The Story of Spontaneity and
Energy Dispersal
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Spontaneity
Spontaneous process are those that occur
naturally.
Hot body cools
A gas expands to fill the available volume
A spontaneous direction of change is
where the direction of change does not
require work to bring it about.
Spontaneity
The reverse of a spontaneous process is a
nonspontaneous process
Confining a gas in a smaller volume
Cooling an already cool object
Nonspontaneous processes require energy
in order to realize them.
Spontaneity
Note:
Spontaneity is often interpreted as a natural
tendency of a process to take place, but it does
not necessarily mean that it can be realized in
practice.
Some spontaneous processes have rates
sooo slow that the tendency is never
realized in practice, while some are
painfully obvious.
Spontaneity
The conversion of diamond to graphite is
spontaneous, but it is joyfully slow.
The expansion of gas into a vacuum is
spontaneous and also instantaneous.
2ND LAW OF
THERMODYNAMICS
Physical Statements of the 2nd Law of
Thermodynamics
Kelvin Statements
“No process is possible in which the sole
result is the absorption of heat from a
reservoir and its complete conversion into
work”
 “It is impossible for a system to undergo a
cyclic process whose sole effects are the
flow of an amount of heat from the
surroundings to the system and the
performance of an equal amount of work
on the surroundings.”
 “It is impossible for a system to undergo a
cyclic process that turns heat completely
into work done on the surroundings.”
Clausius statement
 It is impossible for a process to occur that
has the sole effect of removing a quantity
of heat from an object at a lower
temperature and transferring this quantity
of heat to an object at a higher
temperature.
 Heat cannot flow spontaneously from a
cooler to a hotter object if nothing else
happens
The 2nd Law of Thermodynamics
The 2nd Law of Thermodynamics
recognizes the two classes of processes,
the spontaneous and nonspontaneous
processes.
Implications of the 2nd Law
No heat engine can have an efficiency as
great as unity
No macroscopic process can decrease the
entropy of the universe
Are you kidding me?
Thermodynamic Cat
Hot
Reservoir
Heat
Cold
Reservoir
Engine
Heat
Work
I approve!
What determines the direction of
spontaneous change?
The total internal energy of a system does
NOT determine whether a process is
spontaneous or not.
Per the First Law, energy is conserved in
any process involving an isolated system.
What determines the direction of
spontaneous change?
Instead, it is important to note that the
direction of change is related to the
distribution of energy.
Spontaneous changes are always
accompanied by a dispersal of energy.
Energy Dispersal
 Superheroes with
energy blasts and
similar powers as well
as the Super Saiyans
are impossible
characters.
 They seem to violate
the Second Law of
Thermodynamics!
Power
Genki
dama
Energy Dispersal
A ball on a warm
floor can never be
observed to
spontaneously
bounce as a result
of the energy from
the warm floor
Energy Dispersal
 In order for this to
happen, the thermal
energy represented by
the random motion
and vibrations of the
floor atoms would
have to be
spontaneously
diverted to
accumulate into the
ball.
Energy Dispersal
 It will also require the
random thermal motion to
be redirected to move in a
single direction in order
for the ball to jump
upwards.
 This redirection or
localization of random,
disorderly thermal motion
into a concerted, ordered
motion is so unlikely as to
be virtually impossible.
Energy Dispersal and Spontaneity
Spontaneous change can now be
interpreted as the direction of change that
leads to the dispersal of the total energy
of an isolated system!
INDEED!
Entropy
A state function, denoted by S.
While the First Law can be associated with
U, the Second Law may be expressed in
terms of the S
Entropy and the Second Law
The Second Law can be expressed in
terms of the entropy:
The entropy of an isolated system increases
over the course of a spontaneous change:
ΔStot > 0
Where Stot is the total entropy of the
system and its surroundings.
Entropy: A molecular look
 Boltzmann formula:
 𝑆 = 𝑘 ln 𝑊
 Entropy is a reflection of the
microstates, the ways in
which the molecules of a
system can be arranged
while keeping the total
energy constant.
Entropy
A simple definition of entropy is that it is a
measure of the energy dispersed in a
process.
For the thermodynamic definition, it is
based on the expression:
Entropy
 For a measurable change between two states,
 In order to calculate the difference in entropy
between two states, we find a reversible
pathway between them and integrate the
energy supplied as heat at each stage, divided
by the temperature.
Example
Change in entropy of the
surroundings: ΔSsur
 If we consider a transfer of heat dqsur to the surroundings, which
can be assumed to be a reservoir of constant volume.
 The energy transferred can be identified with the change in internal
energy
 dUsur is independent of how change brought about (U is state function
 Can assume process is reversible, dUsur= dUsur,rev
 Since dUsur = dqsur and dUsur= dUsur,rev,
 ∴ dqsur must equal dqsur,rev
 That is, regardless of how the change is brought about in the
system, reversibly or irreversibly, we can calculate the change of
entropy of the surroundings by dividing the heat transferred by the
temperature at which the transfer takes place.
Change in entropy of the
surroundings: ΔSsur
 For adiabatic change, qsur = 0, so DSsur = 0
Entropy as a State Function
 To prove entropy is a state function we must show that ∫dS is path
independent
 Sufficient to show that the integral around a cycle is zero or

dS 
 Sadi Carnot (1824) devised cycle to represent idealized engine
Hot Th
Reservoir
Step 1: Isothermal reversible
expansion @ Th
qh
w3
-w1
Step 2:Adiabatic expansion
Th to Tc
-w2
Step 3:Isothermal reversible
compression @ Tc (sign of q
negative)
Engine
w4
qc
Cold Tc
Reservoir
Step 4: Adiabatic
compression Tc to Th

dq
T
0
Carnot Engine
How is that
possible?
Carnot Cycle
Carnot Cycle
Step 1: ΔU=0
Step 2: ΔU=w
Step 3: ΔU=0
Step 4: ΔU=-w
Carnot Cycle - Thermodynamic
Temperature Scale
 The efficiency of a heat engine is
the ratio of the work performed to
the heat of the hot reservoir
e=|w|/qh
 The greater the work the greater the
efficiency
 Work is the difference between the
heat supplied to the engine and the
heat returned to the cold reservoir
w = qh -(-qc) = qh + qc
 Therefore, e = |w|/qh = ( qh +
qc)/qh = 1 + (qc/qh )
Hot
Reservoir
qh
w
Heat Engine Work
Cold
Reservoir
Heat
-qc
Efficiency of Heat Engines
 Efficiency is the ratio of the work done by an
engine in comparison to the energy invested in
the form of heat for all reversible engines
𝑤 𝑞ℎ − 𝑞𝑐 𝑇ℎ − 𝑇𝑐
𝑇𝑐
e 𝑜𝑟 η = =
=
=1−
𝑞ℎ
𝑞ℎ
𝑇ℎ
𝑇ℎ
 All reversible engines have the same efficiency
irrespective of their construction.
Carnot Cycle - Thermodynamic
Temperature Scale
Hot

William Thomson (Lord Kelvin) defined a
substance-independent temperature scale
based on the heat transferred between two
Carnot cycles sharing an isotherm

Reservoir
qh
He defined a temperature scale such that qc/qh = Tc/Th




Or as Tc approaches 0 e approaches 1
Efficiency can be used as a measure of
temperature regardless of the working
fluid
 Applies directly to the power
required to maintain a low
temperature in refrigerators
Heat Engine Work
Cold
Reservoir
e = 1 - (Tc/Th )
Zero point on the scale is that
temperature where e = 1
w
Efficiency
Heat
-qc
is maximized
Greater
temperature
difference between reservoirs
The
lower Tc, the greater
the efficiency
Refrigeration
Coefficient of performance (COP or β or c)
𝑞𝑐
𝑞𝑐
𝑇𝑐
𝐶𝑂𝑃 = =
=
𝑤 𝑞ℎ − 𝑞 𝑐
𝑇ℎ − 𝑇𝑐
 COP describes the qc in this case as the
minimum energy to be supplied to a
refrigeration-like system in order to
generate the required entropy to make the
system work.
Entropy changes: Expansion
Entropy changes in a system are
independent of the path taken by the
process
ΔS = 𝑛𝑅
𝑉2
𝑙𝑛
𝑉1
Total change in entropy however depend
on the path:
Reversible process: ΔStot = 0
Irreversible process: ΔStot > 0
Isothermal
Isochoric
Isobaric
Adiabatic
ΔU
0
nCvΔT
q+w
w
q
nRT ln 𝑓
𝑉𝑖
or -wirrev
nCvΔT
nCpΔT or –wirrev
0
𝑉
wrev
-nRT ln
𝑉𝑓
𝑉𝑖
0
-nRT ln
𝑉𝑓
𝑉𝑖
wirrev
-pextΔV
0
-pextΔV
ΔH
0 (for ideal gas)
ΔU
=ΔU + pΔV
=nCpΔT
ΔS
= 𝑛𝑅 𝑙𝑛
𝑉𝑓
𝑉𝑖
=
𝑇2 𝐶𝑣𝑑𝑇
𝑇1 𝑇
=
𝑇2 𝐶𝑝𝑑𝑇
𝑇1 𝑇
𝑛𝑅∆𝑇
1−γ
=-nCvΔT
=-pextΔV
0
Entropy changes: Phase Transitions
ΔtransS =
ΔtransH
Ttrans
Trouton’s rule: An empirical observation
about a wide range of liquids providing
approximately the same standard entropy
of vaporization, around 85 J/mol K.
General equations for entropy during
a heating process
 S as a function of T and V, at constant P
𝑇𝑓
𝑉𝑓
Δ𝑆 = 𝑛𝐶𝑝 𝑙𝑛 + 𝑛𝑅 ln
𝑇𝑖
𝑉𝑖
 S as a function of T and P, at constant V
𝑇𝑓
𝑃𝑓
Δ𝑆 = 𝑛𝐶𝑣 𝑙𝑛 − 𝑛𝑅 ln
𝑇𝑖
𝑃𝑖
Measurement of Entropy (or molar
entropy)
Measurement of Entropy (or molar
entropy)
The terms in the previous equation can be
calculated or determined experimentally
The difficult part is assessing heat
capacities near T = 0.
Such heat capacities can be evaluated via
the Debye extrapolation
Measurement of Entropy (or molar
entropy)
In the Debye extrapolation, the expression
below is assumed to be valid down to
T=0.
𝐶𝑝 =
3
𝑎𝑇
Third Law of Thermodynamics
 At T = 0, all energy of thermal motion has been
quenched, and in a perfect crystal all the atoms or ions
are in a regular, uniform array.
 The localization of matter and the absence of thermal
motion suggest that such materials also have zero
entropy.
 This conclusion is consistent with the molecular
interpretation of entropy, because S = 0 if there is only
one way of arranging the molecules and only one
microstate is accessible (the ground state).
Third Law of Thermodynamics
The entropy of all perfect crystalline
substances is zero at T = 0.
Nernst heat theorem
The entropy change accompanying any
physical or chemical transformation
approaches zero as the temperature
approaches zero: ΔS  0 as T  0
provided all the substances involved are
perfectly crystalline.
Third-Law entropies
These are entropies reported on the basis
that S(0) = 0.
Exercises
HELMHOLTZ AND GIBBS
ENERGIES
Clausius inequality
𝑑𝑞
𝑑𝑆 ≥
𝑇
The Clausius inequality implies that dS ≥
0.
“In an isolated system, the entropy cannot
decrease when a spontaneous change
takes place.”
Criteria for spontaneity
𝑑𝑞
𝑑𝑆 −
≥0
𝑇
In a system in thermal equilibrium with its
surroundings at a temperature T, there is
a transfer of energy as heat when a
change in the system occurs and the
Clausius inequality will read as above:
Criteria for spontaneity
When energy is transferred as heat at constant volume:
𝑑𝑞
𝑑𝑆 −
≥0
𝑇
*dq = dU
𝑇𝑑𝑆 ≥ 𝑑𝑈
 At either constant U or constant S:
 Which leads to
𝑑𝑆𝑈, 𝑉 ≥ 0
𝑑𝑈𝑆, 𝑉 ≤ 0
𝑑𝑈 − 𝑇𝑑𝑆 ≤ 0
Criteria for spontaneity
 When energy is transferred as heat at constant
pressure, the work done is only expansion work
and we can obtain
𝑇𝑑𝑆 ≥ 𝑑𝐻
 At either constant H or constant S:
𝑑𝑆𝐻 , 𝑝 ≥ 0
𝑑𝐻𝑆, 𝑝 ≤ 0
 Which leads to
𝑑𝐻 − 𝑇𝑑𝑆 ≤ 0
Criteria for spontaneity
We can introduce new thermodynamic
quantities in order to more simply
express 𝑑𝑈 − 𝑇𝑑𝑆 ≤ 0 and 𝑑𝐻 − 𝑇𝑑𝑆 ≤ 0
Helmholtz and Gibbs energy
 Helmholtz energy, A:
 Gibbs energy, G:
A = U - TS
G = H - TS
dA = dU – TdS
dG = dH – TdS
dAT,V ≤ 0
dGT,p ≤ 0
Helmholtz energy
A change in a system at constant
temperature and volume is spontaneous if
it corresponds to a decrease in the
Helmholtz energy.
Aside from an indicator of spontaneity, the
change in the Helmholtz function is equal
to the maximum work accompanying a
process.
Helmholtz energy
, useful
Variation of the Gibbs free energy
with temperature
Variation of the Gibbs free energy
with pressure
Variation of the Gibbs free energy
with pressure
Homework
1.
When 1.000 mol C6H12O6 (glucose) is oxidized to carbon dioxide and
water at 25°C according to the equation C6H12O6(s) + 6 O2(g)  6
CO2(g) + 6 H2O(l), calorimetric measurements give ΔrHθ= -2808 kJ
mol-1 and ΔrSθ = +182.4 J K-1 mol-1 at 25°C. How much of this energy
change can be extracted as (a) heat at constant pressure, (b) work?
2.
How much energy is available for sustaining muscular and nervous
activity from the combustion of 1.00 mol of glucose molecules under
standard conditions at 37°C (blood temperature)? The standard
entropy of reaction is +182.4 J K-1 mol-1.
3.
Calculate the standard reaction Gibbs energies of the following
reactions given the Gibbs energies of formation of their components
a) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
b) C12H22O11(s) + 12 O2(g)  12 CO2(s) + 11 H2O(l)
One for the road
Life requires the assembly of a large
number of simple molecules into more
complex but very ordered
macromolecules. Does life violate the
Second Law of Thermodynamics? Why or
why not?