Chapter 17 Free Energy and Thermodynamics

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Chapter 17Free Energy and Thermodynamics
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Goals
Entropy (S, ?S) and spontaneity
Free energy ?G, ?Go
?G, K, product- or reactant-favored
Review H and the 1st Law of Thermodynamics
Chemical Equilibria (ch. 14, etc)
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First Law of Thermodynamics

• First Law of Thermodynamics Energy cannot be
  created or destroyed
• the total energy of the universe cannot change
• it can be transfered from one place to another
• DEuniverse 0 DEsystem DEsurroundings
• system reactants products
• surroundings everything else
• (the transfer of energy from one to the other
  does not change the energy of the universe)

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First Law of Thermodynamics

• For an exothermic reaction, heat from the system
  goes into the surroundings
• two ways energy can be lost from a system,
• converted to heat, q
• used to do work, w
• Energy conservation requires that the energy
  change in the system heat exchanged work
  done
• DE q w (?E internal energy change)
• DE DH PDV (qp DH, enthalpy change)
• DE is a state function (H, P, V)
• internal energy change independent of how done

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Enthalpy, H

• related to the internal energy
• DH generally kJ/mol
• stronger bonds more stable molecules
• if products more stable than reactants, energy
  released exothermic
• DH negative
• if reactants more stable than products, energy
  absorbed endothermic
• DH positive
• The enthalpy is favorable for exothermic
  reactions and unfavorable for endothermic
  reactions.
• Hess Law DHrxn S(DHprod) - S(DHreact)

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Thermodynamics and Spontaneity

• thermodynamics predicts whether a process will
  proceed (occur) under the given conditions
• spontaneous process
• nonspontaneous process does not occur under
  specific conditions.
• spontaneity is determined by comparing the free
  energy of the system before the reaction with the
  free energy of the system after reaction.
• if the system after reaction has less free energy
  than before the reaction, the reaction is
  thermodynamically favorable.
• spontaneity ? fast or slow (rate)

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Spontaneous
Nonspontaneous
ice melts _at_ 25oC water freezes _at_
25oC
water freezes _at_ -10 oC ice melts _at_
-10oC
ball rolls downhill ball rolls
uphill
2Na(s) 2H2O(l) ? H2(g) 2NaOH(aq) ?
H2(g) 2NaOH(aq) 2Na(s) 2H2O(l)
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Diamond ? Graphite
kinetics how fast
Spontaneity direction extent
kinetics
Graphite is thermodynamically more stable than
diamond, so the conversion of diamond into
graphite is spontaneous but its kinetically
too slow (inert) it will never happen in many,
many generations.
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Factors Affecting Whether a Reaction Is
Spontaneous

• The two factors that determine the thermodynamic
  favorability are the enthalpy and the entropy.
• The enthalpy is a comparison of the bond energy
  of the reactants to the products.
• bond energy amount needed to break a bond.
• DH
• The entropy factors relate to the
  randomness/orderliness of a system
• DS
• The enthalpy factor is generally more important
  than the entropy factor

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Entropy, S
Entropy is usually described as a measure of the
randomness or disorder the greater the disorder
of a system, the greater its S. The greater the
order ? the smaller its S.

• Entropy is a thermodynamic function that
  increases as the number of energetically
  equivalent ways of arranging the components
  increases.
• S generally J/K
• S k ln W
• k Boltzmann Constant (R/NA) 1.38 ? 10-23 J/K
• W is the number of energetically equivalent ways,
  (microstates) unitless.

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Entropy Microstates, W
S k ln W
Energetically Equivalent States for the Expansion
of a Gas (4 gas molecules)
1 microstate
1 microstate
?S k ln Wf - k ln Wi
if Wf gt Wi , ?S gt 0 entropy increases.
6 microstates (most probable
distribution)
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Changes in Entropy, DS

• entropy change is favorable when the result is a
  more random system (State C higher entropy).
• DS is positive (DS gt 0)

Some changes that increase the entropy are rxns
where products are in a more disordered state.
(solid gt liquid gt gas) less order (solid lt
liquid lt gas)? larger S reactions which have
larger numbers of product molecules than reactant
molecules. increase in temperature (more
movement) solids dissociating into ions upon
dissolving
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Changes in Entropy in a System(melting)
Particles fixed in space
Particles can occupy many positions
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Changes in Entropy in a System(vaporization)
Particles occupy more space (larger volume)
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Changes in Entropy in a System (solution process)
Structure of solute and solvent disrupted
(also more solute particles)
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Predict entropy change for a process/reaction
For which process/reaction is ?S negative?
Freezing ethanol
? entropy dec
Mixing CCl4 with C6H6
Condensing bromine vapor
? entropy dec
2O3(g) ? 3O2(g)
4Fe(s) 3O2(g) ? 2Fe2O3(s)
? entropy dec
2H2O2(aq) ? 2H2O(l) O2(g)
2Li(s) 2H2O(l) ? 2LiOH(aq) H2(g)
2NH3(g) ? N2(g) 3H2(g)
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The 2nd Law of Thermodynamics

• The entropy of the universe increases in a
  spontaneous process.
• DSuniverse DSsystem DSsurroundings gt 0
• DSuniverse DSsystem DSsurroundings 0
  (equilibrium)
• If DSsystem gtgt 0, DSsurroundings lt 0 for
  DSuniverse gt 0!
• If DSsystem lt 0, DSsurroundings gtgt 0 for
  DSuniverse gt 0!
• the increase in DSsurroundings often comes from
  the heat released in an exothermic reaction,
  DHsystem lt 0.

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The 3rd Law of Thermodynamics
-allows determination of entropy of substances.
(W 1, there is only one way to arrange the
particles to form a perfect crystal)
S k ln W k ln 1 0

• the 3rd Law states that for a perfect crystal at
  absolute zero, the absolute entropy 0 J/molK

?S Sf Si where Si 0 _at_ 0 K the absolute
entropy of a substance is always () positive
at the new T
S k ln W
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Standard Entropies

• S
• entropies for 1 mole at 298 K for a particular
  state, a particular allotrope, particular
  molecular complexity, a particular molar mass,
  and a particular degree of dissolution

Values can be used to calculate the standard
entropy change for a reaction, ?Sorxn ( ?Sosys)
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Trends Standard EntropiesMolar Mass

• For monatomic species, the larger the molar mass,
  the larger the entropy
• available energy states more closely spaced,
  allowing more dispersal of energy through the
  states

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Trends Standard EntropiesStates

• the standard entropy of a substance in the gas
  phase is greater than the standard entropy of the
  same substance in the solid or liquid phase at a
  particular temperature

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Trends Standard EntropiesAllotropes
-different forms of an element

• the more highly ordered form has the smaller
  entropy

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Trends Standard EntropiesMolecular Complexity
(inc of atoms)

• larger, more complex molecules generally have
  larger entropy
• more available energy states, allowing more
  dispersal of energy through the states

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Trends Standard EntropiesDissolution

• dissolved solids generally have larger entropy
• distributing particles throughout the mixture

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Q. Arrange the following in order of increasing
entropy _at_ 25oC! (lowest to highest)
Ne(g), SO2(g), Na(s), NaCl(s) and H2(g)
Na(s) lt NaCl(s) lt H2(g) lt Ne(g) lt SO2(g)
Q . Which has the larger entropy in each pair?

• Li(s) or Li(l)
• b) C2H5OH(l) or CH3OCH3 (l)
• c) Ar(g) or Xe(g)
• d) O2(g) or O3(g)

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Calculate DS? for the reaction4 NH3(g) 5 O2(g)
? 4 NO(g) 6 H2O(g)
standard entropies from Appendix IIB DS, J/K
Given Find
Concept Plan Relationships
Solution
DS is , as you would expect for a reaction with
more gas product molecules than reactant molecules
Check
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Calculating ?So for a Reaction
?So ? So (products) - ? So (reactants)

• Consider 2 H2(g) O2(g) ? 2 H2O(liq) _at_ 25oC
• ?So 2 So (H2O) - 2 So (H2) So (O2)
• ?So 2 mol (69.9 J/Kmol) - 2 mol (130.6
  J/Kmol) 1
  mol (205.0 J/Kmol)
• ?So -326.4 J/K
• Note that there is a decrease in S because 3 mol
  of gas give 2 mol of liquid.

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Temperature Dependence of DSsurroundings

• DHsystem lt 0 (exothermic), it adds heat to the
  surroundings, increasing the entropy of the
  surroundings (DSsurroundings gt 0 )
• DHsystem gt 0 (endothermic), it takes heat from
  the surroundings, decreasing the entropy of the
  surroundings (DSsurroundings lt 0 )

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• 2 H2(g) O2(g) ? 2 H2O(liq) _at_ 25 oC
• ?Sosystem -326.4 J/K

Can calculate ?Horxn ?Hosystem -571.7 kJ
(also from tabulated data)
?Sosurroundings 1917 J/K
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Given DSosurr , DSosys and T, determine DSouniv
and predict if the reaction will be spontaneous.

• 2 H2(g) O2(g) ? 2 H2O(liq) _at_ 25oC
• ?Sosystem -326.4 J/K
• ?Sosurroundings 1917 J/K
• ?Souniverse 1591 J/K

• The entropy of the universe is increasing, so the
  reaction is spontaneous ( product-favored).

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The reaction C3H8(g) 5 O2(g) ? 3 CO2(g) 4
H2O(g) has DHrxn -2044 kJ at 25C. Calculate
the entropy change of the surroundings.
DHsystem -2044 kJ, T 298 K DSsurroundings, J/K
Given Find
Concept Plan Relationships
Solution
Check
combustion is largely exothermic, so the entropy
of the surroundings should increase (inc in
gas mol)
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Spontaneous or Not?
DSuniverse DSsystem DHsystem/T
originally DSuniverse DSsystem
DSsurroundings but
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Without doing any calculations, determine the
sign of Ssys and Ssurr for each reaction.
Predict under what temperatures (all T, low T, or
high T) the reaction will be spontaneous.
DSuniverse DSsystem (-DHsys/T)
?Ssurr (-DHsys/T)
2CO(g) O2(g) ? 2CO2(g) DHrxn
-566.0 kJ
DSsystem (-) 3 mol gas form 2 mol gas DSsurr
() spontaneous _at_ low T
2NO2(g) ? O2(g) 2NO(g) DHrxn 113.1
kJ
DSsystem () 2 mol gas form 3 mol gas DSsurr
(-) spontaneous _at_ high T
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Without doing any calculations, determine the
sign of Ssys and Ssurr for each reaction.
Predict under what temperatures (all T, low T, or
high T) the reaction will be spontaneous.
DSuniverse DSsystem (-DHsys/T)
2H2(g) O2(g) ? 2H2O(g) DHrxn
-483.6 kJ
DSsystem (-) 3 mol gas form 2 mol gas DSsurr
() spontaneous _at_ low T
CO2(g) ? C(s) O2(g) DHrxn 393.5 kJ

DSsystem (-) complicated gas forms a solid
gas DSsurr (-) nonspontaneous _at_ all T
?Ssurr (-DHsys/T)
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At what temperature is the change in entropy for
the reaction equal to the change in entropy for
the surroundings, if ?Horxn -127 kJ and ?Sorxn
314 J/K.
Plan set ?Sorxn ?Sosurr and solve for
T convert kJ to J
rxn implies system!!!
Ans T 404 K
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Gibbs Free Energy, G H -TS
?Suniv ?Ssurr ?Ssys
J. Willard Gibbs1839-1903

• Multiply through by -T
• -T?Suniv ?Hsys - T?Ssys
• -T?Suniv change in Gibbs free energy for the
  system ?Gsystem
• hence, ?G ?H -T?S
• Under standard conditions
• ?Gosys ?Hosys - T?Sosys

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?G ?H - T?S

• Gibbs free energy change total energy change
  for system
• - energy lost in disordering the system
• If the reaction is
• exothermic (negative ?H)
• and entropy increases (positive ?So)
• then ?G must be NEGATIVE
• the reaction is spontaneous (and
  product-favored) at ALL temperatures.

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?G ?H - T?S

• DG will be positive (?G gt 0) when DH is positive
  (endothermic) and DS is negative (more ordered).
  So the change in free energy will be positive at
  all temperatures.
• The reaction will therefore be nonspontaneous at
  ALL temperatures
• When DG 0 the reaction is at equilibrium

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DG DH TDSSpontaneous or Not?
A decrease in Gibbs free energy (DG lt 0)
corresponds to a spontaneous process
An increase in Gibbs free energy (DG gt 0)
corresponds to a nonspontaneous process
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Calculating ?Go ?Go ?H -T?So

• Combustion of acetylene _at_ 25 oC
• C2H2(g) 5/2 O2(g) ? 2 CO2(g) H2O(g)
• Use enthalpies of formation to calculate
• ?Horxn -1238 kJ
• Use standard molar entropies to calculate
• ?Sorxn -97.4 J/K or -0.0974 kJ/K
• ?Gorxn -1238 kJ - (298 K)(-0.0974 J/K)
• -1209 kJ (spontaneous)
• Reaction is product-favored in spite of negative
  ?Sorxn.
• Reaction is enthalpy driven

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Calculating ?Go ?Go ?H -T?So
Is the dissolution of ammonium nitrate
product-favored? If so, is it enthalpy- or
entropy-driven?
NH4NO3(s) heat ? NH4NO3(aq)

• From tables of thermodynamic data we find
• ?Horxn 25.7 kJ (endothermic)
• ?Sorxn 108.7 J/K or 0.1087 kJ/K
  (disorder)
• ?Gorxn 25.7 kJ - (298 K)(0.1087 kJ/K)
• -6.7 kJ (spontaneous)
• Reaction is product-favored in spite of positive
  ?Horxn.
• Reaction is entropy driven

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The reaction CCl4(g) ? C(s, graphite) 2 Cl2(g)
has DH 95.7 kJ and DS 142.2 J/K at 25C.
Calculate DG and determine if it is spontaneous.
DH 95.7 kJ, DS 142.2 J/K, T 298 K DG, kJ
Given Find
Concept Plan Relationships
Solution
Since DG is , the reaction is not spontaneous at
this temperature. To make it spontaneous, we
need to increase the temperature.
Answer
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The reaction CCl4(g) ? C(s, graphite) 2 Cl2(g)
has DH 95.7 kJ and DS 142.2 J/K.
Calculate the minimum T _at_ which it will be
spontaneous.
DH 95.7 kJ, DS 142.2 J/K, DG lt 0 T, K
Given Find
Concept Plan Relationships
Solution
The temperature must be higher than 673K for the
reaction to be spontaneous (i.e. 674 K)
Answer
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Gibbs Free Energy, G

• Calculating ?Go (two ways)
• a) Determine ?Horxn and ?Sorxn and use Gibbs
  equation (at various temps).
• b) Use tabulated values of free energies of
  formation, ?Gfo _at_ 25oC

?Gorxn ? ?Gfo (products) - ? ?Gfo (reactants)
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Calculate DG? at 25?C for the reactionCH4(g) 8
O2(g) ? CO2(g) 2 H2O(g) 4 O3(g)
standard free energies of formation from Appendix
IIB DG?, kJ
Given Find
Concept Plan Relationships
Solution
(spontaneous)
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The reaction SO2(g) ½ O2(g) ? SO3(g) has DH?
-98.9 kJ and DS? -94.0 J/K at 25C. Calculate
DG? at 125?C and determine if it is spontaneous.
DH? -98.9 kJ, DS? -94.0 J/K, T 398 K DG?, kJ
Given Find
(PRACTICE PROBLEM)
Concept Plan Relationships
Solution
Since DG is -ve, the rxn is spontaneous at this
temperature, but less spontaneous than at 25?C
(-127 kJ)
Answer
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?G, ?G, and Keq

• ?G is the change in free energy at non-standard
  conditions.
• ?G is related to ?G
• ?G ?G RT ln Q where Q reaction
  quotient
• When Q lt K or Q gt K, the reaction is spontaneous.
• When Q K reaction is at equilibrium
• When ?G 0 reaction is at equilibrium
• Therefore, ?G - RT ln K

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Thermodynamics and Keq

• FACT ?Gorxn is the change in free energy when
  pure reactants convert COMPLETELY to pure
  products.
• FACT Product-favored systems have Keq gt 1
  (?Grxn lt 0).
• Therefore, both ?Grxn and Keq are related to
  reaction favorability.

Summary ?G - RT ln K
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?Gorxn - RT lnK

• Calculate K for the reaction _at_ 25 oC
• N2O4 ? 2 NO2 ?Gorxn 4.8 kJ
• ?Gorxn 4800 J - (8.31 J/K)(298 K) ln K

K e1.94 0.14 (reactant favored) When
?Gorxn gt 0 (nonspontaneous), then K lt 1!!
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• Estimate the equilibrium constant and position of
  equilibrium for the following reaction at 427C
• N2(g) 3 H2(g) ? 2 NH3(g)

DH 2(-46.19) - 0 3( 0) -92.38 kJ
-92380 J
DS 2 (192.5) - (191.50) 3(130.58)
-198.2 J/K
DG -92380 J - (700 K)(-198.2 J/K)
DG 46400 J (nonspontaneous)
DG -RT lnK
46400 J -(8.314 J/K)(700 K) lnK
ln K -7.97
K e-7.97 3.45 ? 10-4
since K is ltlt 1, the position of equilibrium
favors reactants
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• Calculate DG at 427C for the reaction below if
  the PN2 33.0 atm, PH2 99.0 atm, and PNH3 2.0
  atm
• N2(g) 3 H2(g) ? 2 NH3(g)

DH 2(-46.19) - 0 3( 0) -92.38 kJ
-92380 J
DS 2 (192.5) - (191.50) 3(130.58)
-198.2 J/K
DG -92380 J - (700 K)(-198.2 J/K)
DG 46400 J (nonspontaneous)
DG -46300 J -46 kJ ? DG
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Q. Rank the following in order of increasing
molar entropy (So) _at_ 25oC!
a) Cl2(g), I2(g), Br2(g), and F2(g)
F2(g) lt Cl2(g) lt Br2(g) lt I2(g)
b) H2O(g), H2O2 (g), H2S(g)
H2O(g) lt H2S(g) lt H2O2(g)
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Use only the information provided here to
determine the value of ?Ssurr _at_ 355 K. Predict
whether this reaction shown will be spontaneous _at_
this temperature, if ?Hrxn -114 kJ.

A) ?Ssurr 114 kJ/K, reaction is not
spontaneous
B) ?Ssurr 321 J/K, reaction is spontaneous
C) ?Ssurr 114 kJ/K, reaction is spontaneous
D) ?Ssurr -355 J/K, reaction is not
spontaneous
E) ?Ssurr 321 J/K, it is not possible to
predict the spontaneity of this reaction
without more information.