# Chapter13

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```					            Chapter 13

Thermodynamics: Spontaneous Processes,
Entropy, and Free Energy
Chapter 13: Entropy and Free Energy
In chapter 13 the question of why some chemical
(or physical) processes are spontaneous is asked.
R→P
The answer is found within the 2nd Law of
thermodynamics. Spontaneous processes occur
due to and increase in the total entropy (S), i.e.
DS > 0 for the system plus the surrounding.
For example, exothermic chemical reactions are
often spontaneous because of increases in
disorder of the surrounding caused by the release
of heat.
DSuniverse = DSsystem + DSsurrounding

Entropy is also a measure
of the unavailability of
the system’s energy to do
work.

Consider that heat and
work are both forms of
energy. Which is the
more ordered form?
Entropy (S) is defined quantitatively by the amount
of heat absorbed (isothermally) by the system
divided by the absolute temperature

DS = Dq/T (Joule/K)
For a glass of ice melting the
heat from the room is
transferred to the ice until the
temperatures are equal.
The dispersal of
energy…warmer to cooler
result always results in an
increase in entropy.
Maxwell’s Demon…a gedanken experiment.

Two containers are filled with the same gas at
equal temperatures. When a faster than average
molecule flies near the door the demon opens it.
Over time the average temperature will increase in
one side and decrease in the other…driving the
system out of equilibrium.
ENTROPY AND WHY ENDOTHERMIC PROCESSES
OCCUR.
Entropy (S°) is a measure of randomness or disorder in a
system or its surrounding.
Spontaneous processes are accompanied by a net
increase in the entropy of the universe. This statement is
know as the 2nd Law of thermodynamics.
DSuniverse = DSsystem + DSsurrounding
So that…
DSuniverse = DSsystem + DSsurrounding > O
Third Law of Thermodynamics
• Third Law of Thermodynamics - the entropy
of a perfect crystal is zero at absolute zero.
 S is explicitly known (=0) at 0 K, S values at other
temps can be calculated.
• Absolute entropy is the entropy change of a
substance taken from S = 0 (at T = 0 K) to
some other temperature.
• Standard molar entropy (So) is the absolute
entropy of 1 mole of a substance in its
standard state.
The 3rd Law of Thermodynamics defines the entropy of a
perfect crystal, at zero Kelvin to be equal to zero.
Graphite has a higher standard molar entropy than diamond.
Why might that be?

S° = 2.4 J/mol K                S° = 5.7 J/mol K
The entropy of a substance like water is linked to the
motion (i.e. the temperature) of the molecules
Select Standard Molar Entropy Values (1 atm, 298 K)
So                                          So
Formula                          Formula         Name
(J/(mol•K)                                  (J/(mol•K)
Br2(g)          245.5          CH4(g)         methane          186.2
Br2(l)          152.2        CH3CH3(g)         Ethane          229.5
Cdiamond(s)        2.4          CH3OH(g)        Methanol         239.7
Cgraphite(s)       5.7          CH3OH(l)                         126.8
CO(g)           197.7       CH3CH2OH(g)        Ethanol         282.6
CO2(g)           213.8       CH3CH2OH(l)                        160.7
H2(g)           130.6      CH3CH2CH3(g)       Propane          269.9
N2(g)           191.5      CH3(CH2)2CH3(g)    n-Butane         310.0
O2(g)           205.0      CH3(CH2)2CH3(l)                     231.0
H2O(g)           188.8          C6H6(g)        Benzene          269.2
H2O(l)          69.9           C6H6(l)                         172.8
NH3(g)           192.3        C12H22O11(s)     Sucrose          360.2
Molecular entropy is also linked to the complexity of
molecular structure, i.e. the internal motions of the
molecules and molar mass.

Standard Molar Entropies (298K, 1 bar)

Compound (g)              S° (J/mol۰K)

He                       126

Ar                      155

CH4                      186

C2H6                      230

C3H8                      270

C4H10                     310
Standard Molar Entropies values increase with mass
(number of electrons) which leads to more accessible
energy levels (microstates)
Problem
Rank the compounds in each of the following
groups in order of increasing standard molar
entropy, S°.
a. CH4(g), C2H6(g), and C2H8(g)

b. CCl4(l), CHCl3(l), and CH2Cl2(l)

c. CO2(s), CO2(g), and CS2(g)
Find the compounds in Appendix 4 and compare the
actual values.
Problem 36
Predict the sign of DS for each of the following
processes.
a. Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq)

b. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

c. N2O5(g) → NO2(g) + NO3(g)
Entropy Calculations
We can calculate the entropy of a chemical reaction
using standard molar entropies and the stoichiometry
of the balanced chemical equation.

DSrxn = SnS°[products] – SmS°[reactants]

Use the standard molar entropies in Appendix 4 to
calculate the DS° value for the following reaction.
H2S(g) + 3/2 O2(g) ↔ H2O(g) + SO2(g)
Free Energy
The magnitudes and signs
of DSsys and DSsurr
determine the magnitude
and sign of DSuni.
If DSuni > 0 then the
reaction is spontaneous
Free Energy (DGsys) is the maximum amount of energy
available to do work.
For
DGsys < 0 (i.e. negative) the reaction is spontaneous
(exergonic).
DGsys > 0 (i.e. positive) the reaction is not spontaneous
(endergonic)
DGsys = 0 ; the system is at Equilibrium
Since DGsys depends on temperature, some reactions will
become spontaneous as the temperature of the system
increases or decreases.
Gibbs Free Energy Function
DGsys = DHsys - TDSsys

Changes in entropy, enthalpy and free energy for ice melting

DG<0
Calculate and compare values for DSsys and Dssurr for the
reaction:

H2(g) + ½ O2(g) → H2O(g)
Free energy (we now see) is the maximum amount
of energy available to do work. It can also be
calculated using Standard Free Energy of Formation
Values (Appendix 4)

DGrxn = SnGf°[products] – SmGf°[reactants]

Use the values in Appendix 4 to calculate the DG°rxn
for the following reaction:
H2S(g) + 3/2 O2(g) ↔ H2O(g) + SO2(g)
Driving Forces for Spontaneous
Chemical Processes
1. The formation of low energy products
(exothermic processes; DH < 0)
2. The formation of products that have greater
entropy than the reactants (DS > 0).
•   The free energy (G) relates enthalpy,
entropy, and temperature for a process.
   G = H - TS or DG = DH - TDS
Effects of DH, DS, and T on DG*

DH   DS          DG               Spontaneity

<0   >0       Always <0       Always Spontaneous

Spontaneous at Low
<0   <0     <0 at low temp
Temperatures
Spontaneous at High
>0   >0     <0 at high temp
Temperatures

>0   <0       Always >0       Never Spontaneous

DG = DH - TDS
Show that hydrogen cyanide (HCN) is a gas at 25°C by
estimating its normal boiling point from the following data:

DH°f, kJ/mol   S°, J/(mol · K)
HCN(l)           108.9          113
HCN(g)           135.4          202
Spontaneous Processes
• In thermodynamics, a spontaneous
process is one that proceeds in a given
direction without outside intervention.
Spontaneous process can take forever!

• A non-spontaneous process only occurs
for as long as energy is continually
Thermodynamics
• The second law of thermodynamics
states that the total entropy of the
universe increases in any spontaneous
process.
• Entropy (S) is a measure of the
distribution of energy in a system at a
specific temperature.
Entropy and Microstates
• Quantum mechanics teaches that energy is
not continuous at the atomic scale; only
certain levels of energy are possible.
• The motion of molecules is quantized, which
means different states are separated by
specific energies.
• An energy state, also called an energy level,
is an allowed value of energy.
• A microstate is a unique distribution of
particles among energy levels.
Motion
• Three types of
motion.
 Translational
 Rotational
 Vibrational
• As the temperature
of a sample
increases, the
amount of motion
increases.
Energy States
Statistical Entropy
• Entropy is related to the number of
microstates by the following equation:
S = k ln(W)
 S is entropy
 W is the number of microstates*
 k is the Boltzmann constant (k = 1.38 x
10-23 J/K)
 *For an ideal gas, W can be counted as
permutations in a range of positions and
momenta.
 W = N!/PNi!
Ludwig Boltzmann (1844-
1906) founded the field of
statistical mechanics and
statistical thermodynamics.
These ideas pre-suppose
the existence of (or the
reality of) atoms…which
physicists of the day, in
particular Ernst Mach.
Boltzmann turned to
philosophy to refute
objections to his’ theory, but
ultimately committed suicide.
Driving the Human Engine
• Exergonic reactions are spontaneous
(DG < 0).
• Endergonic reactions are
nonspontaneous (DG > 0).
• The laws of thermodynamics describe
the chemical reactions that power the
human engine.
Glycolysis
The conversion of glucose to
glucose 6-phosphate is the first
step in the catabolism
(physiological combustion) of
starches.
The energy for this process is
provided by the hydrolysis of
ATP
ChemTour: Entropy

Click to launch animation
PC | Mac

This ChemTour includes an “Entropy Battle” game that
challenges students to maintain order within a system as
the temperature rises and the phase level moves from solid
to gas.
ChemTour: Dissolution of
Ammonium Nitrate

Click to launch animation
PC | Mac
ChemTour: Gibbs Free Energy

Click to launch animation
PC | Mac

Students learn to calculate the maximum potential energy
available to do work in a system. An interactive “Gibbs free
energy calculator” allows students to manipulate variables
entropy, enthalpy, and temperature to explore the effect on
DG of a reaction.
Shown to the left are three
possible configurations (A, B,
and C) for placing 4 atoms in two
boxes. Which of the following
processes is accompanied by the
largest increase in entropy, ΔS?

A) A → B             B) B → C        C) C → A

Entropy of Four Atoms in Two Boxes
Consider the following arguments for each answer
and vote again:

A. The entropy change from state A, which has 1
microstate, to state B, which has 4 microstates, is the
greatest.

B. State C is the most probable equilibrium state, so the
B → C transition has the largest entropy.

C. The entropy of the C → A transition is equal to the
sum of the entropies for the A → B and B → C
transitions.

Entropy of Four Atoms in Two Boxes
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?

A) ΔSsurr > 0         B) ΔSsurr = 0   C) ΔSsurr < 0

Isothermal Expansion of an Ideal Gas
Consider the following arguments for each answer
and vote again:

A. The gas is doing work, thereby increasing the entropy
of the surroundings.

B. For a reversible expansion, entropy is constant, so
ΔSsys = ΔSsurr = ΔSuniv = 0.

C. The expansion is isothermal, ΔSsys > 0, and reversible,
ΔSuniv = 0. Therefore, ΔSsurr < 0.

Isothermal Expansion of an Ideal Gas
An ideal gas is expanded
adiabatically (q = 0) into a
vacuum. Which of the following
statements is true for this process?

A) ΔEsys < 0        B) ΔGsys < 0      C) ΔSsys < 0

Gas Expansion into a Vacuum
Consider the following arguments for each answer
and vote again:

A. During the expansion, the gas performs work, so the
energy decreases.

B. The expansion of a gas into a vacuum is a
spontaneous process, so ΔGsys is negative.

C. Although the gas volume increases, the temperature
decreases dramatically, thereby reducing the entropy.

Gas Expansion into a Vacuum
Consider the following possible gas phase reaction:

Which of the following is probably true for this reaction?

A) ΔH > 0        B) ΔS > 0   C) ΔG > 0

Formation of CH Cl from CH and CCl
Consider the following arguments for each answer
and vote again:

A. It is energetically more favorable to have all of the
same type of bonds in a molecule than it is to mix and
match.

B. There are more microstates for the arrangements of H
and Cl atoms on 2 CH2Cl2 molecules than on a CH4
and a CCl4 molecule.

C. Free energy is required to accommodate formation of
the electric dipole moment on CH2Cl2.

Formation of CH Cl from CH and CCl
What can be said of ΔG° for the
condensation of water vapor,
H2O(g) → H2O( ),
at 25°C if the partial pressure of
H2O(g) is 1.0 atm?

A) ΔG° > 0           B) ΔG° = 0     C) ΔG° < 0

ΔG° for Condensation of Water at 25° C
Consider the following arguments for each answer
and vote again:
A. The vaporization of water is spontaneous when the
partial pressure of H2O(g) is 1.0 atm at 25°C.
Therefore, ΔG° > 0 for the condensation of water
vapor.
B. At 25°C, condensation will occur spontaneously
only when the partial pressure of H2O(g) rises above
the equilibrium partial pressure of 1.0 atm.
C. The equilibrium partial pressure of H2O(g) is less
than 1.0 atm at 25°C, so water vapor will condense
spontaneously.

ΔG° for Condensation of Water at 25° C
To the left is a plot of vapor
pressure versus temperature for the
vaporization of ethanol.

C2H5OH(λ)      →    C2H5OH(g).

At which temperature is ΔG° = 0
for the vaporization of ethanol at
1.0 atm?

A) > 100°C          B) 100°C     C) < 100°C

ΔG° of Vaporization of Ethanol
Consider the following arguments for each answer
and vote again:

A. Ethanol has a higher molecular mass than water and
so requires more heat for vaporization.

B. The temperature at which ΔG° = 0 for vaporization is
100°C for all liquids.

C. For ethanol, ΔG° = 0 when the vapor pressure equals
1.0 atm, which occurs at a temperature lower than
100°C.

ΔG° of Vaporization of Ethanol
Which of the following plots shows the
correct relationship between ΔG° (y-
axis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?

A)                      B)              C)

ΔG° Versus T for the Sublimation of I (s)
Consider the following arguments for each answer
and vote again:

A. Once the temperature becomes high enough that the
equilibrium partial pressure of I2(g) is greater than 1.0
atm, the reaction will be spontaneous.

B. This process becomes less spontaneous at higher
temperatures because more iodine must be vaporized.

C. As the temperature increases and the system
approaches equilibrium, ΔG° will decrease. When the
system moves past equilibrium, ΔG° will begin to
increase.

ΔG° Versus T for the Sublimation of I (s)
The formation of ozone, O3(g),
from molecular oxygen is an
endothermic process, with
ΔH° = 85 J/mole.

3 O2(g)       2 O3(g)

At what temperatures will the
reaction proceed spontaneously if
PO2 = PO3 = 1.0 atm?

A) High temperatures B) Low temperatures C) No temperatures

Spontaneity of Ozone Formation
Consider the following arguments for each answer
and vote again:
A. Because the formation of ozone is an endothermic
reaction, it will be spontaneous only at high
temperatures.
B. The formation of ozone will occur only at low
temperatures, where the O2(g) molecules will begin to
condense and form O3(g) molecules.
C. The reaction is endothermic and ΔS° < 0, so at no
temperature can this reaction be spontaneous at 1.0
atm.

Spontaneity of Ozone Formation

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