Free Energy and Thermodynamics

flinkexistenceMechanics

Oct 27, 2013 (4 years and 2 months ago)

57 views

Free Energy and Thermodynamics

1
st

Law of Thermodynamics


Chapter 6: Energy is conserved.


Energy cannot be created or destroyed, rather it is
transferred from one place to another.


Energy can be transferred in two ways.


__________ or __________


Internal

energy of a system = KE + PE


Enthalpy


the first thermodynamic quantity


H = E + P
D
V

Spontaneity


The first law does not tell us the

extent to which

a
reaction will happen (or not).



Reactions are said to be spontaneous in one direction

(and non
-
spontaneous in the opposite direction).



Ex) Releasing a ball


which direction will it go


up or
down?

Spontaneity


Which reaction is
spontaneous?



H
2
O
(l)



H
2(g)

+ ½ O
2(g)



H
2(g)

+ ½ O
2(g)


H
2
O
(l)


Spontaneity


Which reaction is
spontaneous?



2Fe
(s)

+ 3/2O
2(g)


Fe
2
O
3(s)



Fe
2
O
3(s)


2Fe
(s)

+ 3/2O
2(g)

Spontaneity


Often, it is dependent on
the temperature.



H
2
O
(s)



H
2
O
(l)



Spontaneous above or
below 0
o
C?

Spontaneity


Scientists first thought that the criteria for spontaneity
was based solely on whether a reaction was exothermic
or endothermic.


However, ice melting at room temperature is
endothermic as is the dissolution of some salts like
NH
4
Cl.


Clearly, a second criteria is needed to predict
spontaneity.

Carnot Cycle


Sadi

Carnot theorized about
an ideal steam engine


one
that worked at 100%
efficiency.


No heat energy would be lost


all energy is converted to
work.

Two Types of Processes


Reversible


a system is changed in such a method that
BOTH the system and surroundings can be returned
to their former states by EXACTLY reversing the
change.


Irreversible


is one that cannot be reversed without
altering the system or surroundings permanently.

Reversible Processes


Phase changes at their melting or boiling point
temperature are always reversible.


Equilibrium reactions when they reach a steady state
are reversible.

Irreversible Processes


In (a), we have a gas occupying the right half of the
container.


In (b), the partition is removed and the gas spontaneously
expands to fill the container.


In (c), the system is restored by compressing the gas with a
piston. But, this requires work done by the surroundings
changing it permanently!!!

Entropy


A second quantity in thermodynamics.


A measurement of the randomness of a system.


Also a state function just like internal energy (E) and
enthalpy (H).


Thus,
D
S =
S
final



S
initial


For any reversible process,
D
S =
q
rev

/ T


Units for
D
S = J/K mol


LEP #1

Entropy


When a system undergoes a change, both the system
and

the surroundings are affected.


D
S
universe

=
D
S
system

+
D
S
surroundings


When
D
S
universe

> 0 J/K, process is
spontaneous
.


When
D
S
universe

< 0 J/K, process is
non
-
spontaneous
.


When
D
S
universe

= 0 J/K, process is
reversible
.


2
nd

Law of Thermodynamics = the entropy of the
universe always increase for a spontaneous process.


LEP #2

Molecular Interpretation


Molecules can undergo three basic types of motions.


Translational


Vibrational


Rotational

Molecular Interpretation


As any gas is heated, its average KE increases


KE is
proportional to temperature.


This additional KE can be split up among the three
types of motion.


Ludwig Boltzmann


decided to look at entropy from a
statistical viewpoint.


S = k
ln
(W)


k = Boltzmann’s constant (1.38 x 10
-
23

J/K)


W = number of possible microstates

Molecular Interpretation


Number of microstates depends on the relevant
numbers of particles and the positions they can
occupy.

Molecular Interpretation


The number of microstates is akin to playing cards.

Molecular Interpretation


An increase in the entropy means that the randomness
(or disorder) of the system has increased.


Or


an increase in the number of microstates.


More or less microstates if we have four molecules of
gas rather than two?


More or less microstates if we have two decks of cards
rather than one?

Entropy and Life


Human beings (and all life
forms) are highly ordered.


Does this violate the 2
nd

Law
of Thermodynamics?

Entropy and Life


You can’t break even!


To recharge a battery with 100 kJ of
useful energy will require more
than 100 kJ


because of the Second Law of
Thermo!


Every energy transition results in a
“loss” of energy


Its an “Energy Tax” demanded by
nature!

Predicting
D
S


Solids are rigid and ordered = low entropy


Liquids are confined to a specific volume, but are free to
move = more entropy


Gases are free to move anywhere = high entropy


In general, we can predict an increase in the entropy if:


More molecules or particles are produced.


More gases are produced.


Temperature is increased.


Volume is increased.


LEP #3

3
rd

Law of Thermodynamics


The entropy of a pure
crystalline substance at
absolute zero is zero.


As the substance is
warmed, its entropy
increases.


Large increases are seen
for phase changes.

Entropy Change


Absolute entropies, under
standard conditions, can be
determined for all substances.


Values are found in Appendix C.


S
o
solid

< S
0
liquid

<
S
o
gas


S
o

increases with molar mass


S
o

increases with more atoms in
formula

Entropy Change


The change in entropy for any reaction can be
calculated just like
D
H was in Chapter 5.


D
S
o

=
S

nS
o
(products)


S

nS
o
(reactants)


n = coefficients in chemical reaction


LEP #5

Entropy Change


What is
D
S
o

for:



N
2(g)

+ 3 H
2(g)



2 NH
3(g)



Given S
o

for N
2(g)
=191.5 J/K mol, H
2(g)
=130.6 J/k mol,
and NH
3(g)
=192.5 J/K mol?


This value is for the
system
.


How does the entropy of the
surroundings

change?


D
S
surr
.

=
-
D
H
sys
.

/ T


If
D
H
sys
.

=
-
92.38 kJ, then what is
D
S
surr
.
?


What is
D
S universe?

Gibbs Free Energy


D
S
univ
.

=
D
S
sys
.

+
D
S
surr
.


We have just seen that
D
S
surr
.

=
-
D
H
sys
.

/ T.


So,
D
S
univ
.

=
D
S
sys
.

+
-
D
H
sys
.

/ T.


Multiplying both sides by

T yields:


-
T
D
S
univ
.

=
-
T
D
S
sys
.

+
D
H
sys
.


Josiah Gibbs decided to label
-
T
D
S
univ
.
As
D
G.


D
G =
D
H


T
D
S.


Signs for
D
G and their interpretation.

Free Energy and Reactions

Why is energy “Free”?


The change in free energy (
D
G)
represents the maximum amount of
energy available to do work.


Consider the reaction:


C
(
s, graphite
)

+

2 H
2(
g
)
→ CH
4(
g
)


D
H
°
rxn

=

74.6 kJ = exothermic


D
S
°
rxn

=

80.8 J/K = unfavorable


D
G
°
rxn

=

50.5 kJ = spontaneous


D
G
°

is less than
D
H
°

because some of
the released heat energy is lost to
increase the entropy of the
surroundings

Standard Free Energies


Like
D
H
f
o
, there is also a standard free energy of
formation for substances


D
G
f
o
.


These can then be used to calculate the
D
G
o

for any
reaction using the values in Appendix C.


D
G
o

=
S

n
D
G
f
o
(products)


S

n
D
G
f
o
(reactants).


LEP #6

Free Energy and Temperature


Some reactions are
ALWAYS

spontaneous whereas
some are
ALWAYS

non
-
spontaneous.


Some reactions are
DEPENDENT

on the temperature

Free Energy and Temperature


When a reaction becomes just spontaneous (or non
-
spontaneous), the
D
G = 0.


D
G =
D
H


T
D
S.


0 =
D
H


T
D
S.


D
H = T
D
S.


T =
D
H /
D
S.


Warning


D
H is in kJ and
D
S is in J.


LEP #7

Applying to a Reaction


Once we can predict
D
S based on looking at the
reaction AND knowing our relationship between
D
H
and
D
S, we can also predict the outcome on
D
G.


Ex) 2 SO
2(g)

+ O
2(g)



2 SO
3(g)

;
D
H
o

=
-
196.6 kJ


What would we predict for
D
S?


What effect does this have on
D
G?


LEP #8

Free Energy and Equilibrium


The change in free energy along
the reaction path is given by the
equation:
D
G =
D
G
o

+ RT
lnQ
.


At equilibrium,
D
G = 0 and Q = K.


0 =
D
G
o

+ RT
lnK


D
G
o

=
-
RT
lnK


LEP #9, 10, 11

Free Energy and Equilibrium

Equilibrium and Temperature


From Ch 14, we saw that the equilibrium constant is
temperature dependent. We can show why with:


Combining these two equations


D
G
°

=
D
H
°



T
D
S
°


D
G
°

=

RTln
(
K
)


It can be shown that


This equation is in the form
y

=
m
x

+ b