1. Chapter 17 Free Energy and Thermodynamics
Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 17 Free Energy and Thermodynamics
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall

2. First Law of Thermodynamics
you can’t win!
First Law of Thermodynamics: Energy cannot be Created or Destroyed
the total energy of the universe cannot change
though you can transfer it from one place to another
DEuniverse = 0 = DEsystem + DEsurroundings
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3. First Law of Thermodynamics
Conservation of Energy
For an exothermic reaction, “lost” heat from the system goes into the surroundings
two ways energy “lost” from a system,
converted to heat, q
used to do work, w
Energy conservation requires that the energy change in the system equal the heat released + work done
DE = q + w
DE = DH + PDV
DE is a state function
internal energy change independent of how done
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4. Energy Tax you can’t break even!
to recharge a battery with 100 kJ of useful energy will require more than 100 kJ
every energy transition results in a “loss” of energy
conversion of energy to heat which is “lost” by heating up the surroundings
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5. fewer steps generally results in a lower total heat tax
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6. Thermodynamics and Spontaneity
thermodynamics predicts whether a process will proceed under the given conditions
spontaneous process
nonspontaneous processes require energy input to go
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
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7. Comparing Potential Energy
The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.
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8. Reversibility of Process
any spontaneous process is irreversible
it will proceed in only one direction
a reversible process will proceed back and forth between the two end conditions
equilibrium
results in no change in free energy
if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction
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9. Thermodynamics vs. Kinetics
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10. Diamond → Graphite
Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations).
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11. 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 relates to the randomness/orderliness of a system DS
The enthalpy factor is generally more important than the entropy factor
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12. Enthalpy 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 DH°rxn = S(DH°prod) - S(DH°react)
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14. Entropy
entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S
S generally J/mol
S = k ln W
k = Boltzmann Constant = 1.38 x J/K
W is the number of energetically equivalent ways, unitless
Random systems require less energy than ordered systems
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15. W Energetically Equivalent States for the Expansion of a Gas
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16. Macrostates → Microstates
These microstates all have the same macrostate
So there are 6 different particle arrangements that result in the same macrostate
This macrostate can be achieved through
several different arrangements of the particles
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17. Macrostates and Probability
There is only one possible arrangement that gives State A and one that gives State C
There are 6 possible arrangements that give State B
Therefore State B has higher entropy than either State A or State B
The macrostate with the highest entropy also has the greatest dispersal of energy
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18. Changes in Entropy, DS
entropy change is favorable when the result is a more random system.
DS is positive
Some changes that increase the entropy are:
reactions whose products are in a more disordered state.
(solid > liquid > gas)
reactions which have larger numbers of product molecules than reactant molecules.
increase in temperature
solids dissociating into ions upon dissolving
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19. Increases in Entropy
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20. The 2nd Law of Thermodynamics
the total entropy change of the universe must be positive for a process to be spontaneous
for reversible process DSuniv = 0,
for irreversible (spontaneous) process DSuniv > 0
DSuniverse = DSsystem + DSsurroundings
if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount
when DSsystem is negative, DSsurroundings is positive
the increase in DSsurroundings often comes from the heat released in an exothermic reaction
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21. Entropy Change in State Change
when materials change state, the number of macrostates it can have changes as well
for entropy: solid < liquid < gas
because the degrees of freedom of motion increases solid → liquid → gas
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22. Entropy Change and State Change
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23. Heat Flow, Entropy, and the 2nd Law
Heat must flow from water to ice in order for the entropy of the universe to increase
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24. Temperature Dependence of DSsurroundings
when a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings
when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings
the amount the entropy of the surroundings changes depends on the temperature it is at originally
the higher the original temperature, the less effect addition or removal of heat has
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25. DGreaction = S nDGprod – S nDGreact
Gibbs Free Energy, DG
maximum amount of energy from the system available to do work on the surroundings
G = H – T∙S
DGsys = DHsys – TDSsys
DGsys = – TDSuniverse
DGreaction = S nDGprod – S nDGreact
when DG < 0, there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when DG is negative
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26. Ex. 17.2a – The reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) has DHrxn = kJ at 25°C. Calculate the entropy change of the surroundings.
Given:
Find:
DHsystem = kJ, T = 298 K
DSsurroundings, J/K
Concept Plan:
Relationships: DS
T, DH
Solution:
Check:
combustion is largely exothermic, so the entropy of the surrounding should increase significantly

27. Free Energy Change and Spontaneity
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28. Gibbs Free Energy, DG process will be spontaneous when DG is negative
DG will be negative when
DH is negative and DS is positive
exothermic and more random
DH is negative and large and DS is negative but small
DH is positive but small and DS is positive and large
or high temperature
DG will be positive when DH is + and DS is −
never spontaneous at any temperature
when DG = 0 the reaction is at equilibrium
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29. DG, DH, and DS
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30. Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = kJ and DS = J/K at 25°C. Calculate DG and determine if it is spontaneous.
Given:
Find:
DH = kJ, DS = J/K, T = 298 K
DG, kJ
Concept Plan:
Relationships: DG
T, DH, DS
Solution:
Since DG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.
Answer:

31. Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = kJ and DS = J/K. Calculate the minimum temperature it will be spontaneous.
Given:
Find:
DH = kJ, DS = J/K, DG < 0
T, K
Concept Plan:
Relationships: T
DG, DH, DS
Solution:
The temperature must be higher than 673K for the reaction to be spontaneous
Answer:

32. The 3rd Law of Thermodynamics Absolute Entropy
the absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles
the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K
therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy
therefore, the absolute entropy of substances is always +
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33. Standard Entropies S° extensive
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
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35. Relative Standard Entropies States
the gas state has a larger entropy than the liquid state at a particular temperature
the liquid state has a larger entropy than the solid state at a particular temperature
Substance
S°,
(J/mol∙K)
H2O (g)
70.0
H2O (l)
188.8
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36. Relative Standard Entropies Molar Mass
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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37. Relative Standard Entropies Allotropes
the less constrained the structure of an allotrope is, the larger its entropy
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38. Relative Standard Entropies Molecular Complexity
larger, more complex molecules generally have larger entropy
more available energy states, allowing more dispersal of energy through the states
Substance
Molar
Mass
S°,
(J/mol∙K)
Ar (g)
39.948
154.8
NO (g)
30.006
210.8
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39. Relative Standard Entropies Dissolution
dissolved solids generally have larger entropy
distributing particles throughout the mixture
Substance
S°,
(J/mol∙K)
KClO3(s)
143.1
KClO3(aq)
265.7
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40. Substance
S, J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
188.8
Ex –Calculate DS for the reaction 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
Given:
Find:
standard entropies from Appendix IIB
DS, J/K
Concept Plan:
Relationships: DS
SNH3, SO2, SNO, SH2O,
Solution:
Check:
DS is +, as you would expect for a reaction with more gas product molecules than reactant molecules

41. DGreaction = DHreaction – TDSreaction
Calculating DG
at 25C:
DGoreaction = SnGof(products) - SnGof(reactants)
at temperatures other than 25C:
assuming the change in DHoreaction and DSoreaction is negligible
DGreaction = DHreaction – TDSreaction
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43. Substance
DGf, kJ/mol
CH4(g)
-50.5
O2(g)
0.0
CO2(g)
-394.4
H2O(g)
-228.6
O3(g)
163.2
Ex –Calculate DG at 25C for the reaction CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g)
Given:
Find:
standard free energies of formation from Appendix IIB
DG, kJ
Concept Plan:
Relationships:
DG
DGf of prod & react
Solution:

44. Ex. 17. 6 – The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = -98
Ex – The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = kJ and DS = J/K at 25°C. Calculate DG at 125C and determine if it is spontaneous.
Given:
Find:
DH = kJ, DS = J/K, T = 398 K
DG, kJ
Concept Plan:
Relationships:
DG
T, DH, DS
Solution:
Since DG is -, the reaction is spontaneous at this temperature, though less so than at 25C
Answer:

45. DG Relationships
if a reaction can be expressed as a series of reactions, the sum of the DG values of the individual reaction is the DG of the total reaction
DG is a state function
if a reaction is reversed, the sign of its DG value reverses
if the amounts of materials is multiplied by a factor, the value of the DG is multiplied by the same factor
the value of DG of a reaction is extensive
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46. Free Energy and Reversible Reactions
the change in free energy is a theoretical limit as to the amount of work that can be done
if the reaction achieves its theoretical limit, it is a reversible reaction
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47. Real Reactions
in a real reaction, some of the free energy is “lost” as heat
if not most
therefore, real reactions are irreversible
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48. DG under Nonstandard Conditions
DG = DG only when the reactants and products are in their standard states
there normal state at that temperature
partial pressure of gas = 1 atm
concentration = 1 M
under nonstandard conditions, DG = DG + RTlnQ
Q is the reaction quotient
at equilibrium DG = 0
DG = ─RTlnK
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50. DG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
Example - DG
Calculate DG at 427°C 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)
Q =
PNH32
PN21 x PH23
(2.0 atm)2
(33.0 atm)1 (99.0)3 =
= 1.2 x 10-7
DH° = [ 2(-46.19)] - [0 +3( 0)] = kJ = J
DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K
DG° = J - (700 K)( J/K)
DG° = J
DG = DG° + RTlnQ
DG = J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
DG = J = -46 kJ

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52. Example - K
Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C
N2(g) + 3 H2(g) Û 2 NH3(g)
DH° = [ 2(-46.19)] - [0 +3( 0)] = kJ = J
DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K
DG° = J - (700 K)( J/K)
DG° = J
DG° = -RT lnK
J = -(8.314 J/K)(700 K) lnK
lnK = -7.97
K = e-7.97 = 3.45 x 10-4
since K is << 1, the position of equilibrium favors reactants

53. Temperature Dependence of K
for an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant
for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant
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