Download Chapter 10: Entropy and the Second Law of Thermodynamics

Entropy and the Second Law of Thermodynamics
Chapters 10
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Spontaneity
Entropy
The Second Law of Thermodynamics
The Third Law of Thermodynamics
Gibbs Free Energy
Free Energy and Chemical Reactions
Spontaneous Process
Are the following spontaneous or nonspontaneous processes
a) The reaction between NaOH(aq) and HCl(aq)
b) The decomposition of water to give H2(g) and O2(g)
c) The melting of a ice cube
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Spontaneity: Natures Arrow
• Some processes or reactions proceed in only one direction.
– Gasoline reacts spontaneously with oxygen to form carbon dioxide and
water, but water and carbon dioxide never spontaneously react to reform
gasoline.
• A spontaneous process takes place without continuous
intervention, according to thermodynamics.
– Spontaneous processes are not necessarily rapid processes.
– The combustion of diamond is thermodynamically spontaneous, but
diamonds are considered to last forever.
• Some spontaneous reactions only occur once they are initiated.
– The combustion of gasoline is a spontaneous reaction but only occurs
when the reaction is initiated with a spark.
• Nonspontaneous reactions only occur with a continual input of
energy.
Enthalpy and Spontaneity
• Exothermic reactions are generally preferred over
endothermic reactions.
– Melting ice is an endothermic process but occurs spontaneously at
room temperature.
– Enthalpy is not the exclusive determinant of spontaneity.
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Thermodynamic Example
Consider 2.4 moles of a gas contained in a 4.0 L bulb at
a constant temperature of 32°C. This bulb is connected
to an evacuated 4.0 L sealed bulb with a valve. Assume
the temperature is constant.
a) What should happen to the gas when you open the
valve?
b) Calculate H, E, q, and w for the process
c) Given your answer to part b, what is the driving force
for the process?
Entropy
• Entropy is a state function and
was first introduced in
considering the efficiency of
steam engines.
– The Carnot cycle uses a
combination of adiabatic
processes (no heat is exchanged)
and isothermal processes
(temperature is constant).
• For the Carnot cycle, the sum
of q/T around the closed path
is equal to zero and therefore
defines the state function.
– The state function is entropy.
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Spontaneous Processes and Entropy
• Thermodynamics lets us predict whether a process
will occur but gives no information about the
amount of time required for the process.
• The driving force for a spontaneous process is an
increase in the entropy of the universe.
• Entropy, S, can be viewed as a measure of
randomness, or disorder.
Definition of Entropy
• Entropy can be tentatively defined as a measurement of the
randomness, or disorder, of a system.
– For large numbers of particles, probability favors random
arrangements.
– Statistical mechanics or statistical thermodynamics provides a
quantitative basis and molecular perspective to entropy using
probability.
• For entropy, the probability for the number of ways in
which particles can achieve the same energy is used.
– The way by which a collection of particles can assume a given
energy is a microstate.
– The number of possible microstates is designated by .
– The number of microstates increases as the “randomness” of the
system increases.
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Defining Entropy
• The Maxwell-Boltzmann
distribution indicates the
overall collection of molecular
speeds but not the speed of
individual particles.
• Energy is exchanged during
molecular collisions, without
disrupting the overall
distribution of speeds.
The Boltzmann Equation for Entropy
• As the number of microstates for a system
increases, the entropy of the system increases.
This relationship is defined by the equation:
S = kb ln 
– S is entropy
– kb is the Boltzmann constant.
• The ideal gas constant per molecule = R/NA
–  is the number of microstates.
• The particular way in which particles are distributed amongst
the states. Number of microstates = 
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Probability and Spontaneous Change
• Probability has important uses in
chemistry.
• Consider 2 connected bulbs
containing two molecules.
• Each arrangement (microstate) has
an equally probability of occurring
– 25% chance of both molecules being
in the left bulb
– 50% chance of one molecules being in
the each bulb
– 25% chance of both molecules being
in the right bulb
Possible arrangements (states)
of four molecules in a twobulbed flask.
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Chance of an Arrangement in Two Flask System
40%
35%
% Occurrence
30%
25%
N=4
N=10
N=100
N=1000
20%
15%
10%
5%
0%
0%
20%
40%
60%
% of Molecules in Left Flask
80%
100%
Judging Entropy Changes in Processes
• Certain types of changes will result in increased
entropy.
– Certain phase changes.
• Ssolid < Sliquid << Sgas
– An increase in the number of particles present.
– An increase in the temperature of a substance.
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Judging Entropy Changes in Processes
• When a solid melts to form a liquid, entropy increases.
– In solids, the particles are held in place rigidly, limiting the number
of ways a specific energy can be obtained.
– In liquids, the particles move past each other, increasing the
number of ways a specific energy can be obtained.
– The number of microstates increases during melting.
• A chemical reaction that generates two moles of gas when
only one mole of gas was initially present will increase the
entropy of a sample.
– The number of possible microstates increases as the number of
particles increases.
• When a sample is heated, the temperature of the sample
increases.
– As temperature increases, the number of possible velocities
increases.
Entropy Change
ΔS =
qrev
T
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Entropy Changes
• The greater the degree of randomness or disorder
in a system, the greater the entropy of the system
• Entropy increases are expected to accompany
processes in which
– Pure liquids or liquids solutions are formed from solids
– Gases are formed, either from solids or liquids
– The number of molecules of gases increase during the
reaction
– The temperature of a substance is increased (increased
temperature means increased molecular motion)
Entropy Change Examples
Predict whether each of the following processes
involves an increase, decrease or uncertain
change in entropy
1.
2.
3.
4.
5.
2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g)
2SO2(g) + O2(g)  2SO3(g)
NaCl(aq)  NaCl(s)
CO(g) + H2O(g)  CO2(g) + H2(g)
CO(g) + H2O(l)  CO2(g) + H2(g)
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Criteria for Spontaneous Change:
The Second Law of Thermodynamics.
ΔStotal = ΔSuniverse = ΔSsystem + ΔSsurroundings
The Second Law of Thermodynamics:
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
All spontaneous processes produce an
increase in the entropy of the universe.
Criteria for Spontaneous Change:
The Second Law of Thermodynamics.
• The second law of thermodynamics: in any
spontaneous process, the total entropy of the
universe is positive
– (Su > 0).
Su = Ssys + Ssurr > 0
• Su = entropy of the universe
• Ssys = entropy of the system
• Ssurr = entropy of the surroundings
– It is impossible to convert heat completely to work,
since work is a process that involves moving random
motions into more ordered ones.
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Implications and Applications
• The entropy change for the surroundings can be
calculated from the heat flow from the system,
which is equal to –H.
Ssurr  
H
T
– For an exothermic reaction, the entropy of the
surroundings increases.
The Third Law of Thermodynamics
• The third law of thermodynamics states that the entropy of
a perfect crystal of any pure substance approaches zero as
the temperature approaches absolute zero.
• Because S is explicitly known (= 0) at 0 K, S values at
other temps can be calculated.
• The entropy of one mole of a chemical substance under
standard conditions is the standard molar entropy, So.
• The entropy change for a reaction can be calculated from
the standard molar entropies of the reactants and products.
S    i Sproducts i    j Sreactantsj
i
j
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(a) A perfect crystal of hydrogen chloride at 0 K.
(b) As the temperature rises above 0 K, lattice vibrations allow some
dipoles to change their orientations, producing some disorder and
an increase in entropy.
The H2O molecule
can vibrate and rotate
in several ways, some
of which are
shown here.
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Entropy as a Function of Temperature
The Third Law of Thermodynamics
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Implication of the Third Law of Thermodynamics
• Entropy is a state function, so the value of S must be independent of the path
taken from reactants to products.
• The entropy change for a reaction can be calculated from the standard molar
entropies of the reactants and products.
S    i Sproducts i    j Sreactantsj
i
j
S° Examples
• What is S° for the reaction
2NO(g) + O2(g)  2NO2(g)
Given: S°NO = 211 J mol-1 K-1
S°O₂ = 205 J mol-1 K-1
S°NO₂ = 240 J mol-1 K-1
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Free Energy and Free Energy Change
• Hypothetical process:
– only pressure-volume work, at constant T and P.
qsur = -qp = -ΔHsys
• Make the enthalpy change reversible.
– large surroundings, infinitesimal change in
temperature.
• Under these conditions we can calculate entropy.
S sur 
H sys
qsur

T
T
or TS sur   H sys
Free Energy and Free Energy Change
For the universe:
TΔSu. = TΔSsys + TΔSsur
TΔSu. = TΔSsys – ΔHsys = -(ΔHsys – TΔSsys)
-TΔSu. = ΔHsys – TΔSsys
For the system:
G = H – TS (Gibbs Free Energy)
ΔG = ΔH – TΔS (at constant T)
ΔGsys = - TΔSu
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Criteria for Spontaneous Change
at Constant Temperature and Pressure
ΔGsys < 0 (negative), the process is spontaneous.
ΔGsys = 0 (zero), the process is at equilibrium.
ΔGsys > 0 (positive), the process is non-spontaneous.
Free Energy and Spontaneous Change
G  H  TS
• For the last two cases, the temperature at which a reaction changes from
spontaneous to nonspontaneous can be calculated.
T
H
S
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Free Energy and Spontaneous Change
• For a spontaneous reaction with a negative H and a
negative S, the reaction only occurs at low temperature.
– These reactions are referred to as enthalpy driven because the
negative value for the enthalpy is responsible for the negative
value of the Gibbs free energy change.
• For a spontaneous reaction with a positive H and a
positive S, the reaction only occurs at high temperature.
– These reactions are referred to as entropy driven because the
product of the positive entropy change and the absolute
temperature is responsible for the negative value of the Gibbs free
energy change.
Spontaneous Change Predictions
Using the enthalpy and entropy changes to
predict under what conditions would the
following reactions occur spontaneously
1.
2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g) H = -236kJ/mol
2.
I2(g)  2I(g)
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Melting of Ice
Free Energy and Work
• It can be shown that the Gibbs free energy change
is equal to the maximum useful work done by the
system.
G = -wmax
– Work is not a state function.
– Maximum work realized only if the reaction or process
is carried out along a very specific path.
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Free Energy and Work
• A reversible path is the specific path required for
maximum useful work.
– For systems with a reversible path, the system is near
equilibrium and a small incremental change in a
variable will bring the system back to its initial state.
• In an irreversible change, a small incremental
change in any variable does not restore the initial
state.
– The amount of work available is always less than
maximum for an irreversible change.
Free Energy and Work
• The Gibbs free energy change establishes the
upper bound to the amount of work obtained from
a given process.
– Actual work produced in any real application may be
considerably less.
• Reactant mixtures are generally very far from
equilibrium.
– Systems that are far from equilibrium often change
rapidly, and rapid changes tend to be irreversible.
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Standard Free Energy Change, ΔG°
• The standard free energy of formation, ΔGf°.
– The free energy change for a reaction in which one mole of a
substance in its standard state is formed from its elements in their
standard states.
– Gf° = 0 for elements in their free standard state.
• The standard Gibbs free energy change for a reaction, G°,
can be calculated from Gibbs free energies of formation,
Gf°.
G    i Gf productsi    j Gf reactantsj
i
j
– This equation provides an alternative method to
calculate G°, without entropy or enthalpy.
G° = H° - TS°
Free Energy and Chemical Reactions
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Characteristics of G
 G is an extensive property
 G changes sign when a process is reversed
 G for a net or overall process can be obtained
by summing the G values for the individual
steps
4. For a chemical reaction at constant T and P
G = H - TS
5. For a chemical reaction
ΔG° = [ npΔGf°(products) - nrΔGf°(reactants)]
G° Examples
1.
Given that at 400°C the reaction
4Cu(s) + O2(g)  2Cu2O(s) G° = -250 kJ/mol
What is G° at 400°C for the following reactions
a)
b)
2Cu(s) + ½O2(g)  Cu2O(s)
Cu2O(s)  2Cu(s) + ½O2(g)
2.
Will Cu2O(s) spontaneously decompose at 400°C with
the compounds in their standard states?
3.
What happens if the Cu2O(s) decomposition reaction is
made to run simultaneously with
C(s) + ½O2(g)  CO(g) G° = -175 kJ/mol
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Coupled Reactions
• In order to drive a non-spontaneous reactions
we changed the conditions (i.e. temperature or
electrolysis)
• Another method is to couple two reactions.
– One with a positive ΔG and one with a negative ΔG.
– Overall spontaneous process.
G° Examples
4.
Consider the reaction
2NO(g) + O2(g)  2NO2(g)
Given at 25°C H° = -114.1 kJ/mol and S° = -146.3 J mol-1 K-1
What is G°?
5.
Given the Gf° of NO(g) is 86.57 kJ/mol and Gf° of NO2(g) is
51.30kJ/mol
What is G°?
6.
Given for water at 25°C Hvap° = 42 kJ/mol and
Svap° = 119 J mol-1 K-1
What is Gvap° at 25°C?
Will water spontaneously evaporate at 25°C?
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Free Energy Change and Equilibrium
Relationship of ΔG° to ΔG for Nonstandard
Conditions
2 N2(g) + 3 H2(g)  2 NH3(g)
ΔG = ΔH - TΔS
ΔG° = ΔH° - TΔS°
For ideal gases ΔH = ΔH°
ΔG = ΔH° - TΔS
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Relationship Between S and S°
qrev = -w = RT ln
ΔS =
Vf
Vi
V
qrev
= R ln f
T
Vi
ΔS = Sf – Si = R ln
Pf
P
Vf
= R ln i = -R ln
Pi
Pf
Vi
Let i be the standard state
Sf = S° - R ln
P
P
= S° - R ln P
= S° - R ln
1
P°
2 N2(g) + 3 H2(g)  2 NH3(g)
SN2 = S°N2 – Rln PN2
SH2 = S°H2 – Rln PH2
SNH3 = S°NH3 – Rln PNH3
ΔSrxn = 2(S°NH3 – Rln PNH3) – 2(S°N2 – Rln PN2) –3(S°H2 – Rln PH2)
ΔSrxn = 2 S°NH3 – 2S°N2 –3S°H2+ Rln
ΔSrxn = ΔS°rxn + Rln
PN22PH32
2
PNH
3
PN22PH3 2
2
PNH
3
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ΔG Under Non-standard Conditions
ΔG = ΔH° - TΔS
ΔSrxn = ΔS°rxn + Rln
ΔG = ΔH° - TΔS°rxn – TR ln
PN22PH3 2
2
PNH
3
PN22PH3 2
2
PNH
3
2
ΔG = ΔG° + RT ln
PNH3
PN22PH3 2
ΔG = ΔG° + RT ln Q
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