Chapter 17: Thermodynamics Entropy, Free Energy, and Equilibrium

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Chapter 17 ThermodynamicsEntropy, Free
Energy, and Equilibrium
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• Thermodynamics The study of interconversion of
  heat and other forms of energy
• Every chemical reaction is combined with an
  energy change
• Spontaneous process Process that proceeds on
  its own without external influence (-?G).
  Spontaneous does not mean fast
• Non-spontaneous Needs continuous external
  influence to take place (?G)

A spontaneous reaction always moves a reaction
mixture toward equilibrium!!!
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• Which of the following processes are spontaneous?
• Diffusion of perfume molecules from one side of a
  room
• to the other?
• 2) Heat flow from a cold object to a hot object
• 3) Decomposition of rust to iron metal, oxygen,
  and water

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1st law of thermodynamics The energy of the
universe is constant 2nd law of
thermodynamics In any spontaneous process, the
total entropy of a system and its surroundings
always increases 3rd law of thermodynamics The
entropy of a perfectly ordered crystalline
substance at 0 K is zero
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Thermodynamics

• Enthalpy (H) Heat flow
• In to the system ?H endothermic
• Out of the system -?H exothermic
• Are all the spontaneous reactions exothermic?

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Answer No A spontaneous process does not need to
be exothermic! Examples for endothermic
processes which are spontaneous Melting of ice
above OoC , DHfusion Evaporating of water,
DHvap N2O4 ? 2 NO2 , DH0 We need a
second term which will explain that
even endothermic processes are spontaneous
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Entropy (S)

• Entropy (S) Disorder, molecular randomness
• ?S Sfinal Sinitial
• When disorder increases ?S
• When disorder decreases -?S
• Molecular systems tend to move spontaneously
  to a state of maximum randomness or disorder!!!

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Examples 1) NaCl(s) Na(aq)
Cl-(aq) DH 3.88kJ But disorder increases
from a well organized crystal to free movement in
water. 2) Solid liquid gas
randomness increases but melting and
evaporation are endothermic processes
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• Predict the sign of DS for the following
  processes
• sublimation of dry ice CO2(s) ? CO2(g)
• CaSO4(s) ? CaO(s) SO3(g)
• N2(g) 3H2(g) ? 2NH3(g)
• dissolution of iodine in water I2(s) ? I2(aq)
• formation of rain drops H2O(g) ? H2O(l)
• f) Ag(aq) Cl-(aq) ? AgCl(s)

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Why Prefer ?Sgt0

• 1. Entropy and probability
• 2. Entropy and Temperature

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Probability theory the disordered state can be
achieved in more ways than the ordered state. If
you drop 20 coins you will get a mixture of heads
and tails ( disordered state) It is most
unlikely that you get all 20 coins heads or all
20 coins tails ( ordered state)
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Entropy and temperature

• 3rd Law of Thermodynamics
• The entropy of a perfectly ordered crystalline
  substance
• at 0 K is zero
• As the temperature increases, the kinetic energy
  increases,
• molecular motion increases, entropy increases.
• So we can say
• Entropy is associated with molecular motion

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How can we measure the entropy ? How can we
measure the disorder or randomness? Answer
thats not easy and much more knowledge about
physical chemistry is required This is far
beyond this lectures scope Standard molar
entropies are known for many subtances and listed
in tables
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Standard Molar Entropies and Standard Entropies
of Reaction

• Standard Molar Entropy, S?
• The entropy of one mole of the pure substance at
  1 atm pressure and a specific temperature usually
  25?C (STP, standard temperature and pressure)
• Standard Entropy of Reaction, ?S?
• Entropy change for a chemical reaction
• ?S? ?S?products - ?S?reactants
• Based on 1 mole of substance so you have to
  multiply S? by the number of moles present

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Standard Entropy of Reaction, ?S?

• aA bB ? cC dD
• ?S? c S?(C) d S?(D) a S?(A) b S?(B)
• Units coefficients are moles
• S? J/K mol
• ?S? J/K

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Example 2

• Calculate the standard entropy of reaction at
  25?C for the decomposition of calcium carbonate
• CaCO3(s) ? CaO(s) CO2(g)
• Substance S? (J/K mol)
• CaCO3(s) 92.9
• CaO(s) 39.7
• CO2(g) 213.6

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DS0 S0(CaO) S0(CO2) - S0(CaCO3) DS0
1mol x (39.7J/K mol) 1mol(213.6J/K mol) -
mol(92.9J/K mol) 160.4 J/K Entropy is
positive Disorder increases
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A system tends to move spontaneously toward a
state of minimum enthalpy enthalpy and maximum
entropy
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• 1st Law of Thermodynamics
• In any process, spontaneous or non-spontaneous,
  the total energy of a system and its surroundings
  is constant
• Endothermic or exothermic, no explanation if
  process is spontaneous or non-spontaneous, simply
  a statement of conservation of energy
• 2nd Law of Thermodynamics
• In any spontaneous process, the total entropy of
  a system and its surroundings always increases

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Entropy and the Second Law of Thermodynamics

• ?Stotal ?Ssystem ?Ssurroundings
• if ?S (total) gt 0 spontaneous
• if ?S (total) lt 0 non spontaneous
• if ?S (total) 0 equilibrium
• A reaction that is non-spontaneous in the forward
  
• direction is spontaneous in the reverse direction
• because
• ?S (total) reverse reaction - ?S (total)
  forward reaction

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• a)  Exothermic reaction ?Hlt0, because the
  surroundings gain heat (entropy increases
  ?Ssurrgt0 ), heat is lost from the system
• b) Endothermic reaction ?Hgt0, surroundings
  lose heat (entropy decreases ?Ssurr lt0), and
  system gains the heat

• DSsurr - DH / T

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• We are more interested in the reaction mixture
  (the system)
• than in the surrounding
• Gibbs Free Energy, G
• G H TS enthalpy temp (in K) x
  entropy
• ?G ?H - T?S
• if ?G lt 0 spontaneous
• if DG gt 0 non-spontaneous
• if ?G 0 equilibrium
• Free energy has two terms, one is independent on
  temp
• the other is dependent on the temperature

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DGDH-TDS
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• Consider the decomposition of gaseous N2O4
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• N2O4(g) ? 2 NO2(g) ?H 57.1 kJ
• ?S 175.8 J/K
• Is this reaction spontaneous under standard-state
  conditions at 25?C?
• Estimate the temperature at which the reaction
  becomes spontaneous

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DG DH - TDS 57.1 kJ 298K x 0.1758
kJ/K 4.7 kJ DG gt 0 reaction is
non-spontaneous Set DG 0 and solve for
T 57.1 kJ 0 DH - TDS T DH /
DS ----------------- 325K 0.1758 kJ/K
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Free energy of a substance (its enthalpy and
entropy) depends on temperature pressure physica
l state (solid, liquid, gas) concentration (in
case of solutions) Free energy changes for
chemical reactions must be compared under
well-defined standard state conditions
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Standard Free-Energy Changes for Reactions

• Standard State Conditions Solids, liquids, and
  gases in pure form at 1 atm pressure, Solutes at
  1M concentration, specified temperature, usually
  25 celsius
• Standard Free Energy Change, ?G? The change in
  free energy that occurs when reactants in their
  SS are converted to products in their SS
• ?G ?H - T?S ?G? ?H? - T?S?

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Example 4

• Consider the thermal decomposition of calcium
  carbonate
• CaCO3(s) ? CaO(s) CO2(g) ?H? 178.3 kJ
• ?S? 160.4 J/K
• a) Calculate the standard free energy change
  for this
• reaction at 25?C
• b)    Will a mixture of solid CaCO3, CaO, and
  gaseous
• CO2 at 1 atm pressure react spontaneously at
  25?C?
• c) Assuming that ?H? and ?S? are independent of
  temperature, estimate the temperature at which
  the reaction becomes spontaneous

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• Standard Free Energy of Formation, ?G?f
• The free energy change for formation of one mole
  of the substance in its standard state from the
  most stable form of its constituent elements in
  their standard states
• ?G?f measures the thermodynamic stability with
  respect to its constituent elements
• ?G?f lt 0, the substances are stable and do not
  decompose to their constituent elements under
  standard state conditions

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• ?G?f gt0 the substances are thermodynamically
  unstable with respect to their constituents
  elements
• a) There is no point in trying to synthesize a
  substance that has a ?G?f because it would
  degrade into its constituents
• b) You would need to synthesize it at different
  temperatures and or pressures or start with
  different starting materials that has a reaction
  with a -?G?f
• ?G? ??G?f(products) - ??G?f(reactants)
• 6. General reaction aA bB ? cC dD
• ?G? c?G?f(C) d?G?f(D) a?G?f(A)
  b?G?f(B)

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• Calculate the standard free energy change for the
  reaction of calcium carbide with water. Might
  this reaction be used for the synthesis of
  acetylene (C2H2)?
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• CaC2(s) 2 H2O(l) ? C2H2(g) Ca(OH)2(s)
• ?G?f (CaC2) -64.8 kJ/mol
• ?G?f (H2O(l)) -237.2 kJ/mol
• ?G?f (C2H2) 209.2 kJ/mol
• ?G?f (Ca(OH)2) -898.6 kJ/mol

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• ?G?f (CaC2) -64.8 kJ/mol
• ?G?f (H2O(l)) -237.2 kJ/mol
• ?G?f (C2H2) 209.2 kJ/mol
• ?G?f (Ca(OH)2) -898.6 kJ/mol
• DGof (209.2 (-898.6) - (-64.8 2 x 237.2)
• DGof - 149.2 kJ/mol

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Free Energy changes and Composition of the
Reaction Mixture The sign ( or -) of the free
energy DG0 tells us the direction of spontaneous
reaction. DG0 free energy change under
standard-state conditions What happens if the
reaction mixture is not in STP condition? Nonstan
dard state condition DG DG0 RT lnQ Q
reaction quotient (for gases Qp, for solutions
Qc) Reaction quotient same form as the
equilibrium constant K
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Calculate DG for the formation of ethylene C2H4
from carbon and hydrogen at 250C when the partial
pressures are 100 atm H2 and 0.10 atm C2H4 2
C(s) 2 H2(g) ? C2H4(g) DG0 68.1
kJ Is the reaction spontaneous in the forward or
the reverse direction? pC2H4 0.10 Qp
-------- ------- 1 x 10-5 (pH2)2
1002 DG DG0 RT lnQp DG 68.1 kJ/mol
(8.314 x 10-3 x 298) x ln(1 x 10-5) kJ/mol DG
39.6 kJ/mol Because DG gt 0, the reaction is
spontaneous in the reverse direction
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Free Energy and Chemical Equilibrium

• For ?G ?G? RT ln Q
• When the reaction mixture is mostly reactants
  
• Qltlt1 and RT lnQ ltlt0
• ?Glt0 no matter if DG is positive or negative
• forward reaction is always spontaneous when the
• concentration of products is very small
• When the reaction mixture is mostly products
• Qgtgt1 and RT lnQ gtgt0
• ?Ggt0 no matter if DG is positive or negative
• reverse reaction is always spontaneous when the
• concentration of reactants is very small

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Relationship between free energy and the
equilibrium constant DG DG0 RT ln Q At
equilibrium DG 0 Q K (reaction
quotient equilibrium constant) DG0 RT ln K
0 DG0 - RT ln K K is either Kp for gas
reactions or Kc for solutions
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• DG0 - RT ln K DG0 DHo TDSo
• Setting the two equations equal to each other
• RT ln K DG0 DHo TDSo
• Solving for ln K
• DHo TDSo - DHo TDSo -
  DHo DSo
• ln K ----------------- ------- -------
  ----- -------
• -RT RT RT RT
  R
• - DHo 1 DSo
• ln K ------- -- ------
• R T R

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DG0 - RT ln K allows us to calculate
either the equilibrium constant for a reaction
from the standard free energy change or vice
versa the free energy change for a reaction from
the equilibrium constant
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At 250C, Kw 1.0 x 10-14 Calculate DG0 for the
reaction 2 H2O ? H3O OH- DG0 - RT ln K
K 8.314 x 10-3 kJ/K mol
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At 250C, Kw 1.0 x 10-14 Calculate DG0 for the
reaction 2 H2O ? H3O OH- DG0 - RT ln K
DGo - (8.314 x 10-3 x 298 x ln(1.0 x
10-14) 80 kJ/mol
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Calculate the vapor pressure of water at
25oC H2O (l) ? H2O(g) equilibrium equation
for this process Kp pH2O DGo - RT ln
Kp Go to appendix B in your text book and
find DGof for H2O (l) -237.2 kJ/mol DGof for
H2O (g) -228.6 kJ/mol
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DGo DGof(H2O g) - DGof(H2O l) -228.6
(-237.2) 8.6 kJ -8.6 ln Kp
ln pH2O - DGo / RT -----------------------
---- 8.314 x 10-3 x 298 ln pH2O -
3.471 pH2O 0.0311 atm or 23.6 mm
Hg