The Second Law of Thermodynamics

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Chapter 4 The Second Law of Thermodynamics The
Concepts
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OUTLINE SECTION 4.1 - Dispersal of
energy SECTION 4.2 - Entropy SECTION 4.3 -
Entropy changes for certain processes SECTION 4.4
- The Third Law of Thermodynamics SECTION 4.5 -
Helmholtz and Gibbs energies SECTION 4.6 -
Standard molar Gibbs energies
HOMEWORK EXERCISES 4.4 26 Parts(a)
only. PROBLEMS 4.5, 4.7
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The Dispersal of Energy

• From the First Law we have learnt that we may
  inter-convert heat and work, DU q w.
• It is found from experiment that
• Work can be completely transformed into heat.
• However,
• 2) Heat cannot be completely transformed into
  work.

A transformation whose only final result is to
transform into work heat extracted from a
reservoir is impossible. Kelvins Postulate
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The Dispersal of Energy
Kelvin considered the possibility of building a
perpetuum engine of the second kind by
considering the Carnot cycle.
The Reversible Carnot Cycle
Total amount of heat absorbed by the system is qh
qc. DU 0 ? w -(qh qc)
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The Dispersal of Energy
Expressions for the heat transfer involved in the
Reversible Carnot Cycle
AB Isothermal Expansion qh -w nRTh
ln(VA/VB) BC Adiabatic expansion q
0 CD Isothermal Compression qc -w nRTh
ln(VD/VC) DA Adiabatic Compression q 0
Adiabatic process TV (? -1) constant, can show
VA/VB VD/VC
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The Dispersal of Energy
The Carnot Cycle Continued. The efficiency, ?,
of a heat engine or Carnot cycle
Or
When converting heat into work, some work, qc, is
lost by the system to the surroundings and the
temperature of the system drops. Consistent
with the Second Law, a perpetuum engine of the
second kind cannot be built.
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The Dispersal of Energy
Example What is the efficiency of a reversible
cyclic engine that operates between reservoirs at
10oC and 300oC?
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The Dispersal of Energy
Refridgeration A Carnot Cycle working in the
reverse direction will extract heat, qc, from a
source at the low temperature, Tc, by absorbing
an amount of work, w.
w -(qh qc)
Heat cannot pass from a colder to a hotter body
without some other change occuring at the same
time. Clausius. In the absence of any other
change, the heat transfer during a process can
only occur in one direction.
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The Dispersal of Energy
The direction of spontaneous change A
spontaneous change is one that does not require
work to be done to bring that change about. As
we have just discovered spontaneous change occurs
in one direction. It is in the chemists
interest to find out the criterion for a
spontaneous change. We shall see that the
entropy, S, lets us assess whether one state is
accessible from another by spontaneous change.
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The Dispersal of Energy

• The direction of spontaneous change
• It is in the chemists interest to find out the
  criterion for a spontaneous change.
• Our first guess for this criterion might be that
  the system goes in the direction of lower kinetic
  and/or potential energy (ball rolling down a
  hill, exothermic reaction etc.)
• More to it. Consider some other spontaneous
  processes
• Some endothermic reactions
• Transfer of heat to cold body from a hot body in
  thermal contact.
• Expansion of gas into a vacuum, or mixing of two
  gases.
• We note that
• Spontaneous changes are accompanied by an
  increase in randomness or chaos in the system.

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Entropy, S

• Entropy, S, is a measure of degree of chaos.
• It is a state function i.e.
• For a change of state A ? B DS SB SA
• For our system

To move forward we need to quantify entropy.
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(From Chapter 2).
Types of changes If the system is kept
continually in a state of equilibrium while a
change is carried out then that process is called
REVERSIBLE. Example Compressing a gas in a
cylinder (the system) by pushing in the piston
very slowly. Truly reversible processes are
ideal they are carried out infinitely slowly and
will be capable of doing the maximum amount of
work. If the piston were pushed in fast the gas
near the piston would heat up and/or currents
would form and the system would no longer be in
equilibrium. This would be an IRREVERSIBLE
process. All spontaneous processes are
irreversible.
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Entropy, S
Recall from the reversible Carnot cycle the
efficiency of an engine
So
Thus when we go round the reversible cycle
The Carnot cycle is a relatively simple idealized
cycle. The whole treatment can be expanded to
treat any reversible cycle to give
This is characteristic for a state function. We
define ENTROPY by
For measureable changes we use
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Example 1 In an isolated system at 273 K and 10
atm, 5 moles of a perfect gas expands reversibly
to a final pressure of 1 atm. Keeping the gas in
thermal contact with a heat reservoir means the
process is performed isothermally. Calculate (i)
the entropy change of the gas, (ii) the entropy
change of the surroundings and (iii) the total
entropy change.
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Example 2 Following Example 1 find the three
entropy changes if the expansion is carried out
irreversibly against constant external pressure
of 1 atm.
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Entropy, S
Examples 1 and 2 demonstrate qrev gt qirrev
We know that
Thus
Let us now consider an adiabatic expansion of gas.
Recall all spontaneous processes are irreversible
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Entropy, S
For any spontaneous change
This is the Clausius Inequality.
The Second Law of Thermodynamics The entropy of
the universe strives toward a maximum
(Clausius) Or The entropy of an isolated system
increases in the course of a spontaneous change
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Combining the first and second laws
First law dU dq dw
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Summary of Statements about Entropy and Entropy
Change

• A positive value of DS implies an increase in
  randomness. It also implies that there is less
  heat available to do useful work. A positive DS
  accompanies all spontaneous processes.
• DS is defined as , or simply
• DS is greater than (see examples 1
  and 2).
• DS can be calculated only from reversible paths.
• The entropy, S, is the property of a system that
  is conserved in an adiabatic reversible process.
• The entropy, S, of the universe will increase in
  any process that isnt carried out reversibly.

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AN INTERPRETATION OF ENTROPY AT THE MICROSCOPIC
LEVEL S is a measure of the degree of disorder
in a system. In fact Boltzmann derived S kB
ln W where W is the number of different
microscopic arrangements that correspond to the
specified state of the macroscopic system. kB is
Boltzmann's constant R/NA. The more
"restrictions" there are on the particles, the
smaller S. e.g. 2 H has a greater entropy than
H2. e.g. Ssolid lt Sliquid lt Sgas. e.g. gas
expands from V to 2V. DS is often gt0 for
reactions where the number of particles
increases. However, it is negative for reactions
of the type AB (aq) ? A(aq) B-(aq). Why?
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Entropy Changes During Physical Changes.
Volume, Pressure and Temperature changes
Fundamental Thermodynamic Equation
dU TdS - pdV
Assuming fixed chemical content
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Example Calculate the entropy change in 5 mol
of an ideal gas when it expands isothermally and
reversibly from 100 L to 200 L
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Example Calculate the entropy change when argon
at 25o C and 1 atm in a container of volume 500
cm3 is allowed to expand to 1000 cm3 and is
simultaneously heated to 100o C. Cp,m (Ar)
20.786 J K-1 mol-1. Assume the gas behaves
perfectly.
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Entropy Changes During Physical Changes.
The entropy of a phase transition at the
transition temperature
What is a transition temperature?
Ttrs 273 K for the melting transition of
water. At the transition temp. any transfer of
heat between the system and its surroundings is
reversible. At constant pressure qp DH and so
from our definition for the entropy (DS qrev /
T ) we may write
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Entropy Changes During Physical Changes.
Entropies of vaporization of liquids
The enthalpy of vaporization of CCl4 is 30 kJ
mol-1 and the vaporization transition temp. is
349.85 K calc DvapS.
Using Troutons rule estimate the molar enthalpy
of vaporization of Br2 given that it boils at
332.35 K ?
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Entropy Changes During Physical Changes.
The measurement of entropy. We may relate the
entropy of a system at temperature T to its
entropy at T 0 K. We need to measure the
systems heat capacity Cp at different
temperatures and add in the entropy of a
transition change. The expression is
Problem difficult to measure Cp near T 0. In
practice we do an extrapolation. At low T we find
that the expression Cp aT 3 works well. (Debye
Extrapolation)
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The Third Law of Thermodynamics.
In a perfect crystal all atoms or ions are in a
regular, uniform array. Consider such a crystal
at T 0. No thermal motion perfect order
suggests that S 0. The entropy change
accompanying any physical or chemical
transformation approaches zero as the temperature
approaches zero. Nernst Heat Theorem The Third
Law of Thermodynamics The entropy of all perfect
crystalline substances is zero at T 0
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The Third Law of Thermodynamics.
Standard Reaction Entropy
Example Determine the change in entropy at 298
K for the reaction H2(g) Cl2(g) ? 2HCl(g)
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The dependence of Dr S on T
Recall Kirchoffs Law
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Self-test 4.6 Calculate the standard reaction
entropy for the combustion of methane to carbon
dioxide and liquid water at 25oC.
Negative entropy change?
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Example 1 In an isolated system at 273 K and 10
atm, 5 moles of a perfect gas expands reversibly
to a final pressure of 1 atm. Keeping the gas in
thermal contact with a heat reservoir means the
process is performed isothermally. Calculate (i)
the entropy change of the gas, (ii) the entropy
change of the surroundings and (iii) the total
entropy change.
DStotal DSsysDssurr 0
DSsys -DSsurr
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Example 2 Following Example 1 find the three
entropy changes if the expansion is carried out
irreversibly against constant external pressure
of 1 atm.
DStotal DSsysDSsurr gt 0
DSsys gt -DSsurr
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Examples 1 and 2 produced the Clausius Inequality
Reversible DStotal 0 i.e DSsys -DSsurr
Irrevesible DStotal gt 0 i.e DSsys gt -DSsurr
dSsys ? -dSsurr
Switching to infinitesimals
dSsurr -dq / T
This IS our signpost for spontaneous change. It
involves knowledge of what both the system AND
surroundings are doing
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The Helmholtz Energy and Gibbs Energy
In certain special cases we can avoid calculating
DSsurr and still maintain a criterion for
spontaneous change. We consider here one such
case, an isothermal process. We introduce two
new state functions
The Helmholtz (free) energy A U TS
The Gibbs (free) energy G H - TS
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The Helmholtz Energy and Gibbs Energy
The Helmholtz (free) energy A U TS
The Gibbs (free) energy G H - TS
For an isothermal process
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1) The Helmholtz Energy for an isothermal process
Reversible process
Irreversible process
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1) The Helmholtz Energy for an isothermal process
Reversible process
Irreversible process
Work done BY the system loss in A
(reversible) Something less than the loss
in A (irrevesible)
Therefore -DA is the MAXIMUM WORK THAT THE SYSTEM
CAN DO for a given isothermal change of state
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Example 4.5 When 1 mol of glucose is oxidized to
carbon dioxide and water at 298 K according
to C6H12O6(s) 6O2(g) ? 6CO2(g)
6H2O(l) calorimetric experiments show DrU?
-2808 kJ mol-1 and DrS? 182.4 J K-1 mol-1. How
much of this energy change can be extracted as a)
heat at constant p and b) as work?
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1) The Gibbs Energy for an isothermal and
isobaric process (const. T and p)
w? non-pV work
Reversible process
Irreversible process
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1) The Gibbs Energy for an isothermal and
isobaric process (const. T and p)
Reversible process
Irreversible process
Non-pV work done BY the system loss in G
(reversible) Something
less than the loss in G (irrevesible)
Therefore -DG is the MAXIMUM NON-pV WORK THAT THE
SYSTEM CAN DO for a given isothermal change of
state
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Example 4.6 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 37o C? The standard
entropy of reaction is 182.4 J K-1 mol-1.
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The Helmholtz Energy and Gibbs Energy
We have two new criterion for spontaneity for
isothermal processes
Const. T no work dA 0 DA 0
0 reversible lt 0 irreversible
(spontaneous)
Const. T no pV work dG 0 DG 0

• Notes
• This treatment describes A and G as criteria for
  spontaneity. However, in any change of state
  there can be be a change in A or G.
• All these parameters refer to the system

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The Helmholtz Energy and Gibbs Energy
Some notes on the Helmholtz energy The change
in the Helmholtz energy is equal to the maximum
work accompanying a process. DA DU TDS
(isothermal). If a decrease in entropy
accompanies a process i.e. DS lt 0 consequently
the max work DA is less than DU.
Some notes on the Gibbs energy Commonly used in
chemistry. At constant T and p, chemical
reactions are spontaneous in the direction of
decreasing Gibbs energy. DG DH TDS
(isothermal). Consider a spontaneous endothermic
reaction
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Relationships between state functions Be
prepared!
U and S are defined by the first and second laws
of thermodynamics, but H, A and G are defined
using U and S.
The four relationships are
We can write the fundamental thermodynamic
equation in several forms with these equations
dU TdS PdV dH TdS VdP dA -SdT -
PdV dG -SdT VdP
Gibbs Equations
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Standard Molar Gibbs Energy
Standard Gibbs Energy of Reaction
Or
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A Case Study

• CH4(g) 2O2(g) ? CO2(g) H2O(l)
• (2) CO2(g) H2O(l) ? CH4(g) 2O2(g)
• In which direction does the spontaneous change
  occur?

Our first criterion for the direction of
spontaneous change was in terms of entropy dSsys
? -dSsurr or dS ? dq/T or for an isolated system
dS ? 0 .
(self-test 4.6) The standard reaction entropy,
DrS?, for reaction (1) 243 J K-1 mol-1.
At first glance the result suggests that the
spontaneous reaction is reaction (2). This is
counter-intuitive. However, we have only
calculated the change in entropy accompanying the
reaction, i.e. we havent considered the
surroundings.
For the reaction to be spontaneous DSsys ?
-DSsurr , so DSsurr ? 243 J K-1 mol-1
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A Case Study
We have developed new criteria for spontaneous
change. These are dAT,V ? 0 and dGT,p ? 0
These criteria do not require knowledge of what
the surroundings are doing.
Self-test 4.9 Calculate the standard reaction
Gibbs Energy for the combustion of methane to
carbon dioxide and liquid water at 25oC.
CH4(g) 2O2(g) ? CO2(g) H2O(l)
From Atkins
DfG? / kJ
mol-1 CH4(g) -50.7
O2(g) 0 CO2(g) -394.4
H2O(l) -237.1
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A Case Study
DrS?, for reaction (1) 243 J K-1 mol-1
DrG?, for reaction (1) 818 J K-1 mol-1
Because DrG? 0 we know that reaction (1) is
spontaneous
Recall For the reaction to be spontaneous DSsys ?
-DSsurr , so DSsurr ? 243 J K-1 mol-1
For the reaction to overcome the decrease in
entropy it must cause an increase in entropy in
the surroundings.
How is this achieved?
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A Case Study
CH4(g) 2O2(g) ? CO2(g) H2O(l)
From Atkins
DfH? / kJ
mol-1 CH4(g) -74.81
O2(g) 0 CO2(g) -393.51
H2O(l) -285.83
DfH? -890 kJ mol-1.
DfU? DfH? - RTDng -890 kJ mol-1 -
8.314 J K-1 mol-1 . 298 K . (-2 mol)
-885 kJ mol-1
Exothermic reaction. Energy is lost from the
system to the surroundings. This energy is used
to bring about the required entropy increase in
the surroundings.
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A Case Study
So we have shown that if for a reaction DrG?
0 and DrS? 0
Then DrH? DrG?
DG DH TDS
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SUMMARY ENTROPY S defined via the Carnot cycle
dS dqrev/T. Thermodynamic efficiency of
engines, refrigerators and heat-pumps. For an
IRREVERSIBLE or SPONTANEOUS change DSuniverse gt
0. For a REVERSIBLE change DSuniverse 0. At
the microscopic level, S kB ln W. Disorder
versus order. We have seen how to work out DS
for phase changes, temperature changes and gas
expansions and mixing. The Gibbs energy G H -
TS DG (at const T and p) DH - TDS.
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SUMMARY
At constant T and p, for a SPONTANEOUS change
DGsystem lt 0. For a REVERSIBLE change DGsystem
0. At constant T and V it is useful to consider
the Helmholtz energy A U - TS, where the
criterion for spontaneity is dAsystem lt 0. For a
reversible process at constant T and p, DG
wnon-pV. For any chemical reaction (where "DfG?"
is the Gibbs energy of formation)