# Spontaneity, Entropy and Free Energy

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```					 Thermodynamics:
Spontaneity, Entropy
and Free Energy
Thermodynamics
Thermodynamics studies how changes in
energy, entropy and temperature affect the
spontaneity of a process or chemical reaction.
Using thermodynamics we can predict the
direction a reaction will go, and also the driving
force of a reaction or system to go to equilibrium.
Spontaneity
A spontaneous process is one that occurs
without outside intervention. Examples include:
- a ball rolling downhill
- ice melting at temperatures above 0oC
- gases expanding to fill their container
- iron rusts in the presence of air and water
- two gases mixing
Spontaneity
Spontaneous processes can release energy (a
ball rolling downhill), require energy (ice
melting at temperatures above 0oC), or involve
no energy change at all (two gases mixing) .

Spontaneity is independent of the speed or
rate of a reaction. A spontaneous process may
proceed very slowly.
Spontaneity
There are three factors that combine to
predict spontaneity. They are:
1. Energy Change
2. Temperature
3. Entropy Change
Entropy

A measure of
randomness or
disorder
Entropy
Entropy, S, is a measure of randomness or
disorder. The natural tendency of things is to
tend toward greater disorder. This is because
there are many ways (or positions) that lead to
disorder, but very few that lead to an ordered
state.
Entropy
The entropy of a system is defined by the
Boltzmann equation:
S = k ln W

k is the Boltzmann constant, and W is the
number of energetically equivalent ways to
arrange the components of the system.
Entropy
Gases will
spontaneously and
uniformly mix
because the mixed
state has more
possible arrangements
(a larger value of W
and higher entropy)
than the unmixed
state.
Entropy

The driving force for a spontaneous process is an
increase in the entropy of the universe.
ΔSo and       Phase Changes

Gases have more entropy than liquids or solids.
ΔSo and       Mixtures

Mixtures have more entropy than pure substances.
Entropy Values of Common
Substances
The   2nd   Law of Thermodynamics

In any spontaneous process there is always
an increase in the entropy of the universe.
The   2nd   Law of Thermodynamics
Water spontaneously freezes at a
temperature below 0oC. Therefore, the process
increases the entropy of the universe.
The water molecules become much more
ordered as they freeze, and experience a
decrease in entropy. The process also releases
heat, and this heat warms gaseous molecules in
air, and increases the entropy of the
surroundings.
The   2nd   Law of Thermodynamics
Since the process
is spontaneous
below 0oC, ΔSsurr,
which is positive,
must be greater in
magnitude than ΔS
of the water
molecules.
Entropy
Entropy can be viewed as the dispersal or
randomization of energy. The freezing of water (an
exothermic process) releases heat to the
surroundings, and thus increases the entropy of
the surroundings. The process is spontaneous at
or below 0oC because the increase in entropy of
the surroundings is greater than the decrease in
entropy of the water as it freezes.
Δ S and Spontaneity
Spontaneity
Entropy, temperature and heat flow all play a
role in spontaneity. A thermodynamic quantity,
the Gibbs Free Energy (G), combines these factors
to predict the spontaneity of a process.

ΔG = ΔH - TΔS
Spontaneity
ΔG = ΔH - TΔS

If a process releases heat (ΔH is negative)
and has an increase in entropy (ΔS is positive), it
will always be spontaneous.
The value of ΔG for spontaneous processes
is negative.
Spontaneity
ΔG = ΔH - TΔS
Spontaneity and ΔG
If ΔG is negative, the process is
spontaneous (and the reverse process is non-
spontaneous).
If ΔG is positive, the process is non-
spontaneous, and the reverse process is
spontaneous.
If ΔG = 0, the system is at equilibrium.
ΔG
Although ΔG can be used to predict in
which direction a reaction will proceed, it does
not predict the rate of the reaction.
For example, the conversion of diamond to
graphite has a ΔGo = -3 kJ, so diamonds should
spontaneously change to graphite at standard
conditions. However, kinetics shows that the
reaction is extremely slow.
The Significance of ΔG
ΔG represents the driving force for the
reaction to proceed to equilibrium.
The Significance of ΔG
If negative, the value of ΔG in KJ is the
maximum possible useful work that can be
obtained from a process or reaction at constant
temperature and pressure.
In practice, some energy is always lost, so the
actual work produced will be less than the
calculated value.
The Significance of ΔG
If positive, the value of ΔG in KJ is the
minimum work that must be done to make the
non-spontaneous process or reaction proceed.

In practice, some additional work is required
to make the non-spontaneous process or
reaction proceed.
Reversibility
A reversible reaction is a reaction that achieves
the theoretical limit with respect to free energy.
That is, there is no loss of energy (usually as
heat) to the surroundings.
All real reactions are irreversible, and do not
achieve he theoretical limit of available free
energy.
Predicting the sign of             ΔSo

For many chemical reactions or physical
changes, it is relatively easy to predict if the
entropy of the system is increasing or
decreasing.
If a substance goes from a more ordered
phase (solid) to a less ordered phase (liquid or
gas), its entropy increases.
Predicting the sign of           ΔSo

For chemical reactions, it is sometimes
possible to compare the randomness of
products versus reactants.
2 KClO3(s)  2 KCl(s) + 3 O2(g)
The production of a gaseous product from a
solid reactant will have a positive value of ΔSo.
Calculating Entropy Changes
Since entropy is a measure of randomness, it
is possible to calculate absolute entropy values.
This is in contrast to enthalpy values, where we
can only calculate changes in enthalpy.
A perfect crystal at absolute zero has an
entropy value (S) =0. All other substances have
positive values of entropy due to some degree of
disorder.
Calculating Entropy Changes
Fortunately, the entropy values of most
common elements and compounds have been
tabulated. Most thermodynamic tables,
including the appendix in the textbook, include
standard entropy values, So.
Entropy Values of Common
Substances
Entropy Values
For comparable
structures, the
entropy increases
with increasing mass
Entropy Values
For molecules with similar masses, the more
complex molecule has greater entropy. The
molecule with more bonds has additional ways
to absorb energy, and thus greater entropy.
Calculating Entropy Changes
For any chemical reaction,

Δ Soreaction= Σmolprod Soproducts- Σmolreact Soreactants

The units of entropy are joules/K-mol.
Calculation of       ∆Go

∆Go, the standard free energy change, can be
calculated in several ways.

∆Go = ∆Ho - T ∆So

It can be calculated directly, using the
standard enthalpy change and entropy change
for the process.
Calculation of          ∆Go

∆Go = ∆Ho - T ∆So

∆Ho is usually calculated by using standard
enthalpies of formation, ∆Hfo.

∆Horxn = Σnprod ∆Hoproducts- Σnreact ∆Horeactants
Calculation of       ∆Go

∆Go = ∆Ho - T ∆So

Once ∆Ho and ∆So have been calculated, the
value of ∆Go can be calculated, using the
temperature in Kelvins.
Calculation of         ∆Go

∆Go can also be calculated by combining
the free energy changes of related reactions.
This is the same method used in Hess’ Law to
calculate enthalpy changes. If the sum of the
reactions gives the reaction of interest, the sum
of the ∆Go values gives ∆Go for the reaction.
Calculation of         ∆Go

Lastly, ∆Go can be calculated using standard
free energies of formation, ∆Gfo. Some tables
of thermodynamic data, including the appendix
of your textbook, include values of ∆Gfo.

∆Gorxn = Σmolprod ∆Gfo prod - Σmolreact ∆Gfo react
Calculation of       ∆Go

When calculating ∆Go from standard free
energies of formation, keep in mind that ∆Gfo
for any element in its standard state is zero.
As with enthalpies of formation, the formation
reaction is the reaction of elements in their
standard states to make compounds (or
allotropes).
Calculation of   ∆Go
Calculation of          ∆Go

Note the values of zero for nitrogen, hydrogen and
graphite.
Spontaneity Problem
   Consider the reaction:
CaCO3(s) ↔CaO(s) + CO2(g) at 25oC.
Calculate ∆Go using the tables in the appendix
of your textbook. Is the process spontaneous at
this temperature? Is it spontaneous at all
temperatures? If not, at what temperature does
it become spontaneous?
Spontaneity Problem
   Consider the reaction:
CaCO3(s) ↔CaO(s) + CO2(g) at 25oC.
Calculate ∆Go using the tables in the appendix
of your textbook. Is the process spontaneous at
this temperature?
Calculation of ∆Grxno will indicate
spontaneity at 25oC. It can be calculated using
∆Gfo values or from ∆Hfo and ∆So values.
Calculation of          ∆Go

CaCO3(s) ↔CaO(s) + CO2(g)
∆Grxno = Σnprod ∆Gfo prod - Σnreact ∆Gfo react
Calculation of           ∆Go

CaCO3(s) ↔CaO(s) + CO2(g)
∆Grxno =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]
–[1 mol(-1128.8 kJ/mol)]
Calculation of           ∆Go

CaCO3(s) ↔CaO(s) + CO2(g)
∆Grxno =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]
–[1 mol(-1128.8 kJ/mol)] = +130.4 kJ
Spontaneity Problem
   Consider the reaction:
CaCO3(s) ↔CaO(s) + CO2(g) at 25oC.
Calculate ∆Go using the tables in the appendix
of your textbook. Is the process spontaneous at
this temperature?
Since ∆Grxno =+130.4 kJ, the reaction is not
spontaneous at 25oC.
Spontaneity Problem
   Consider the reaction:
CaCO3(s) ↔CaO(s) + CO2(g) at 25oC.
Is it spontaneous at all temperatures? If not, at
what temperature does it become spontaneous?
Spontaneity Problem
   Consider the reaction:
CaCO3(s) ↔CaO(s) + CO2(g) at 25oC.
Is it spontaneous at all temperatures? If not, at
what temperature does it become spontaneous?

At 25oC, ∆Grxno is positive, and the reaction
is not spontaneous in the forward direction.
Spontaneity Problem
   Consider the reaction:
CaCO3(s) ↔CaO(s) + CO2(g) at 25oC.
Is it spontaneous at all temperatures? If not, at
what temperature does it become spontaneous?
Inspection of the reaction shows that it
involves an increase in entropy due to
production of a gas from a solid.
Spontaneity Problem
   Consider the reaction:
CaCO3(s) ↔CaO(s) + CO2(g) at 25oC.
Is it spontaneous at all temperatures? If not, at
what temperature does it become spontaneous?
We can calculate the entropy change and the
enthalpy change, and then determine the
temperature at which spontaneity will occur.
CaCO3(s) ↔CaO(s) + CO2(g)
Since ∆Go = ∆Ho - T∆So, and there is an
increase in entropy, the reaction will become
spontaneous at higher temperatures.
To calculate ∆So, use the thermodynamic
tables in the appendix.
CaCO3(s) ↔CaO(s) + CO2(g)

∆Srxno =[1mol(213.6J/K-mol)+1mol(39.7J/K-mol)]
-[1mol(92.9J/K-mol)] = 160.4 J/K
CaCO3(s) ↔CaO(s) + CO2(g)
∆Go = ∆Ho - T∆So

Since we know the value of ∆Go (+130.4 kJ)
and ∆So (160.4 J/K), we can calculate the value
of ∆Ho at 25oC.
130.4 kJ = ∆Ho –(298K) (160.4 J/K)
∆Ho = 130.4 kJ + (298K) (.1604 kJ/K)
∆Ho = + 178.2 kJ
CaCO3(s) ↔CaO(s) + CO2(g)
∆Go = ∆Ho - T∆So

If we assume that the values of ∆Ho and
∆So don’t change much with temperature, we
can estimate the temperature at which the
reaction will become spontaneous.
CaCO3(s) ↔CaO(s) + CO2(g)
∆Go = ∆Ho - T∆So

∆Go is positive at lower temperatures, and
will be negative at higher temperatures. Set ∆Go
equal to zero, and solve for temperature.
0 = ∆Ho - T∆So
T = ∆Ho
∆So
CaCO3(s) ↔CaO(s) + CO2(g)
∆Go = ∆Ho - T∆So
0 = ∆Ho - T∆So
T = ∆Ho
∆So
T = (178.2 kJ)/(160.4 J/K)(10-3kJ/J)
=1111K or 838oC
The reaction will be spontaneous in the forward
direction at temperatures above 838oC.
∆G for Non-Standard Conditions
The thermodynamic tables are for standard
conditions. This includes having all reactants
and products present initially at a temperature of
25oC. All gases are at a pressure of 1 atm, and
all solutions are 1 M.
∆G for Non-Standard Conditions
For non-standard temperature, concentrations
or gas pressures:

∆G = ∆Go + RTlnQ

Where R = 8.314 J/K-mol
T is temperature in Kelvins
Q is the reaction quotient
∆G for Non-Standard Conditions
For non-standard temperature, concentrations
or gas pressures:

∆G = ∆Go + RTlnQ

For Q, gas pressures are in atmospheres, and
concentrations of solutions are in molarity, M.
∆G o   and Equilibrium
A large negative value of ∆Go indicates that
the forward reaction or process is spontaneous.
That is, there is a large driving force for the
forward reaction. This also means that the
equilibrium constant for the reaction will be
large.
∆G o    and Equilibrium
A large positive value of ∆Go indicates that
the reverse reaction or process is spontaneous.
That is, there is a large driving force for the reverse
reaction. This also means that the equilibrium
constant for the reaction will be small.
When a reaction or process is at equilibrium,
∆Go = zero.
∆G o   and Equilibrium
∆G o    and Equilibrium
∆G = ∆Go + RT lnQ

At equilibrium, ∆G is equal to zero, and
Q = K.
0 = ∆Go + RT lnK
∆Go = - RT lnK
∆G o     and Equilibrium
   Calculate, ∆Go and K at 25oC for:
C (s, diamond) ↔ C (s, graphite)
∆G o    and Equilibrium
Calculate, ∆Go and K at 25oC for:
C (s, diamond) ↔ C (s, graphite)
∆Go = (1 mol) ∆Gof (graphite) - (1 mol) ∆Gof (diamond)
= 0 -(1 mol)(2.900 kJ/mol)
= -2.900 kJ
The reaction is spontaneous at 25oC.
∆G o   and Equilibrium
   Calculate, ∆Go and K at 25oC for:
C (s, diamond) ↔ C (s, graphite)
∆Go = -2.900 kJ
∆Go = -2.900 kJ = -RT ln K
-2.900 kJ = -(8.314J/mol-K) (298.2K)ln K
ln K = 1.170
K= e1.170 = 3.22
C (s, diamond) ↔ C (s, graphite)

The negative value of ∆Go and the
equilibrium constant >1 suggest that diamonds
can spontaneously react to form graphite.
Although the reaction is thermodynamically
favored, the rate constant is extremely small due
to a huge activation energy. The disruption of
the bonding in the diamond to form planar sp2
hybridized carbon atoms is kinetically unfavorable.

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