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George Mason University
General Chemistry 212
Chapter 20
Thermodynamics
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and
Change, 7th edition, 2011, McGraw-Hill
Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George
Mason are intended for those students enrolled in a science
/engineering oriented curricula, with particular emphasis on
chemistry, biochemistry, and biology The material on these slides is
taken primarily from the course text but the instructor has modified,
condensed, or otherwise reorganized selected material.
Additional material from other sources may also be included.
Interpretation of course material to clarify concepts and solutions to
problems is the sole responsibility of this instructor.
2/6/2015
1
Thermodynamics

Thermodynamics: Enthalpy, Entropy, Free Energy
The Direction of Chemical Reactions

The First Law of Thermodynamics
Conservation of Energy



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Limitations of the First Law

The Sign of H and Spontaneous Change

Freedom of Motion and Disposal of Energy
The Second Law of Thermodynamics

Predicting Spontaneous Change

Entropy and the Number of Microstates

Entropy and the Second Law
The Third Law of thermodynamics

Standard Molar Entropies
2
Thermodynamics



Calculating the Change in Entropy of a Reaction

The Standard Entropy of Reaction

Entropy Changes in the Surroundings

Entropy Change and the Equilibrium State

Spontaneous Exothermic and Endothermic Reactions
Entropy, Free Energy, and Work

Free Energy Change (∆G) and Reaction Spontaneity

Standard Free Energy Changes

G and Work

Temperature and Reaction Spontaneity

Coupling of Reactions
Free Energy, Equilibrium, and Reaction Direction
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3
Thermodynamics
Enthalpy (∆H)
Sum of Internal Energy (E) plus
Product of Pressure & Volume
(Endothermic vs. Exothermic)
( Hrxn = Hf prod - Hf react)
Entropy (S)
Measure of system order/disorder &
the number of ways energy can
can be dispersed through
the motion of its particles
All real processes occur spontaneously
in the direction that increases the
Entropy of the universe
(universe = system + surroundings)
Gibbs Free Energy (∆G)
Difference between Enthalpy and
the product of absolute temperature
and the Entropy
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w  P Δ V
ΔE  q p  w
qp  ΔE  P ΔV
 H  E  P V
ΔH  qp
(Constant Pressure)
Euniv = Esys + Esurr
(Spontaneous)
ΔSuniv > 0
ΔSsys = - Ssurr
ΔSuniv = 0
(At Equilibrium)
ΔSsys = R  ln
o
ΔSrxn
=
Vfinal
P
C
= R  ln initial = R  ln final
Vinitial
pfinal
Cinitial
o
 m Sproducts
-
o
 n Sreactants
o
ΔG osys  H sys
- TΔSosys
ΔGorxn =
o
 mΔG f(products)
-
o
 nΔG f(reactants)
4
Thermodynamics

Thermodynamics - study of relationships between heat
and other forms of energy in chemical reactions

The direction and extent of chemical reactions can be
predicted through thermodynamics (i.e., feasibility)

In thermodynamics, a state variable is also called a state
function

Examples include:
Temperature (T), Pressure (P), Volume (V),
Internal Energy (E), Enthalpy (H), and Entropy (S)

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In contrast Heat (q) and Work (W) are not state
functions, but process functions
5
Thermodynamics

Chemical reactions are driven by heat (Enthalpy) and/or
randomness (Entropy)

A measure of randomness (disorder) is Entropy (S)

An increase in disorder is spontaneous

Spontaneous reactions are moving toward equilibrium

Spontaneous reactions move in the direction where energy
is lowered, and move to Q/K = 1 (equilibrium)

Thermodynamics is used to determine spontaneity
(a process which occurs by itself) and the natural forces
that determine the extent of a chemical reaction (i.e., Kc)
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6
Thermodynamics

For a reaction to be useful it must be spontaneous
(i.e., goes to near completion, i.e., far to the right)


Spontaneity of a reaction depends on:

Enthalpy - heat flow in chemical reactions

Entropy - measure of the order or randomness of
a system (Entropy units - J/ oK)

Entropy is a state function; S = Sfinal - Sinitial

Higher disorder equates to an increase in Entropy

Entropy has positional and thermal disorder
There are three principal laws of thermodynamics, each of
which leads to the definition of thermodynamic properties
(state variables) which help us to understand and predict
the operation of a physical system
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7
Laws of Thermodynamics

The Laws of Thermodynamics define fundamental
physical quantities (temperature, energy, and entropy that
characterize thermodynamic systems. The laws describe
how these quantities behave under various circumstances,
and forbid certain phenomena (such as perpetual motion)

The First Law of Thermodynamics is a statement of the
conservation of energy

The Second Law is a statement about the direction of that
conservation

The Third Law is a statement about reaching Absolute
Zero (0° K)
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8
Laws of Thermodynamics

First law of thermodynamics:

The first law, also known as the Law of Conservation
of Energy, states that energy cannot be created or
destroyed in a chemical reaction

Energy can only be transferred or changed from one
form to another. For example, turning on a light would
seem to produce energy; however, it is electrical
energy taken from another source that is converted

It relates the various forms of kinetic and
potential energy in a system to the work
(W = -PΔV) which a system can perform and to the
transfer of heat

It applies to the changes in internal energy (ΔE) when
energy passes, as work (W), as heat (q), or with
matter, into or out from a system
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9
Laws of Thermodynamics

The first law is usually formulated by stating that the
change in the Internal Energy (E) of a closed system
is equal to the amount of Heat (q) supplied to the
system, minus the amount of Work (W = -PV)
performed by the system on its surroundings

The law of conservation of energy can be stated
The Energy of an Isolated System is Constant
ΔE = q + w = q - P V
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10
Laws of Thermodynamics

First Law of system Thermodynamics

Conservation of Energy, E (or U in some texts)
Any change in the energy of the system
corresponds to the interchange of “heat” (work) with
an “External” surrounding

Total Internal Energy (E) - The sum of the kinetic
and potential energies of the particles making up a
substance

Kinetic Energy (Ek) - The energy associated with an
object by virtue of its motion,
Ek = ½mv2

Potential Energy (Ep) - The energy an object has by
virtue of its position in a field of force,
Ep = mgh
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(kgm2/s2) (joules)
(kg  m/s2 m = kgm2/s2)
11
Laws of Thermodynamics

Work – The energy transferred is moved by a force,
such as the expansion of a gas in an open system under
constant pressure
Pressure
= kg/(ms2)
Volume
= m3
Work (W) = kg/(ms2)  m3 = kgm2/s2 = joules (J)

Internal Energy
● The Internal Energy of a system, E, is precisely
defined as the heat at constant pressure (qp)
plus the work (w) done by the system:
ΔE = qp + w
ΔE = qp + (-P ΔV)
qp  ΔE  P ΔV
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12
Laws of Thermodynamics

Enthalpy is defined as the internal energy plus the
product of the pressure and volume – work

The change in Enthalpy is the change in internal
energy plus the product of constant pressure and the
change in Volume
Recall


ΔH = qp
(At Constant Pressure)
The change in Enthalpy equals the heat gained or lost
(heat of reaction) at constant pressure – the entire change
in “internal energy” (E), minus any expansion “work”
done by the system (PV) would have negative sign
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13
Laws of Thermodynamics





E – total internal energy; the sum of kinetic and
potential energies in the system
q – heat flow between system and surroundings
(-q indicates that heat is lost to surroundings)
w – work (-w indicates work is lost to surroundings)
H – Enthalpy – extensive property dependent on
quantity of substance and represents the heat
energy tied up in the chemical bonds (heat of
reaction)
Useful Units in Energy expressions
 1 J (joule) = 1 kgm2/s2

1 Pa (pascal) = 1 kg/ms2

1 atm = 1.01325 x 105 Pa
1 atm = 760 torr = 760 mm Hg

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14
Exchanges of Heat and Work
with the Surroundings
Pressure x
Volume
Work = expansion
of volume due to
forming a gas
q>0
w>0
on system
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q<0
w<0
by system
15
Practice Problem
Consider the combustion of Methane (CH4) in Oxygen
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
The heat of reaction (q) at 25 oC and 1.00 atm is -890.2 kJ. What
is E for the change indicated by the chemical equation at 1 atm?
n = 3 mol converted to 1 mol = -2 mol
@ 25 oC and 1 atm, 1 mol of gas = 24.5 L, thus
V = -2(24.5) = -49 L  (1m3/1000 L) = -0.049 m3
E = q - PV
E = -890.2 kJ – 1 atm x (-0.049 m3)
E = -890.2 kJ – (1.01 x 105 Pa)(-0.049 m3)
E = -890.2 kJ – (1.01 x 105 kg/ms2)(-0.049m3)
E = -890.2 kJ + (4949 J x 1 kJ/1000 J)
E = -890.2 kJ + 4.949 kJ = -885 kJ
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16
Laws of Thermodynamics

Second Law of Thermodynamics:

The second law introduces a new state variable,
Entropy (S)

Entropy is a measure of the number of specific ways in
which the energy of a thermodynamic system can be
dispersed through the motions of its particles

In a natural thermodynamic process, the sum of the
entropies of the participating thermodynamic
systems increases

The total entropy of a system plus its environment
(surroundings) can not decrease; it can remain
constant for a reversible process but must always
increase for an irreversible process
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17
Laws of Thermodynamics

Second Law of Thermodynamics:

According to the second law the entropy of an isolated
system not in thermal equilibrium never decreases;
such a system will spontaneously evolve toward
thermodynamic equilibrium, the state of maximum
entropy of the system
More simply put: the entropy of the world only
increases and never decreases

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A simple application of the second law of
thermodynamics is that a room, if not cleaned and
tidied, will invariably become more messy and
disorderly with time - regardless of how careful one is
to keep it clean. When the room is cleaned, its entropy
decreases, but the effort to clean it has resulted in an
increase in entropy outside the room that exceeds the
entropy lost
18
Laws of Thermodynamics

The 2nd Law of Thermodynamics

Entropy is a state function; S = Sf - Si

Higher disorder equates to an increase in Entropy

Entropy has positional and thermal disorder

The Entropy, S, is conserved for a reversible process

The disorder of the system and thermal surroundings
must increase for a spontaneous process

The total Entropy of a system and its surroundings
always increases for a “Spontaneous” process

A process occurs spontaneously in the direction that
increases the Entropy of the universe
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19
Laws of Thermodynamics

A spontaneous change, whether a chemical or physical
change, or just a change in location is one that:

Occurs by itself under specified conditions

Occurs without a continuous input of energy from
outside the system

In a non-spontaneous change, the surroundings must
supply the system with a continuous input of energy

Under a given set of conditions, a spontaneous change in
one direction is not spontaneous in the “other” direction
A limitation of the 1st Law of Thermodynamics

Spontaneous does not equate to “Instantaneous”

The first and second laws make it impossible to construct
a perpetual motion machine.
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20
Laws of Thermodynamics

Limitations of the 1st law of Thermodynamics

The 1st Law accounts for the energy involved in a
chemical process (reaction)

The internal energy (E) of a system, the sum of the
kinetic and potential energy of all its particles,
changes when heat (q) and/or work (w= -PV) are
gained or lost by the system
ΔE = q + w = q - P V

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Energy not part of the system is part of the
surroundings
21
Laws of Thermodynamics

The surroundings (sur) and the system (sys)
together constitute the “Universe (univ)”
Euniv = Esys + Esur

Heat and/or work gained by system is lost by
surroundings
(q + w)sys = - (q + w)surr

The “total” energy of the Universe is constant
ΔEsys = - ΔEsur
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 ΔEsys + ΔEsur = 0 = ΔEuniv
22
Laws of Thermodynamics

The first Law, however, does not account for the
“direction” of the change in energy
Ex. The burning of gas in your car

Potential energy difference between chemical bonds in
fuel mixture and those in exhaust is converted to kinetic
energy to move the car

Some of the converted energy is released to the
environmental surroundings as heat (q)

Energy (E) is converted from one form to another, but
there is a “net” conservation of energy

1st law does not explain why the exhaust gas does not
convert back into gasoline and oxygen

1st law does not account for the “direction” of a
spontaneous change
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23
Laws of Thermodynamics

Spontaneous Change and Change in Enthalpy (H)

It was originally proposed (19th Century) that the “sign” of
the Enthalpy change (H) – the heat lost or gained at
constant pressure (qp) – was the criterion of spontaneity

Exothermic processes (H < 0) were “spontaneous”

Endothermic processes (H > 0) were “nonspontaneous”

Ex. Combustion (burning) of Methane in Oxygen is
CH 4 (g) + 2O2 (g)  CO2 (g) + 2H 2O(g)
ΔH = - 802 kJ
“Spontaneous” and “Exothermic” (H < 0)
When Methane burns in your furnace, heat is released
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24
Laws of Thermodynamics

The sign of the change in Enthalpy (H), however, does
not indicate spontaneity in all cases

An Exothermic process can occur spontaneously under
certain conditions and the opposite Endothermic process
can also occur spontaneously under other conditions
Ex. Water freezes below 0oC and melts above 0oC
Both changes are spontaneous
Freezing is Exothermic
Melting (& Evaporation) is Endothermic
Most Water-Soluble Salts have a positive Hsoln
yet they dissolve spontaneously
The decomposition of N2O5 is Endothermic and
spontaneous
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N 2O5 (s)  2NO2 (g) + 1 2O2 (g)
ΔHrxn = +109.5 kJ
25
Entropy

Freedom of motion & energy dispersion

Endothermic processes result in more
particles (atoms, ions, molecules) with more
freedom of motion – Entropy increases

During an Endothermic phase change, “fewer”
moles of reactant produce “more” moles of
product

The energy of the particles is dispersed over
more quantized energy levels
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26
Entropy

Endothermic Spontaneous Process
Less freedom of particle motion  more freedom of motion
Localized energy of motion  dispersed energy of motion
Phase Change:
Solid  Liquid  Gas
Dissolving of Salt: Crystalline Solid + Liquid  Ions in Solution
Chemical Change: Crystalline Solids  Gases + Ions in
Solution

In thermodynamic terms, a change in the freedom of motion
of particles in a system, that is, in the dispersal of their
energy of motion, is a key factor determining the direction of
a spontaneous process
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27
Entropy


Quantized Energy Levels

Electronic

Kinetic - vibrational, rotational, translational
Microstate

A single quantized state at any instant

The total energy of the system is dispersed throughout
the microstate

New microstates are created when system conditions
change

At a given set of conditions, each microstate has the
same total energy as any other

A given microstate is just as likely to occur as any other
microstate
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28
Entropy

Microstates vs Entropy (Positional Disorder)

Boltzmann Equation
S = k lnW
where k – Boltzmann Constant
R
8.31447J / mol • K
-23
k =
=
=
1.38

10
J/K
23
NA
6.02214  10 / mole
where R = Universal Gas Constant
NA = Avogadro’s Number
where W = No. of Microstates
R
S =
ln W = 1.38 x 10-23  lnW  J / K
NA
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29
Entropy

The number of microstates (W) possible for a given
number of particles (n) as the volume changes is a
function of the nth power of 2:
Wfinal
n
= 2
Winitial
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30
Entropy

Compute Ssys
ΔSsys = Sfinal - Sinitial = k  ln Wfinal - k  ln Winitial
ΔSsys = k  ln

When n becomes NA , i.e. 1 mole
ΔSsys
R
R
N
A
=
 ln 2 =
 N A  ln 2 = R  ln 2
NA
NA
ΔSsys = R  ln 2 



2/6/2015
Wfinal
R
= k × ln 2n =
× ln 2n
Winitial
NA
 8.314 J / mol • K  0.693 
= 5.76 J / mol • K
The Boltzman constant “k = R/NA” has become “R”
A system with fewer microstates (smaller Wfinal) has
lower Entropy (Lower S)
A system with more microstates (larger Wfinal) has
higher Entropy (higher S)
31
Entropy

Entropy change – Volume, Pressure, Concentration
S = R ln(V, P, C)
ΔSsys = Sfinal - Sinitial = R x ln (V, P, C)final - R x ln (V, P, C)initial
Recall : V  T
or
V = const × T
1
V
or PV = const
P
V  [conc] or V  n or V = const  [conc] or V = const  n
ΔSsys = R  ln
ΔSsys
2/6/2015
Vfinal
Vinitial
1
P
C ( n)
= R  ln final = R  ln final
1
Cinitial (n)
p initial
Vfinal
p initial
Cfinal
= R  ln
= R  ln
= R  ln
Vinitial
Pfinal
Cinitial
32
Entropy

Changes in Entropy

The change in Entropy of the system (Ssys) depends
only on the difference between its final and initial values
ΔSsys = Sfinal - Sinitial

(Ssys) > 0 when its value increases during a change
Ex. Sublimation of dry ice to gaseous CO2
CO2 (s)  CO2 (g)
ΔSsys = Sfinal - Sinitial = SgaseousCO2 - SsolidCO2

(Ssys) < 0 when its value decreases during a change
Ex. Condensation of Water
H 2O(g)  H 2O(l)
ΔSsys = Sliquid H 2O - Sgaseous H2O
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33
Entropy

Entropy Changes based on Heat Changes
 The 2nd Law of Thermodynamics states that the change
in Entropy for a gas expanding into a vacuum is related
to the heat absorbed (qrev) and the temperature (T) at
which the exchange occurs
qrev
ΔSsys =



2/6/2015
T
Qrev refers to a “Reversible” process where the
expansion of the gas can be reversed by the application
of pressure (work, PV)
The heat absorbed by the expanding gas increases the
dispersal of energy in the system, increasing the
Entropy
If the change in Entropy, Ssys, is greater than the heat
absorbed divided by the absolute temperature, the
process occurs spontaneously
34
Laws of Thermodynamics

Determination of the Direction of a Spontaneous Process
Second Law Restated
All real processes occur spontaneously in the
direction that increases the Entropy of the universe
(system + surroundings)

When changes in both the system and the surroundings
occur, the universe must be considered

Some spontaneous processes end up with higher
Entropy

Other spontaneous processes end up with lower
Entropy
2/6/2015
35
Laws of Thermodynamics

The Entropy change in the system or surroundings can
be positive or negative

For a spontaneous process, the “sum” of the Entropy
changes must be positive

If the Entropy of the system decreases, the Entropy of
the surroundings must increase, making the net
increase to the universe positive
ΔSuniv = ΔSsys + ΔSsurr > 0
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(spontaneous process )
36
Laws of Thermodynamics

The 3rd Law of Thermodynamics:

Entropy & Enthalpy are both “state” functions

Absolute Enthalpies cannot be determined, only
changes
i.e., No reference point

Absolute Entropy of a substance provides a reference
point and can be determined
2/6/2015

The Entropy of a system approaches a constant value
as the temperature approaches zero

The entropy of a system at absolute zero is typically
zero, and in all cases is determined only by the
number of different ground states it has
37
Laws of Thermodynamics

Specifically, the entropy of a pure crystalline
substance (perfect order, where all particles are
perfectly aligned with no defects of any kind) at
absolute zero temperature is zero

This statement holds true if the perfect crystal has
only one state with minimum energy
Ssys = 0 at 0oK
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38
Entropy

Entropy values for substances are compared to “standard”
states


2/6/2015
Standard States

Gases
– 1 atmosphere (atm)

Concentrations – Molarity (M)

Solids
– Pure Substance
Standard Molar Entropy

So (Units – J/molK @ 298oK)

Values available in Reference Tables (Appendix “B”)
39
Entropy

Predicting Relative So Values of a System

2/6/2015
Temperature Changes

So increases as temperature increases

Temperature increases as “heat” is absorbed
(q > 0)

As temperature increases, the Kinetic
Energies of gases, liquids, and solids increase
and are dispersed over larger areas
increasing the number of microstates
available, which increases Entropy
40
Entropy
 At any T > 0o K, each particle
moves about its lattice
position
 As temperature increases
through the addition of
“heat”, movement is greater
 Total energy is increased
giving particles greater
freedom of movement
 Energy is more dispersed
 Entropy is increased
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41
Entropy

Predicting Relative So Values of a System (Con’t)
 Physical States and Phase Changes
 So increases for a substance as it changes from a
solid to a liquid to a gas
 Heat must be absorbed (q>0) for a change in phase
to occur
 Increase in Entropy from liquid to gas is much larger
than from solid to liquid Svapo >> Sfuso
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42
Entropy

Predicting Relative So Values of a System (Con’t)

Dissolving a solid or liquid

Entropy of a dissolved solid or liquid is greater than
the Entropy of the “pure” solute

As the crystals breakdown, the ions have increased
freedom of movement

Particle energy is more dispersed into more
“microstates”
Entropy is increased
2/6/2015

Entropy increase is “greater” for ionic solutes than
“molecular” solutes – more particles are produced

The slight increase in Entropy for “molecular” solutes
in solution arises from the separation of molecules
from one another when mixed with the solvent
43
Entropy

Predicting Relative So Values of a System (Con’t)

Dissolving a Gas

Gases have considerable freedom of motion and
highly dispersed energy in the gaseous state

Dissolving a gas in a solvent results in diminished
freedom of motion
Entropy is “Decreased”

Mixing (dissolving) a gas in another gas

Molecules separate and mix increasing microstates
and dispersion of energy
Entropy “Increases”
2/6/2015
44
Entropy

Predicting Relative So Values of a System (Con’t)

Atomic Size

2/6/2015
Multiple substances in a given phase will have
different Entropies based on Atomic Size and
Molecular Complexity

Down a “Periodic” group energy levels become
“closer” together as the atoms get “Heavier”

No. of microstates and molar Entropy increase
45
Entropy

Predicting Relative So Values of a System (Con’t)

Molecular Complexity

Allotropes – Elements that occur in different forms
have higher Entropy in the form that allows more
freedom of motion
Ex. Diamond vs Graphite
Diamond bonds extend the 3 dimensions, allowing
limited movement – lower Entropy
Graphite bonds extend only within twodimensional sheets, which move relatively easy to
each other – higher Entropy
2/6/2015
46
Entropy

Predicting Relative So Values of a System (Con’t)

Molecular Complexity (Con’t)

2/6/2015
Compounds

Entropy increases as the number of atoms (or
ions) in a formula unit of a molecule increases

The trend is based on types of movement and the
number of microstates possible

NO (Nitrous Oxide) in the chart below can vibrate
only toward and away from each other

The 3 atoms of the NO2 molecule have more
virbrational motions
47
Entropy

Predicting Relative So Values of a System (Con’t)
 Molecular Complexity (Con’t)
 Compounds of large molecules
2/6/2015

A long organic hydrocarbon chain can rotate and
vibrate in more ways than a short chain

Entropy increase with “Chain Length”

A ring compound with the same molecular formula
as a corresponding chain compound has lower
Entropy because a ring structure inhibits freedom
of motion
cyclopentane (C5H10) vs pentene (C5H10)
Scyclopentane <
Spentene
48
Entropy

Predicting Relative So Values of a System (Con’t)

Physical State vs Molecular Complexity
When gases are compared to liquids:
The effect of physical state (g, l, s) usually
dominates that of molecular complexity, i.e.,
the No. atoms in a formula unit or chain
length
2/6/2015
49
Practice Problem
Choose the member with the higher Entropy in each
of the following pairs, and justify the choice

1 mol of SO2(g) or 1 mol SO3(g)
SO3 has more types of atoms in the same
state, i.e., more types of motion available
More Entropy

1 mol CO2(s) or 1 mol CO2(g)
Entropy increases in the sequence:
2/6/2015
s < l < g
50
Practice Problem

3 mol of O2(g) or 2 mol of O3(g)
The two samples contain the same number of
oxygen atoms (6), but different numbers of
molecules
O3 is more complex, but the greater number of
molecules of O2 dominates – more moles of
particles produces more microstates
2/6/2015
51
Practice Problem (Con’t)

1 mol of KBr(s) or 1 mol KBr(aq)
Both molecules have the same number of ions (2)
Motion in a crystal is more restricted and energy is
less dispersed
 KBr(aq) has higher Entropy

Sea Water at 2oC or at 23oC
Entropy increases with increasing temperature
Seawater at 23oC has higher Entropy

1 mol CF4(g) or 1 mol CCl4(g)
For similar compounds Entropy increases with
increasing molar moss
2/6/2015
S(CF4)(g)
<
S(CCl4)(g)
52
Practice Problem

Predict the sign of S for each process:
 Alcohol Evaporates
ΔSsys positive, the process described is liquid alcohol
becoming gaseous alcohol
The gas molecules have greater Entropy than the liquid
molecules
 A solid explosive converts to a gas
ΔSsys positive, the process described is a change from
solid to gas, an increase in possible energy states for the
system
 Perfume vapors diffuse through a room
ΔSsys positive, the perfume molecules have more possible
locations in the larger volume of the room than inside the
bottle
A system that has more possible arrangements has
greater Entropy
2/6/2015
53
Practice Problem
Without using Appendix B predict the sign of S for:
2K(s) + F2(g) → 2KF(s)
ΔSsys negative – reaction involves a gaseous reactant
and no gaseous products, so Entropy decreases
The number of particles also decreases, indicating a
decrease in Entropy
NH3(g) + HBr(g) → NH4Br(s)
ΔSsys negative – gaseous reactants form solid product
and number of particles decreases, so Entropy
decreases
NaClO3(s) → Na+(aq) + ClO3ΔSsys positive – when a solid salt dissolves in water,
Entropy generally increases
2/6/2015
54
Entropy

Calculating Change in Entropy

Gases


2/6/2015
The sign of the Standard Entropy of Reaction (Sorxn)
of a reaction involving gases can often be predicted
when the reaction involves a change in the number
of moles that occurs and all the reactants and
products are in their “standard” states
Gases have great freedom of motion and high molar
Entropies

If the number of moles of gas increases, Sorxn is
usually positive

If the number of moles of gas decreases, Sorxn is
usually negative
o
ΔSrxn
=
o
 m Sproducts
-
o
 n Sreactants
55
Practice Problem
Calculate Sorxn for the combustion of 1 mol of Propane
at 25oC C H (g) + 5O (g)  3CO (g) + 4H O(l)
3
8
2
2
2
ΔSorxn =  mSo products -  nSoreactants
Calculate Δn to determine if the change in moles from
reactant to product indicates increased or decreased Entropy
Δn = 3 mol (product) - 6 mol (reactant) = - 3
Entropy should decrease (ΔSorxn < 0)
Calculate Sorxn , using values from Appendix B
ΔSorxn = [(3 mol CO 2 )(So of CO 2 ) + (4 mol H 2O)(So of H 2O)]
- [(1 mol C3 H 8 )(So of C3H 8 ) + (5 mol O2 )(So of O 2 )]
ΔSorxn = [(3 mol )(213.7 J / mol  K) + (4 mol)(69.9 J / mol  K)]
- [(1 mol)(269.9 J / mol  K) + (5 mol)(205.0 J / mol  K)]
2/6/2015
ΔSorxn = - 374 J / K < 0
56
Entropy

Entropy Changes in the Surroundings

2nd Law – For a spontaneous process, a decrease in
Entropy in the system, Ssys, can only occur if there is
an increase in Entropy in the surroundings, Ssys
ΔSuniv = ΔSsys + ΔSsur > 0
(spontaneous process)

Essential role of the surroundings is to either add heat
to the system or remove heat from the system –
surroundings act as a “Heat Sink”

Surroundings are generally considered so large that its
temperature essentially remains constant even though
its Entropy will change through the loss or gain of heat
2/6/2015
57
Thermodynamics

Surroundings participate in two (2) types of Enthalpy
changes

Exothermic Change

Heat lost by system is gained by surroundings

Increased freedom of motion from temperature
increase in surroundings leads to Entropy increase
Exothermic Change :

qsur > 0
ΔSsur > 0
Endothermic Change

Heat gained by system is lost by surroundings

Heat lost reduces freedom of motion in surroundings,
energy dispersal is less, and Entropy decreases
Endothermic Change :
2/6/2015
qsys < 0
qsys > 0
qsur < 0
ΔSsur < 0
58
Thermodynamics
Temperature of the Surroundings
 At Lower Temperatures
 Little random motion
 Little energy
 Fewer energy levels
 Fewer microstates
 Transfer of heat from system has larger effect on
how much energy is dispersed
 At Higher Temperatures
 Surroundings already have relatively large quantity of
energy dispersal
 More energy levels
 More available microstates
 Transfer of heat from system has much smaller effect
on the total dispersion of energy
59
2/6/2015

Thermodynamics

Temperature of the Surroundings

The change in Entropy of the surroundings is “greater”
when heat is added at lower temperatures

Recall 2nd Law – The change in Entropy of the
surroundings is directly related to an “opposite” change
in the heat (q) of the system and “inversely” related to
the temperature at which the heat is transferred
ΔSsur = 
T
Recall that for a process at “Constant Pressure”, the
heat (qp) = H
ΔSsur = 2/6/2015
qsys
H sys
T
60
Practice Problem
How does the Entropy of the surroundings change during an
Exothermic reaction?
Ans: In an Exothermic process, the system releases heat
to its surroundings. The Entropy of the surroundings
increases because the temperature of the
surroundings increases (ΔSsur > 0)
How does the Entropy of the surroundings change during an
Endothermic reaction?
Ans: In an Endothermic process, the system absorbs heat
from the surroundings and the surroundings become
cooler. Thus, the Entropy of the surroundings
decreases (ΔSsur < 0)
2/6/2015
61
Practice Problem
What is the Entropy of a perfect crystal at 0 oK
Ans: According to the Third Law, the Entropy is zero
Does the Entropy increase or decrease as the temperature rises?
Ans: Entropy will increase with temperature
Why is ∆Hof = 0 but So > 0 for an element?
Ans: The third law states that the Entropy of a pure, perfectly
crystalline element or compound may be taken as zero at
zero Kelvin. Since the standard state temperature is 25°C
and Entropy increases with temperature, S° must be
greater than zero for an element in its standard state
Why does Appendix B list ∆Hof values but not ∆Sof
Ans: Since Entropy values have a reference point (0 Entropy at
0 K), actual Entropy values can be determined, not just
Entropy changes
2/6/2015
62
Practice Problem
Predict the spontaneity of the following:
 Water evaporates from a puddle
Spontaneous, evaporation occurs because a few of the
liquid molecules have enough energy to break away
from the intermolecular forces of the other liquid
molecules and move spontaneously into the gas phase
 A lion chases an antelope
Spontaneous, a lion spontaneously chases an antelope
without added force
 An isotope undergoes radioactive decay
Spontaneous, an unstable substance decays
spontaneously to a more stable substance
2/6/2015
63
Practice Problem
 Earth moves around the sun
Spontaneous
 A boulder rolls up a hill
The movement of a boulder against gravity is
nonspontaneous
 Sodium metal and Chlorine gas form solid Sodium Chloride
The reaction of an active metal (Sodium) with an active
nonmetal (Chlorine) is spontaneous
 Methane burns in air
Spontaneous, with a small amount of energy input,
Methane will continue to burn without additional energy
(the reaction itself provides the necessary energy) until it
is used up
2/6/2015
64
Practice Problem
 A teaspoon of sugar dissolve in hot coffee
Spontaneous, the dissolved sugar molecules have more
states they can occupy than the crystalline sugar, so the
reaction proceeds in the direction of dissolution
 A soft-boiled egg become raw
Not spontaneous, a cooked egg will not become raw
again, no matter how long it sits or how many times it is
mixed
 Water decomposes to H2 & O2 at 298 oK
Water is a very stable compound; its decomposition at
298 K and 1 atm is not spontaneous
2/6/2015
65
Practice Problem
Calculate Suniv and state whether the reaction occurs
spontaneously at 298oK for the following reaction
N2 (g) + 3H2 (g)
2NH3 (g)
o
ΔSsys
= - 197 J / K
For the reaction to react spontaneously :
ΔSuniv > 0 and
o
To find ΔSsurr , determine ΔHsys
o
ΔHosys = ΔHrxn
Ssurr > + 197 J / K
(From ΔHrxn values in Appendix B)
o
o
o
Hrxn
= ΔHprod
- ΔHreact
ΔHosys = [(2 mol NH3 )(-45.9 kJ / mol)]-[(3 mol H2 )(0 kJ / mol) + (1 mol N2 (0 kJ / mol)]
ΔSsur
ΔHosys = - 91.8 kJ
1000 J
-91.8
kJ
x
ΔH sys
kJ = 308 J / K
= =T
298o K
ΔSuniv = ΔSsys + ΔSsur = -197 J / K + 308 J / K = 111 J / K
ΔSuniv > 0
2/6/2015
 Reaction proceeds spontaneously at 298o K
66
Thermodynamics

Entropy Change and the Equilibrium State

For a process “spontaneously” approaching equilibrium,
the change in Entropy is positive
ΔSuniv > 0

At equilibrium, there is no net change in the flow or
energy to either the system or the surroundings
ΔSuniv = 0

Any change in Entropy in the system is exactly balanced
by an opposite Entropy change in the surroundings
At Equilibrium : ΔSuniv = ΔSsys + ΔSsur = 0
ΔSsys = - Ssur
2/6/2015
67
Practice Problem
Calculate Suniv for the vaporization of 1 mol water at 100oC
(373oK)
H2O(l)
H2O(g)
@ 373o K)
Entropy of System is increasing as Heat is absorbed from
surroundings changing the liquid to a gas
Compute Ssys from Standard Molar Entropies (from Appendix B)
ΔSosys =
o
 mSprod
-
o
 nSreac
= Soof H 2 0(g, 373o K) - Soof H2O(l, 373o K)
ΔSosys = 195.9 J / K - 86.8 J / K = 109.1 J / K
Compute Ssurr from Hosys and Temperature (T = 373oK)
ΔH osys
ΔSsur = T
where ΔHosys = ΔHovap at 373o K = 40.7 kJ / mol = 40.7 x 103 J / mol
ΔSsur = -
2/6/2015
ΔHosys
T
40.7 x 103 J
= = -109 J / K
373 K
ΔSuniv = ΔSsys + ΔSsurr = 109.1 - 109  0
68
Thermodynamics

Summary – Spontaneous Exothermic &
Endothermic Reactions
 Exothermic Reaction (Hsys < 0)
 Heat, released from system, is absorbed by
surroundings
 Increased freedom of motion and energy
dispersal in surroundings (Ssurr > 0)
 Ex. Exothermic where Entropy change:
(Ssys) > 0
C6 H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H 2O(g) + Heat
6 moles gas yields 12 moles gas and heat
o
ΔSrxn
=
ΔSsys > 0 ΔSsur > 0
2/6/2015
o
 m Sproducts
-
o
 n Sreactants
> 0
then ΔSuniv = ΔSsys + ΔSsur > 0
69
Thermodynamics

Summary – Spontaneous Exothermic &
Endothermic Reactions (Con’t)

Exothermic Reaction (Hsys < 0)

Ex. Exothermic where Entropy change (Ssys) < 0

Entropy in surroundings must increase even more
(Ssurr > > 0) to make the total S positive
CaO(s) + CO2 (g)  CaCO3 + Heat
Entropy of system decreases because :
The amount (mol) of gas decreases
Heat released increases Entropy of surroundings even more
ΔSsys < 0 ΔSsur > > 0
2/6/2015
then ΔSuniv = ΔSsys + ΔSsur > 0
70
Thermodynamics
Summary – Spontaneous Exothermic &
Endothermic Reactions (Con’t)
 Endothermic Reaction (Hsys > 0)
 Heat lost by surroundings decreases the molecular
freedom of motion and dispersal of energy
 Entropy of surroundings decreases (Ssurr) < 0
 Only way an Endothermic reaction can occur
spontaneously is if (Ssys) > 0 and large enough to
outweigh the negative Ssurr
 Ex. Solution Process for many ionic compounds
 Heat is absorbed to form solution
 Entropy of surroundings decreases
 However, when crystalline solids become freemoving ions, the Entropy increase in the system is
quite large (Ssys) > > 0
 Ssys increase far outweighs negative Ssurr
2/6/2015

71
Practice Problem
Acetone, CH3COCH3, is a volatile liquid solvent (it is used in
nail polish, for example). The standard Enthalpy of
formation of the liquid at 25 oC is -247.6 kJ/mol; the same
quantity for the vapor is -216.6 kJ/mol. What is the Entropy
change when 1.00 mol of liquid acetone vaporizes at 25 oC?
CH3COCH 3 (l)
ΔHosys =
o
 mHprod
-
o
CH 3COCH 3 (g)
 nHreac
@ 298o K)
= Hoof Acetone(g, 298o K) - Hoof Acetone(l, 298o K)
ΔHosys = - 216.6 J - (-247.6 J) = 31.0 J
ΔSsur = -
ΔH osys
T
= -
31.0 J
= - 0.104 J / K
o
298 K
Endothermic reaction - Energy from surroundings is input to
system to vaporize acetone (∆Hosys is positive)
Energy (temperature) of surroundings is decreased, decreasing
Entropy
2/6/2015
72
Gibbs Free Energy

Entropy, Free Energy and Work

Gibbs Free Energy (G)

Using Hsys & Ssurr , it can be predicted whether
a reaction will be “Spontaneous” at a particular
temperature

J. Willard Gibbs developed a single criterion for
spontaneity

The Gibbs “Free Energy” (G) is a function that
combines the system’s Enthalpy (H) and
Entropy (S)
G = H - TS
ΔG sys  H sys - TΔSsys
2/6/2015
73
Gibbs Free Energy

Gibbs Free Energy Change and Reaction Spontaneity
The Free Energy Change (G) is a measure of the
spontaneity of a process and of the useful energy
available from it
At Equilibrium : ΔSuniv = ΔSsys + ΔSsur = 0
ΔSsur = -
H sys
T
ΔSuniv = ΔSsys + ΔSsur = ΔSsys -
TΔS univ  TΔSsys - H sys
ΔHsys
T
-TΔSuniv = ΔHsys - TΔSsys
Recall :
ΔGosys = ΔHosys - TΔSosys
2/6/2015

"Standard Free Energy Change"
-TΔSuniv = ΔG sys = ΔHsys - TΔSsys
74
Gibbs Free Energy

Gibbs Free Energy Change and Reaction Spontaneity
-TΔS univ  H sys - TΔSsys = G sys

The sign of G tells if a reaction is spontaneous

From the 2nd Law of Thermodynamics


Suniv > 0 for spontaneous reaction

Suniv < 0 for nonspontaneous reaction

Suniv = 0 for process in “Equilibrium”
Absolute Temperature (K) is “always positive”
TΔSuniv > 0
2/6/2015
or - TΔSuniv < 0 for spontaneous process

G < 0 for a spontaneous process

G > 0 for a nonspontaneous process

G = 0 for a process in equilibrium
75
Practice Problem
Calculate Gsyso (Grxno) at 25oC for the following reaction
Δ
4KClO3 (s) 
 3KClO4 (s) + KCL(s)
Calculate Hsyso from Hfo values from tables
ΔHosys = ΔHorxn =
o
 m ΔHf(products)
-
o
 n ΔHf(reactants)
ΔHosys = [(3 mol KClO4 )(ΔHfo of KClO4 ) + (1 mol KCl)(ΔHfo of KCl)]
-[(4 mol KClO 3 )(ΔHfo of KClO 3 )]
ΔHosys = [(3 mol)(- 432.8 kJ / mol) + (1 mol)( - 436.7 kJ / mol)]
- [(4 mol)( - 397.7 kJ / mol)]
ΔHosys = -144 kJ
2/6/2015
Con’t
76
Practice Problem (Con’t)
Calculate Ssyso from So values from tables
ΔSosys = ΔSorxn =
o
 mΔSproducts
-
o
 nΔSreactants
ΔSosys = ΔSrxn = [(3 mol KClO4 )(So of KClO4 ) + (1 mol KCl)(So of KCl)]
- [(4 mol KClO3 )(So of KClO3 )]
o
ΔSosys = ΔSrxn
= [(3 mol)(151.0 J / mol  K) + (1 mol)(82.6 J / mol  K )]
- [(4 mol)(143.1 J / mol  K)]
o
ΔSosys = ΔSrxn
= - 36.8 J / K
Calculate Gsyso at 298oK
o
ΔG osys  H sys
- TΔSosys
J
1 kJ 

ΔGosys = - 144 kJ - (298 K)(-36.8

)  = - 133 kJ
K
1000 J 

2/6/2015
77
Gibbs Free Energy

Standard Free Energy of Formation (Gfo)

Gfo is the free energy change that occurs when 1
mole of compound is made from its “elements” and
all of the components are in their “standard” states
ΔGorxn =

o
 mΔG f(products)
-
o
 nΔG f(reactants)
Gfo values have properties similar to Hfo values

Gfo of an element in its standard form is “zero”

An equation coefficient (m or n) multiplies Gfo by
that number

Reversing a reaction changes the sign of Gfo

Gfo values are obtained from tables
2/6/2015
78
Thermodynamics

G and the Work (w) a System Can Do
 For a Spontaneous process (G < 0) at constant
Temperature (T) and Pressure (P), G is the
“Maximum” of useful work obtainable from the
system as the process takes place
ΔG = Wmax



2/6/2015
For a Nonspontaneous process (G > 0) at
constant T & P, G is the “Minimum” work that
must be done to the system to make the process
take place
In any process, neither the “maximum” or the
“minimum” work is achieved because some “Heat”
is lost
A reaction at equilibrium, which includes phase
changes (G = 0), can no longer do “any work”
79
Thermodynamics

The Effect of Temperature on Reaction Spontaneity

When the signs of H & S are the same, some
reactions that are non-spontaneous at one temperature
become spontaneous at another, and vice versa

The temperature at which a reaction becomes
spontaneous is the temperature at which a
“Positive” G switches to a “Negative” G

This occurs because of the changing magnitude of the
-T S term


2/6/2015
This cross-over temperature (reaction at equilibrium)
occurs when G = 0
ΔG = 0 = ΔH - TΔS
Thus:
ΔH = TΔS
ΔH
T =
ΔS
80
Thermodynamics

The Effect of Temperature on Reaction
Spontaneity
ΔG sys  H sys - TΔSsys

2/6/2015
Reactions Independent of Temperature
 Spontaneous Reaction at all Temperatures
 H < 0 (Exothermic)
S > 0
 (- TS) term is always negative
 G is always “negative”
 Nonspontaneous Reaction at all
Temperatures
 H > 0 (Endothermic)
S < 0
 Both oppose spontaneity
 - TS is positive
 G is always positive
81
Practice Problem
Predict spontaneity of the following reactions
2H 2O2 (l)  2H 2O(l) + O2 (g)
ΔH = - 196 kJ
ΔH = < 0
ΔS = 125 J / K
ΔS > 0
- TΔS < 0
ΔG = ΔH - TΔS < 0
Reaction is spontaneous at all temperatures
3O2 (g)  2O3 (g)
ΔH = 286 kJ
ΔH = > 0
ΔS = - 137 J / K
ΔS < 0
- TΔS > 0
ΔG = ΔH - TΔS > 0
Reaction is not spontaneous at any temperature
2/6/2015
82
Thermondynamics

Temperature Dependent Reactions
When H & S have the same sign, the
relative magnitudes of the –TS and H
terms determine the sign of G

2/6/2015
Reaction is spontaneous at high Temperatures

H > 0
S > 0

S favors spontaneity (-TS) < 0)

H does not favor spontaneity

Spontaneity will occur only when -TS
(generally high temperature) is large enough
to make G negative
83
Practice Problem

Predict spontaneity of the following reaction
2N2O(g)
+
∆H = 197.1 kJ
O2(g)  4NO(g)
and
∆S = 198.2 J/K
With a Positive ∆H, the reaction will be
spontaneous only when - T∆S is large enough to
make ∆G negative
ΔG = ΔH - TΔS < 0
This would occur at “Higher” temperatures
The oxidation of N2O occurs spontaneously at
2/6/2015
T > 994 K
84
Thermodynamics

Temperature Dependent Reactions (Con’t)
When H & S have the same sign, the
relative magnitudes of the (– TS) and H
terms determine the sign of G

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Reaction is spontaneous at lower Temperatures

H < 0
S < 0

H favors spontaneity

S does not favor spontaneity (- TS) > 0)

G will only be negative when -TS is smaller
the H term, usually at a lower temperature
85
Practice Problem

Predict spontaneity of the following reaction
4Fe(s) + 3O2(g)  2Fe2O3(s)
∆H = -1651 kJ
and ∆S = -549.4 J/K
∆H favors spontaneity, but ∆S does not
With a negative ∆H, the reaction will occur spontaneously
only if the -T ∆S term is smaller than the ∆H term.
ΔG = ΔH - TΔS > 0
This happens only at lower temperatures
The production of Iron(III) oxide occurs spontaneously at
any T < 3005 K
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86
Thermodynamics

Summary – Reaction Spontaneity and the Sign of
ΔH, ΔS, and ΔG
o
ΔG osys  H sys
- TΔSosys
ΔG sys  H sys - TΔSsys
Recall : At G = 0
ΔH
T=
ΔS
∆H
∆S
-T∆S
∆G
Description
—
+
—
—
Spontaneous at all Temperatures
+
—
+
+
Nonspontaneous at all Temperatures
+
+
—
—
+
Spontaneous at Higher Temperature
Nonspontaneous at Lower Temperatures
—
—
+
—
+
Spontaneous at Lower Temperatures
Nonspontaneous at Higher Temperatures
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87
Practice Problem
Predict the spontaneity of the following reaction
2N 2O(g) + O2 (g)  4NO(g)
ΔH = 197.1 kJ
ΔS = 198.2 J / K
ΔH = > 0 ΔS > 0
- TΔS = ?
ΔG = ΔH - TΔS = ?
Reaction will be spontaneous when Temperature
is high enough to make ΔG negative
4Fe(s) + 3O2 (g)  2Fe2O3 (s)
ΔH = - 1651 kJ
ΔS = - 549.4 J / K
ΔH = < 0 ΔS < 0
- TΔS = ?
ΔG = ΔH - TΔS = ?
With the negative H, the reaction will be spontaneous
only if - TS is smaller than the H to make G negative
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This would have to occur at lower temperatures
88
Thermodynamics

Free Energy, Equilibrium, and Reaction Direction



2/6/2015
From Chapter 17

Q < K (Q/K < 1) – reaction proceeds “Right”

Q > K (Q/K > 1) – reaction proceeds “Left”

Q = K (Q/K = 1) – Reaction has reached “Equilibrium”
Energy & Spontaneity

Exothermic (H < 0) – reaction proceeds “Right”

Endothermic (H > 0) – reaction proceeds “Left”
Free Energy & Spontaneity

G < 0 for a spontaneous process

G > 0 for a nonspontaneous process

G = 0 for a process in equilibrium
89
Thermodynamics

Relationship between Q/K and G

If Q/K < 1, then ln(Q/K) < 0 and if G < 0
Then: Reaction is Exothermic and spontaneous

If Q/K > 1, then ln(Q/K) > 0 and if G > 0
Then: Reaction is Endothermic and nonspontaneous

If Q/K = 1, then ln(Q/K) = 0 and if G = 0
Then: Reaction has reached equilibrium

In each case the signs of ln(Q/K) and G are the same
for a given reaction direction

Gibbs noted that ln(Q/K) and G are proportional to
each other and are related (made equal) by the
proportionality constant “RT”
Q
ΔG = RT ln
= RT ln Q - RT ln K
K
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90
Thermodynamics

Recall: Q represents the concentrations (or pressures) of
systems components at any time during the reaction,
whereas, K represents the concentrations when the
reaction has reached “equilibrium”

G depends on how the Q ratio of the concentrations
differs from the equilibrium ratio, K

Expressing G when “Q” is at standard state conditions


All concentrations are = 1 M (pressures = 1 atm)
o
ΔG
= RT ln 1 - RT ln K
Q=1
ΔGo = RT* 0 - RT ln K = - RT ln K


2/6/2015
Standard Free Energy (Go) can be computed from the
Equilibrium constant (K)
Logarithmic relationship means a “small” change in Go
has a large effect on the value of K
91
Thermodynamics

For expressing the free energy for nonstandard initial
conditions

Substitute Go equation into G equation
ΔG = RT ln Q - RT ln K = RT ln Q - (-ΔG o )
ΔG = ΔGo  RT ln Q
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92
Practice Problem
If the partial pressures of all species in the reaction below
are 0.50 atm, what is G (kJ) for the reaction at 25oC?
Kp = 0.16
PCl 3 (g) + Cl 2 (g)  PCl 5 (g)
Qp =
PPCl
PCl PCl
3
0.50
0.50
=
=
= 2.0
(0.50)(0.50)
0.25
5
2
Q
 2.0 
ΔG = RT ln
= (8.314 J / mol  K)(298 K)  ln 

K
 0.16 
1 kJ
J
ΔG = 6.3  103 1000 J = 6.3 kJ / mol
mol
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93
Practice Problem
Calculate the value of the thermodynamic equilibrium
constant (K) at 25 oC for the reaction given below:
The values of standard free energy of formation of the
substances in kJ/mol at 25 oC are NO2, 51.30; N2O4, 97.82)
N 2O4 (g)  2NO2 (g)
ΔGorxn =
 mΔG f(products)
o
ΔG orxn =
 mΔG f(NO )
2
o
-
-
o
 nΔG f(reactants)
o
 nΔG f(N O )
2 4
= (2  51.30) - (97.82)
ΔGorxn = 4.78 kJ
ΔGo = - RT ln K
1000 J
/ mol
ΔG o
kJ
ln K = = = - 1.93
J
RT
8.314
 298K
mol • K
4.78 kJ
K = e-1.93 = 0.15
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94
Practice Problem
Obtain the Kp at 35oC for the reaction in the previous problem
The standard enthalpies of formation of the substances in kJ/mol
at 25oC are:
N2O4 9.16 J/mol-K
NO2 33.2 J/mol-K
The standard molar entropies at 25 oC are:
N2O4 304.3 J/mol-K
NO2 239.9 J/mol-K
N 2O4 (g)  2NO2 (g)
ΔHosys =
o
 mHprod
-
o
 nHreac
= 2Hoof NO2 ) - Hoof N 2O4 )
ΔHosys = (2  33.2) - (9.16) = 57.24 kJ / mol
ΔSorxn =
o
 m Sproducts
-
o
 n Sreactants
ΔSosys = ΔSorxn = 2ΔSoNO - ΔSoN O
2
2
4
ΔSosys = ΔSorxn = (2  239.9) - (304.3) = 175.5 J / mol  K
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Con’t
95
Practice Problem (Con’t)
o
ΔG osys  H sys
- TΔSosys
ΔGosys = 57.24 kJ / mol - 308 K  175.5 J / mol • K
kJ
ΔG osys = 57.24
1000 J
kJ - 308 K x 175.5 J / mol • K = 3.19 x 103 J / mol
mol
ΔGo = - RT ln K
ΔG o
ln K = =RT
3.19 x 103 J / mol
= - 1.25
J
o
8.314
x
308
K
o
mol • K
K = e-1.25 = 0.29
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96
Summary

Laws of Thermodynamics

1st Law – The change in the Internal Energy of a closed
system (E) is equal to the amount of Heat (q) supplied
to the system, minus the amount of Work (w)
performed by the system on its surroundings
ΔE = q + w = q - P V

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Limitations of the 1st Law of Thermodynamics

The 1st Law accounts for the energy involved in a
chemical process (reaction)

The first Law, however, does not account for the
“direction” of the change in energy
97
Summary

2nd Law of Thermodynamics

The total Entropy of a system and its surroundings
always increases for a “Spontaneous” process

All real processes occur spontaneously in the
direction that increases the Entropy of the universe
(system + surroundings)

The 2nd Law of Thermodynamics states that the
change in Entropy for a gas expanding into a vacuum
is related to the heat absorbed (qrev) and the
temperature (T) at which the exchange occurs
ΔSsys
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qrev
=
T
98
Summary

2/6/2015
3rd Law of Thermodynamics

The Entropy of a system approaches a constant
value as the temperature approaches zero

The Entropy of a perfect crystal at “absolute zero” is
zero
99
Summary Equations
w  P Δ V
ΔE  qp  w
ΔE  qp  (-P ΔV)
qp  ΔE  P ΔV
H  E  PV
ΔH = qp
(at constant Pressure)
Euniv = Esys + Esurr
(q + w)sys = - (q + w)surr
ΔEsys = - ΔEsurr
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 ΔEsys + ΔEsurr = 0 = ΔEuniv
100
Summary Equations
S = k lnW
k =
R
8.31447J / mol • K
-23
=
=
1.38
x
10
J/K
23
NA
6.02214 x 10
R
S =
ln W = 1.38 x 10-23 x lnW
NA
S = R ln(V, P, C)
ΔSsys = Sfinal - Sinitial = R x ln (V, P, C)final - R x ln (V, P, C)initial
ΔSsys = R x ln
Vfinal
P
C
= R x ln final = R x ln final
Vinitial
p initial
Cinitial
ΔSsys
ΔSsurr
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ΔSsurr
qrev
=
T
qsys
= T
H sys
= T
101
Summary Equations
o
ΔSrxn
=
 m Sproducts
o
-
 n Sreactants
ΔHosys =
 mHproducts
o
-
 nHreactants
ΔGorxn =
o
 mΔG f(products)
Exothermic Change :
qsys < 0
Endothermic Change :
qsys > 0
ΔSuniv = ΔSsys + ΔSsurr > 0
o
o
-
o
 nΔG f(reactants)
qsurr > 0
ΔSsurr > 0
qsurr < 0
ΔSsurr < 0
(spontaneous process)
At Equilibrium : ΔSuniv = ΔSsys + ΔSsurr = 0
ΔSsys = - Ssurr
ΔG sys  H sys - TΔSsys
ΔG = RT ln
Q
= RT ln Q - RT log K
K
ΔG o = - RT ln K
ΔG = ΔG o + RT ln Q
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(Q = 1 at standard state)
(at any non - standard state)
102
Summary Equat
Free Energy, Equilibrium, and Reaction Direction

Q < K (Q/K < 1) – reaction proceeds
“Right”

Q > K (Q/K > 1) – reaction proceeds
“Left”

Q = K (Q/K = 1) – reaction has reached “Equilibrium”
Energy & Spontaneity


Exothermic (H < 0) – reaction proceeds “Right”

Endothermic (H > 0) – reaction proceeds “Left”
Free Energy & Spontaneity


G < 0 for a spontaneous process

G > 0 for a nonspontaneous process

G = 0 for a process in equilibrium
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103