Chapter 17
1
FREE ENERGY AND THERMODYNAMICS
SUROVIEC
SPRING 2014
I. Spontaneous Change and Equilibrium
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 Chemical changes –
reactions
 Physical changes –
expansions, heat,
precipitation
 Spontaneous change –
occurs without outside
intervention


Does not tell us about rate
of change
Does indicate that we are
headed to equilibrium
II. Heat and Spontaneity
3
HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq)
DH = ?

CH4 (g) + O2 (g)  CO2 (g) + H2O (g)
DH = ?

NH4NO3 (s)  NH4+ (aq) + NO3- (aq)
DH = ?


gas expanding : ?

Melting of ice : DH = ?
III. Dispersal of Energy
4
 Entropy – S
 Second law of thermodynamics
 In a spontaneous process the entropy of the universe increases
DS = S final - Sinitial
A. Dispersal of Energy
 The dispersal of energy over as
many different energy states is the
KEY to entropy
 Thermal energy has been
transferred
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7
B. Dispersal of Matter
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 Since it is difficult to calculate
to exact energy levels and
dispersal of total energy among
them lets initially look at the
dispersal of matter among a
system.
B. Dispersal of Matter
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 Looking at the flask
again, the entropy of a
state increases with the
number of energetically
equivalent ways to
arrange the components
of the system.
 2nd law states that for any
spontaneous process, the
entropy of the universe
increases.
C. Energy change and change in state
10
 Entropy of a sample of
matter changes as it
changes state.
IV. Heat transfer
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 There can also be
spontaneous processes
that occur even when
entropy seems to
decrease.
B. Temperature dependence of DSsurr
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 If freezing increases
entropy of surroundings,
why would water not
freeze at all temps?
C. Quantifying entropy changes in the
surrounding
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 When a system
exchanges heat with the
surroundings, it changes
the entropy of the
surroundings
 At constant pressure we
can use qsys to quantify
the change in entropy for
the surroundings.
C. Quantifying entropy changes in the
surrounding
14
 We also know that the
higher the temperature
the lower the magnitude
of DSsurr
Example
15
Consider the combustion reaction below:
C3H8 (g) + 5O2(g)  3CO2(g) + 4H2O(l) ΔHrxn = -2044 kJ
a)Calculate DSsurr at 25oC
b)Determine the sign of DSsys
c)Determine the sign of DSuniv
Example
16
 Given DH°rxn = -125kJ and ΔS°rxn = 253J/K at
25°C, what is the ΔSuniv?
V. Gibbs Free Energy
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 Knowing that:
DSuniv = DSsys -
DH sys
T
 We can use that to rearrange and solve for Gibbs free
energy.
V. Gibbs Free Energy
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 The change in Gibbs Free Energy for a process
occurring at a constant temperature and pressure is
proportional to the negative of ΔSuniv
A. Effect of ΔH, ΔS and T
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 These terms are all related, but how?
VI. So how do we know if a reaction is going to be
spontaneous?
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Example
21
Given:
C2H4(g) + H2(g)  C2H6(g)
Where ΔH = -137.5kJ and ΔS = -120.5 J/K
a)Calculate ΔG at 25°C
b)How does T affect ΔG
Example
22
Given:
2Ca(s) + o2(g)  2CaO(s)
Where ΔH°rxn = -1269.8 kJ and ΔS°rxn = -364.6 J/K
a)Calculate ΔG°rxn at 25°C
b)Is it spontaneous?
VI. Entropy changes in chemical reactions
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 As a reminder: ΔH°rxn is change in enthalpy for a
process in which reactants and products are in their
std. states.
A. S° and the 3rd Law of Thermodynamics
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 Remember that we said that ΔH°f for an element in
its std. state is zero.
 However since S is temperature dependent, we can
better define the zero of entropy.
B. Relative standard entropies
25
 As we have seen, the entropy of a substance depends
on its state.
C. Calculating ΔS°rxn
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 To calculate ΔS°rxn we use the following equation:
DS orxn = ånp S o (products) -ånr S o (reactants)
Example
27
Compute ΔS°rxn for the following reaction:
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
188.8
Example
28
Compute ΔS°rxn for the following reaction:
2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)
J/molK
H2S(g)
205.8
O2(g)
205.2
SO2(g)
248.2
H2O(g)
188.8
VI. Calculating ΔG°rxn
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A. Free energy changes
 We can use ΔH°rxn and ΔS°rxn
Example
30
 Calculate the ΔG°rxn for the following reaction at 25°C
2SO2(g) + O2(g)  2SO3(g)
ΔH°rxn
ΔS°rxn
kJ/mol
J/molK
SO2(g)
-296.8
248.2
O2(g)
0
205.2
SO3(g)
-395.7
256.8
Example
31
 What is the previous reaction was not at 25°C but
instead was at 125°C?
B. ΔG°rxn using free energies
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 Since ΔG°rxn is the change in free energy and fee
energy is a state function, we can solve ΔG°rxn as
we did for ΔH°rxn and ΔS°rxn
Example
33
 Use standard free energies of formation to determine
ΔG°rxn at 25°C
CH4(g) + 8O2(g)  CO2(g) + 2H2O(g) + 4O3(g)
kJ/mol
CH4(g)
-50.5
O2(g)
0
CO2(g)
-394.4
H2O(g)
-228.6
O3(g)
163.2
C. Determining ΔG°rxn for a series of
reactions
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 To determine ΔG°rxn for a series of reactions, we use the
same rules that we did with ΔG°rxn
Example
35
 Determine the ΔG°rxn for the following reaction:
2NO(g) + O2(g)  2NO2(g)
N2(g) + O2(g)  2NO(g)
2N2O(g)  2N2(g) + O2(g)
ΔG°rxn = -71.2 kJ
ΔG°rxn = +175.2 kJ
ΔG°rxn = -207.4 kJ
VII. DGo, K and product favorability
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 K = equilibrium constant
 How does K relate to ΔG°rxn ?
 ΔG°rxn is the increase or decrease in free energy as
the reactants are completely converted to products
 Reactants are not always converted completely to
products.
VII. DGo, K and product favorability
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 When reactants are mixed, they proceed to position
of lowest ree energy and then reach equilibrium.
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Fig. 19-13, p.929
Combining Everything that we have learned
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1.
2.
3.
The free energy at equilibrium is lower than the free
energy of reactants or products
DGorxn gives the position of the equilibrium
DGorxn describes the direction of the reaction
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