

Spontaneous reactions are reactions, that
once started, continues by itself without
further input of energy from the outside.
If a reaction is spontaneous under a given set
of conditions, then the reverse reaction is
considered non-spontaneous.
•
Identify whether the reactions below are
spontaneous or not:
– H2O (s)

H2O (l)
– 2H2 (g) + O2 (g)

2H2O (l)
*requires a spark to begin
– 2H2O (l)  2H2 (g) + O2 (g)
*requires consistent electric current
•
Almost all exothermic reactions are considered to
be spontaneous (at 25 *C and 1 atm).
– ΔH for a spontaneous reactions tends to be negative.
•
•
•
However, some endothermic reactions at specific
temperatures may be considered spontaneous,
for example the melting of ice at 1 atm above 0
*C:
H2O (s)

H2O (l)
ΔH = 6.0 kJ
Endothermic reactions that are non-spontaneous
at room temperature often become spontaneous
when the temperature is raised.
The Randomness Factor: In general, nature tends
to move spontaneously from more ordered to
more random states (order to disorder).


The randomness factor discussed is treated
quantitatively as entropy (S). Basically the
greater the disorder (more random the
distribution of molecules) the greater the
entropy.
-Entropy, like enthalpy is a state property so
that ΔS = Sfinal - Sinitial
◦ unit for S = J/mol K
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•
•
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A liquid has higher entropy than the solid
from which it is formed.
A gas has a higher entropy than the liquid
from which it is formed.
Increasing the temperature of a substance
increases its entropy.
A completely ordered pure crystalline solid
has an entropy of 0 K (3rd law of
thermodynamics).
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•
•
•
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Elements have nonzero standard entropies.
Standard molar entropies of pure substances are
always positive quantities.
Aqueous ions may have negative entropy values.
As a group, gases tend to have higher entropies
than liquids. An increase in the number of moles
of a gas also leads to a higher entropy and vice
versa.
As a molecule becomes more complex, the
higher the entropy (more ways for the atoms to
move about with respect to one another).

Predict whether ΔS is positive or negative for
each of the following processes:
◦ Taking dry ice from a freezer where its temperature
is -80°C and allowing it to warm to room
temperature
◦ Dissolving bromine in hexane
◦ Condensing gaseous bromine to liquid bromine
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



Which of the following
reactions results in the
largest increase in entropy?
(A) CO2(s) CO2(g)
(B) H2(g) + Cl2(g)  2HCl(g)
(C) KNO3(s)  KNO3(l)
(D) C(diamond)  C(graphite)


To calculate the standard entropy change, ΔS,
use the following relation:
ΔS = ∑ ΔS products - ∑ ΔS reactants
The 2nd Law of Thermodynamics: In a
spontaneous process, there is a net increase
in entropy, taking into account both the
system and surroundings, ΔS > 0

Calculate the entropy change at 25ºC in J/K
for
2SO2(g) + O2(g)  2SO3(g)
given the following data:
◦ SO2(g): 248.1 J/mol-K
◦ O2(g): 205.3 J/mol-K
◦ SO3(g): 256.6 J/mol-K

[2(256.6)] - [2(248.1) + 1(205.3)] = -188.3 J/K

Calculate ΔS for
◦ (1) dissolving one mole of calcium hydroxide in
water
◦ (2) the combustion of one mole of methane, CH4,
to form carbon dioxide and liquid water
•
Gibbs free energy is a quantity that basically
allows us to put the two quantities, enthalpy and
entropy, together in such a way as to arrive at a
single function whose sign will determine
whether the reaction is spontaneous. The basic
definition of Gibbs free energy is:
ΔG = ΔH - TΔS
(The Gibbs-Helmholtz Equation)
– G = Gibbs Free Energy (J)
– H = Enthalpy (J/mol)
– T = Temperature (K)
– S = Entropy (J/mol K -- please be aware that the mole
in the unit tends to cancel out from the ΔS equation)

The Gibb’s free energy equation combines all
the information that we have learned thus far.
But what does the Gibb’s free energy value
tell us about a reaction? It tells us the
following:
◦ If ΔG is negative, the reaction is spontaneous in the
forward direction.
◦ If ΔG is equal to zero, the reaction is at equilibrium.
◦ If ΔG is positive, then the reaction is
nonspontaneous in the forward direction, but the
reverse reaction will be spontaneous.
◦ For elements at standard state (pure elements at
25ºC and 1 atm are assigned a value of zero).

For the reaction CaSO4 (s)  Ca2+
calculate:
◦ ΔH°
◦ ΔS°
◦ ΔG° at 25°C
(aq)
+ SO42- (aq)


The standard free energy of formation, ΔGf,
can also be used to solve for the free energy
of a reaction (reference Appendix 1 in text):
ΔGf rxn = ∑ ΔGf products - ∑ ΔGf reactants
If it is a negative quantity then the compound
can be formed spontaneously from the
elements, like in the formation of water:
H2 (g)

+
½ O2 (g)

H2O (l)
ΔGf = -237.2 kJ
Elements in their elemental state will have a
ΔGf = 0


Using ΔG°f values from Appendix 1, calculate
the standard free energy change at 25°C for
the reaction CaSO4 (s)  Ca2+ (aq) + SO42- (aq)
Using ΔG°f values from Appendix 1, calculate
the standard free energy change at 25°C for
the dissolution of 1 mole of calcium chloride.
ΔS+
ΔG =
ΔG = ??
negative
spontaneous
at high
temperatures
spontaneous
at all
temperatures
ΔH–
0
K=1
ΔH+
ΔG = ??
ΔG = positive
spontaneous
at low
temperatures
non-spontaneous
at all
temperatures
ΔS–

At what temperature does ΔG° become zero
for the reaction
Fe2O3 (s) + 3H2 (g)  2Fe (s) + 3H2O (g)