Chapter 14: Thermodynamics: Spontaneous Processes, Entropy, and Free Energy

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CHEM 162: Gilbert Chapter 14 page 1
Chapter 14: Thermodynamics: Spontaneous Processes,
Entropy, and Free Energy

Problems: 14.1-14.62, 14.69-14.73, 14.75-14.83, 14.86-14.87, 14.91-14.92


Why do some reactions occur but others don’t?

We can answer these questions using thermodynamics, the study of energy transformations.

1
st
Law of Thermodynamics: Energy is neither created nor destroyed.
– Essentially the law of conservation of energy
– While energy is converted from one form to another, the energy of the universe is constant.

system: that part of the universe being studied

surroundings: the rest of the universe outside the system


In an exothermic reaction like methane burning, CH
4
(g) + 2 O
2
(g)

CO
2
(g) + 2 H
2
O(g)
the bonds formed in the products are
stronger than the bonds broken in
the reactants.

The difference in energy between
the bonds broken and the bonds
formed is released to the
surroundings.

Heat flows from the system to the
surroundings.



In an endothermic reaction like the following, N
2
(g) + O
2
(g)

2 NO(g)
more energy is required to break the
stronger bonds in the reactants
compared to the bonds formed in the
reactants.

The difference in energy must be
absorbed from the surroundings
for the reaction to occur.

Heat flows from the surroundings
to the system when the reaction
occurs.


CHEM 162: Gilbert Chapter 14 page 2
14.1 SPONTANEOUS PROCESSES AND ENTROPY

spontaneous process: takes place naturally, without continuous outside intervention
– e.g., a drop of food coloring will spread in a glass of water or
once ignited H
2
gas burns in O
2
gas.

nonspontaneous process: only takes place as a result of continuous intervention
– e.g., electricity is required to convert water to H
2
gas and O
2
gas



Keep in mind that spontaneous processes may be fast or slow.


a. An explosion is fast and spontaneous.


b. The process of rusting is slow and spontaneous.




Consider the following spontaneous processes at room temperature:
• A drop of food coloring will spread in a glass of water.
• Methane (CH
4
) burns in O
2
gas.
• Ice melts in your hand.
• Ammonium chloride dissolves in a test tube with water, making the test tube colder.


Example: What do these processes or reactions have in common? Are they all exothermic?
Are they all endothermic?



CHEM 162: Gilbert Chapter 14 page 3
The result of all of these processes/reactions is an increase in entropy.

Entropy (S): a measure of the molecular randomness or disorder of a system
– A thermodynamic function that describes the number of arrangements (positions and/or
energy levels) available to a system at a given instant.

The more ways a particular state can be achieved, the more likely or the greater the probability
of finding the system in that state.

For example, imagine a brand new deck of cards with the jokers removed.
– If the deck is divided and shuffled, the deck goes from being ordered to disordered; the
more it’s shuffled, the more disordered it becomes. But why?
– In 1877 Ludwig Boltzmann, an Austrian physicist, introduced the concept of entropy to
explain why cards become more disordered the more they’re shuffled.

– He determined there were 8.066×10
67
ways a deck of cards can be organized
compared to only one way for it to be perfectly ordered like a new pack of cards.


There are more states for a disordered deck of cards.


When shuffled a deck of cards is more likely to be disordered.

Nature proceeds towards the states that have the highest probabilities of existing.

Thus, the universe tends towards the most probable disordered states.


Ex. Consider the ideal gas in the following
image. Explain what will happen when
the valve between the bulbs is opened.





Now, consider why the gas spreads.
– Consider the many arrangements resulting
in about the same number of atoms in
each of the two bulbs compared to all of
the atoms gathering in one bulb or many
more atoms in one bulb versus the other.

If we consider a particular arrangement
of particles a microstate, then many
more microstates exist where particles are evenly distributed in both bulbs and a much
smaller number of microstates exist where the particles are in only one bulb or mostly
concentrated in one bulb.


Thus, if a gas is placed in one bulb of an empty container, and the connecting valve
is opened, the gas will spontaneously expand to fill the entire vessel.


However, the opposite process (the gaseous particles filling a vessel all moving to
one bulb leaving the other bulb completely or mostly empty), while not completely
impossible, is highly improbable.
CHEM 162: Gilbert Chapter 14 page 4
The expansion of a gas also demonstrates the idea of positional probability, a type of
probability that depends on the number of arrangements in space that yield a particular state.
– The greater the space or volume available to particles, the higher the positional entropy.

For example, positional entropy increases from solid to liquid and from liquid to gas.
– Solids have the smallest volume and relatively few positions available for particles.
– Gases have much greater volume with many positions available for gas particles.





Positional Entropy and Solutions
– The formation of a solution—when a solid dissolves (e.g., NaCl dissolving in water) or two
liquids mix—increases the positional entropy of a system.


The overall volume available to each component increases.


Thus, the number of positions available to the particles in the system increases.





CHEM 162: Gilbert Chapter 14 page 5
Entropy and Energy States
– Positional entropy results from the different energy levels available to a system.
– To understand entropy we must recognize how atoms and molecules behave at the
molecular level.

For example, the kinetic energy of a molecule can be in any of the following forms:
– translational: motion through space
– rotational: motion about its center of mass
– vibrational: stretching, compression, bending and twisting of chemical bonds



The more complex the molecule, the more energy levels available to it.

The higher its entropy.
– Compare the limited vibrational energy available for the NO molecule compared to the
various forms of vibrational energy available for the NO
2
molecule below.




Consider an O
2
molecule and the different kinds of motion available to it:



CHEM 162: Gilbert Chapter 14 page 6
Quantum mechanics teaches that energy at the atomic level is not continuous but discrete.
– Only certain energy levels are available for electrons within an atom.
– The distribution of electrons in those energy levels is given by an electron configuration.

In the same way, the motion of atoms and molecules or their energy is also quantized.
– Translational energy levels are very close together.

The quantized difference between energy levels is so small that for temperatures above
absolute zero, especially for room temperature and above, the energy levels for
translational energy form a continuum.

Very little energy is required for a gaseous atom or molecule to move through
space (i.e., experience translational motion).

Consider where vibrational and rotational energy levels fall on the Electromagnetic Spectrum:



The rotational and vibrational energy levels for atoms and molecules are also quantized.
– An O
2
molecule absorbing energy in the microwave region

rotational energy
– An O
2
molecule absorbing more energy in the infrared region

vibrational energy
– The difference between rotational and vibrational energy levels is also small but
greater than the difference between different translational energy levels.

At low temperatures, O
2
molecules may not have enough energy to rotate and vibrate.

At higher temperature like room temperature O
2
molecules are rotating, vibrating, and
moving in space.


Note that at a given instant, an O
2
molecule occupies a specific energy state or energy level.
– At lower temperatures, only certain energy states are available or accessible.
– As temperature increases, more energy states become accessible.
– The O
2
molecules moves in space and also rotates and vibrates.


It can even gain enough energy for its electrons to be excited to higher energy levels.


The

molecule reaches its first electronic excited state.

At very high temperatures, the molecule vibrates enough to dissociate into two O atoms.
– This is the dissociation energy for the O
2
molecule.
CHEM 162: Gilbert Chapter 14 page 7
We can represent these different energy states or levels in the following way:



Note that rotational energy levels (orange) are much closer together than the vibrational
energy levels (green).
– Each energy level corresponds to an O
2
molecule at a given temperature and state.
– 1. An O
2
molecule may be in the electronic ground state (all electrons in the lowest
energy levels) with only certain vibrational modes available (v
1
).
– 2. An O
2
molecule may be in an electronic excited state (with one or more
electrons in a higher energy level even if the lowest levels are not full) with more
vibrational modes available (v
2
).


Example: Consider the differences between an O
2
molecule and an O
3
molecule.

a. Explain the difference between v
3
for the Electronic ground state O
2
molecule and v
3
for
the Electronic excited state O
2
molecule.








b. Would the O
3
molecule have more or fewer vibrational energy levels? Explain why.





CHEM 162: Gilbert Chapter 14 page 8
Now imagine a system consisting of a mole of O
2
molecules at the same temperature.

Not only do the molecules have these available energy levels, but now they’re also colliding
with one another, so all of these energy levels are superimposed on one another.

More energy levels are available for all of the O
2
molecules in the system.

More microstates are available for the system.

While the average speed for a molecule at a given temperature is constant, many gas
molecules can move at a range of speeds characterized by Maxwell-Boltzmann distribution
curves.

Consider H
2
at two different temperatures.
– Consider the distribution of
molecular speeds for equal
numbers of H
2
gas molecules at
(a) 300K and (b) 373K at the right.
– Instead of using bars, a curve
shows the distribution of speeds.
– These curves are generally called
Maxwell-Boltzmann distribution
curves, as Maxwell was the first to
deduce the pattern theoretically
and Ludwig Boltzmann was the first to effectively substantiate them.


Next, consider O
2
at three different temperatures.
– Consider the Maxwell-Boltzmann distribution for O
2

at 300K, 1000K, and 2000K.
– Again these curves actually correspond to the
distribution of speeds that would be given by the sets
of bars indicating the number of molecules with each
speed at each temperature.
– Consider many O
2
molecules moving and colliding
at a given temperature.
– They experience translational motion while also
rotating and vibrating.
– Because kinetic energy is KE= ½ mv
2
, a particle’s
velocity (speed and directionality) is directly related
to its kinetic energy.

The different energy states (the combination of the

translational, rotational, and vibrational energies) available to an O
2
molecule also have a
Maxwell-Boltzmann distribution.


The higher the system’s temperature, the more energy states available to the system.


More energy states available

more microstates available to a system.


The higher the entropy of the system
CHEM 162: Gilbert Chapter 14 page 9
Ex. 1: Explain which one of the following has the greater entropy: 1 mole of H
2
O at room
temperature or 1 mole of O
2
gas at room temperature.







14.2 THERMODYNAMIC ENTROPY

Isothermal and Nonisothermial Processes

Isothermal processes occur at a constant temperature (e.g. melting and boiling).

Entropy is a state function (or state property), so it depends only on the present state.

The change in state functions depends only on the initial state and the final state, not
the pathway from the initial to the final state.
– Thus, unlike reaction rates that depend on the reaction pathway (i.e., whether or not
a catalyst is used), state functions like entropy are the same regardless of pathway.

The entropy change (ΔS) can be determined using only initial and final states:

ΔS = S
final
− S
initial



Example: Check all of the following for which the entropy change (ΔS) is positive:
 a. Dry ice, CO
2
(s), undergoes sublimation at room temperature.
 b. Sugar dissolves in water.
 c. Water freezes at 0°C.
 d. A supersaturated solution of sodium acetate recrystallizes.


Entropy Changes in Chemical Reactions with Gases

The change in positional entropy is dominated by the relative number of gaseous molecules in
a chemical reaction.

If a reaction increases the number of gaseous molecules, ΔS is positive.

If a reaction decreases the number of gaseous molecules, ΔS is negative.


Example: Check all of the following for which the entropy change (ΔS) is positive:
 a. N
2
(g) + 3 H
2
(g)

2 NH
3
(g)
 b. C
3
H
8
(g) + 5 O
2
(g)

3 CO
2
(g) + 4 H
2
O(g)
 c. (NH
4
)
2
CO
3
(s)

2 NH
3
(g) + CO
2
(g) + H
2
O(g)
 d. N
2
O
4
(g)

2 NO
2
(g)
CHEM 162: Gilbert Chapter 14 page 10
Spontaneous Combustion: http://www.youtube.com/watch?v=dvo6Q5nbI9Q


Early studies in thermodynamics led scientists to propose that a process would be
spontaneous if it were exothermic.

A system tends to undergo changes that lower its potential energy (i.e., make it more stable).

Exothermicity is a driving force for spontaneous reactions and processes.


Example: Does ice melt at room temperature? Is this process endothermic or exothermic?





– Thus, some endothermic processes also occur spontaneously.


ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

2
nd
Law of Thermodynamics: In any spontaneous process, there is always an increase
in the entropy of the universe.
– Essentially, the entropy of the universe is increasing.

Increasing the entropy of the universe (S
univ
) is the driving force for a spontaneous process.

ΔS
univ
= ΔS
sys
+ ΔS
surr


where
Δ
S
sys
and
Δ
S
surr
represent the changes in the entropy of the system and
surroundings, respectively.

Thus, if
ΔS
universe
is positive for a process or reaction, it is spontaneous.


14.4 CALCULATING ENTROPY CHANGES

The Effect of Temperature on Spontaneity

Consider the vaporization of 1 mole of water to steam, H
2
O(l)

H
2
O(g)

a. If 1 mole of water is the system, is ΔS
sys
for the vaporization of water positive or negative?
Explain why.






CHEM 162: Gilbert Chapter 14 page 11
Consider that the entropy changes for the surroundings (ΔS
surr
) are determined primarily
by heat flow.

We can determine the heat flow and hence, the ΔS
surr
, if we know whether the reaction is
endothermic or exothermic.

b. Is the process of water vaporizing endothermic or exothermic? Explain how heat flows
to/from the system and the surroundings.





c. Does the heat flow described in part b increase or decrease the entropy of the particles in
the surroundings?






d. Given the ΔS
sys
from part a and the ΔS
surr
from part b, is the vaporization of water
spontaneous or not—i.e., is ΔS
univ
positive or negative? From practical experience, does
water always vaporize?






From practical experience, we know water only vaporizes spontaneously for temperatures
above water’s boiling point (100°C) at normal atmospheric pressure (about 1 atm).

Temperature is another key factor in determining the significance of ΔS
surr
and the
overall sign for ΔS
univ
—i.e., whether or not a process or reaction is spontaneous.



14.5 FREE ENERGY (G) AND FREE-ENERGY CHANGE (ΔG)

Thus, two driving forces causing chemical reactions to occur are:

1. The formation of more stable (low-energy) products from less stable (high-energy)
reactants in exothermic reactions (ΔH<0).

2. The formation of products that have more entropy than the reactants (ΔS>0).



CHEM 162: Gilbert Chapter 14 page 12
Determining the Entropy Change in the Universe
– The American mathematical physicist J. Willard Gibbs (1839-1903) is credited with
establishing the mathematical foundation of modern thermodynamics.
– He developed a way to use ΔH
sys
and ΔS
sys
to predict whether or not a reaction would
be spontaneous at constant temperature and pressure.


Gibbs defined a new state function, free energy (G), which is the energy available to
do useful work—i.e., that portion of energy that can be dispersed into the universe.


We can express ΔS
surr
in terms of the enthalpy change for the system, ΔH, as follows:

ΔS
surr
= −
T

with ΔH in joules (J), and absolute T in Kelvins

Note, because ΔH is the enthalpy change for the system, the sign convention is critical.


In exothermic reactions, the system releases heat to the surroundings.
– Because ΔH is negative

ΔS
surr
is positive.


In endothermic reactions, the system absorbs heat from the surroundings.
– Because ΔH is positive

ΔS
surr
is negative.


Ex. 1 a. Calculate ΔS
surr
and ΔS
univ
for the vaporization of 1.00 mole of water at 125°C.
(ΔH=40.7 kJ/mol, and ΔS
sys
=119 J/K for 1 mole of water vaporizing.)






b. Calculate ΔS
surr
and ΔS
univ
for the vaporization of 1.00 mole of water at 25°C.







Thus, endothermic reactions will only be spontaneous at
high temperatures
.

Exothermicity
as a driving force is only relevant at low temperatures.

Thus, if ΔS
surr
= −
T
H Δ
, then ΔS
univ
is ΔS
univ
= ΔS
sys
+ ΔS
surr
= ΔS
sys

T
H
sys
Δ


Then multiplying both sides by T yields: T ΔS
univ
= T ΔS
sys
− ΔH
sys
= − (ΔH
sys
− T ΔS
sys
)

Then multiplying both sides by -1 yields: − T ΔS
univ
= ΔH
sys
− T ΔS
sys


CHEM 162: Gilbert Chapter 14 page 13
This leads to
Gibb’s free energy (G)
,
G = H

T S


and
Gibb’s free energy change (
Δ
G)
: ΔG = − T ΔS
univ


Gibb’s equation: ΔG = ΔH

T ΔS

where the “sys” subscript can be dropped since each term is based on the system.


Because T is always positive and the sign of ΔS
univ
indicates if a process/reaction is
spontaneous, for any process/reaction occurring at a constant temperature,
the
sign of
Δ
G indicates if the process/reaction is spontaneous
.
– If ΔS
univ
>0, the process is spontaneous

ΔG = −T(+) = −ve (a negative value).
– If ΔS
univ
<0, the process is nonspontaneous

ΔG = −T(−) = +ve (a positive value).

If ΔG<0 (negative), the
reaction/process
is spontaneous.
If ΔG>0 (positive), the
reaction/process
is nonspontaneous (or the reverse
reaction/process is spontaneous).
If ΔG=0, t
he reaction/process is at equilibrium
(
neither the forward nor the
reverse process is favored
).

Combining this with what we determined previously yields the following temperature-
dependence for the Gibb’s equation and spontaneous reactions:

Δ
S
Δ
H < 0
Δ
S > 0

Spontaneous at
all

temperatures

Δ
H > 0
Δ
S > 0


Spontaneous at
high
temperatures




Δ
H

Δ
H < 0
Δ
S < 0


Spontaneous at
low
temperatures


Δ
H > 0
Δ
S < 0

Nonspontaneous




Example: Use Gibb’s equation to calculate the entropy change (ΔS) for one mole of water
vaporizing. Use water’s boiling point and its molar heat of vaporization, ΔH
vapor
(40.7 kJ/mol).





CHEM 162: Gilbert Chapter 14 page 14
14.3 ABSOLUTE ENTROPY, THE 3
rd
LAW OF THERMODYNAMICS, AND STRUCTURE

In Chapter 5 on Thermochemistry, because
enthalpy (H)
is also a
state function
, the
enthalpy

change
(
Δ
H)
was determined using the
initial and final states
for a reaction.
– The
enthalpy

change

H
) for a reaction was calculated using the
enthalpy

change

H
)
for the formation of reactants and products from their elements in their standard states.
– However, since enthalpy is “heat flow”, there is no inherent heat flow of a substance, only
the heat flow associated with forming the substance from its component elements.


The absolute enthalpy of a substance cannot be determined.


The 3
rd
Law of Thermodynamics:
At absolute zero
(T= 0 K), the
entropy is zero
(S=0) for a
perfect crystal
(in which all particles are perfectly aligned, with no gaps, in a crystal structure).

Consider the images below showing a perfect crystal with S=0 versus imperfect crystals
with higher entropy (S>0).



Example: Explain why a crystal would become more disordered as temperature increases.






By setting a value of zero for entropy at absolute zero to be the reference point, scientists can
estimate the entropy value for pure substances at 25°C.


The
absolute entropy (S
°
) of a compound or element can be determined
.
– If we consider that
entropy
is a thermodynamic
function
that describes the
number of


microstates
(positions and energy levels)
available to a system
in a given state
.


The
more microstates
available to an atom or molecule, the
higher its entropy
.

The
more microstates

available to a system
with those atoms or molecules


The
higher the entropy of the system


CHEM 162: Gilbert Chapter 14 page 15
Consider again that the
kinetic energy
of a molecule can be in any of the following forms:

translational
(motion through space),
rotational
(motion about its center of mass), and

vibrational
(stretching, compression, bending and twisting of chemical bonds)



The
more complex the molecule
, the more energy levels available to it.

The
higher its entropy
.
– The additional vibrational energy levels in NO
2
molecules increase its entropy .


S°=211 J/mol∙K for NO and S°=240 J/mol∙K for NO
2






Ex. 1 The absolute entropy would be higher for
1 mole of nitrogen
or
1 mole of neon
at
25°C? Explain.





Ex. 2 Use the element symbols and chemical formulas to rank
1 mole
of each of the following
substances in terms of
increasing entropy
at 25°C: C
diamond
, water, ice, oxygen gas,
ammonia, mercury, and helium.




_________ < _________ < _________ < _________ < _________ < _________ < _________
lowest entropy highest entropy
CHEM 162: Gilbert Chapter 14 page 16
14.4 CALCULATING ENTROPY CHANGES

Standard Entropy Values (S°) are given for substances at 25°C and 1 atm
.

superscript °
denotes at
P=1 atm and T=25°C
– Just like the calculation of ΔH
°
from standard enthalpies of formation,
the
standard entropy change

(
Δ
S°)
can be calculated

for a general reaction,


a A + b B

c C + d D
where a,b,c,d = stoichiometric coefficients


ΔS°

=

Σ

n



(products) –
Σ

m


(reactants)
= [c

(C) + d

(D)] – [a

(A) + b

(B)]


C
2
H
5
OH
(l)
O
2
(g)

CO
2
(g)
H
2
O
(g)
H
2
O
(l)

N
2
(g)

NO
(g)

161 J/mol∙K

205 J/mol∙K

214 J/mol∙K 189 J/mol∙K 70 J/mol∙K

192 J/mol∙K

211 J/mol∙K


Ex. 1: For each of the following, calculate the standard entropy change (ΔS°)

and the free
energy change (ΔG°), then indicate if the reaction is spontaneous at 25°C. If the
reaction is not spontaneous at 25°C, calculate at what temperatures the reaction would
be spontaneous. (Assume the given standard entropy values, S°, are valid over the
required temperature ranges for these reactions.)

a. N
2
(g)
+ O
2
(g)


2 NO
(g)
ΔH°= +180.6 kJ











b. 2 H
2
(g)
+ O
2
(g)


2 H
2
O
(l)
ΔH°= −572 kJ











CHEM 162: Gilbert Chapter 14 page 17
c. C
2
H
5
OH
(l)
+ 3 O
2
(g)


2 CO
2
(g)
+ 3 H
2
O
(g)
ΔH°= −1791 kJ


















14.5 FREE ENERGY AND FREE ENERGY CHANGE

The Meaning of Free Energy

Let’s consider that the energy released by a spontaneous chemical reaction at constant
temperature and pressure is the maximum amount of energy available to do useful work.

i.e., this is the
free energy
from the chemical reaction.

Rearranging
ΔG = ΔH

T ΔS
and solving for ΔH yields:
ΔH = ΔG
+
T ΔS


Remember that
ΔH
results from bonds being broken and made in chemical reactions.
– The
ΔG
part can be used for motion, light, heat, etc.—ie., to do useful work.

Why
ΔG
is called free energy and is considered the energy to do useful work.
– However, keep in mind that the work obtained from
ΔG
is never 100% efficient.
– Next, the
TΔS
part is the temperature-dependent change in entropy, that part of the energy
change that simply increases the entropy of the universe

It is not usable!


Calculating Free Energy Changes

Unlike the standard enthalpy change (ΔH°), which is heat flow accompanying a reaction and
can be determined experimentally using a calorimeter, the standard free energy change (ΔG°)
cannot be measured directly.
– However, ΔG° can be determined from other measurements and values.

This is important because knowing ΔG° for different conditions will allow us to compare
the relative tendency for reactions to occur.
CHEM 162: Gilbert Chapter 14 page 18
Note:
°
Δ
f
G
=
0 for elements in their standard states
(their naturally occurring form at 25°C)
just like
°
Δ
f
H
=0 for elements in their standard states
.


Consider again the following definitions for
standard states
:

Definitions of standard state (or reference form)
1. A gasesous substance with P=1 atm
2. An aqueous solution with a concentration of 1M at a pressure of 1 atm
3. Pure liquids and solids
4. The most stable form of elements at 1 atm and 25°C


standard free energy of formation (
°
Δ
f
G
) of a substance
is the free-energy change for the
formation of 1 mole of the substance from its elements in their standard states,
– Thus, the
°
Δ
f
G
of a substance is a measure of its thermodynamic stability with respect to its
constituent elements.
– Substances with negative values for
°
f
ΔG
are stable and do not decompose into their
constituent elements.
– e.g. water’s
°
f
ΔG
=-237.2 kJ/mol while CO
2
’s
°
f
ΔG
=-394.4 kJ/mol


Consider that the
standard free energy change


) is the free-energy change for a
reaction when the reactants in their standard states are converted to the products in their
standard states.

Thus, we can use a table of standard free energy of formation values (
°
Δ
f
G
) for reactants
and products to calculate
standard free energy changes


) for various chemical
reactions, just like we did for Δ
H
°

and Δ
S
°.



Thus, the
standard free energy change

(
Δ
G°)
can be calculated as


ΔG°

=
(sum of the
°
Δ
f
G
for all products) – (sum of the
°
Δ
f
G
for all reactants)

so for a general reaction,
a A + b B

c C + d D
where a,b,c,d =
stoichiometric coefficients


ΔG°

=

Σ

n

°
Δ
f
G

(products) –
Σ

m
ΔG°

(reactants)


= [c
°
Δ
f
G
(C) + d
°
Δ
f
G
(D)] – [a
°
Δ
f
G
(A) + b
°
Δ
f
G
(B)]





CHEM 162: Gilbert Chapter 14 page 19
Ex. 1: CaCl
2
dissolves in water as follows: CaCl
2
(s)


Ca
2+
(aq)
+ 2 Cl

(aq)

a. Calculate ΔG° for this process at 25°C using the following:

CaCl
2
(s)’s
°
f
ΔG
=−748.1 kJ/mol; Ca
+2
(aq)’s
°
f
ΔG
=−553.6 kJ/mol; Cl

(aq)’s
°
f
ΔG
=−131.2 kJ/mol.












b. Does this occur spontaneously? Explain.






Ex. 2: Calculate the free energy change (ΔG°) for the following reactions using the standard
free energy of formation values given below, then indicate if the reaction is
spontaneous.
°
f
ΔG (NO)=+90.3 kJ/mol and
°
f
ΔG (NO
2
)=+33.2 kJ/mol

a.
N
2
(g)
+ O
2
(g)


2 NO
(g)









b.
2 NO
(g)
+ O
2
(g)


2 NO
2
(g)











CHEM 162: Gilbert Chapter 14 page 20
CHAPTER 16: CHEMICAL EQUILIBRIUM

Problems: 16.99-16.118

16.9 EQUILIBRIUM AND THERMODYNAMICS

The standard free energy change (
Δ
G°) only applies to standard state conditions, where
temperature is 25°C and all gas pressures are 1 atm.

Thus,
Δ
G° only allows us to predict if a reaction is spontaneous at these conditions.

However, many reactions do not occur at these conditions

A more general
free energy change at nonstandard conditions (
Δ
G)
allows us to predict
if a reaction/process is spontaneous for all other conditions:

ΔG = ΔG°
+
RT ln

Q

where R=8.3145 J/mol

K, T=absolute temperature in Kelvins, and
Q=reaction quotient indicating the state of a system at a given instant.


Ex. 3 Consider again the vaporization of 1 mole of water:
H
2
O
(l)

H
2
O
(g)

Δ
H°=+40.7

kJ

a. Calculate the standard free energy

G°) for the process given
Δ
S°= 118.9 J/mol

K,
then explain whether or not the vaporization of water spontaneous under standard
state conditions?








Then how do we explain the presence of water vapor in the atmosphere at 25°C?

b. Write the expression for the reaction quotient (Q) for this process then calculate the
free energy for the process when the partial pressure of water vapor in the air is
5.0
×
10
-3
atm at 25°C.








CHEM 162: Gilbert Chapter 14 page 21
c. Calculate the free energy for the process when the partial pressure of water vapor in
the air is 5.0
×
10
-3
atm at 95°C.








d. Compare the free energy values for the vaporization of water at 25°C and 95°C and
explain the difference. What does the difference indicate about the spontaneity of
the process as temperature increases? Is this expected?







Ex. 4: a. Use the standard free energy change (
Δ
G°) you calculated on Ex. 2 part b on p. 19
to calculate
Δ
G for the reaction below at 95°C when the partial pressure of each
reactant is 5.00 atm and the partial pressure of the NO
2
is 0.50 atm.

2 NO
(g)
+ O
2
(g)


2 NO
2
(g)













b. Is the reaction more or less spontaneous under these conditions? Explain if this is
what you would expect based on Le Châtelier’s Principle and the change from
standard state conditions.








CHEM 162: Gilbert Chapter 14 page 22
The Meaning of
Δ
G for a Chemical Reaction

Do the results from Ex. 4 on the previous page indicate that given partial pressures of 5.00 atm
for NO and O
2
and a partial pressure of 0.50 atm for NO
2
at 95°C will insure the reaction will go
to completion? Not necessarily.

Although the reaction is more spontaneous at these conditions, indicating reactants are more
likely to be converted to products, the lowest free energy for the system may not be one where
only products are present.

The system achieves the
lowest possible free energy
by going to
equilibrium
,
not completion.

We can represent this using the analogy of a ball rolling down a hill.



The first example, showing a ball will roll from point A to point B, is analogous to a phase
change. For example, at room temperature, ice will melt completely into liquid water which has
a lower free energy than ice at room temperature.

The second example show that the ball will roll from point A towards point B since point A is
higher than point B. This is similar to reactants being converted to products in a chemical
reaction.
– However, because the lowest point is at C, the ball will stop before it reaches point B,
similar to a chemical reaction where the equilibrium lies to the right, favoring products.


FREE ENERGY AND EQUILIBRIUM


Previously we defined equilibrium to be when the forward and reverse reactions are occurring
at equal rates.

We can now consider equilibrium from a thermodynamic point of view.

The
equilibrium point

occurs at the
lowest value of free energy
available to a
reaction system.
CHEM 162: Gilbert Chapter 14 page 23
Consider the following hypothetical reaction: A
(g)

B
(g)


The figure below shows the stages when 1.00 mole of A is placed in a vessel with P=2.00 atm.
(a) The initial free energy of A (G
A
) is much higher than the initial free energy of B (G
B
).
(b) G
A
decreases and G
B
increases as reactant A is converted to product B.
(c) Given enough time, the system reaches equilibrium where G
A
=G
B
.

The plot of G versus Fraction of A reacted on the left shows G for the system changing as it
approaches equilibrium.



The second set shows the stages when 1.00 mole of B is placed in a vessel with P=2.00 atm.
(d) The initial free energy of A (G
A
) is much lower than the initial free energy of B (G
B
).
(e) G
A
increases and G
B
decreases as reactant B is converted to product A.
(f) Given enough time, the system reaches equilibrium where G
A
=G
B
.

Notice that in both examples where total number of moles in the container is 1.00 and the total
pressure is 2.00 atm, and they both reach the same equilibrium point—regardless of the initial
amounts consisting of only reactants, only products, or a mixture of reactants and products.
CHEM 162: Gilbert Chapter 14 page 24
Starting with
ΔG = ΔG°
+
RT ln

Q
we can solve for
ΔG°
at equilibrium.


(Hint: Consider the value for
ΔG
at equilibrium and Q=K at equilibrium.)






Thus,
at equilibrium
:
ΔG° =



Ex. 1 Consider again the vaporization of 1 mole of water :
H
2
O
(l)

H
2
O
(g)

Δ
H°=+40.7

kJ


a. Solve for the equilibrium constant for this process at 25°C using the
Δ
G° you
calculated on Ex. 3a on p. 20.








b. Note that the vapor pressure for water is about 0.031 atm at 25°C. What is the
meaning of “vapor pressure of water”? Explain how it is related to the K calculated in
part a.





c. Given the formula for K, explain how the equilibrium constant changes as
temperature increases for this process? Is this expected?





d. Explain the relationship between the standard free energy change and the
equilibrium constant K for an endothermic process like the vaporization of water.




CHEM 162: Gilbert Chapter 14 page 25
Ex. 2 Consider the following reaction:
CO(g) + H
2
O(g)
CO
2
(g)
+ H
2
(g)

a. Calculate
Δ

for the reaction at 25
°
C using the standard values below:

Substance CO(g) CO
2
(g) H
2
(g) H
2
O(g)
°
Δ
f
H
(kJ/mol)
-110.5 -393.5

-241.8
Δ

(J/mol⋅K)

197.9 213.6 130.6 188.8














b. Calculate the equilibrium constant for this reaction.







c. Calculate the free energy change for the reaction at 345
°
C where the partial
pressure of the each reactant is 2.50 atm and the partial pressure of each product is
1.50 atm. Is the reaction spontaneous at these conditions?
















CHEM 162: Gilbert Chapter 14 page 26
16.10 Changing K with Changing Temperatures

Combining the following equations:
ΔG°
= −
RT ln

K
and
ΔG° = ΔH°

T ΔS°


yields

RT ln

K = ΔH°

T ΔS°


and dividing both sides by –RT yields
R
S
RT
H
K ln
oo
Δ
+
Δ
=
-


which can be written in the form of a line equation:
R
ΔS

T
1
R
ΔH-
K ln
oo
+=









y
=
m x + b

This plot of
ln K versus

T
1
is a
van’t Hoff plot
, where
slope=
R
ΔH-
o
and
y-intercept=
R
ΔS
o
.


Example: Consider the van’t plot for the reaction,
CO(g) + H
2
O(g)
CO
2
(g)
+ H
2
(g)
,
for temperatures ranging from 478K to 643K.


Calculate
Δ
H° and
Δ

for the reaction using
the data above.
Compare these values
of
Δ
H° and
Δ
S° with
those calculated based
on
Δ
H
f
° and S° values
on the previous page.

















van't Hoff plot for CO reacting with Steam
y = 4810.1x - 4.7262
2.00
2.50
3.00
3.50
4.00
4.50
5.00
5.50
6.00
1.50E-03 1.60E-03 1.70E-03 1.80E-03 1.90E-03 2.00E-03 2.10E-03 2.20E-03
1/ T (1/K)
ln K
CHEM 162: Gilbert Chapter 14 page 27
Next, consider equilibrium constants K
1
and K
2
at temperatures T
1
and T
2
, respectively:

R
ΔS

T
1
R
ΔH-
K ln
1
1
oo
+=








and
R
ΔS

T
1
R
ΔH-
K ln
2
2
oo
+=











Now, solve for
ln K
2
– ln K
1
=


















+
R
ΔS

T
1
R
ΔH-
2
oo

















+
R
ΔS

T
1
R
ΔH-
1
oo



=



















T
1
R
ΔH-
2
o



















T
1
R
ΔH-
1
o

=



















T
1
-
T
1
R
ΔH-
12
o




van’t Hoff Equation:

















=
121
2
T
1
-
T
1

R
ΔH-

K
K
ln
o



Example: Again consider the following reaction,
CO(g) + H
2
O(g)
CO
2
(g)
+ H
2
(g)
,
a. Calculate the equilibrium constant for the reaction at 725K using the equilibrium
constant for the reaction at 25°C you calculated using
Δ
H
f
° values on page 25.













b. How does the equilibrium constant at 725K compare to the equilibrium constant
at 25°C? Is this expected given the nature of the reaction? Explain.