Unit 5 – Part 1: Thermodynamics
 Entropy and the Second Law of
Thermodynamics
 Gibbs Free Energy
 Free Energy and Equilibrium Constants
Entropy and the 2nd Law of
Thermodynamics
 Industrial chemists are responsible for
designing cost-effective manufacturing
processes.
2 NH3 (g) + CO2 (g)
NH2CONH2 (aq) + H2O (l)
urea
 Commercial uses of urea:
 nitrogen fertilizer for plants
 used to manufacture certain plastics and
adhesives
Entropy and the 2nd Law of
Thermodynamics
 Some of the questions a chemist must
consider:
 Does the reaction need to be heated?
 How much product will be present at
equilibrium?
 Does the reaction naturally proceed in this
direction?
Entropy and the 2nd Law of
Thermodynamics
 In order to determine if a reaction proceeds
naturally in the direction written, we need to
know if it is spontaneous.
 Capable of proceeding in the direction
written without needing to be driven by an
outside source of energy
Entropy and the 2nd Law of
Thermodynamics
 Examples of spontaneous processes:
 an egg breaking when dropped
 a ball rolling down a hill
 ice  water at room temperature
 Examples of non-spontaneous processes:
 a ball rolling up a hill
 water  ice at room temperature
Entropy and the 2nd Law of
Thermodynamics
 Examples of spontaneous reactions:
 CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
 2 N (g)  N2 (g)
 Examples of non-spontaneous reactions:
 2 H2O (g)  2 H2 (g) + O2 (g)
 O2 (g)  2 O (g)
Entropy and the 2nd Law of
Thermodynamics
 Many but not all spontaneous processes are
exothermic.
 Enthalpy (DH) alone cannot be used to predict
whether or not a reaction is spontaneous.
 To predict whether a reaction is spontaneous,
we need to use the second law of
thermodynamics and a thermodynamic quantity
called entropy.
Entropy and the 2nd Law of
Thermodynamics
 Entropy (S):
 a thermodynamic quantity related to the
disorder or randomness of a system.
 The more disordered or random the system
is, the larger its entropy is.
 A state function
 not path dependent
Entropy and the 2nd Law of
Thermodynamics
 Every chemical has an entropy associated with it
that depends on its physical state, temperature,
and pressure:
 H2O (l)
69.91 J/mol.K at 25oC/1 atm
 H2O (g)
188.83 J/mol.K at 100oC/1 atm
 Appendix C:
 table of thermodynamic properties including S
Entropy and the 2nd Law of
Thermodynamics
 The entropy change (DS) can be calculated for
any process:
 DS = Sfinal - Sinitial
 DS = Sproducts - Sreactants
 Sign conventions for DS:
 DS = positive  more disordered
 DS = negative  less disordered
Entropy and the 2nd Law of
Thermodynamics
Example: The following process occurs at 0oC
and 1 atm pressure. Does the system become
more ordered or more disordered?
H2O (s)
H2O (l)
DS = 22 J/K
Entropy and the 2nd Law of
Thermodynamics
 The sign of DS can be predicted:
 In general, any change that increases the
overall disorder or randomness will result in
a positive value for DS.
 In general, the overall entropy increases when:
 a molecule (or anything else) is broken into
two or more smaller molecules
 there is an increase in the number of moles
of a gas
 a solid changes to a liquid or gas
 a liquid changes to a gas
Entropy and the 2nd Law of
Thermodynamics
Example: Without doing any calculations, predict
whether DS will be positive or negative.
Breaking an egg
N2 (g) + 3 H2 (g)
2 NH3 (g)
2NH3(g) + CO2(g)
NH2CONH2 (aq) + H2O (l)
Entropy and the 2nd Law of
Thermodynamics
 The entropy change (DS) for a reaction or
process can be calculated using the following
equation:
DSo = S n Soproducts - S m Soreactants
where So = the standard molar entropy
 Note: This is similar to the method used to
calculate DHo for a reaction!
Entropy and the 2nd Law of
Thermodynamics
 Standard molar entropy (So) :
 the entropy value for one mole of a chemical
species in its standard state
 1 atm pressure
 1 M (for those in solution)
 NOTE:
Unlike DHof, the standard molar
entropy of a pure element is NOT zero.
Entropy and the 2nd Law of
Thermodynamics
Example: Predict whether the entropy will
increase or decrease for the following reaction.
Calculate DSo.
C6H12O6 (s)
2 C2H5OH (l) + 2 CO2 (g)
Entropy and the 2nd Law of
Thermodynamics
 How does the change in entropy relate to
the spontaneity of a chemical reaction or
process?
 The Second Law of Thermodynamics can be
used to predict whether a reaction will
occur spontaneously.
 The total entropy of a system and its
surroundings always increases for a
spontaneous process.
Gibbs Free Energy
 Simply looking at the sign of DS for a chemical
reaction or process does not tell you if the
reaction is spontaneous.
 Spontaneous reactions involve an overall
increase in the entropy of the universe.
 Reactions that have a large, negative DH tend to
be spontaneous:
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
DH = -802 kJ
Gibbs Free Energy
 The Gibbs free energy (G) is used to relate
both the enthalpy change and the entropy
change of a reaction to its spontaneity.
G = H - TS
where G = Gibbs free energy (“free energy”)
H = enthalpy
T = temperature (K)
S = entropy
Gibbs Free Energy
 Free energy is a state function.
 The change in free energy (DG) of a system
can be used to determine the spontaneity of a
process or reaction.
 For a process occurring at constant
temperature:
DG = DH - TDS
Gibbs Free Energy
 For a reaction occurring at constant temperature
and pressure, the sign of DG can be used to
determine if a reaction is spontaneous in the
direction written:
 DG = negative
 reaction is spontaneous in the forward
direction
 DG = zero
 reaction is at equilibrium
Gibbs Free Energy
 The sign of DG can be used to determine if a
reaction is spontaneous in the direction written
(cont):
 DG = positive
 reaction is not spontaneous in the direction
written
 work must be supplied by the surroundings
to make the reaction occur in the direction
written
 reaction is spontaneous in the reverse
direction
Gibbs Free Energy
Example: Using the definition of DG, calculate
the DG for the following reaction at 35oC:
2 H+ (aq) + S2- (aq)
 H2S (g)
DH = -61.9 kJ
DS = + 183.6 J/K
Gibbs Free Energy
 The standard free energy of formation (DGof)
has been tabulated for many different
substances. (see Appendix C)
 the change in free energy associated with the
formation of 1 mole of a substance from its
elements under standard conditions
 pure solid
 pure liquid
 gas at 1 atm pressure
 solution with 1 M concentration
Gibbs Free Energy
 There is not a standard temperature for
determining DGof.
 25oC is often used for tables of data
 values can be calculated at other
temperatures as well
 DGof for an element in its standard state is
zero.
Gibbs Free Energy
 The standard free energy change for a
chemical process can be calculated using the
following expression:
DGo = S n DGof (products) - S m DGof (reactants)
 Note:
This is similar to the way we
calculated DHo and DSo
Gibbs Free Energy
Example: Calculate DGo for the following
reaction using the standard free energies of
formation.
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)