Chapter 19' Chemical Thermodynamics'

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Chapter 19. Chemical Thermodynamics.

• ?G ?H T?S

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Enthalpy (H) and entropy (S)

• In Chapter 5 we discussed the role of enthalpy (
  ?H) in driving reactions. We saw that highly
  exothermic reactions such as the burning of
  gasoline proceeded very strongly, while the
  reverse reaction was highly endothermic, and did
  not proceed spontaneously at all. (CO2 and H2O do
  not react spontaneously to form gasoline and
  oxygen). However, enthalpy is not the only
  driving force in reactions. There is a second
  contribution, the entropy. Entropy relates to
  probability, and the increase in disorder of a
  system.
• We can think about probability in terms of a gas
  moving to occupy two flasks instead of only one

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We connect two flasks as shown at the right by a
closed stopcock, one of which is
evacuated. When the stopcock is opened, the air
will rush into the evacu- ated flask to
equalize the pressure. What is the driving force
for this? It is in fact probability, or entropy.
4
19.1 Spontaneous Processes.

• A spontaneous process is one that proceeds on
  its own without any outside assistance.
• A gas will expand from one flask into a second
  as a spontaneous process, but the reverse will
  never occur.
• Processes that are spontaneous in one direction
  are non-spontaneous in the opposite direction.

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The rusting of nails in air a spontaneous
process
Fresh nails
rusted nails
spontaneous
not spontaneous
2 Fe (s) 3 O2 (g) ? Fe2O3 (s) (rust)
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Examples of spontaneous processes
water
Ice at 25 oC will spontaneously melt
ice
spontaneous
not spontaneous
The reverse will never happen. Water at 25
oC will never freeze.
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Examples of spontaneous processes
But water at -10 oC will spontaneously freeze
ice
spontaneous
water
not spontaneous
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Equilibrium

• A mixture of water and ice at 0 oC is at
  equilibrium.

The relative quantity of ice and water
remains the same if we dont add or remove heat.
There is no spontaneous process in either
direction. If we remove some heat, the
temperature remains at 0 oC, but some water
turns to ice.
ice
water (at 0 oC)
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Reversible processes

• These occur with infinitesmal changes in the
  system.
• The melting and freezing of ice at 0 oC is
  reversible. Thus, if we add some heat, some of
  the ice melts, and if we remove some heat, some
  of the water freezes, but the temperature remains
  at 0 oC. At all other temperatures an
  irreversible process occurs.
• Melting and freezing of a substance at its
  melting point are reversible.
• Boiling and condensation of a substance at its
  boiling point are also reversible.
• Reversible processes are not spontaneous.

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Example

• When benzene vapor converts to benzene liquid at
  80.1 oC (B.Pt. of benzene) and 1 atm is this a
  spontaneous process?
• The condensation of a gas to a liquid at the
  boiling point of a substance is not a spontaneous
  process. To move the equilibrium in either
  direction we have to add or subtract energy.

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19.2 Entropy and the second Law of Thermodynamics.

• Entropy is associated with randomness. The
  entropy, S, of a system is a state function just
  like the internal energy. The change in entropy
  is given by
• ?S Sfinal - Sinitial

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Relating entropy to heat transfer and temperature

• For a process that occurs at constant
  temperature such as melting of a solid at its
  melting point
• ?Ssystem qrev/T
• To calculate the entropy change of a system we
  need to know the heat change, which is qrev, and
  divide it by the temperature in K. The term qrev
  is the reversible flow of heat in a reversible
  process such as enthalpy of melting of a solid at
  its melting point, or of boiling of a liquid at
  its boiling point.

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Example

• The molar enthalpy of fusion of Hg is 2.29
  kJ/mol. What is the entropy change when 50 g of
  Hg freezes at the normal melting point of -38.9
  oC? (fusion melting)
• The important thing to realize here is that qrev
  ?H(fusion)
• Hg(s) ? Hg(l) ?H 2.29 kJ/mol
• So for freezing
• Hg(l) ? Hg(s) ?H -2.29 kJ/mol

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• Moles Hg 50.0 g x 1mole 0.249 mol
• 200.59 g
• kJ 0.249 mole x -2.29 kJ -0.571 kJ -571 J
• 1 mole
• T 273.15 38.9 243.3 K
• ?S qrev/T -571/243.3 -2.44 J/K
• Note that units of entropy are J/K. A positive
  change in entropy means an increase in entropy of
  the system.

?H (freezing) Hg(l)
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The Second Law of Thermodynamics

• The total entropy of the universe increases in
  any spontaneous process.
• The sum of the entropy change of the system and
  the surroundings is always positive.
• Reversible ?Suniverse ?Ssystem ?Ssurround
  0
• Irreversible ?Suniverse ?Ssystem ?Ssurround
  gt 0

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Example

• What happens to the entropy of the universe when
  a lesser spotted thrush begins to do the
  makarena?
• Answer It increases, ?Suniverse gt 0
• For any spontaneous process, and thats any
  process where anything actually happens, the
  entropy of the universe increases. You may get
  questions like the above, and if you bear in mind
  that (except for a reversible process) when
  anything happens, the entropy of the universe
  increases.

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19.3 The Molecular Interpretation of Entropy.

• Molecules can undergo translational, rotational,
  and vibrational motion.
• The number of ways a system can be arranged is
  the number of microstates it has. The entropy of
  a system is given by
• S k ln W
• where W is the number of possible microstates,
  and k
• is Boltsmanns constant
• 1.38 x 10-23 J/K.

Ludwig Boltzmann
(1844-1906) Boltzmann worked out the nature of
entropy.
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Probability and gas molecules in interconnected
flasks

• Consider a pair of flasks connected by a
  stop-cock. The probability that all the molecules
  will be in one flask is (½)n, where n is the
  number of molecules.

one molecule, probability (1/2)1 1/2
two molecules, probability (1/2)2 1/4
four molecules, probability (1/2)4 1/16
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The relationship between probability and entropy

• So if n 6.02 x 1023, we have a very small
  probability of finding all
• the molecules in one flask!
• Entropy relates to probability. The gas
• molecules in the apparatus at right are moving
  randomly, but the probability aspect keeps them
  evenly distributed between the two flasks.

6.02 x 1023 molecules, probability (1/2)6.02 x
10
23
There is thus a chance that all the air
molecules in this room would rush into one
corner, and we would all die, but it is so small
that it would never happen.
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The standard molar entropy of a substance (Sº)

• The entropies of substances may be measured
  experimentally. The entropy of one mole of a
  substance in its standard state is known as the
  standard molar entropy (Sº). The units of Sº are
  J/mol.K. Some values of standard molar entropies
  are given overleaf. Note that So values increase
  solid lt liquid lt gas, e.g. C6H6 (s) lt C6H6 (l) lt
  C6H6 (g) increase with the size of molecules,
  e.g. HF lt HCl lt HBr lt HI or CH4 lt C2H6 lt C3H8
  and with being a solid or liquid dissolved in
  solution in another liquid e.g. CH3COOH (l) lt
  CH3COOH (aq).

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Standard molar entropies of selected substances
at 298 K (So, J/mol-K)

• Gases
• H2(g) 130.6 N2(g) 191.5
• O2(g) 205.0 C6H6(g) 269.2
• Liquids
• H2O(l) 69.9 C6H6(l) 172.8
• Solids
• Li(s) 29.1 Fe(s) 27.23
• FeCl3(s) 142.3 NaCl(s) 72.3
• C(diamond) 2.43 C(graphite) 5.69

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Making Qualitative predictions about ?S.

• Entropy increases with temperature, since the
  particles vibrate and move about more rapidly,
  and so there is greater disorder. It also
  increases with increasing volume, as with a an
  expanding gas, and with the number of
  independ-ently moving particles. Thus, entropy
  increases solid lt liquid lt gas.

Figure shows increase in entropy of a
substance with increasing temp. as we pass
solid?liquid?gas.
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Of all phase states, gases have the highest
entropy
Ssolid
lt Sliquid
lt Sgas
gas
liquid
solid
In a gas, molecules are more randomly distributed
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Larger Molecules generally have a
larger entropy
Ssmall
lt Smedium
lt Slarge
Larger molecules have more internal motion
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size of molecules increases

Sgas gt Sliquid
dissolving a gas in a liquid is accompanied by a
lowering of the entropy

dissolving a liquid in another liquid is
accompanied by an increase in entropy

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Often, dissolving a solid or liquid will
increase the entropy
solution
liquid
dissolves
lower entropy
more disordered arrangement higher entropy
solid
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Dissolving a gas in a liquid decreases the
entropy
gas
dissolves
overall more disordered arrangement higher
entropy
solution of gas in liquid, lower S
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Entropy increases with increasing temperature
along the series solid lt liquid lt gas
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Note S increases with increase in number of
moles of gas in a reaction.
What is the sign of DS for the following
reactions?
FeCl2 (s) H2 (g) ? Fe (s) 2 HCl (g)
DS
solid
solid
gas
gas
1 mol
2 mol
Ba(OH)2 (s) ? BaO (s) H2O (g)
DS
solid
solid
gas
2 SO2 (g) O2 (g) ? 2 SO3 (g)
DS -
gas
gas
gas
2 mol
1 mol
2 mol
DS -
Ag (aq) Cl- (aq) ? AgCl (s)
in solution
insoluble
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For each of the following pairs of substances,
which substance has a higher molar entropy at
25oC ?
HCl (l) HCl (s)
Li (s) Cs (s)
C2H2 (g) C2H6 (g)
Pb2 (aq) Pb (s)
O2 (g) O2 (aq)
HCl (l) HBr (l)
CH3OH (l) CH3OH (aq)
N2 (l) N2 (g)
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19.4 Entropy Changes in Chemical Reactions.

• This parallels the calculation of standard molar
  enthalpies of reaction.
• ?So SnSo(products) - SmSo(reactants
  )
• Note that So is not zero for elements.
• Recall that So is a state function.

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Standard molar entropies of selected substances
at 298 K (So, J/mol-K)

• Gases
• H2(g) 130.6 N2(g) 191.5
• O2(g) 205.0 C6H6(g) 269.2
• Liquids
• H2O(l) 69.9 C6H6(l) 172.8
• Solids
• Li(s) 29.1 Fe(s) 27.23
• FeCl3(s) 142.3 NaCl(s) 72.3
• C(diamond) 2.43 C(graphite) 5.69

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Example

• Calculate ?So for the following reaction
• C2H4(g) H2(g) ? C2H6(g)
• So products So C2H6(g) 229.5 J/mol-K
• So reactants So C2H4(g) So H2(g)
• 219.4 130.58 J/mol-K
• 349.98 J/mol-K
• So(products) So(reactants)
• 229.5 349.98 J/mol-K
• -120.5 J/mol-K

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Entropy Changes in the Surroundings.

• The change in entropy in the surroundings is
  caused entirely by heat transferred from the
  system to the surroundings, and so we have q
  ?H.
• ?S(surroundings) -qsys /T -(?H/T)
• Thus, if asked to calculate the change in
  entropy of the surroundings, just use the above
  equation.

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Section 19.5. Gibbs Free Energy.

• We have seen that both enthalpy changes (?H) and
  entropy changes (?S) contribute to the overall
  energy that drives a reaction. Josiah Willard
  Gibbs (1839-1903) proposed a new state function,
  now called G in his honor, known as the free
  energy.
• G H - TS

J. Willard Gibbs (1839 1903)
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• Again, we work with changes in G, so we have
• ?G ?H - T?S
• increase in increase in
• entropy of entropy of
• surroundings system
• -?G is in fact the increase in entropy in the
  universe.

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• ?H is the increase in entropy of the
  surroundings from ?S(surroundings) -q/T.
• Note
• 1) If ?G is negative, the reaction is
  spontaneous.
• 2) If ?G is zero, the reaction is at
  equilibrium.
• 3) If ?G is positive, the reaction is non-
• spontaneous , but reverse reaction is
• spontaneous.

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19.6. Free Energy and temperature. (p. 827)

• Temperature helps determine whether ?G will be
  negative, and the process spontaneous.
• ?H ?S -T?S ?G ( ?H T?S)
• __________________________________________
• - - - (spontaneous)
• - (non-spontaneous)
• - - or depends on T
• - or - depends on T
• __________________________________________

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DG lt 0 ? reaction is spontaneous (product
favored)
exergonic
DG gt 0 ? reaction is non-spontaneous
endergonic
DG 0 ? reaction is at equilibrium
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19.4 Free Energy Changes (?G) in Chemical
Reactions.

• This parallels the calculation of standard molar
  enthalpies of reaction.
• ?Go(rxn) Sn?Gfo(products) -
  Sm?Gfo(reactants)
• Note that ?Gof is zero for elements.
• Recall that ?Gof is a state function.

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Table of ?Gof , ?Hof, (kJ/mol) and So (J/mol-K)
for some substances (p. 1123)

• Substance ?Hof ?Gof So
• _________________________________________________
  _______
• Al(s) 0 0 28.32
• C(s, graphite) 0 0 5.69
• C2H6(g) -84.68 -32.89 229.5
• O2(g) 0 0 205.0
• Ni(s) 0 0 29.9
• NiCl2(s) -305.3 -259.0
  97.65
• Cl2(g) 0 0 222.96

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• Example Calculate ?Go, ?Ho, and ?So for the
  following (problem 19.54)
• Ni(s) Cl2(g) NiCl2(s)
• ?Ho -305.3 0 0 -305.3 kJ
• ?So 97.65 29.9 222.96 -155.21 J/K
• ?Go -269.0 0 0 -259.0 kJ
• Alternatively, from ?G ?H T?S
• ?Go -305.3 kJ (-155.21 J x 1 kJ x 298 K)
• 1 K 1000 J
• -259.0 kJ

Note units
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The reaction of sodium metal with water
2 Na (s) 2 H2O (l) ? 2 NaOH (aq) H2 (g)
Is the reaction spontaneous?
What is the sign of DG?
What is the sign of DH?
What is the sign of DS?
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The reaction of sodium metal with water
2 Na (s) 2 H2O (l) ? 2 NaOH (aq) H2 (g)
Is the reaction spontaneous?
Yes
What is the sign of DG?
DG negative
What is the sign of DH?
DH negative (exothermic!)
What is the sign of DS?
DS positive
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