Chemistry 202
Chapter 9
Thermochemistry the Second and Third Laws
Thermodynamics
Thermodynamics: study of energy
The First Law of thermodynamics:
Law of conservation of energy: energy of the universe is constant.
Internal energy (U): sum of the kinetic and potential energies.
 “delta”: change
U = q + w
Heat
Work
Thermodynamics
U = q + w
Energy flows into system via heat (endothermic): q = +x
Energy flows out of system via heat (exothermic): q = -x
Surroundings
Surroundings
Energy
Energy
System
System
U  0
U  0
Endothermic
Unfavorable
Exothermic
Favorable
Thermodynamics
There are two ways energy is “lost” from a system:
– Converted to heat, q
– Used to do work, w
Energy conservation requires that the energy change in the system is
equal to the heat released plus work done.
U = q + w
U = H + PV
U is a state function:
State function: a property of system that changes independently of its pathways.
(independent of how done)
Driving forces
Energy spread: concentrated energy is dispersed widely.
(Exothermic process)
heat
Matter spread: molecules of a substance are spread out and occupy
a larger volume.
Dissolving is endothermic process,
but because of matter spread, it occurs.
Spontaneous Change
If no external influence drives a change (no work done on the
system) the change is spontaneous.
A hot block of metal
spontaneously cools to
the temperature of its
surroundings.
Spontaneously
Cold
Hot
Not spontaneously
The reverse
process does not
occur.
Spontaneous Change
A spontaneous change occur
naturally without external
influences.
A gas expands
spontaneously into the
lower evacuated flask.
Spontaneously
Not spontaneously
The reverse process
does not occur.
Nonspontaneous changes can be
made to happen in an
"unnatural" direction. processes
require energy input to go.
we have to force it to happen; a
nonspontaneous change is done with
work.
Spontaneous Change
Spontaneity is determined by comparing the chemical potential energy of
the system before the reaction with the free energy of the system after the
reaction.
– If the system after reaction has less potential energy than before the
reaction, the reaction is thermodynamically favorable.
The direction of
spontaneity can be
determined by comparing
the potential energy of the
system at the start and
the end.
Spontaneous Change
Graphite is more stable than diamond, so the conversion of diamond into
graphite is spontaneous. But don’t worry, it’s so slow that your ring won’t
turn into pencil lead in your lifetime (or through many of your generations).
Spontaneous Change
A reversible process will proceed back and forth between the two end
conditions:
1. Any reversible process is at equilibrium.
2. This results in no change in free energy.
If a process is spontaneous in one direction, it must be nonspontaneous
in the opposite direction.
Entropy (S)
A measure of disorder or randomness.
Energy spread  Faster random motions of the molecules in surroundings.
Matter spread  Components of matter are dispersed (occupy a larger volume).
The Second Law of thermodynamics:
The entropy (S) of the universe is always increasing.
We run towards a disorder (heat death of universe).
A Spontaneous process is one that happens in nature on its own.
(because of increasing entropy)
Dissolving
Energy Tax
• To recharge a battery with 100 kJ of useful energy
will require more than 100 kJ because of the
second law of thermodynamics.
• Every energy transition results in a “loss” of
energy.
– An “energy tax” demanded by nature
– Conversion of energy to heat, which is “lost” by
heating up the surroundings
Heat Tax
Fewer steps
generally results in
a lower total heat
tax.
Entropy (S)
When a solid melts, the particles
have more freedom of
movement.
More freedom of
motion increases the randomness
of the system. When systems
become more random, energy is
released. We call this energy,
entropy.
Melting is an endothermic process, yet ice will spontaneously melt above 0 °C.
An endothermic reaction can be spontaneous.
Entropy (S)
Evaporation is an endothermic process as well.
An endothermic reaction can be spontaneous.
Entropy (S)
The natural progression of a system and its surroundings is from order to
disorder, from lower to higher entropy.
To quantify entropy we take a thermally insulated, sealed flask or a calorimeter,
to measure and make precise predictions about disorder.
qrev
ΔS =
T
q: is energy transferred, “rev” means energy must be transferred reversibly (J)
T: is the absolute temperature (K)
S: Change in entropy (J·K-1)
Example 1: A large flask of water is placed on a heater and 100. J of
energy is transferred reversibly to the water at 25°C. What is the change
in entropy of the water?
qrev
ΔS =
T
ΔS =
qrev
100. J
100. J
=
=
= +0.336 J·K-1
T
(273.15 + 25°C)K 298K
As heat flows into the water the entropy of the water increases.
We refer to a "large flask of water" to ensure that the
temperature remains virtually constant as the heat is
transferred: This equation applies only at constant temperature.
Deriving a change in Entropy
A dispersion of molecules is a positional disorder.
These changes can occur by either a changes in temperature, volume or
pressure.
Below are the expressions derived over the next few slides.
ΔS = C ln
T2
T1
C = heat capacity
changes in
temperature
ΔS = nR ln
V2
V1
n = moles
R = universal
changes in volume
ΔS = nR ln
P1
P2
n = moles
R = universal
Note P1 and P2
changes in pressure
Example 2: A sample of nitrogen gas of volume 20.0 L at 5.00 kPa is
heated from 20.°C to 400.°C at constant volume (CV). What is the change
in the entropy of the nitrogen? The molar heat capacity of nitrogen at
constant volume, CV,m, is 20.81 J·K-1·mol-1. Assume ideal behavior.
A rise in temperature means ΔS = C ln
T2
will be positive. (C = CV)
T1
CV = nCV,m so to find CV we’ll first have to find moles of gas using PV = nRT
or n =PV/RT
n=
5.00 kPa × 20.0 L
T2
substitute
this
into
ΔS
=
nC
ln
v,m
T1
8.314 kPa·L·K−1 ×293K
ΔS = nC v,m ln
T2
=
T1
5.00 kPa × 20.0 L
× 20.81 J·K-1·mol-1 ln 673K = +0.710 J·K-1
−1
8.314 kPa·L·K ×293K
293K
Example 3: The temperature of 5.5 g of stainless steel is increased from
20.°C to 100.°C. What is the change in the entropy of the stainless steel?
The specific heat capacity of stainless steel is 0.51 J·(°C)-1·g-1 .
A rise in temperature means ΔS = C ln
T2
will be positive.
T1
C = g Cg
ΔS = g Cg ln
T2
= 5.5 g × 0.51 J·(°C)-1·g-1 ln 373K = +0.68 J·K-1
T1
293K
The change is positive, an increase in entropy.
Example 4: What is the change in entropy of the gas when 1.00 mol N2(g)
expands isothermally (constant T) from 22.0 L to 44.0 L?
ΔS = nR ln
V2
44.0 L
= 1.00 mol × 8.314 J·K−1·mol−1 × ln
= + 5.76 J·K−1
V1
22.0 L
As expected, the entropy increases as the space
available to the gas expands.
Note: the change in entropy of 1 mole of substance has units of J·K−1 .
Molar entropy is the change of entropy per mole has units of J·K−1·mol-1.
Calculating entropy for an irreversible process
Example 5: In an experiment, 1.00 mol Ar(g) was compressed suddenly
(and irreversibly) from 5.00 L to 1.00 L by driving in a piston (like a big
bicycle pump), and in the process, its temperature increased from 20.0°C to
25.2°C. What is the change in entropy of the gas? (Ar C v,m is 12.74 J·K1·mol-1)
This problem is solved in 2 steps:
Step 1: find the effect from change in volume,
Step 2: find the effect from the change in temperature,
Combine both steps.
Calculating entropy for an irreversible process
Example 5: In an experiment, 1.00 mol Ar(g) was compressed suddenly
(and irreversibly) from 5.00 L to 1.00 L by driving in a piston (like a big
bicycle pump), and in the process, its temperature increased from 20.0°C to
25.2°C. What is the change in entropy of the gas? (Ar C v,m is 12.74 J·K1·mol-1)
Step 1: find the effect from change in volume,
ΔS = nR ln
V2
1.00 L
= 1.00 mol × 8.314 J·K−1·mol−1 × ln
= −13.4 J·K−1
V1
5.00 L
Step 2: find the effect from the change in temperature,
ΔS = nC v,m ln
T2
= 1.00 mol × 12.74 J·K-1·mol-1 ln 298.4 K =
T1
293.2 K
+ 0.22 J·K-1
Calculating entropy for an irreversible process
Example 5: In an experiment, 1.00 mol Ar(g) was compressed suddenly
(and irreversibly) from 5.00 L to 1.00 L by driving in a piston (like a big
bicycle pump), and in the process, its temperature increased from 20.0°C to
25.2°C. What is the change in entropy of the gas? (Ar C v,m is 12.74 J·K1·mol-1)
Combine both steps.
ΔS = −13.4 J·K−1 + 0.22 J·K-1 = −13.2 J·K−1
Effect of compression is greater then a small rise in temperature.
Entropy Changes Accompanying Changes in Physical State
ΔHvap
ΔHfus
Phase changes occur at constant temperature.
To calculate entropy ΔS =
qrev
we note:
T
(1) the temperature remains constant as heat is added;
(2) the heat transfer is reversible;
(3) all at constant pressure meaning: qrev = ΔH.
This means: ΔS° =
ΔH°
is either ΔHfus° or ΔHvap° under standard conditions.
T
ΔS°: Standard entropy of vaporization or fusion at standard states (pure, 1 atm.)
Example 6: What is the standard entropy of vaporization of acetone at its
normal boiling point of 56.2°C?
The boiling point corresponds to 329.4 K, and we note from Table 7.3 that
the standard enthalpy of vaporization of acetone at its boiling point is 29.1
kJ·mol-1.
Converting 29.1 kJ·mol-1 to 29,100 J·mol−1 is standard, ΔS is typically
reported in J·K−1 since heat changes in ΔS tend to be small.
ΔHvap°
29,100 J·mol−1
ΔS° =
=
= 88.3 J·K−1·mol−1
T
329.4 K
Third Law of Thermodynamics
A perfectly orderly state of matter with neither positional nor, thermal disorder,
this is a state where entropy is zero, a state of perfect order.
This is the absolute scale of entropy; just as Kelvin is the absolute scale of
temperature. This is the third law of thermodynamics:
The entropies of all perfect crystals approach zero as the
absolute temperature approaches zero.
A "perfect crystal" is where S  0 as T  0K, implies a perfect orderly array,
and so no positional disorder.
We expect the entropy of any substance to be greater than zero above T = 0K.
Entropy (S)
Boltzmann Equation:
Statistical entropy
S = k ln W
S = Entropy J/mol
k = Boltzmann constant = 1.38 × 10−23 J/K (gas constant divided by Avogadro’s number)
W = is the number of positions atoms or molecules can arrange into and
are microstates (unitless).
Random systems require less energy than ordered systems.
Entropy (S)
Every time we change the position of a molecule it represents a different
microstates. Each microstate has the same total energy. This collection is
called an ensemble.
A microstate lasts only for an instant so, when we measure the bulk properties,
we are measuring an average taken over many microstates that’s why the
Boltzmann formula is called statistical entropy.
If there is only one microstate, then all molecules are in exactly the
same energy level, then W = 1. In this case the total system has zero
entropy (In 1 = 0) and zero disorder.
Example: Calculate the entropy of a tiny solid made up of four diatomic
molecules of a compound such as carbon monoxide, CO, at T = 0 when (a) the
four molecules have formed a perfectly ordered crystal in which all molecules
are aligned with their C atoms on the left (top-left image) and (b) the four
molecules lie in random orientations (but parallel).
If all four molecules are
pointing in one direction (as in
the top left image or bottom
right), the entropy of the solid is
zero.
The entropy is greater than zero when the molecules become disordered
and adopt different orientations. S = k ln W, where W = 1,
S = k ln 1 = 0
Example 7: Calculate the entropy of a tiny solid made up of four diatomic
molecules of a compound such as carbon monoxide, CO, at T = 0 when (a) the
four molecules have formed a perfectly ordered crystal in which all molecules
are aligned with their C atoms on the left (top-left image) and (b) the four
molecules lie in random orientations (but parallel).
S = k lnW
(b) Because each molecules has 2 different orientations and there are 4 sets
of molecules
W = 2 × 2 × 2 × 2 = 24 = 16 possible combinations.
S = (1.381 × 10-23 J·K−1 ) × ln 16 = 3.8281 × 10-23 J·K−1
Example 8: Calculate the entropy of a solid in which it is supposed that a
substituted benzene molecule, C6H5F, can take anyone of six orientations
with the same energy. Suppose there is 1.0 mol of molecules in the sample.
Here there are 6 different orientations for 6.022×1023 sets of molecules.
W = 6 6.022×10^23
Note:
ln ax = x ln a
S = k lnW
W = k ln (6 6.022×10^23)
so W = 6.022 × 1023 ln 6
S = (1.381 × 10-23 J·K−1 ) × 6.022 × 1023 ln 6 = 15 J·K−1
Example 9: The entropy of 1.00 mol FCIO3 (s) at T = 0K is 10.1 J·K−1.
Suggest an interpretation.
An FCIO3 (s) molecule is tetrahedral, and so we can expect it to be able to
take up any of 4 orientations in a crystal
W = 4 6.022 × 10^23
S = k lnW
S = (1.381 × 10-23 J·K−1 ) × 6.022 × 1023 ln 4 = 11.5 J·K−1
Equivalencies of Statistical and Thermodynamic Entropies
q
ΔS = rev
T
Bulk samples
S = k ln W
Boltzmann’s formula for
Statistical behavior
These are two different expressions for entropy, but are they the same?
q
For ΔS = rev we’ve seen how this expression changes with volume (ΔS =
T
V
T
P
nR ln 2 ) , Temperature (ΔS = C ln 2 ), and pressure (ΔS = nR ln 1 ).
V1
T1
P2
The question is does Boltzmann’s equation change with either
volume or pressure?
Equivalencies of Statistical and Thermodynamic Entropies
Temperature
Each level is a onedimensional box .
For example at T = 0, if only
the lowest energy level is
occupied than W = 1 and the
entropy is zero. There is no
"disorder," because we know
which state each molecule
occupies.
At higher
temperatures,
more molecules
occupy more
energy levels or
microstates, so
the entropy
increases.
Equivalencies of Statistical and Thermodynamic Entropies
Volume
We know gases expand spontaneous when the volume is allowed to expand.
Going from (a) to (b) is an isothermal expansion
(constant temp).
Expanding the volume allows molecules more
microstate orientations.
Still maintaining our same range of energy, the
levels start to bunch up to accommodate the
added microstates.
Expansion creates more microstates or W ∝ volume.
Equivalencies of Statistical and Thermodynamic Entropies
These are identical.
ΔS =
qrev
T
Bulk samples
S = k ln W
Boltzmann’s Statistical behavior
1 Both are state functions: W depends only on the current state of the system (not the past).
2 Both are extensive properties (depends on the number of atoms i.e.
volume and mass): Doubling the number of molecules increases the
number of microstates from W to W2, or k InW to k InW2, or 2k InW.
Equivalencies of Statistical and Thermodynamic Entropies
These are identical.
ΔS =
qrev
T
Bulk samples
S = k ln W
Boltzmann’s Statistical behavior
3 Both increase in a spontaneous change: the overall disorder of the
system and its surroundings increases, which means that the number of
microstates increases.
4 Both increase with temperature. When the temperature of the system
increases, more microstates become accessible, and so the statistical
entropy increases.
Standard Molar Entropies (Sm°)
ΔSfus° =
ΔHfus°
for either fusion (or vapor).
T
Gases have the most freedom (more W’s). Liquids have a greater
freedom of movement than solids.
More complex
Standard Molar Entropies (Sm°)
heavier
CH3OH
CH3CH2CH3
CH3CH3
127.2
171
127
Diamond’s low entropy is due to it’s rigid bonds: the atoms do not jiggle
around as much as the atoms of lead.
Lead atoms are heavier than carbon atoms, and heavier (more protons,
neutrons and electrons) atoms have more vibrational energy levels available
than do lighter ones.
Standard Molar Entropies (Sm°)
heavier
CH3OH
CH3CH2CH3
CH3CH3
127.2
171
127
Polar molecules have lower entropy values.
Heavier molecules have greater entropy values.
Standard Reaction Entropies: changes to the system
It is possible to predict the sign of the entropy change of the system?
We know a gas has a greater entropy value than either solids or liquids.
A change in the amount of gas molecules normally dominates any other
entropy change in a reaction.
Example 10: Without doing any calculations, estimate the sign of the
entropy change for the reaction N2(g) + 3 H2(g) → 2 NH3(g) and explain
your answer.
Answer: negative, because there is a net decrease in the number of
moles of gas molecules.
Standard Reaction Entropies: changes to the system
To calculate the change in entropy in a reaction, we calculate
the difference between the entropies of the products and
those of the reactants.
ΔS (system) = ΣnSm° (products) - ΣnSm° (reactants)
Σ = sum up
n = stoichiometric coefficients, moles
Sm° = standard molar entropies
Example 11: Calculating the standard reaction entropy Calculate the
standard reaction entropy of
N2 (g) + 3 H2 (g) → 2 NH3 (g) at 25°C.
Sm° N2 (g) = 191.6 J·K−1
Sm° H2 (g) = 130.7 J·K−1
Sm° NH3 (g) = 192.4 J·K−1
We expect this to have a negative entropy, because there is a net decrease
in the number of moles of product.
ΔS (system) = ΣnSm° (products) - ΣnSm° (reactants)
ΔS = {2 moles Sm° (NH3 (g))} - { 1 mole Sm° (N2 (g)) + 3 moles Sm° (H2 (g))}
ΔS = {2 (192.4 J·K−1) } - {(191.6 J·K−1) + 3(130.7 J·K−1)}
ΔS = -198.9 J·K−1
Global Changes in Entropy: Total
Ssystem = Sreaction
ΔS (total) = ΔS (system) + ΔS (surroundings)
A process is spontaneous only if
the total change in entropy, the
sum of the changes in the
system and the surroundings, is
positive.
45
Global Changes in Entropy: Total
Positive Total Entropy = Spontaneous Process
– For reversible process Suniv = 0
– For irreversible (spontaneous) process Suniv > 0
Suniverse = Ssystem + Ssurroundings
If the entropy of the system decreases, then the entropy of the
surroundings must increase by a larger amount.
When Ssystem is negative, Ssurroundings must be positive and big for a
spontaneous process.
ΔS (total) “+” = ΔS (system) + ΔS (surroundings)
small “–”
large “+”
Global Changes in Entropy: Surroundings
At low temperatures, T, there is a greater effect on ΔS.
ΔH
ΔS = T
a sneeze in a quiet library will attract attention
then sneezing in a noisy street.
In (a) a more chaotic system, adding or removing heat has less of an effect than
in (b) a less chaotic system.
Example 12: Calculate the entropy change of the surroundings when 2.00
mol NH3(g) is formed from the elements in their most stable forms at 298K.
ΔHreaction for 1/2 N2 (g) + 3/2 H2 (g) → NH3(g) (25°C) is - 46.11 kJ·mol−1
- 46.11 kJ·mol−1 ×
1 kJ
= − 4.611 × 103 J·mol−1
1000 J
3
ΔHsys
2 mol × − 4.611 × 10 J·mol−1
ΔSsurr = or = + 309 J·K−1
T
298 K
Since we’ve already calculated ΔSsystem = -198.9 J·K−1 (from standard molar
enthalpy’s Ex. 11) for this reaction, we can find the total ΔS,
Therefore, ΔS (total) = ΔS (system) + ΔS (surroundings)
- 198.9 J·K−1 + (- 4.611 × 103 J·mol−1)
ΔS (total) = - 4.810 × 103 J·mol−1 (unfavorable)
Overall Change in Entropy
ΔS (total) is a combination of both the system and surroundings:
ΔS (total) = ΔS (system) + ΔS (surroundings)
ΔS (system) = ΣnSm° (products) - ΣnSm° (reactants)
• If ΔS (total) is positive (an increase), the process is spontaneous.
• If ΔS (total) is negative (a decrease), the reverse process is
spontaneous.
• If ΔS (total) = 0, the process has no tendency to proceed in either
direction (phase change is the most common).
Example 13: Assess whether the combustion of magnesium is spontaneous
at 25°C under standard conditions, given:
2 Mg(s) + O2(g) → 2 MgO (s) , K = 273.15 + (25°C)K = 298 K
ΔS°(system) = -217 J·K−1 , ΔHsys° = -1202 kJ (exothermic),
ΔS° (total) = ΔS° (system) + ΔS°(surroundings)
ΔS° (system) = -217 J·K−1 (unfavorable)
ΔS° (surroundings) = ΔSsurr° = + 4.03 ×103 J·K−1 (favorable)
ΔS° (total) = -217 J·K−1 + -
ΔHsys°
−1,202,000 J
= =
T
298 K
−1,202,000 J
= + 3.82 × 103 J·K−1
298 K
Because ΔS°(total) is positive, the reaction is spontaneous under standard
conditions even though the entropy of the system decreases. The large
amount of heat produced drives the reaction.
Example 14: Is the formation of iodine chloride from its elements in their
most stable forms spontaneous under standard conditions at 25°C? For
the reaction:
Cl2(g) + I2(g) → 2ICl(g)
ΔS°(system) = + 155.9 J·K−1, ΔHsys° = + 35.0 kJ (Endothermic).
ΔS° (total) = ΔS° (system) + ΔS°(surroundings)
ΔS° (system) = + 155.9 J·K−1 (favorable)
ΔS° (surroundings) = ΔSsurr° = - 117.45 J·K−1 (ufavorable)
ΔHsys°
35000 J
==
T
298 K
ΔS° (total) = + 155.9 J·K−1 - 117.45 J·K−1 = + 38.45 J·K−1
Yes, this is spontaneous even though the reaction is endothermic, the
reason: entropy is driving the reaction (dipole-dipole).
Example 15: Is the formation of benzene from its elements in their most
stable forms spontaneous at 25°C? For the reaction
6 C(s) + 3 H2(g) → C6H6 (l), K = 273.15 + (25°C)K = 298 K
ΔS°(system) = -253.18 J·K−1, ΔHsys° = +49.0 kJ (endothermic).
Here both the system and the surround are both unfavorable.
ΔS° (total) = ΔS° (system) + ΔS°(surroundings)
ΔS° (system) = -253.18 J·K−1 (unfavorable)
ΔS° (surroundings) = ΔSsurr° = - 160 J·K−1 (unfavorable)
ΔHsys°
49,000 J
==
T
298 K
ΔS° (total) = -253.18 J·K−1 + (- 160 J·K−1) = -410 J·K−1
Because ΔS°(total) is negative, the reaction is nonspontaneous because
both the system (6 moles → 1 mole, and gas → liquid) and surroundings
(endothermic) are decreasing.
Summery of Three Points
(1) As just seen, endothermic reactions are sometimes
driven by the dominating increase in disorder of the system.
(2) Work is related to heat by ΔU = q + w (state functions) and we have
q
shown wrev > wirrev, it follows that qrev > qirrev , which means ΔS ≥
…
T
In an isolated system q = 0 so ΔS ≥ 0 meaning the entropy cannot
decrease in an isolated system or in other words, the entropy of the
universe is steadily increasing (Clausius inequality).
Summery of Three Points
(3) A reversible and irreversible path between two given states of the system
leave the surroundings in different states.
ΔS (system) is a state function so it only depends on the difference between
the final and initial states.
ΔS (surroundings) is not a state function. That’s because work can be
q
calculated either wrev or wirrev so ΔS ≥
is different for different process.
T
Let’s calculate the total entropy change for the expansion of an ideal gas to
see how this works.
Example 16: Calculate ΔS (system), ΔS (surroundings), then ΔS (total) for
(a) the isothermal, reversible expansion and (b) the isothermal irreversible,
free expansion (no work, w = 0) of 1.00 mol of ideal gas molecules from
8.00L to 20.00L at 292K. Explain any differences between the two paths.
(a) Isothermal reversible expansion changes ΔU = 0
(so q = -w), the surroundings is the negative of the
system and the overall change is zero Ssys = - Ssurr
(because for reversible process Stot = 0).
(b) Irreversible changes, the final state of the
surroundings is different and entropy increases (w = 0
because of free expansion) so U = q.
Part (a) we use ΔU = w + q, ΔU = 0 for isothermal an expansion and q = -w or
Ssys = -Ssurr.
w = -nRT ln
Ssys =
V2
V
(Eq. 4 Ch. 8), So q = +nRT ln 2 (because q = -w)
V1
V1
q
so,
T
ΔSsys = nR ln
V2
20.00 L
, 1.00 mol × 8.314 J·K−1·mol−1 × ln 8.00 L = +7.6 J·K−1
V1
Part (a) a reversible path, ΔStot = ΔSsys + ΔSsurr = 0
ΔSsurr = -Ssys = - nR ln
V2
or −7.6 J·K−1
V1
So ΔStot = +7.6 J·K−1 − 7.6 J·K−1 = 0
Part (b) U = q + w we use w = 0 for an nonexpansion process (Eq. 8 & 9
Ch. 8)
U = q
and U = 0 (because of isothermal irreversible process) so q = 0.
qsurr = 0
Ssurr = −
qsurr
T
So Ssurr = 0
ΔSsys = +7.6 Jk-1 (it is a state function it’s change is the same).
Stot = +7.6 J.K-1
Equilibrium and Entropy
The system remains in its current state until it is disturbed by changes in
temperature, volume, or adding more reactants.
Walking on a sunny day, your body
temperature is about the same as
ambient temperatures, so your body
and the air are at equilibrium.
When you get in, or out of a pool, you
cool off, the water is extracting heat
from your body, you are not at
equilibrium.
Equilibrium and Entropy
The equilibrium state is a dynamic equilibrium, where the forward and
reverse processes are continually at matching rates.
A block of metal at the same temperature as its surroundings, is thermal
equilibrium.
Energy has no net tendency to flow into or out of the block as heat. Energy
continues to flow in both directions, but there is no net flow.
Example 17: Confirm that liquid benzene and benzene vapor are in
equilibrium at the normal boiling point of benzene, 80.1°C, and 1atm
pressure. The enthalpy of vaporization of benzene at its boiling point is
30.8 KJ·mol−1 and its entropy of vaporization is 87.2 J·K−1·mol−1 .
For a reversible path a reversible path, ΔStot = ΔSsys + ΔSsurr = 0
ΔSsys = + 87.2 J·K−1·mol−1
K = 273.15 + (80.1°C)K = 353.3 K
ΔSsurr =
ΔStot = 0
ΔH
30.8 KJ·mol−1
q=−
=−
= -0.0872 kJ·K−1·mol−1
T
353.3 K
or -87.2 J·K−1·mol−1
Gibbs Free Energy
The Gibbs free energy, G, is the
maximum amount of work energy
that can be released to the
surroundings by a system for a
constant temperature and pressure
system.
It tells us how much nonexpansion work (work under free expansion
when w usually = 0) we can get from the system.
Gibbs free energy is often called the chemical potential because it
is analogous to the storing of energy in a mechanical system.
Nonexpansion Work
Nonexpansion work includes electrical work and mechanical work (like
stretching a spring or carrying a weight up a slope ).
Electrical work-the work of pushing electrons through an electrical circuitis the basis of generating chemical electrical power (Ch. 14).
Nonexpansion work also includes the work of muscular activity, the work
of linking amino acids together to form protein molecules, and the work
of sending nerve signals through neurons; key ideas in bioenergetics.
Gibbs Free Energy
ΔStot = ΔSsys + ΔSsurr
ΔStot = ΔSsys - ΔHsurr/T
-TΔStot = -TΔSsys + ΔHsurr to get the familiar form:
ΔG = ΔHsys – TΔSsys
Where G, for Gibbs, is a state function, at constant temperature.
Gibbs Free Energy
• A process will be spontaneous when G is negative.
 G will be negative when
– H is negative and S is positive.
• Exothermic and more random
– H is negative and large and S is negative but small.
• Or low temperature
– H is positive but small and S is positive and large.
• Or high temperature
•
G will be positive when H is positive and S is negative.
– Never spontaneous at any temperature.
•
When G = 0 the reaction is at equilibrium.
Gibbs Free Energy
Example 18: Can a nonspontaneous process with a negative ΔS become
spontaneous if heat is absorbed by the system (endo) ?
[Answer: No, the criteria is for a –ΔG, so for any +ΔH and – ΔS, ΔG is always
“+.” ]
Example 19: Can a nonspontaneous process with a positive ΔS become
spontaneous if the temperature is increased ?
[Answer: Yes, only when +ΔH is small, and the temperature is increased. ]
Example 20: Calculate the change in molar Gibbs free energy, ΔGm, for the
process H2O (s) → H2O (l) at 1 atm and (a) 10.°C; (b) 0.0°C. Decide for each
temperature whether melting is spontaneous or not. Treat ΔHfus and ΔSfus as
independent of temperature.
The enthalpy of fusion (ΔHfus) is 6.01 kJ·mol−1 and the entropy of fusion (ΔSfus )
is 22.0 J·K−1·mol−1. These values are almost independent of temperature over
the temperature range considered.
ΔG = ΔH – TΔS
(a) 10.°C
ΔH in kJ
TΔS also converted to kJ
ΔG = ΔH – TΔS , = 6.01 kJ·mol−1 – (273.15 + 10.°C)K × 22.0 J·K−1·mol−1 =
-0.22 kJ·mol−1
Example 20: Calculate the change in molar Gibbs free energy, ΔGm, for the
process H2O (s) → H2O (l) at 1 atm and (a) 10.°C; (b) 0.0°C. Decide for each
temperature whether melting is spontaneous or not. Treat ΔHfus and ΔSfus as
independent of temperature.
The enthalpy of fusion (ΔHfus) is 6.01 kJ·mol−1 and the entropy of fusion
(ΔSfus) is 22.0 J·K−1·mol−1. These values are almost independent of
temperature over the temperature range considered.
ΔG = ΔH – TΔS
(b) 0.0°C
ΔG = ΔH – TΔS , = 6.01 kJ·mol−1 – (273.15 + 0.0°C)K × 22.0 J·K−1·mol−1
= 0.0 kJ·mol−1
In (a) ΔG is “-” so ice melting above freezing is spontaneous, in (b) at 0.0°C
it’s at equilibrium, this makes sense since it’s in a phase change where we’ve
already calculated fusion ΔStot = 0.
Gibbs Free Energy & Temperature
G = H - TS
Gibbs free energy decreases as the
temperature is raised.
As T increases, TS increases, too, and a larger
quantity is subtracted from H.
Gibbs free energy decreases more sharply
with temperature for gases than for liquids.
Because:
S°m,gas > S°m,liquid > S°m,solid
Gibbs Free Energy & Temperature
ΔG° does depend on temperature ΔG° = ΔH° – TΔS° . There are four cases
to consider:
Spontaneous if:
(a) T is low
a)
b)
c)
d)
(b) T is high
(c) never
(d) always
Exothermic ΔH° < 0, unfavorable ΔS° < 0, as long as T remains low, ΔG° is “-”
Endothermic ΔH° > 0, favorable
ΔS° > 0, if the T is high enough, ΔG° will be “-”
Endothermic ΔH° > 0, unfavorable ΔS° < 0, this is never spontaneous at any T
Exothermic ΔH° < 0, favorable
ΔS° > 0, spontaneous at any T.
Example 21: Estimate the temperature at which it is thermodynamically
possible for carbon to reduce iron(III) oxide to iron under standard
conditions by the endothermic reaction
2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g)
ΔS° J·K−1 :
87.4
ΔH° kJ:
-824.2
5.7
27.3
213.7
0
0
-393.5
For an endothermic ΔH° > 0, because gas is produced ΔS° > 0 (favorable),
if the T is high enough, ΔG° will be “-” and the reaction will be spontaneous.
The crossover point is where ΔG° goes from “+” to “-” so we can take
ΔG° = ΔH° – TΔS° set it to 0,
ΔH° = TΔS° , then solve for T ,
T=
ΔH°
ΔS°
Example 21: Estimate the temperature at which it is thermodynamically
possible for carbon to reduce iron(III) oxide to iron under standard
conditions by the endothermic reaction
2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g)
ΔS° J·K−1 :
87.4
ΔH° kJ:
-824.2
5.7
27.3
213.7
0
0
-393.5
use ΔH° = ΣnΔHf° (products) - Σn ΔHf° (reactants)
= {4 mol ΔHf° (Fe, s) + 3 mol ΔHf° (CO2, g) }kJ {2 mol ΔHf° (Fe2O3,s) + 3 mol ΔHf° (C, s)}kJ
= (4 × 0 + 3 × -393.5)kJ - (2 × -824.2 + 3 × 0 )kJ
= +467.9 kJ
use ΔS° = ΣnΔSf° (products) - Σn ΔSf° (reactants)
= {4 mol ΔSf° (Fe, s) + 3 mol ΔSf° (CO2, g) } J·K−1 {2 mol ΔSf° (Fe2O3,s) + 3 mol ΔSf° (C, s)} J·K−1
= (4 × 27.3 + 3 × 213.7) J·K−1 - (2 × 87.4 + 3 × 5.7 )J·K−1
= +558.4 J·K−1 or + 0.5584 kJ·K−1
T=
ΔH°
,
ΔS°
+467.9 kJ
= 838K,
+0.5584 kJ·K−1
or anything temperature above 565 °C
Gibbs Free Energy of Reaction
Our principal interest in chemistry is to calculate ΔG for a reaction.
Gibbs free energy of reaction, ΔG (which is commonly referred to as the
"reaction free energy").
ΔG = ΣnG (products) - ΣnG (reactants)
Σ = sum up
n = stoichiometric coefficients, moles
G = molar Gibbs Free Energy
Gibbs Free Energy of Reaction
For standard conditions of a chemical reaction, ΔG°
ΔG° = ΣnG° (products) - ΣnG° (reactants)
For standard ΔG°, meaning formed from it’s elements or ΔGf° at 1 bar,
298K (25°C) per mol.
So ΔG is under variable condition, but ΔGf° means fixed conditions.
Example 22: Calculate the standard Gibbs free energy of formation of
C3H6 (g), cyclopropane, at 25°C.
3C (graphite) + 3 H2 (g) → C3H6(g)
We can use ΔG° = ΔH° – TΔS° , where ΔG° = ΔGf°
ΔHf° (C3H6, g) = +53.30 kJ
ΔS°= ΣnS (products) - ΣnS (reactants),
ΔSm° (C3H6, g) - {3ΔSm° (C, gra) + 3ΔSm° (H2, g)}
= 237.4 J·K−1·mol−1 – {3(130.68) J·K−1·mol−1 + 3(5.740) J·K−1·mol−1}
= -171.86 J·K−1 or -0.17186 kJ·K−1
ΔG° = +53.30 kJ - 298 K × -0.17186 kJ·K−1 = +104.5 kJ
When ΔGf° is “large +” the reaction is nonspontaneous, and in this case,
at any temperature (because S0 and H>0).
Labile verses Inert
Standard Gibbs free energy of formation is an
indication of a compound's stability.
If ΔGf° < 0 than elements are poised to change
spontaneously into the compound.
If ΔGf° > 0, then the reverse is true, the compound is
poised to change spontaneously into the pure
elements.
The ΔGf° for benzene is + 124 kJ·mol−1 at 25°C, so
benzene is unstable and is poised to decompose
spontaneously into its elements.
Labile verses Inert
That tendency may not be realized because the
decomposition may be very slow. Benzene can, in
fact, be kept indefinitely without decomposing at all.
Substances that are thermodynamically unstable
but survive for long periods are called nonlabile
or even inert. So benzene is thermodynamically
unstable but nonlabile.
Substances that decompose or react rapidly are
called labile. Most radicals are labile.
Labile verses Inert
Question 23: Is glucose stable relative to its elements at 25°C and under
standard conditions ?
glucose, ΔGf° = -910. kJ·mol−1
Answer: Yes; ΔGf° is a negative value.
Question 24: Is methylamine, CH3NH2, stable relative to its elements at
25°C and under standard conditions ?
CH3NH2, ΔGf° = +32.16 kJ·mol−1
Answer: No; ΔGf° is a positive value.
Practice
Practice 1: Calculate the entropy change associated with the isothermal
expansion of 5.25mol of ideal gas atoms from 24.252L to 34.058L.
Practice
Practice 2: Which would you expect to have a higher molar entropy single
crystals of BF3 or of COF2 (a) at T = 0K, (b) at 25ºC? Why?.
Practice
Practice 3: Calculate the change in molar Gibbs energy for the process
NH3(l)  NH3(g) at 1atm and (a) -15.0 ºC; (b) -45 ºC (see Table 8.3 and 9.1).
In each case, indicate whether vaporization would be spontaneous.
Practice
Practice 4: Initially a sample of ideal gas at 323K occupies 1.67L at 4.95atm.
The gas is allowed to expand to 7.33L by two pathways: (a) isothermal,
reversible expansion; (b) isothermal irreversible free expansion. Calculate
Stot, S and Ssurr for each pathway.
Practice
Practice 5: Calculate the standard reaction entropy, enthalpy, and Gibbs free
energy for the following reaction from data in Appendix 2A.
2H2O2(l)  2H2O(l) + O2(g)
Practice
Practice 6: Calculate the standard entropy of vaporization of ammonia at
210.0K, given that the molar heat capacities at constant pressure of liquid
ammonia and ammonia vapor are 80.8 J.K-1.mol-1 and 35.1 J.K-1mol-1,
respectively, in this range (see Table 8.3).