Ch 20. Thermodynamics

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Ch 20. Thermodynamics

• Brady Senese, 4th Ed.

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Chapter 20 Thermodynamics

• The First Law of Thermodynamics internal energy
  may be transferred as heat or work, but cannot be
  created or destroyed was originally discussed
  in Chapter 7

• For an isolated system the internal energy is
  constant
• ? E Ef - Ei 0
• We cant measure the internal energy of anything,
  so we measure the changes in energy
• DE work (w) heat (q)

3

• The change in internal energy is defined in terms
  of heat (q) and work (w)
• Internal energy is a state function which means
  that its value does not depend on how the change
  from one state to another was carried out

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State Function

• Any characteristic of a system that is
  independent of pathway, designated with capital
  letters e.g. H, T, V, P
• A system is frequently characterized by changes
  in the state function, ?X
• ?XrxnS(n?Xfinal)- S(n?Xinitial) where nthe
  number of moles of that substance in the reaction

DH S(nDHprod) S(nDHreact)
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Standard State

• Most Stable form of the pure substance at
• 1 atm pressure
• Stated temperature. If temp is not specified,
  assume 25 C
• Solutions are 1M in concentration.
• Measurements made under standard state conditions
  have the mark ?H

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Learning Check

•  Calculate ?H for the following reactions.
• H2O(l)   CO2(g) ? H2CO3(aq) kJ/mol
• -285.9 -393.5  -698.7 ?Hf0298K
• NH3(g) HCl(g) ? NH4Cl(s)
• -46.19   -92.30   -315.4  ?Hf0298K

-19.3 kJ/mol
DH SDHprod SDHreact -698.7 -
(-285.9)(-393.5)
-176.9 kJ/mol
-315.4 - (-46.19)(-92.30)
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• There are two types of work that we are
  interested in electrical and P-V work which
  will be discussed now
• The work done by a system depends on the volume
  change and the external pressure
• For a reaction at constant volume

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• qv is called the heat at constant volume
• Usually reactions are carried out at fixed
  pressure
• For these reactions, enthalpy is more convenient

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• qp is called the heat at constant pressure
• The internal energy and enthalpy changes are
  different whenever a volume change occurs for the
  system
• Only when the volume change is large is the
  difference significant
• Large volume changes can occur when gases are
  involved in the reaction

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• Treating the gases as ideal

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Work (w) Force Distance -Patm?V

• In reactions, work is most often due to the
  expansion or contraction of a system due to
  changing moles of gas.
• The deployment of an airbag is one example of
  this process.
• 1 C3H8(g) 5 O2(g) ? 3 CO2(g) 4 H2O(g)
• 6 moles of gas ? 7 moles of gas
• Since PVnRT, P?V ?nRT.
• Thus, we can predict the work w - ?nRT

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Enthalpy (H) and Internal Energy (E)

• Enthalpy is the heat transferred at constant
  pressure, thus, ?H qp
• ?E q w
• If there is no change in the moles of gas, then
  ?E qp
• If the reaction occurs in a fixed volume
  container, ?E qv
• If the moles of gas change, ?E qvqp ?nRT ?H
  ?nRT
• Generally ?H is very close in value to ?E, unless
  work term is huge like in an explosion

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Learning Check

• Consider the nitrogen triiodide decomposition,
  2NI3(s)?N2(g)3I2(g) . Is ?? ??? Why?
• For the following reaction for picric acid,
  calculate ??, ??

?nRT 17.3 kJ
?H0 -13298.8 kJ
DE0 -13316.2 kJ
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Spontaneous Reactions

• A spontaneous change is a change the occurs by
  itself (without continuous outside assistance)
• Examples A rock falling off a ledge and water
  flowing down-hill
• Nonspontaneous events occur at the expense of
  spontaneous ones
• Often, but not always accompanied by exothermic
  processes

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• Most, but not all, exothermic reactions are
  spontaneous
• Some endothermic reactions are spontaneous
• Heat changes alone are not sufficient to predict
  if a process will proceed spontaneously
• We can begin to explore this by considering the
  heat transfer between a hot and cold object

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• Consider a system of six objects that are either
  hot (designated 1) or cold (designated 0)
• If three are hot and three are cold, the system
  has a total of three units of energy
• The hot system could transfer 0, 1, 2, or 3 units
  of energy to the cold objects
• There are 20 possible states of the system,
  starting with all hot to the left (111000)

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Energy transfer possibilities between six objects
with a total system energy of three units. The
allowed values of energies for an object are 0 or
1 unit.
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• There are four possible outcomes the transfer of
  0, 1, 2, or 3 units of energy
• The probability of a state being produced is
  proportional to the number of ways it can be made
• Probabilities must be normalized

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• For our heat transfer problem

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• The results can be stated in a number of ways
• There is a 5 chance no energy will be
  transferred
• There is a 95 chance some energy will be
  transferred
• There is a 5 chance that all the energy will be
  transferred
• There is a 90 chance that 1 or 2 units of energy
  will be transferred
• This simple model demonstrates the role of
  probability in determining the direction of a
  spontaneous process

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• Spontaneous processes tend to proceed from states
  of low probability to states of higher
  probability
• The entropy (symbol S) is used in thermodynamics
  to describe the number of equivalent ways that
  the energy can be distributed
• The greater the statistical probability of a
  particular state, the greater the entropy

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• Like internal energy and the enthalpy, entropy is
  also a state function
• An event that is accompanied by an increase in
  the entropy of the system will have a tendency to
  occur spontaneously
• A positive change in entropy means the final
  state (products) are more probable that the
  initial state (reactants)

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Consider the reaction 3A?3B, where A molecules
can take on energies that are multiples of 10
energy units and B molecules can take on energies
in multiples of 5 energy units. Suppose the total
energy of the system in 20 energy units. (a)
There are two ways to distribute all the energy
between 3 A molecules. (b) There are four ways to
distribute all the energy between 3 B molecules.
The entropy of B is higher than A because there
are more ways to distribute the same amount of
energy.
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• It is often possible to predict whether the
  entropy change will be positive or negative
• Volume For gases, the entropy increases as
  volume increases

(a) A gas in a container separated from a vacuum
by a partition. (b) The partition is removed. (c)
The gas expands to achieve a more probably
(higher entropy) particle distribution.
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• Temperature As temperature increases, the
  entropy increases

(a) At absolute zero, the particles (black dots)
are in their equilibrium lattice positions and
entropy is relatively low. (b) At higher
temperatures, the molecules vibrate. Entropy
increases. (c) As the temperature increases
further, more violent vibrations can occur and
the entropy is higher than in (b).
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• Physical state Entropy increases as a substance
  goes from the solid to liquid to vapor

The crystalline solid has very low entropy. The
liquid has higher entropy because the molecules
can move freely and there are more ways to
distribute kinetic energy among them. The gas has
the highest entropy because the particles are
randomly distributed throughout the container and
there are many, many was to distribute the
kinetic energy.
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• For chemical reactions involving gases, the sign
  of the entropy change is easy to predict
• The number of particles is a major factor that
  affects the sign of the entropy change for
  chemical reactions
• When all other things are equal, the reaction
  that increases the number of particles in the
  system tend to have a positive entropy change

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Adding additional particles to a system increases
the number of ways that the energy can be
distributed in the system. With all other things
being equal, a reaction that produces more
particles will have a positive entropy change.
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Predict The Sign Of ? S In The Following

• Dry ice ?carbon dioxide gas
• Moisture condenses on a cool window
• AB ?A B
• A drop of food coloring added to a glass of water
  disperses
• 2Al(s) 3Br2(l) ?2AlBr3(s)

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Factors Affecting Entropy (?S) (Cont.)

• Number of particles
• Number of bonds
• The Entropy of a pure crystalline solid at 0 K is
  defined as 0.

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• According to the Second Law of Thermodynamics
  whenever a spontaneous event takes place in our
  universe, the total entropy of the universe
  increases
• In thermodynamics, the universe is the sum or
  total of the system and surroundings
• It can be shown that the entropy change of the
  surroundings is equal to the heat transferred to
  the surroundings from the system divided by the
  Kelvin temperature

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The Second Law of Thermodynamics

• ?Stotal ?Ssystem ?Ssurroundings
• since ?Ssurroundingsqsurroundings /T
• and since qsurroundings-qsystem
• since ?Ssurroundings -qsystem /T -?Hsystem /T
• ?Stotal ?Ssystem- ?Hsystem /T
• ?Stotal (T?Ssystem- ?Hsystem )/T
• T?Stotal T?Ssystem- ?Hsystem -(?Hsystem
  T?Ssystem)
• Since ?Stotal is () for all spontaneous
  reactions,
• -(?Hsystem T?Ssystem)gt0
• ?Hsystem T?Ssystemlt0 for all spontaneous
  reactions.

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• Temperature must be considered when determining
  if a process is spontaneous
• Liquid water spontaneously freezes when the
  temperature is below 0oC to form solid water
  (ice)
• Solid water (ice) spontaneously melts when the
  temperature is above 0oC
• The three factors that can influence spontaneity
  are the enthalpy change, the entropy change, and
  the temperature

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• A new function, called the Gibbbs free energy , G
  is used to determine if events are spontaneous
• It is also a state function

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• At constant temperature and pressure, a change
  can only be spontaneous if it is accompanied by a
  decrease in the free energy of the system
• General comments can be made about the
  possibility of a spontaneous process and the
  signs of the enthalpy and entropy changes

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The Driving Forces Of A Reaction

• Reactions typically occur spontaneously when the
  ?H is (-) or if ?S is ()
• If the ?H gt T?S term, the reaction is said to be
  enthalpy driven
• If the T?S term gt ?H term, the reaction is said
  to be entropy driven
• Free energy was derived to give the overall
  picture of both of the driving forces

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Component Hfo (kJ/mol) Fe2O3(s) -822.2 Al(s)
0 Al2O3(s) -1,669.8 Fe (s)
0 Fe and Al are zero because, by
definition, the Hfo of elements in their standard
states is zero (OK, technically the iron is in
the liquid state, but in the end it becomes solid
again).
Fe2O3(s) 2 Al(s) Al2O3(s) 2
Fe(l) The H for this reaction is the sum of the
Hfo's of the products - the sum of the Hfo's of
the reactants (multiplying each by their
stoichiometric coefficient in the balanced
reaction equation), i.e. Horxn (1
mol)(HfoAl2O3) (2 mol)(HfoFe) - (1
mol)(HfoFe2O3) - (2 mol)(HfoAl) Horxn (1
mol)(-1,669.8 kJ/mol) (2 mol)(0) - (1
mol)(-822.2 kJ/mol) - (2mol)(0 kJ/mol) Horxn
-847.6 kJ
WOW Big Exothermicity!
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Temperature Controlled Reactions

• Temperature-controlled reactions are spontaneous
  at one temperature and not at another
• They are identified by the fact that their ?H and
  ?S have the same sign
• They change spontaneity at the temperature when
  T ?H/?S

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(Gibbs) Free Energy (G)

• ?G0 ?H0-T?S0
• ?G0 is highly dependent on temperature.
• May be calculated using ?Gf0 data only at 298K.
• Note that ?G0 implies standard conditions. If
  the concentrations and pressures vary, we must
  correct for this to calculate ?G

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Learning Check

• Calculate ?G0 for the following using both
  approaches at 298K
• CaCO3(s) ?CO2(g) CaO(s)
• ?Hf (kJ/mol) ?Gf (kJ/mol) Sf J/molK
• CO2(g) -393.5 -394.4 213.7
• CaCO3(s) -1432.7 -1320.3 107
• CaO(s) -635.5 -604.2 40
• Which approach is needed if you want ?G0 at 500K?

?Hrxn403.7 kJ/mol
?Srxn146.7 J/molK
?G0rxn 403.7 298K x 0.1467 kJ/molK 359 kJ
?G0rxn 321.7
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• The Third Law of Thermodynamics makes the
  experimental determination of absolute entropies
  possible
• At absolute zero the entropy of a perfectly
  ordered crystalline substance is zero
• S 0 at T 0 K (perfect
  crystal)
• The entropy of 1 mol of a substance at a
  temperature of 298 K (25 oC) and a pressure of 1
  atm is called the standard entropy, So
• A number of standard entropies have been
  tabulated

42

• Some of these are reported in Table 20.1 and
  Appendix C
• They can be used to calculate standard entropy
  changes for reactions
• If the reaction under consideration corresponds
  to the formation of 1 mol of compound from its
  elements, then the calculated standard entropy
  change is called the standard entropy of
  formation,

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Learning Check

• Calculate ?S0 for the following
• CO2(s) ? CO2(g)
• 187.6 213.7 S0 (J/molK)
• CaCO3(s) ? CO2(g) CaO(s)
• 92.9   213.7   40   S0 (J/molK)

26.1 J/molK
161 J/molK
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• Some standard entropies of formation are not
  tabulated, so must be calculated when needed
• Like for the entropy change, the standard free
  energy change is determined at 298 K and 1 atm
• Some standard free energies of formation are
  tabulated in Table 20.2 and Appendix C

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• They can be used to find standard free energy
  changes for reactions
• A chief use of spontaneous reactions is to
  produce useful work
• The maximum conversion of chemical energy to work
  occurs if the reaction is carried out under
  thermodynamically reversible conditions

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?G0 for Reversible Reactions

• A process is defined as thermodynamically
  reversible if it can be reversed and
• If it is always very close to equilibrium (the
  change in quantities is very small)
• For reversible reactions, ?G0 represents the
  maximum work output

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A reversible expansion of a gas. As water
molecules evaporate one at a time, the external
pressure gradually decreases and the gas slowly
expands. The process would be reversed if one gas
molecule condensed into the liquid. A truly
reversible process would require an infinite
number of tiny steps, and would take forever to
actually accomplish.
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• The change in free energy provides a limit to the
  amount of available energy in a reaction
• The maximum amount of energy produced by a
  reaction that can be theoretically harnessed as
  work is equal to
• This is the energy that need not be lost to the
  surroundings as heat and it therefore free to be
  used for work
• This allows the efficiency of a system to be
  determined

49

• A system that is neither spontaneous nor
  nonspontaneous is at equilibrium
• This occurs when the free energy change is zero
• Consider the equilibrium between ice and water at
  0oC
• The system will remain at equilibrium as long as
  no heat is added or removed
• Both phases can exist together indefinitely

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• No work can be done by a system at equilibrium
  because the available (free) energy is zero
• Only one temperature is possible for a phases
  change at equilibrium
• This is the melting temperature (point) of the
  substance for a solid-liquid equilibrium
• This is the boiling temperature (point) of the
  substance for a liquid-vapor equilibrium

51

• Free energy change diagrams can be used to
  represent phase changes

The free energy diagram for the conversion of
H2O(l) to H2O(s). At the left of each diagram the
system consists entirely of H2O(l). At the right
is H2O(s). The horizontal axis represents the
extent of conversion between H2O(l) and H2O(s).
52

• Free energy changes for reactions are usually
  more complex

The minimum on the curve indicates the
composition of the reaction mixture at
equilibrium. Because the standard free energy
change is positive, the position of the
equilibrium lies close to the reactants.
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Free energy curve for a reaction having a
negative standard free energy change. The
reaction proceeds to equilibrium by moving
down-hill from left to right. The position of
the equilibrium lies close to the products
because the standard free energy change for the
reaction is negative.

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• At equilibrium
• Given two of the three quantities (enthalpy
  change, entropy change, and temperature) the
  third can be calculated

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?G0 At Equilibrium

• Thus, 0 ?H-T ?S
• T ?H/?S for all equilibrium processes,
  including phase changes.
• What is the expected melting point for Cu?

56

• While the tables of thermodynamic quantities are
  extensive, they are incomplete
• Reasonable estimates of the heat of reaction, for
  example, can be made from atomization energies
  and average bond energies
• Atomization energies can be found in Table 20.3
  and Appendix C
• A number of average bond energies are reported in
  Table 20.4

57

Two paths for the formation of methane from its
elements in their standard states. Steps (1) and
(2) involve the atomization of hydrogen and
carbon, which can be obtained from Table 20.3 or
Appendix C. Step (3) can be estimated by adding
the energies from Table 20.4 for all the bond
formed going from isolated gas phase atoms to the
product.
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Chemical Potential Energy is stored in the bonds
?Hrxn SD(Bonds broken) - S D (Bonds
formed)
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Calculating ?Hrxn Using Bond Dissociation Energies

• ?H SD(Bonds broken) - S D(Bonds formed)
•  
• CH4(g) 3Cl2(g) ? CHCl3(g) 3HCl(g)
• Examine Structures of each and decide what bonds
  must be broken and what bonds must be formed.
• Broken 4C-H, 3Cl-Cl. Formed 1C-H, 3C-Cl, 3H-Cl
• Exothermic reactions form stronger bonds in the
  product than in the reactant

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• Example The standard enthalpy of formation of
  ethane from Appendix C is 84.5 kJ/mol. Estimate
  this value using average bond energies and
  atomization energies.

61

• The estimate differs from the tabulated value by
  7
• This error is remarkably small considering how
  easy the calculation was to perform

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Example

• Given the reaction S8(s) 12O2(g)? 8SO3(g),
  Calculate ??0 from ??0f . Given that S-S bonds
  have D 225 kJ/mol bond, and that OO has D498
  kJ/mol, what is the bond dissociation energy for
  the sulfur to oxygen bonds?

63
Thermodynamics Review

• The First Law of Thermodynamics internal energy
  may be transferred as heat or work, but cannot be
  created or destroyed was originally discussed
  in Chapter 7

• For an isolated system the internal energy is
  constant
• ? E Ef - Ei 0
• We cant measure the internal energy of anything,
  so we measure the changes in energy
• DE work (w) heat (q)

64
Enthalpy (H) and Internal Energy (E)

• Enthalpy is the heat transferred at constant
  pressure, thus, ?H qp
• ?E q w
• If there is no change in the moles of gas, then
  ?E qp
• If the reaction occurs in a fixed volume
  container, ?E qv
• If the moles of gas change, ?E qvqp ?nRT ?H
  ?nRT
• Generally ?H is very close in value to ?E, unless
  work term is huge like in an explosion

65
Entropy (S)

• A measure of randomness
• The more possible arrangement of particles, the
  higher the entropy
• Processes favor high probability (p)

66
Factors Affecting Entropy (?S)

• Mixing
• Volume change of gas
• Temperature- not a significant change if state is
  the same
• Physical State

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The Second Law of Thermodynamics
Spontaneous processes
(Gibbs) Free Energy (G)

• G H-TS, thus
• ?G ?H- T?S
• Note that ?G is highly dependent on temperature.
• Reactions whose ?Glt0 are "spontaneous" or
  "efficient
• ?G0 at equilibrium.
• ?Ggt0 inefficient (non-spontaneous)

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The Third Law of Thermodynamics

• The entropy of a pure crystalline solid at 0 K is
  defined as 0
• All others states and materials have Sgt0.
• ?S ?Sproducts- ?Sreactants
• Note that the units are J/molK