Spontaneity entropy___free_energy

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

Spontaneous Processes and Entropy
Spontaneous processes occurs without
outside intervention
Spontaneous processes can be fast or slow

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

Kinetics
depends on the pathway taken
tells us the speed of the process
depends on
activation energy
temperature
concentration
catalysts

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

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

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

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

Ssolid < Sliquid < Sgas
Choose the substance with the higher
positional entropy:
CO2(s) or CO2(g)?
N2(g) at 1 atm and 25o
C or N2(g) at .010 atm and
25o
C?

Predict the sign of the entropy change
solid sugar is added to water
iodine vapor condenses onto a cold surface
forming crystals

Second Law of Thermodynamics
The entropy of the universe is increasing
The universe is made up of the system and
the surroundings
∆Suniverse = ∆Ssystem + ∆Ssurroundings

A process is spontaneous if the ∆Suniverse is
positive
If the ∆Suniverse is zero, there is no tendency for
the reaction to occur

The effect of temperature on spontaneity
H2O(l) --> H2O(g)
water is the system, everything else is the
surroundings
∆Ssystem increases, i.e. ∆Ssystem is positive, because
there are more positions for the water molecules
in the gas state than in the liquid state

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
∆Ssurroundings is negative

Sign of ∆S depends on the heat flow
Exothermic Rxn: ∆Ssurr >0
Endothermic Rxn: ∆Ssurr< 0
Magnitude of ∆S is determined by the temperature
∆Ssurr = - ∆H
Τ

Signs of Entropy Changes
∆Ssys ∆Ssurr ∆Suniv Spontaneous?
+ +
- -
+ -
- +

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):
∆G = ∆H - T∆S

How does the free energy related to
spontaneity?
∆G = ∆H - T∆S
− ∆G = - ∆H + ∆S (remember, - ∆H = ∆Ssurr )
T T T
−∆G = ∆Ssurr + ∆Ssys (remember, ∆Ssurr + ∆Ssys = ∆Suniv)
T
-∆G = ∆Suniv
T

∆Suniv > 0 for a spontaneous reaction
∆G < 0 for a spontaneous reaction
∆G > 0 for a nonspontaneous reaction
Useful to look at ∆G because many chemical
reactions take place under constant pressure
and temperature

H2O(s) --> H2O(l)
∆Ho
= 6.03 x 103
J/mole
∆So
= 22.1 J/K.
mole
Calculate ∆G, ∆Ssurr, and ∆Suniv at -10o
C, 0o
C,
and 10o
C

For the melting of ice
∆Ssys and ∆Ssurr oppose each other
spontaneity will depend on temperature
∆So
is positive because of the increase in
positional entropy when the ice melts
∆Ssurr is negative because the reaction is
endothermic

At what temperatures is Br2(l) --> Br2(g)
spontaneous?
What is the normal boiling point of Br2?
∆Ho
= 31.0 kJ/mol ∆So
= 93.0 J/K.
mol

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)

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

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, ∆S will be positive

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

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

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)

Entropy is a state function
entropy does not depend on the pathway
taken
∆Srxn = Σn∆So
products- Σn∆So
reactant

Calculate ∆So
at 25o
C for
2NiS(s) + 3 O2(g) --> 2 SO2(g) + 2 NiO(s)
Substance So
(J/K.
mol)
SO2 248
NiO 38
O2 205
NiS 53

Calculate ∆So
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

What did you expect the ∆So
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 ∆So

Free Energy and Chemical Reactions
Standard Free Energy Change
∆Go
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

How do we calculate ∆Go
?
∆Go
= ∆Ho
- T∆So
(for a reaction carried out
at constant temperature)
Use Hess’ Law
Use ∆Go
f (standard free energy of
formation)
∆Go
= Σn∆Go
f (products) - Σn∆Go
f (reactants)

Calculate ∆Go
for the reaction at 25o
C
2SO2(g) + O2(g) --> 2 SO3(g)
Substance ∆Ho
f(kJ/mol) ∆So
(J/K.
mol)
SO2(g) -297 248
SO3 -396 257
O2 0 205

Calculate ∆Go
for the reaction Cdia --> Cgr
using the following data:
Cdia + O2 --> CO2(g) ∆Go
= -397 kJ
Cgr + O2 --> CO2(g) ∆Go
= -394 kJ

Calculate ∆Go
for the reaction
2CH3OH + 3 O2--> 2 CO2 + 4 H2O
Substance ∆Go
f(kJ/mol)
CH3OH -163
O2 0
CO2 -394
H2O -229

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

Given that G = ∆Go
+ RTln(P)
where G is the free energy at some P (not necessarily 1 atm)
where ∆Go
is the free energy at 1 atm
Ex: N2(g) + 3 H2(g) --> 2 NH3(g)
(lots of equations…lots of equations…)
∆G = ∆Go
+ 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

Calculate ∆G 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?

Free Energy and Equilibrium
Equilibrium occurs at the lowest value of
free energy available to the reaction
system, i.e., when ∆G = 0
At equilibrium, ∆G = 0, Q = Keq so
∆G = 0 = ∆Go
+ RT ln Keq
∆Go
= - RT ln Keq
Use this equation to find Keq given ∆Go
, or to
find ∆Go
given Keq

Relationship between ∆Go
and Keq
∆Go
Keq
= 0 1
< 0 >1
> 0 < 1

For N2 + 3 H2 --> 2 NH3, ∆Go
= - 33.3 kJ per mole of N2
consumed at 25o
C. 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

4Fe + 3 O2 <====> 2Fe2O3 Calculate the
equilibrium constant using the following
information:
Substance ∆Ho
f (kJ/mol) So
(J/K.
mol)
Fe2O3 -826 90
Fe 0 27
O2 0 205

Keq and temperature
We used Le Chatelier’s Principle to determine
how Keq would change when temperature
changes
Use ∆G to determine the new Keq at a new
temperature
∆Go
= -RT ln K = ∆Ho
- T∆So
ln K = - ∆Ho
. 1 + ∆So
R T R

Spontaneity entropy___free_energy

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