Spontaneity Entropy and Free Energy

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Spontaneity Entropy and Free Energy

• WHAT DRIVES A REACTION TO BE SPONTANEOUS?

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ENTHALPY (?H)

• heat content
• (exothermic reactions are generally favored)

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1st Law of Thermodynamics

• The total energy in the universe is constant.
• DE q (heat) w (work)
• -heat out -work on surroundings
• heat in work on system
• Calculating DH
• stoichiometryusing info given with equation
• ?H rxn ? n DHf products -- ? n DHf
  reactants
• Hesss Lawadding up equations and DHs
• calorimetry q mCDT or q CDT
• bond energies ? bonds broken -- ? bonds
  formed
• Note pure elements have DHf values of 0

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ENTROPY SThe Second Law of Thermodynamics

• The universe is constantly increasing disorder.
• DSuniv DSsystem DSsurroundings
• ?Ssurroundings based on heat flow
• exothermic DSsurr
• endothermic -DSsurr

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ENTROPY (?S)

• Disorder of a system (more disorder is favored)
  Nature tends toward chaos! Think about your room
  at the end of the week!
• Factors that can indicate entropy changes
• Phase changes
• Temperature changes
• Volume changes
• Mixing substances
• Change in number of particles
• Change in moles of gas

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Spontaneous reactions are those that occur
without outside intervention. They may occur
fast OR slow (that is kinetics). Some
reactions are very fast (like combustion of
hydrogen) other reactions are very slow (like
graphite turning to diamond)
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Predicting the entropy of a system based on
physical evidence

• The greater the disorder or randomness in a
  system, the larger the entropy.
• The entropy of a substance always increases as it
  changes from solid to liquid to gas.
• When a pure solid or liquid dissolves in a
  solvent, the entropy of the substance increases
  (carbonates are an exception! --they interact
  with water and actually bring MORE order to the
  system)

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Predicting the entropy of a system based on
physical evidence

• When a gas molecule escapes from a solvent, the
  entropy increases.
• Entropy generally increases with increasing
  molecular complexity (crystal structure KCl vs
  CaCl2) since there are more MOVING electrons!
• Reactions increasing the number of moles of
  particles often increase entropy.

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Exercise 2 Predicting Entropy Changes

• Predict the sign of the entropy change for
  each of the following processes.
• A Solid sugar is added to water to
• form a solution.
• B Iodine vapor condenses on a cold
• surface to form crystals.

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Solution

• A ?S
• Sugar molecules have been randomly dispersed in
  the water. Greater entropy DS
• B -?S
• Gaseous iodine is forming a solid. Less entropy
  - DS

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Sample Problem A

• Which of the following has the largest
• increase in entropy?
• a) CO2(s) ? CO2(g)
• b) H2(g) Cl2(g) ? 2 HCl(g)
• c) KNO3(s) ? KNO3(l)
• d) C(diamond) ? C(graphite)

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Answer

• the substance changes from a
• highly organized state to a more
• disorganized state.

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• Choose the substance expected to have the greater
  absolute entropy.
• Pb(s) or Cgraphite
• He(g) at 1 atmosphere or He(g) at 0.5 atmosphere
• H2O(l) or CH3CH2OH(l) at the same temperature
• Mg(s) at 0C or Mg(s) at 150 C both at the same
  pressure

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1. Pb has greater molar entropy. Pb, with metallic
   bonding, forms soft crystals with high amplitudes
   of vibration graphite has stronger (covalent)
   bonds is more rigid and thus more ordered.
2. He(g) at 0.05 has greater molar entropy. At
   lower pressure (greater volume) He atoms have
   more space in which to move so are more random.
3. CH3CH2OH has greater molar entropy. Ethanol
   molecules have more atoms and thus more
   vibrations water exhibits stronger hydrogen
   bonding
4. Mg(s) at 150 C has greater molar entropy. At
   the higher temperature the atoms have more
   kinetic energy and vibrate faster and, thus, show
   greater randomness.

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ENTROPYThe Third Law of Thermodynamics

• The entropy of a perfect crystal at 0 K
• is zero.
• not a lot of perfect crystals out there
• so, entropy values are RARELY ever
• zeroeven elements

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So what?

• This means the absolute entropy of a substance
  can then be determined at any temperature higher
  than 0 K.
• (Handy to know if you ever need to defend why
  G H for elements 0. . . BUT S does not!)

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BIG MAMMA, verse 2

• ?Srxn ? ?S (products) - ? ?S (reactants)
• S is when disorder increases (favored)
• S is when disorder decreases
• Units are usually J/K? mol (not kJ ---tricky!)

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Sample Problem B

• Calculate the entropy change at 25?C, in
• J/K for
• 2 SO2(g) O2(g) ? 2 SO3(g)
• Given the following data
• SO2(g) 248.1 J/K? mol
• O2(g) 205.3 J/K?mol
• SO3(g) 256.6 J/K? mol
• Entropy change -188.3 J/K

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ENTROPY CHANGES FOR REVERSIBLE PHASE CHANGES

• (thats a phase change at constant temperature)
• ?S heat transferred
  q
• temperature at which change occurs
  T
• where the heat supplied (endothermic) (q gt 0)
  or evolved (exothermic) (q lt 0) is divided by the
  temperature in Kelvins

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• EX water (l _at_ 100 ? water (g _at_ 100)
• the entropy will increase.
• Taking favored conditions into consideration,
• the equation above rearranges into
• ?S - ?H
• T
• Give signs to ?H following exo/endo
• guidelines! (If reaction is exo. entropy of
• system increasesmakes sense!)

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• Calculate ?Ssurr for each of these reactions at
  25?C and 1 atm.
• Sb2S3(s)3 Fe(s) ? 2 Sb(s)3 FeS(s) ?H -125kJ
• Sb4O6(s)6 C(s)?4 Sb(s)6 CO(g) ?H 778kJ

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Solution

• ?Ssurr 419 J/K
• ?Ssurr -2.61 103 J/K

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SUMMARY

• ENTROPY
• ?S MORE DISORDER
  
• (FAVORED CONDITION)
• ?S - MORE ORDER

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Whether a reaction will occur spontaneously may
be determined by looking at the ?S of the
universe.

• ?S system ?S surroundings ?S universe
• IF ?S universe is , then reaction is
• spontaneous
• IF ?S universe is -, then reaction is
• NONspontaneous

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Consider

• 2 H2 (g) O2 (g) ? H2O (g)
• ignite rxn is fast!
• ?Ssystem -88.9J/K
• Entropy declines
• (due mainly to 3 ? 2 moles of gas!)

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First law of thermodynamics demands that this
energy is transferred from the system to the
surroundings so... -?Hsystem ?Hsurroundings
OR - (- 483.6 kJ) 483.6 kJ
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• ?Ssurroundings
• ?Hsurroundings 483.6 kJ
• T 298 K
• 1620 J/K

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Now we can find ?Suniverse

• ?S system ?S surroundings ?S universe
• (-88.9 J/K) (1620 J/K) 1530 J/K
• Even though the entropy of the system
  declines, the entropy change for the surroundings
  is SOOO large that the overall change for the
  universe is positive.

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Bottom line

• A process is spontaneous in spite of a
  negative entropy change as long as it is
  extremely exothermic.
• Sufficient exothermicity offsets system
  ordering.

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FREE ENERGY

• Calculation of Gibbs free energy is what
  ultimately decides whether a reaction is
  spontaneous or not.
• NEGATIVE ?Gs are spontaneous.

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?G can be calculated one of several ways

• ?Gºrxn ? ?Go (products) - ? ?Go (reactants)
• ?Gf 0 (for elements in standard state)
• ?G ?H - T?S

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?G ?Go RT ln (Q)

• Define terms
• ?G free energy not at
  standard conditions
• ?Go free energy at standard conditions
• R universal gas constant 8.3145 J/mol?K
• T temp. in Kelvin
• ln natural log
• Q reaction quotient (for gases this is the
  partial pressures of the products divided by the
  partial pressures of the reactantsall raised to
  the power of their coefficients)
• Q products
• reactants

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RatLink

• ?G? -RTlnK
• Terms
• Basically the same as above --- however,
  here the system is at equilibrium, so ?G 0 and
  K represents the equilibrium constant under
  standard conditions.
• K products still raised to power of
  coefficients
• reactants

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nFe ?G? - nFE? Remember
this!!

• Terms
• ?Go just like abovestandard free
• energy
• n number of moles of electrons
• transferred (look at ½
  reactions)
• F Faradays constant 96,485
• Coulombs/mole electrons
• Eo standard voltage
• one volt joule/coulomb

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Sample Problem C

• Find the free energy of formation for the
  oxidation of water to produce hydrogen peroxide.
• 2 H2O(l) O2(g) ? 2 H2O2(l)
• Given the following information
• ?Gf
• H2O(l) -56.7 kcal/mol
• O2(g) 0 kcal/mol
• H2O2(l) -27.2 kcal/mol
• Free energy of formation 59.0 kcal/mol

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Calculate ?H?, ?S?, and ?G?
2 SO2 (g) O2 (g) ? 2 SO3 (g) carried out at
25?C and 1 atm.
Substance ?Hof (kJ/mol) So (J/Kmol)
SO2 (g) -297 248
SO3 (g) -396 257
O2 (g) 0 205
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Solution

• ?H? -198 kJ
• ?S? -187 J/K
• ?G? -142 kJ

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Hesss law of summation

• Using the following data _at_ 25?C, calculate ?Go
• Cdiamond (s) ? Cgraphite (s)
• Cdiamond (s) O2 (g) ?CO2 (g) ?G? -397 kJ
  
• 
  
• Cgraphite (s) O2 (g) ? CO2 (g) ?G? -394 kJ
• ?G? -3 kJ

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?G ?Go RT ln(Q)

• One method for synthesizing methanol (CH3OH)
  involves reacting carbon monoxide and hydrogen
  gases
• CO (g) 2 H2 (g) ? CH3OH (l)
• Calculate ?G at 25?C for this reaction where
  carbon monoxide gas at 5.0 atm and hydrogen gas
  at 3.0 atm are converted to liquid methanol.
• ?G at 25?C -38 kJ/mol rxn

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?G? -RTlnK )

• The overall reaction for the corrosion
  (rusting) of iron by oxygen is
• 4 Fe(s) 3 O2(g) ? 2 Fe2O3(s)
• Calculate the equilibrium constant for this
  reaction at 25?C.
• K .548

Substance ?Hof (kJ/mol) So (J/Kmol)
Fe2O3 (s) -826 90
Fe (s) 0 27
O2 (g) 0 205
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• Gibbs equation can also be used to calculate
  the phase change temperature of a substance.
  During the phase change, there is an equilibrium
  between phases so the value of ?G? is zero.
• Really just like what we did earlier in this
  unit with enthalpy and entropy!

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Sample Problem

• Find the thermodynamic boiling point of
• H2O(l) ? H2O(g)
• Given the following information
• Hvap 44 kJ
• Svap 118.8 J/K
• Thermodynamic BP 370K

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SUMMARY OF FREE ENERGY

• ?G NOT SPONTANEOUS
• ?G - SPONTANEOUS

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Conditions of ?G

• ?H ?S Result
• neg pos spontaneous at all temps
• pos pos spontaneous at high temps
• neg neg spontaneous at low temps
• pos neg not spontaneous, ever

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Relationship to K and E

• ?G K E
  
• 0 at equilibrium K 1 0
• negative gt1, products favored positive
• positive lt1, reactants favored negative

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Summary

• Signs are important!
• DS increased disorder
• - decreased disorder
• DH endothermic
• -exothermic
• DG nonspontaneous
• - spontaneous
• 0 _at_ equilibrium

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Summary of Thermodynamic terms
Term Symbol What it measures Interpretation
Enthalpy change ?H Heat flow exothermic endothermic
Entropy change ?S Disorder disorder ? disorder ?
Gibbs Free Energy change ?G Spontaneity process will occur process will not occur
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3 equations

• D?rxn ? ?f products - ? ?f reactants
• DG DH - TDS
• DG -RTlnK

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?H ?S ?G for the process under consideration
( - ) ( ) Spontaneous at any temperature
( ) ( ) Endo non-spontaneous rxn become spontaneous at high temperature
( - ) ( - ) Exo spontaneous rxn become non-spontaneous at high temperature
( ) ( - ) Non-spontaneous at any temperature
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Enthalpy change

• ?H o ? ? Hof, products - ? ?Hof,
  reactants
• ? Hrxn ? bonds broken - ? bonds formed

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Enthalpy

• ?S ? Sproducts - ? Sreactants
• ?G ?H - T?S
• At equilibrium ?G 0, therefore ?S ?H / T

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Linking thermodynamics, Equilibrium and
Electrochemistry

• ?Go -R T ln Keq nFEo