The Gibbs Free Energy Function and Phase Transitions
H2O(l) --> H2O(s)
Normal freezing point of H2O = 273.15K
The change in enthalpy is the enthalpy of freezing - enthalpy
change associated when one mole of liquid water freezes at
1atm and 273.15K
DH = -6007J
Entropy change
-6007 J
DH
=
DS=
= -21.99 J/K
273.15K
T
DG = DH - TDS = -6007J - (273.15K)(-21.99J/K) = 0J
=> at 273.15K and 1 atm, liquid and solid water are at
equilibrium
As water is cooled by 10K below 273.15K to 263.15K
Assume that DH and DS do not change with T
DG = DH - TDS = -6007J - (263.15K)(-21.99J/K) = -220J
Since DG < 0 water spontaneously freezes
At 10K above the freezing point, 283.15K
DG = DH - TDS = -6007J - (283.15K)(-21.99J/K) = 219J
DG > 0 water does not spontaneously freeze at 283.15K, 10K
above the normal freezing point of water
Above 273.15K, the reverse process is spontaneous - ice
melts to liquid water.
H2O(l) --> H2O(s)
Freezing not
spontaneous;
melting
spontaneous
freezing
spontaneous
Plot of DH and TDS vs T for the freezing of water. At 273.15 K
the system is at equilibrium, DG = 0
For phase transitions, at the transition temperature, the
system is at equilibrium between the two phases.
Above or below the transition temperature, the phase that is
the more stable phase is determined by thermodynamics - DG
of the phase transition.
Reaction Free Energy
DGr = SnGm(products) - SnGm(reactants)
Standard Reaction Free Energy
DGor = SnGom (products) - SnGom(reactants)
Difference in free energy of the pure products in their
standard states and pure reactants in their standard states.
Standard Free-Energy of Formation
The standard free energy of formation, DGfo, of a substance is
the standard reaction free energy per mole for the formation
of a compound from its elements in their most stable form
1/2N2 (g) + 3/2 H2 (g) --> NH3(g) DGfo = -16 kJ
O2 (g) --> O2 (g)
DGfo = 0
DGfo is an indication of a compounds
stability relative to its elements
Thermodynamically stable compound
DGfo < 0
Thermodynamically unstable compound
DGfo > 0
Standard Reaction Free Energy
For a reaction: aA + bB --> cC + dD
DGro = c DGfo (C) + d DGfo(D) - a DGfo(A) -b DGfo(B)
The standard free energy of formation of elements in their
most stable form at 298.15K is zero.
Problem: Calculate the standard free-energy change for the
following reaction at 298K:
N2(g) + 3H2 (g) --> 2NH3(g)
Given that DGfo[NH3(g)] = -16.66 kJ/mol.
What is the DGo for the reverse reaction?
Since the reactants N2(g) and H2(g) are in their standard state
at 298 K their standard free energies of formation are defined
to be zero.
DGro = 2 DGfo[NH3(g)] - DGfo[N2(g)] - 3 DGfo[H2(g)]
= 2 x -16.66 = -33.32 kJ/mol
For the reverse reaction: 2NH3(g) --> N2(g) + 3H2 (g)
DGro = +33.32 kJ/mol
C3H8(g) + 5O2(g) --> 3CO2 (g) + 4H2O(l) DHo = - 2220 kJ
a) Without using the information from the tables of
thermodynamic quantities predict whether DGo for this
reaction will be less negative or more negative than DHo.
b) Use data from App. D to calculate DGro for this reaction.
Whether DGro is more negative or less negative than DHo
depends on the sign of DSo since DGro = DHo - T DSo
The reaction involves 6 moles of gaseous reactants and
produces 3 moles of gaseous product
=> DSo < 0 (since fewer moles of gaseous products
than reactants)
Since DSo < 0 and DHo < 0 and DGro = DHo - T DSo
=> DGro is less negative than DHo
DGro = 3DGfo(CO2 (g))+4DGfo(H2O(l)-DGfo(C3H8(g))-5DGfo(O2(g))
= 3(-394.4) + 4(-237.13) - (-23.47) - 5(0)
= -2108 kJ
Effect of temperature on DGo
Values of DGro calculated using the tabulated values of DGfo
apply only at 298.15K.
For other temperatures the relation:
DGro = DHo - T DSo
can be used, assuming DHo and DSo do not vary significantly
with temperature.
When DGr = 0, T =
o
DHo
DSo
Use this expression to determine temperature above or
below which reaction becomes spontaneous.
Problem: The normal boiling point is the temperature at
which a pure liquid is in equilibrium with its vapor at 1 atm.
a) write the chemical equation that defines the normal
boiling point of liquid CCl4
b) what is the value of DGo at equilibrium?
c) Use thermodynamic data to estimate the boiling point of
CCl4.
a) CCl4(l)
CCl4(g)
b) At equilibrium DG = 0. In any equilibrium for a normal
boiling point, both the liquid and gas are in their standard
states. Hence, for this process DG is DGo.
c)
o
DH
T=
DSo
where for this reaction T is the boiling point
Note: To determine the boiling point accurately we need the
values of DHo and DSo for the vaporization process at the
boiling point of CCl4
DHo = (1mol)(-106.7 kJ/mol) - (1mol)(-139.3kJ/mol) = 32.6 kJ
DSo = (1mol)(309.4 J/mol-K) -(1mol)(214.4 J/mol-K) = 95.0 J/K
o
DH
T=
DSo
=
32.6 kJ
0.095 kJ/K
= 343 K
The Gibbs Function and the Equilibrium Constant
DG = DGro + R T ln Q
Q - reaction quotient
Under standard conditions the concentrations of all reactants
and products are set to 1
(standard pressure of gases = 1 atm
Standard pressure of solutions = 1 M)
Under standard conditions, ln Q = 0
=> DG = DGro
At equilibrium DG = 0 and Q = K
At equilibrium
DGro = - R T ln K or K = e- DGr
o
/RT
DGro < 0 => K > 1
DGro > 0 => K < 1
DG = DGro + R T ln Q = - RT ln K + RT ln Q
DG = R T ln
Q
K
when Q = K => DG = 0; equilibrium
DGforward <
0
DGforward >
0
If the reaction mixture has too much N2 or H2 relative to NH3,
Q < K, and reaction moves spontaneously in the direction to
form NH3 till equilibrium is established; Q = K. If there is
too much NH3 in the mixture, Q > K, decomposition of NH3 is
spontaneous, till equilibrium is established; Q = K
Temperature Dependence of K
The temperature dependence of K can be used to determine
DHo and DSo of a reaction
o
o
ln K = -DGr = - DH + DS
RT
RT
R
If the equilibrium constant of a reaction is known at one
temperature and the value of DHo is known, then the
equilibrium constant at another temperature can be
calculated
o
o
DH
DS
ln K1 =
+
R T1
R
o
o
ln K2 = - DH + DS
R T2
R
ln K2 - ln K1 = ln K2 =
K1
DHo
R
(
1 - 1
T1
T2
)
If DHo is negative (exothermic), an increase in T reduces K; if
it is positive (endothermic), an increase in T increases K
For the equilibrium between a pure liquid and its vapor,
the equilibrium constant is simply the equilibrium vapor
pressure (since pure liquids do not appear in the
equilibrium expression)
Liquid
Gas
K = Pgas
Hence
ln p2 = DHvap ( 1 - 1
p1
R
T1
T2
)
Clausius-Clapeyron
equation
The Clausius-Clapeyron equation indicates the variation of
vapor pressure of a liquid with temperature
Driving Non-spontaneous Reactions
The magnitude of DG is a useful tool for evaluating reactions.
For a spontaneous process at constant pressure and
temperature, the change in free energy for a process equals
the maximum amount of work that can be done by the system
on its surroundings
wmax = DGr
For non-spontaneous reactions (DGr > 0), the magnitude of
DGr is a measure of the minimum amount of work that must
be done on the system to cause the process to occur.
A reaction for which DGr is large and negative (like the
combustion of gasoline), is much more capable of doing
work on the surroundings than a reaction for which DGr is
small and negative (like the melting of ice).
Many chemical reactions, including a large number that are
central to biological systems, are non-spontaneous.
For example: Cu can be extracted from the mineral
chalcolite which contains Cu2S.
The decomposition of Cu2S is non-spontaneous
Cu2S --> 2Cu(s) + S(s)
DGro = +86.2 kJ
Since this reaction is not spontaneous Cu cannot be directly
obtained from the above reaction.
Instead, work needs to be done on this reaction, and this is
done by coupling this non-spontaneous reaction with a
spontaneous reaction so that the overall reaction is
spontaneous.
Consider the following reaction:
S(s) + O2(g) --> SO2 (g)
DGro = -300.4kJ
Coupling this reaction with the extraction of Cu from Cu2S
Net:
Cu2S --> 2Cu(s) + S(s)
DGro = +86.2 kJ
S(s) + O2(g) --> SO2 (g)
DGro = -300.4kJ
Cu2S(s) + O2 (g) --> 2Cu(s) + SO2(g)
DGro = -214.2kJ
Biological systems employ the same principle by using
spontaneous reactions to drive non-spontaneous reactions.
Many biochemical reactions are not spontaneous.
These reactions are made to occur by coupling them with
spontaneous reactions which release heat.
The metabolism of food is the usual source of free energy
needed to do the work to maintain biological systems.
C6H12O6(s) + 6O2 (g) --> 6CO2 (g) + 6H2O(l)
DHo = -2803 kJ
A means of transporting the energy released by glucose
metabolism to the reactions that require energy is needed.
The free energy released by the metabolism of glucose is
used to convert lower-energy ADP (adenine diphosphate) to
higher energy ATP (adenine triphosphate).
The energy stored in the ATP molecule is then available to a
biochemical reaction that requires energy and in the
process ATP is converted to ADP.
C6H12O6
ATP
Free energy
released by
oxidation of glucose
converts ADP to
ATP
6CO2 (g) + 6H2O(l)
ADP
Free energy released
by ATP converts
simple molecules to
more complex
molecules