Entropy and Free
Energy
Chapter 19
•
First Law
– Energy is conserved in chemical processes
• neither created nor destroyed
• converted from one form into another
•
Second Law
– For any spontaneous process, the entropy of
the universe increases
• the real criterion for spontaneity
• changes in randomness of the universe is +
Laws of Thermodynamics

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
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
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Free energy- the energy that is available
to do work.
Entropy- a measure of the disorder of a
system.
Enthalpy-at constant pressure, it is the
heat evolved or absorbed in the reaction.
Spontaneous Reaction
Non-spontaneous Reaction
Law of Disorder-processes move in the
direction of maximum disorder or
randomness
Some definitions…
Entropy of a gas is greater than that of
a liquid or a solid.
2. Entropy increases when a substance is
divided into parts.
3. Entropy tends to increase in chemical
reactions in which the total number of
product molecules is greater than the
total number of reactant molecules.
4. Entropy tends to increase when
temperature increases.
1.
Entropy “Rules”
(see p 729 for examples)

The size and direction of heat (enthalpy)
changes and entropy changes together
determine whether a reaction is
spontaneous.
Reaction Spontaneity
How ΔH and ΔS Affect Reaction Spontaneity
ΔH
Decreases
(exothermic)
Increases
(endothermic)
Decreases
(exothermic)
Increases
(endothermic)
ΔS
Increases
Spontaneous?
(more
disorder in products than
in reactants)
Yes
Increases
ΔS > ΔH
Decreases (less
disorder in products than
in reactants)
ΔH > ΔS
Decreases
No
Units for S: J/K
 Usually given as J/K x mol because we are
interested in a specific substance.
 S° signifies entropy at standard
conditions (101.3 kPa and 25°C).
 Theoretical entropy of a perfect crystal at
0 K is zero.

Some background on the standard
conditions for entropy…

Standard Entropy change (ΔS°) can be
calculated using:

ΔS°(reaction) = ΔS°(products)- ΔS°
Entropy Calculations
(reactants)
Practice Problem:





Calculate the standard entropy change
(ΔS°) that occurs when 1 mol H2O(g) at
25°C and 101.3 kPa condenses to 1
mol H2O(l) at the same temperature.
H2O(g) S° = 188.7 J/Kxmol
H2O(l) S° = 69.94 J/Kxmol
ΔS°=69.94 – 188.7 = -118.8 J/Kxmol
The negative sign indicates that entropy
decreases.


Josiah Gibbs formulated the Gibbs free
Energy change (ΔG) equation.
It is the maximum amount of energy that can
be coupled to another process to do useful
work. The change in Gibbs free energy is
related to the change in entropy (ΔS) and the
change in enthalpy (ΔH) of the system by the
following equation:
ΔG = ΔH – TΔS
(T=temperature in Kelvin)
Free Energy Calculations





If –ΔG, reaction is spontaneous in forward
direction
If +ΔG, reaction is non-spontaneous in
forward direction but spontaneous in reverse
direction. Work must be supplied from
surroundings to make it occur.
If ΔG=0, reaction is at equilibrium
All spontaneous processes release free
energy.
In a spontaneous reaction ΔG is negative
because the system loses free energy.
ΔG
Qualitative Prediction of
Spontaneity
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) + 1235 kJ
• What is the sign of H?
• What is the sign of S?
• Plug signs into: ΔG = ΔH – TΔS
• Prediction???
• spontaneous
Free Energy and Keq

ΔG = ΔG° + RT ln Q(where R=8.314 J/mol
K)
At equilibrium ΔG=0 and Q=K, therefore:
 ΔG°=-RT ln K and K= e-ΔG°/RT

(find these formulas on your AP cheat sheet)

Turn to page 740 and try practice exercise
19.12

G negative  reaction proceeds right to
equilibrium
 G positive  reaction proceeds
left to equilibrium
 G = 0  at equilibrium
G is the energy change by a system going
from initial conditions to equilibrium

Infinite number of combinations of
variables
◦ conc, T, P, etc.

Reference values based on standard
conditions
◦ gases at 1 atm
◦ solids and liquids – most stable form at 1
atm and 298 K
◦ solutions at 1 M
But…
Tabulated
Horxn
Hof
Horxn =
npHof (products) - nrHof (reactants)
Standard Enthalpy
S of a pure crystal at 0 K = 0
Third Law of Thermodynamics
Tabulated
Sorxn
So
Sorxn =
npSo (products) - nrSo (reactants)
Standard Entropy
Tabulated
Gorxn
Gof
Gorxn =
npGof (products) - nrGof (reactants)
Standard Free Energy
one method
reaction A
S
U
M
Gorxn A
reaction B
Gorxn B
reaction C
Gorxn C
Standard
Free Energy Go
reaction of interest
rxn
another method
S
U
M
Gorxn
=
Horxn
- T Sorxn
Standard Free Energy
yet another method
Gorxn
Standard states of all
reactants and products
1 M ; 1 atm
Standard Free Energy
ALL THREE METHODS
Equilibrium
Grxn =
Gorxn + RT lnQ
Free Energy related to
Standard Free Energy