, the reaction is an equilibrium process at 98°C. In fact,
this is the melting point of sodium.
Entropy Changes in the Universe
The third law of thermodynamics states that the entropy of a perfect
crystalline substance is zero at absolute zero.
Entropy increases in a
substance as temperature
increases from absolute
zero.
18.5
Predicting Spontaneity
Measurements on the surroundings are seldom made, limiting the use
of the second law of thermodynamics.
Gibbs free energy (G) or simply free energy can be used to express
spontaneity more directly.
G = H – TS
The change in free energy for a system is:
ΔG = ΔH – TΔS
Predicting Spontaneity
Using the Gibbs free energy, it is possible to make predictions on
spontaneity.
ΔG = ΔH – TΔS
ΔG < 0The reaction is spontaneous in the forward direction.
ΔG > 0The reaction is nonspontaneous in the forward direction.
ΔG = 0The system is at equilibrium
Worked Example 18.5
According to Table 18.4, a reaction will be spontaneous only at high temperatures
if both ΔH and ΔS are positive. For a reaction in which ΔH = 199.5 kJ/mol and ΔS
= 476 J/K∙mol, determine the temperature (in °C) above which the reaction is
spontaneous.
Think
About
It Spontaneity
is favored
by alow
release
energy (ΔHat
Strategy
The
temperature
that divides
high from
is theoftemperature
being
negative)
an increase
entropy
(ΔS–being
which ΔH
= TΔS
(ΔG =and
0).by
Therefore,
we in
use
ΔG = ΔH
TΔS,positive).
substituting 0 for
both
are positive,
as in this
case, only
entropy
ΔG andWhen
solving
forquantities
T to determine
temperature
in kelvins;
wethe
then
convert to
favor spontaneity. For an endothermic process such as this,
degreeschange
Celsius.
which requires the input of heat, it should make sense that adding
Solution
more heat by increasing the temperature will shift the equilibrium to
476 J
1 kJ
the right, thusΔS
making
it “more spontaneous.”
=
= 0.476 kJ/K∙mol
K∙mol 1000 J
T=
ΔH
199.5 kJ/mol
=
= 419
ΔS
0.476 kJ/K∙mol
K
= (419 – 273) = 146°C
Predicting Spontaneity
The standard free energy of reaction (ΔG°rxn) is free-energy change
for a reaction when it occurs under standard-state conditions.
The following conditions define the standard states of pure
substances and solutions are:
 Gases
1 atm pressure
 Liquids
pure liquid
 Solids
pure solid
 Elements
the most stable allotropic form at 1 atm and 25°C
 Solutions
1 molar concentration
Entropy Changes in a System
For a chemical reaction
aA + bB → cC + dD
ΔG°rxn = [cΔG°f (C) + dΔG°f (D)] – [aΔG°f (A) + bΔG°f (B)]
Alternatively,
ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f (reactants)
ΔG°f for any element in its most stable allotropic form at 1 atm is
defined as zero.
Worked Example 18.6
Calculate the standard free-energy changes for the following reactions at 25°C:
(a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
(b) 2MgO(s) → 2Mg(s) + O2(g)
Strategy Look up the ΔG°f values for the reactants and products in each
equation, and use ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f (reactants) to solve for
ΔG°rxn.
Solution From Appendix 2, we have the following values: ΔG°f [CH4(g)] =
−50.8 kJ/mol, ΔG°f [CO2(g)] = −394.4 kJ/mol, ΔG°f [H2O(l)] = −237.2 kJ/mol, and
ΔG°f [MgO(s)] = −569.6 kJ/mol. All the other substances are elements in their
standard states and have, by definition, ΔG°f = 0.
(a) ΔG°rxn = (ΔG°f [CO2(g)] + 2ΔG°f [H2O(l)]) – (ΔG°f [CH4(g)] + ΔG°f [O2(g)])
= [(−394.4 kJ/mol) + (2)(−237.2 kJ/mol)] − [(−50.8 kJ/mol) + (2)(0 kJ/mol)]
= −818.0 kJ/mol
Worked Example 18.6 (cont.)
Solution From Appendix 2, we have the following values: ΔG°f [CH4(g)] =
−50.8 kJ/mol, ΔG°f [CO2(g)] = −394.4 kJ/mol, ΔG°f [H2O(l)] = −237.2 kJ/mol, and
ΔG°f [MgO(s)] = −596.6 kJ/mol. All the other substances are elements in their
standard states and have, by definition, ΔG°f = 0.
(b) ΔG°rxn = (2ΔG°f [Mg(s)] + ΔG°f [O2(g)]) – (2ΔG°f [MgO(s)])
= [(2)(0 kJ/mol) + (0 kJ/mol)] − [(2)(−569.6 kJ/mol)]
= 1139 kJ/mol
Think About It Note that, like standard enthalpies of formation (ΔH°f ), standard
free energies of formation (ΔG°f ) depend on the state of matter. Using water as an
example, ΔG°f [H2O(l)] = −237.2 kJ/mol and ΔG°f [H2O(g)] = −228.6 kJ/mol.
Always double-check to make sure you have selected the right value from the
table.
Worked Example 18.7
The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol,
respectively. Calculate the entropy changes for the solid-to-liquid and liquid-tovapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and
boils at 80.1°C.
Strategy The solid-liquid transition at the melting point and the liquid-vapor
transition at the boiling points are equilibrium processes. Therefore, because ΔG
About Itin For
same
is zeroThink
at equilibrium,
eachthe
case
we substance,
can use ΔGΔS=vap
ΔHis–always
TΔS, substituting 0 for
larger
ΔSfus. the
Theentropy
changechange
in number
of
ΔG andsignificantly
solving for ΔS,
to than
determine
associated
with the
arrangements is always bigger in a liquid-to-gas transition than in a
process.
solid-to-liquid transition.
ΔHfus
10.9 kJ/mol
Solution
ΔSfus =
=
Tmelting
278.7 K
= 0.0391 kJ/K∙mol
ΔSvap =
or 39.1 J/K∙mol
ΔHvap
31.0 kJ/mol
=
Tboiling
353.3 K
= 0.0877 kJ/K∙mol
or 87.7 J/K∙mol
18.6
Free Energy and Chemical Equilibrium
It is the sign of ΔG (not ΔG°) that determines spontaneity.
The relationship between ΔG and ΔG° is:
ΔG = ΔG° + RT lnQ
R is the gas constant (8.314 J/K·mol).
T is the kelvin temperature.
Q is the reaction quotient.
Free Energy and Chemical Equilibrium
Consider the following equilibrium:
H2(g) + I2(g) ⇌ 2HI(g)
ΔG° at 25°C = 2.60 kJ/mol
ΔG depends on the partial pressures of each chemical species.
If PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 3.0 atm:
QP 
 PHI 
P P
H2
Then:
2
I2
 3.0   9.0  2.25

  2.0  2.0  4.0
2
2.60 kJ  8.314  103 kJ 
G 

  298 K ln2.25   4.60 kJ/mol
mol
K mol


Free Energy and Chemical Equilibrium
The spontaneity can be manipulated by changing the partial
pressures of the reaction components:
H2(g) + I2(g) ⇌ 2HI(g)
ΔG° at 25°C = 2.60 kJ/mol
If PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 1.0 atm:
QP 
 PHI 
2

1.0 
2
P P   2.0  2.0 
H2
I2

1.0
 0.25
4.0
Then:
2.60 kJ  8.314  103 kJ 
G 

  298 K ln0.25   0.8 kJ/mol
mol
K·mol


Worked Example 18.8
The equilibrium constant, KP, for the reaction
N2O4(g) ⇌ 2NO2(g)
is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.4
kJ/mol. In a certain experiment, the initial pressures are PN2O4 = 0.453 atm and
PNO2 = 0.122 atm. Calculate ΔG for the reaction at these pressures, and predict
the direction in which the reaction will proceed spontaneously to establish
equilibrium.
Strategy Use the partial pressures of N2O4 and NO2 to calculate the reaction
quotient QP, and then use ΔG = ΔG° + RT lnQ to calculate ΔG.
The reaction quotient expression is
(PNO2)2
(0.122)2
QP =
=
= 0.0329
PN2O4
0.453
Worked Example 18.8 (cont.)
Solution
ΔG = ΔG° + RT lnQ
5.4 kJ
8.314×10-3 kJ
=
+
(298 K)(ln 0.0329)
mol
K∙mol
= 5.4 kJ/mol – 8.46 kJ/mol
= –3.1 kJ/mol
Because ΔG is negative, the reaction proceeds spontaneously from left to right to
reach equilibrium.
Think About It Remember, a reaction with a positive ΔG° value can be
spontaneous if the starting concentrations of reactants and products are such that
Q < K.
Free Energy and Chemical Equilibrium
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG° + RT ln K
ΔG° = –RT ln K
Free Energy and Chemical Equilibrium
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG° + RT ln K
ΔG° = –RT ln K
Free Energy and Chemical Equilibrium
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG° + RT ln K
ΔG° = –RT ln K
Worked Example 18.9
Using data from Appendix 2, calculate the equilibrium constant, KP, for the
following reaction at 25°C:
2H2O(l) ⇌ 2H2(g) + O2(g)
Strategy Use data from Appendix 2 and ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f
(reactants) to calculate ΔG° for the reaction. Then use ΔG° = −RT lnK to solve for KP.
Think About It This is an extremely small equilibrium constant,
Solution
with
the large, positive
value of ΔG°. We know
ΔG°= which
(2ΔG°f is
[Hconsistent
2(g)] + ΔG°
f [O2(g)]) – (2ΔG°
f [H2O(l)])
everyday+ experience
water doeskJ/mol)]
not decompose
=from
[2(0 kJ/mol)
(0 kJ/mol)]that
− [(2)(−237.2
=spontaneously
474.4 kJ/mol into its constituent elements at 25°C.
ΔG° = −RT ln KP
474.4 kJ
8.314×10-3 kJ
=
(298 K) ln KP
mol
K∙mol
−191.5 = ln KP
KP = e−191.5 = 7×10-84
Worked Example 18.10
The equilibrium constant, Ksp, for the dissolution of silver chloride in water at
25°C:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
is 1.6×10-10. Calculate ΔG° for the process.
Strategy Use ΔG° = −RT lnK to calculate ΔG°.
Solution
R = 8.314×10-3 kJ/K∙mol and T = (25 + 273) = 298 K.
ΔG° = −RT ln Ksp
8.314×10-3 kJ
=
(298 K) ln (1.6×10-10)
K∙mol
= 55.9 kJ/mol
Think About It The relatively large, positive ΔG°, like the very small K value,
corresponds to a process that lies very far to the left. Note that the K in
ΔG° = −RT lnK can be any type of Kc (Ka, Kb, Ksp, etc.) or KP.
18.7
Thermodynamics of Living Systems
Many biological reactions have positive ΔG° value, making the
reaction nonspontaneous.
None spontaneous reactions can be coupled with spontaneous
reactions in order to drive a process forward:
alanine + glycine → alanylglycine
ATP + H2O → ADP + H3PO4
ΔG° = 29 kJ/mol
ΔG° = –31 kJ/mol
ATP + H2O + alanine + glycine → ADP + H3PO4 + alanylglycine
ΔG° = 29 kJ/mol + –31 kJ/mol = –2 kJ/mol
Thermodynamics of Living Systems
Many biological reactions have positive ΔG° value, making the
reaction nonspontaneous.
18
Key Concepts
A Qualitative Description of Entropy
A Quantitative Description of Entropy
Calculating Δssys
Standard S°
Qualitatively Predicting ΔS°sys
Calculating Δssurr
The Second Law of Thermodynamics
The Third Law of Thermodynamics
Gibbs Free-Energy Change, ΔG
Standard Free-Energy Changes, ΔG°
Using ΔG and ΔG° to Solve Problems
Relationship Between ΔG and ΔG°
Relationship Between ΔG° and K