Spontaneity, Entropy & Free Energy  First Law of Thermodynamics 

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Spontaneity, Entropy & Free Energy
First Law of Thermodynamics
Basically the law of conservation of energy
energy can be neither created nor destroyed
i.e., the energy of the universe is constant
• the total energy is constant
• energy can be interchanged
– e.g. potential energy (stored in chemical bonds) can be
converted to thermal energy in a chemical reaction
– CH4 + O2 --> CO2 + H2O + energy
Doesn’t tell us why a reaction proceeds in a
particular direction
Spontaneity, Entropy & Free Energy
Spontaneous Processes and Entropy
Spontaneous processes occurs without
outside intervention
Spontaneous processes can be fast or slow
Spontaneity, Entropy & Free Energy
Thermodynamics
lets us predict whether a process will
occur
tells us the direction a reaction will go
only considers the initial and final states
does not require knowledge of the pathway
taken for a reaction
Spontaneity, Entropy & Free Energy
Kinetics
depends on the pathway taken
tells us the speed of the process
depends on
activation energy
temperature
concentration
catalysts
Spontaneity, Entropy & Free Energy
Spontaneous Processes
a ball rolls downhill, but the ball never
spontaneously rolls uphill
steel rusts, but the rust never spontaneously
forms iron and oxygen
a gas fills its container, but a gas will never
spontaneously collect in one corner of the
container.
Water spontaneously freezes at temperatures
below 0o C
Spontaneity, Entropy & Free Energy
What thermodynamic principle explains why
these processes occur in one direction?
The driving force for a spontaneous reaction
is an increase in the entropy of the universe
Spontaneity, Entropy & Free Energy
Entropy
Symbol: S
A measure of randomness or disorder
The natural progression is from order to disorder
It is natural for disorder to increase
Entropy is a thermodynamic function
Describes the number of arrangements that are
available to a system in a given state
Spontaneity, Entropy & Free Energy
Entropy
The greater the number of possible
arrangements, the greater the entropy of
a system, i.e., there is a large positional
probability.
The positional probability or the entropy
increases as a solid changes from a liquid
or as a liquid changes to a gas
Spontaneity, Entropy & Free Energy
Ssolid < Sliquid < Sgas
Choose the substance with the higher
positional entropy:
CO2(s) or CO2(g)?
N2(g) at 1 atm and 25oC or N2(g) at .010 atm and
25oC?
Spontaneity, Entropy & Free Energy
Predict the sign of the entropy change
solid sugar is added to water
iodine vapor condenses onto a cold surface
forming crystals
Spontaneity, Entropy & Free Energy
Second Law of Thermodynamics
The entropy of the universe is increasing
The universe is made up of the system and
the surroundings
DSuniverse = DSsystem + DSsurroundings
Spontaneity, Entropy & Free Energy
A process is spontaneous if the DSuniverse is
positive
If the DSuniverse is zero, there is no tendency
for the reaction to occur
Spontaneity, Entropy & Free Energy
 The effect of temperature on spontaneity
H2O(l) --> H2O(g)
water is the system, everything else is the
surroundings
DSsystem increases, i.e. DSsystem is positive, because
there are more positions for the water molecules
in the gas state than in the liquid state
Spontaneity, Entropy & Free Energy
What happens to the surrounding?
Heat leaves the surroundings, entering the system
to cause the liquid molecules to vaporize
When heat leaves the surroundings, the motion of
the molecules of the surroundings decrease, which
results in a decrease in the entropy of the
surroundings
DSsurroundings is negative
Spontaneity, Entropy & Free Energy
Sign of DS depends on the heat flow
Exothermic Rxn: DSsurr >0
Endothermic Rxn: DSsurr< 0
 Magnitude of DS is determined by the temperature
DSsurr = - DH
T
Spontaneity, Entropy & Free Energy
Signs of Entropy Changes
DSsys DSsurr DSuniv
Spontaneous?
+
+
+
+
Spontaneity, Entropy & Free Energy
Free Energy
aka Gibbs Free Energy
G
another thermodynamic function
related to spontaneity
G = H - TS
for a process that occurs at constant
temperature (i.e. for the system):
DG = DH - TDS
Spontaneity, Entropy & Free Energy
How does the free energy related to
spontaneity?
DG = DH - TDS
- DG = - DH + DS
(remember, - DH = DSsurr )
T
T
T
-DG = DSsurr + DSsys (remember, DSsurr + DSsys = DSuniv)
T
-DG = DSuniv
T
Spontaneity, Entropy & Free Energy
DSuniv > 0 for a spontaneous reaction
DG < 0 for a spontaneous reaction
DG > 0 for a nonspontaneous reaction
Useful to look at DG because many chemical
reactions take place under constant pressure
and temperature
Spontaneity, Entropy & Free Energy
H2O(s) --> H2O(l)
DHo = 6.03 x 103J/mole
DSo = 22.1 J/K.mole
Calculate DG, DSsurr, and DSuniv at -10oC,
0oC, and 10oC
Spontaneity, Entropy & Free Energy
For the melting of ice
DSsys and D Ssurr oppose each other
spontaneity will depend on temperature
DSo is positive because of the increase in
positional entropy when the ice melts
DSsurr is negative because the reaction is
endothermic
Spontaneity, Entropy & Free Energy
At what temperatures is Br2(l) -->
Br2(g) spontaneous?
What is the normal boiling point of Br2?
DHo= 31.0 kJ/mol DSo = 93.0 J/K.mol
Spontaneity, Entropy & Free Energy
Entropy Changes in Chemical Reactions
Just like physical changes, entropy
changes in the surroundings are
determined by heat flow
Entropy changes in the system are
determined by positional entropy (the
change in the number of possible
arrangements)
Spontaneity, Entropy & Free Energy
N2 (g) + 3 H2(g) --> 2 NH3 (g)
The entropy of the this system decreases
because
four reactant molecules form two product
molecules
there are less independent units in the system
less positional disorder, i.e. fewer possible
configurations
Spontaneity, Entropy & Free Energy
When a reaction involves gaseous
molecules:
the change in positional entropy is
determined by the relative numbers of
molecules of gaseous reactants and
products
I.e., if you have more product molecules
than reactant molecules, DS will be positive
Spontaneity, Entropy & Free Energy
In thermodynamics, the change in a
function is usually what is important
usually we can’t assign an absolute value to
a function like enthalpy or free energy
we can usually determine the change in
enthalpy and free energy
Spontaneity, Entropy & Free Energy
We can assign absolute entropy values,
i.e., we can find S
A perfect crystal at 0 K, while
unattainable, represents a standard
all molecular motion stops
all particles are in their place
the entropy of a perfect crystal at 0 K
is zero = third law of thermodynamics
Spontaneity, Entropy & Free Energy
Increase the temperature of our
perfect crystal
molecular motion increases
disorder increases
entropy varies with temperature
See thermodynamic tables for So values
(at 298 K and 1 atm)
Spontaneity, Entropy & Free Energy
Entropy is a state function
entropy does not depend on the pathway
taken
DSrxn = SnDSoproducts - SnDSoreactant
Spontaneity, Entropy & Free Energy
Calculate DSo at 25oC for
2NiS(s) +
Substance
SO2
NiO
O2
NiS
3 O2(g) --> 2 SO2(g) + 2 NiO(s)
So(J/K.mol)
248
38
205
53
Spontaneity, Entropy & Free Energy
Calculate DSo for
Al2O3(s) + 3 H2(g) --> 2 Al(s) + 3 H2O(g)
Substance
So (J/K.mol)
Al2O3
51
H2
131
Al
28
H2O
189
Spontaneity, Entropy & Free Energy
What did you expect the DSo to be?
Why is it large and positive?
H2O is nonlinear and triatomic
H2O has many rotational and vibrational
motions
H2 is linear and diatomic
H2 has less rotational and vibrational motions
The more complex the molecule, the
higher the DSo
Spontaneity, Entropy & Free Energy
Free Energy and Chemical Reactions
Standard Free Energy Change
DGo
the change in the free energy that occurs if the
reactants in their standard states are changed to
products in their standard states
can’t be measured directly
calculate from other values
allows us to predict the tendency for a reaction to
go
Spontaneity, Entropy & Free Energy
How do we calculate DGo?
DGo = DHo - TDSo (for a reaction carried
out at constant temperature)
Use Hess’ Law
Use DGof (standard free energy of
formation)
DGo = SnDGof (products) - SnDGof (reactants)
Spontaneity, Entropy & Free Energy
Calculate DGo for the reaction at 25oC
2SO2(g) + O2(g) --> 2 SO3(g)
Substance
DHof(kJ/mol)
SO2(g)
-297
SO3
-396
O2
0
DSo (J/K.mol)
248
257
205
Spontaneity, Entropy & Free Energy
Calculate DGo for the reaction Cdia -->
Cgr using the following data:
Cdia + O2 --> CO2(g)
DGo = -397 kJ
Cgr + O2 --> CO2(g)
DGo = -394 kJ
Spontaneity, Entropy & Free Energy
Calculate DGo for the reaction
2CH3OH + 3
Substance
CH3OH
O2
CO2
H2O
O2--> 2 CO2 + 4 H2O
DGof(kJ/mol)
-163
0
-394
-229
Spontaneity, Entropy & Free Energy
The dependence of free energy on pressure
How does pressure affect enthalpy and entropy?
Pressure does not affect enthalpy
Pressure does affect entropy because pressure
depends on the volume
• 1 mole of a gas at 10.0 L has more positions available
than 1 mole of a gas at 1.0 L
• Slarge volume > Ssmall volume
• Slow pressure > Shigh pressure
Spontaneity, Entropy & Free Energy
 Given that G = DGo + RTln(P)
where G is the free energy at some P (not necessarily 1 atm)
where DGo is the free energy at 1 atm
 Ex: N2(g) + 3 H2(g) --> 2 NH3(g)
(lots of equations…lots of equations…)
 DG = DGo + RT ln Q
Q is the reaction quotient (from the law of mass action)
T is the temperature in K
R is the gas constant, 8.3145 J/mol.K
Spontaneity, Entropy & Free Energy
Calculate DG at 25o C for the reaction
CO(g) + 2 H2(g) --> CH3OH where
carbon monoxide is 5.0 atm and
hydrogen gas at 3.0 atm are converted
to liquid methanol.
What does the answer tell us about this
reaction under these conditions?
Spontaneity, Entropy & Free Energy
Free Energy and Equilibrium
Equilibrium occurs at the lowest value of
free energy available to the reaction
system, i.e., when DG = 0
At equilibrium, DG = 0, Q = Keq so
DG = 0 = DGo + RT ln Keq
DGo = - RT ln Keq
Use this equation to find Keq given DGo, or
to find DGo given Keq
Spontaneity, Entropy & Free Energy
Relationship between DGo and Keq
DGo
Keq
= 0
1
< 0
>1
> 0
<1
Spontaneity, Entropy & Free Energy
 For N2 + 3 H2 --> 2 NH3, DGo = - 33.3 kJ per mole of
N2 consumed at 25oC. Predict the direction in which
the reaction will shift to reach equilibrium
a. PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00 x 10-2 atm
b. PNH3 = 1.00 atm, PN2 = 1.00 atm, PH2 = 1.00 atm
Spontaneity, Entropy & Free Energy
4Fe + 3 O2 <====> 2Fe2O3 Calculate the
equilibrium constant using the following
information:
Substance
Fe2O3
Fe
O2
DHof (kJ/mol)
-826
0
0
So(J/K.mol)
90
27
205
Spontaneity, Entropy & Free Energy
Keq and temperature
We used Le Chatelier’s Principle to determine
how Keq would change when temperature
changes
Use DG to determine the new Keq at a new
temperature
DGo = -RT ln K = DHo - TDSo
ln K = - DHo . 1 + DSo
R T
R
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