The Laws of Thermodynamics in Review
1. The internal energy* of an isolated system is constant. There are
only two ways to change internal energy – heat and work.
U  q  w
2. The entropy of the universe increases in any spontaneous process.
S univ erse  S sy stem  S surroundin gs
3. The entropy of a perfect crystal approaches zero as the temperature
approaches absolute zero.
* Internal energy (U) is the energy stored in a system.
It includes the energy of chemical bonds and intermolecular
forces as well as kinetic energy.
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1st Law: Heat and Work in Review
Heat and work are path functions (not state functions).
Heat at a constant pressure is called Enthalpy. By definition, at
constant pressure:
q  H
Ex) Imagine a reaction done in a balloon. The volume of the
balloon may change over the course of the reaction.
i) If the balloon shrinks, work has been done on the system.
ii) If the balloon expands, the system has done work.
The amount of work done on the system depends on the pressure and on
the change in volume. At constant pressure:
w  PV
2
1st Law: Heat and Work (Review)
Recalling that U  q  w and that q  H at constant
pressure, we can relate enthalpy and internal energy:
q  U  w
Therefore, enthalpy under constant pressure can be defines as:
H  U  PV
3
2nd Law: Gibbs Free Energy
Recall that at constant temperature, the entropy change is:
q rev
S 
T
As q sy stem  H sy stem
at constant P,
Hence at constant P and T:
S sy stem 
H sy stem
T
The heat absorbed by the system will be the same as the heat
released by the surroundings , hence:
S surroundin gs  
H sy stem
T
4
2nd Law: Gibbs Free Energy
Recall that:
S univ erse  S sy stem  S surroundin gs
Then:
Suniverse  S system 
H system
T
Therefore:
T Suniverse  T S system  H system
For any spontaneous reaction, -TSuniverse will be negative.
5
2nd Law: Gibbs Free Energy
-TSuniverse is defined in terms of the system only, and is used to
define a new term, the Gibbs free energy (G), as:
G  H TS
i) At constant temperature and pressure,
G  H TS
since enthalpy is only meaningful at constant pressure.
ii) For any spontaneous reaction, G will be negative.
iii) Gibbs free energy is a state function.
6
Gibbs Free Energy and the Standard State
Changes in Gibbs free energy are determined by reaction conditions.
Hence free energy values are always made in reference to standard
states.
Standard temperature and pressure (STP) are defined as:
i) T = 25 °C
ii) P = 1 bar (100 kPa)
Standard states are defined as:
i) A solid is in the standard state if it is at STP.
ii) A liquid is in the standard state if it is at STP.
iii) A gas is in the standard state if it is at STP and behaves ideally.
iv) A solution is in the standard state if it is has one solute with a
concentration of 1 mol/L, is at STP and behaves ideally.
Note that “a liquid” is not the same as “a solution”.
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Gibbs Free Energy and the Standard State
Standard molar Gibbs free energy of formation (fG° or fGm°) is
generally tabulated alongside standard molar enthalpy of formation
and standard entropy.
It refers to the free energy change when one mole of a compound is
made from pure elements in their standard states.
e.g.
Br2(l)  Br2(g)
Na(s)
1
 Br2(l)  NaBr(s)
2
kJ
Δ f G(Br 2(g) )  82.40
mol
kJ
Δ f G(NaBr (s ) )  -349.0
mol
If a substance is a pure element in its standard state, this means
that its standard molar Gibbs free energy of formation must be 0.
8
Gibbs Free Energy and Predicting Spontaneity
As the Gibbs free energy is a state function it is used to calculate the
free energy change for any reaction (rG°) in the standard state.
e.g.
H2 O(l)  SO3(g)  H2SO 4(aq)
Δ r G  ???
kJ
mol
kJ
 f G (SO 3( g ) )  371.1
mol
 f G (H 2O (l ) )  237.1
 f G (H 2SO 4 (aq ) )  744.5
kJ
9
mol
Gibbs Free Energy and Predicting Spontaneity
r G    f G (products)  f G (reactants )
Note the similarity to Hess’s Law.
Important notes:
The state of matter is important.
i) Only one allotrope of any element can have fG° = 0.
Most often the one that is most stable under STP.
ex) graphite, O2(g), etc.; white phosphorus are exceptions.
ii) As ions cannot exist in isolation hence without counterions;
Therefore the standard state H+(aq) (1 M at STP) is used as the
reference for ions.
ie. fG°(H+(aq)) = 0.
10
Gibbs Free Energy and Predicting Spontaneity
A reaction that we know is spontaneous occurs when potassium metal is
added to water. What is the free energy change for this reaction under
standard conditions?
e.g.
K (s)  H2O(l)  K

(aq)
(aq)
 OH
1
 H2(g)
2
Δ r G  ???
kJ
mol
kJ
 f G (K (aq ) )  283.3
mol
kJ
 f G (OH (aq ) )  157.2
mol
 f G (H 2O (l ) )  237.1
11
Gibbs Free Energy and Predicting Spontaneity
A reaction that we know doesn’t proceed so easily is the reaction of
hydrogen and nitrogen to make ammonia.
What is the free energy change for this reaction under standard
conditions?
f G (NH 3( g ) )  16.45
kJ
mol
12
Gibbs Free Energy and Activity
The Haber-Bosch process isn’t performed under standard
conditions so this free energy value doesn’t apply in this case.
Reaction conditions: 450 oC, 100 atm, catalyst: magnetite (Fe3O4)
In fact, many reactions are run under nonstandard conditions.
To deal with reactions under nonstandard conditions, we need to
consider the activity (a) of the substances involved.
Activity is a measure of a substance’s deviation from the
standard state:
i) Any substance in the standard state will have a = 1.
Recall that the standard state requires that gases and solutes
behave ideally.
ii) Activity is dimensionless and therefore has no units.
13
Gibbs Free Energy and Activity
For ideal substances, activity is relatively straightforward:
i) Pure solids have a = 1
ii) Ideal gases have a = P/P° where P° = 1 bar
iii) Ideal solutes have a = c/c° where c° = 1 mol/L
iv) Ideal solvents have a = X where X is the mole fraction of
solvent molecules.
For nonideal substances, activity must be modified by an activity
coefficient ():
i) Nonideal solutes have a =  c/c° where c° = 1 mol/L
ii) Activity coefficients are concentration-dependent as they
account for such nonideal behaviour arising from intermolecular
forces between different particles of solute. Thus, the activity
coefficient will tend towards 1 with dilution.
14
Gibbs Free Energy & Nonstandard Conditions
To calculate the free energy of a reaction under nonstandard
conditions, we use the free energy under standard conditions and
the deviation from the standard state conditions:
r G m  r G mo  RT lnQ
where rGm is the molar Gibbs free energy change under the
given conditions, R is the ideal gas constant, T is the given
temperature (in K) and Q is the reaction quotient:
Q 
product of activities of products
product of activities of reactants
There is a similarity between the formula for reaction quotient (Q)
and the formula for equilibrium constant (K).
The equilibrium constant only describes the system at equilibrium.
15
Gibbs Free Energy & Nonstandard Conditions
Calculate the molar Gibbs free energy change for the rusting of iron
under atmospheric conditions.
 fG (Fe 2O 3(s ) )  742.2
kJ
mol
PO2  0.21 bar
T  25 C
Q = 10.39
G = -736.4 kJ/mol
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Gibbs Free Energy & Nonstandard Conditions
Calculate the molar Gibbs free energy change for dissolving carbon
dioxide in a soft drink under atmospheric conditions.
kJ
mol
kJ
 fG (H 2O (l ) )  237.1
mol
kJ
 fG (H 2CO3(aq ) )  623.1
mol
 fG (CO2( g ) )  394.4
Q = 2.66
G=10.8 kJ/mol
PC O2  0.038 bar
[H2 CO3 ]  0.1 M
[sugar]  0.6 M
[H2 O]  55.33 M
T  25 C
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Gibbs Free Energy & Nonstandard Conditions
Repeat the calculation for dissolving carbon dioxide in a soft drink
with a higher pressure for carbon dioxide (5.0 bar).
kJ
mol
kJ
 fG (H 2O (l ) )  237.1
mol
kJ
 fG (H 2CO3(aq ) )  623.1
mol
 fG (CO2( g ) )  394.4
PC O2  5.0 bar
Q = 0.0202
G=-1.26 kJ/mol
[H2 CO3 ]  0.1 M
[sugar]  0.6 M
[H2 O]  55.33 M
T  25 C
18
Free Energy Change and Equilibrium
Recall that the free energy change for a reaction under any conditions
(rG) can be calculated from the free energy change of the same
reaction under standard conditions (rGº):
 r Gm   r Gmo  RT ln Q
where Q is the reaction quotient:
product of activities of products
Q
product of activities of reactants
We also know that:
i) The forward reaction is spontaneous if rG < 0.
ii) The reverse reaction is spontaneous if rG > 0.
What if rG = 0?
19
Free Energy Change and Equilibrium
When rG = 0, we can say that:
 r Gm   r Gmo  RT ln Q  0
Which rearranges to give:
 r Gmo   RT ln Q
The forward and reverse reactions are in balance, ie at
equilibrium!
Therefore:
Q=K
Hence:
 r G   RT ln K
o
m
20
Free Energy Change and Equilibrium
I) If we know the equilibium constant for a reaction, we can
calculate its standard molar free energy change:
 r Gmo  RT ln K
II) If we know the standard molar free energy change for a
reaction, we can therefore calculate its equilibrium constant:
K e
r Gmo
RT
Note that these equations refer to rGº (for standard conditions)
NOT to rG (which is ZERO at equilibrium!)
21
Q vs. K
reactants
Q
products
product of activities of products
product of activities of reactants
K
product of activities of products
product of activities of reactants
Consider a very generic equilibrium reaction:
i) When does Q = K?
AB
at equilibrium K =Q = aB/aA
ii) When is Q < K? When there is excess A or a depletion of B w.r.t
equilibrium aA > aA(eq) or aB < aB (eq) => Q < K
Reaction proceeds forward
When is Q > K? When there is excess B or a depletion of A w.r.t
equilibrium aA < aA(eq) or aB > aB (eq) => Q > K
Reaction proceeds in reverse
22
Free Energy Change and Equilibrium
Consider the dimerization of NO2:
2 NO2(g)
N2O4(g)
Calculate the equilibrium constant for this reaction at 25 ºC.
kJ
mol
kJ
 fG(N2O4( g ) )  97.89 23
mol
 fG(NO2( g ) )  51.31
K = 6.74
Free Energy Change and Equilibrium
Is the dimerization of NO2 spontaneous at 25 ºC if the partial pressure
of NO2 is 0.4 bar and the partial pressure of N2O4 is 1.8 bar?
24
Free Energy Change and Equilibrium: Acids
The acid ionization constant, Ka, is the equilibrium constant for the
reaction in which an acid ionizes.
e.g.
HF(aq)
H+(aq) + F-(aq)
K a  6.6  104
pKa is a measure of strength of an acid.
An acid with a low pKa is stronger than an acid with a high pKa.
The pKa value for an acid comes from its acid ionization constant:
pK a   log K a
Also an acid will be weaker than another acid whose Ka value is
smaller. (Even a “big” Ka value is pretty small. Ka values range
from about 10-50 to 1010.)
25
Free Energy Change and Equilibrium: Acids
Calculate the standard molar free energy of formation for HF(aq).
HF(aq)
H+(aq) + F-(aq)
K a  6.6  104
 fG(H(aq ) )  0
kJ
mol
kJ
 fG(F( aq ) )  278.8 26
mol
Free Energy Change and Equilibrium:Solubility
The solubility product, Ksp, is the equilibrium constant for the
reaction in which an ionic compound dissolves in water.
e.g.
PbI2(s)
Pb2+(aq) + 2 I-(aq)
K s p  ???
Ex) Calculate the solubility of lead(II) iodide in water at 25 ºC.
Solubility is reported in #moles of solid soluble in 1L water.
kJ
mol
kJ
 fG(I(aq ) )  51.57
mol
kJ
 fG(PbI2( s ) )  173.6 27
mol
 fG(Pb(2aq ) )  24.43
Free Energy Change and Equilibrium:Solubility
Simple Ksp calculations is that they almost never give good results.
Multiple simultaneous equilibria that must be considered, not just the one for
Ksp.
When PbI2 is dissolved in water, two complex ions also form:
Pb2+(aq) + 4 I-(aq)
Pb2+(aq) + 3 OH-(aq)
[PbI4]2-(aq)
[Pb(OH)3]-(aq)
K f  3.0  104
K f  3.8  1014
In this case the Ksp calculation does gives almost the same answer as a
complete calculation factoring in both complex ions; because, the Kf values for
both complex ions are relatively small (compared to some Kf values in the 1030
range).
Usually the complex ion formation is significant enough that it must be
considered. Otherwise in the solubility calculation will give very poor values.
28
Free Energy Change and Equilibrium:Solubility
Complex ions are not the only factor complicating solubility calculations.
Ex) What equilibria would you have to consider if you wanted to calculate
the solubility of carbon dioxide in water?
How would you increase the solubility?
29