1. Chapter 17 Free Energy and Thermodynamics
Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro
Chapter 17 Free Energy and Thermodynamics
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA

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 is “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. The Energy Tax You can’t break even!
To recharge a battery with 100 kJ of useful energy will require more than 100 kJ
because of the Second Law of Thermodynamics
Every energy transition results in a “loss” of energy
an “Energy Tax” demanded by nature
and 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 occur under the given conditions
processes that will occur are called spontaneous
nonspontaneous processes require energy input to go
Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction
if the system after reaction has less potential 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 because there is a net release of energy when it proceeds in that direction
it will proceed in only one direction
A reversible process will proceed back and forth between the two end conditions
any reversible process is at 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. Spontaneous Processes
Spontaneous processes occur because they release energy from the system
Most spontaneous processes proceed from a system of higher potential energy to a system at lower potential energy
exothermic
But there are some spontaneous processes that proceed from a system of lower potential energy to a system at higher potential energy
endothermic
How can something absorb potential energy, yet have a net release of energy?
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12. Melting Ice
Melting is an Endothermic process, yet ice will spontaneously melt above 0 °C.
When a solid melts, the particles have more freedom of movement.
More freedom of motion increases the randomness of the system. When systems become more random, energy is released. We call this energy, entropy
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13. Factors Affecting Whether a Reaction Is Spontaneous
There are two factors that determine whether a reaction is spontaneous. They are the enthalpy change and the entropy change of the system
The enthalpy change, DH, is the difference in the sum of the internal energy and PV work energy of the reactants to the products
The entropy change, DS, is the difference in randomness of the reactants compared to the products
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14. Enthalpy Change DH generally measured in kJ/mol
Stronger bonds = more stable molecules
A reaction is generally exothermic if the bonds in the products are stronger than the bonds in the reactants
exothermic = energy released, DH is negative
A reaction is generally endothermic if the bonds in the products are weaker than the bonds in the reactants
endothermic = energy absorbed, DH is positive
The enthalpy change is favorable for exothermic reactions and unfavorable for endothermic reactions
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15. 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 10−23 J/K
W is the number of energetically equivalent ways a system can exist
unitless
Random systems require less energy than ordered systems
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16. W
These are energetically equivalent states for the expansion of a gas.
It doesn’t matter, in terms of potential energy, whether the molecules are all in one flask, or evenly distributed
But one of these states is more probable than the other two
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17. Macrostates → Microstates
These microstates all have the same macrostate
So there are six 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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18. Macrostates and Probability
There is only one possible arrangement that gives State A and one that gives State B
There are six possible arrangements that give State C
The macrostate with the highest entropy also has the greatest dispersal of energy
Therefore State C has higher entropy than either State A or State B
There is six times the probability of having the State C macrostate than either State A or State B
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19. Changes in Entropy, DS DS = Sfinal − Sinitial
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 random state
solid more ordered than liquid more ordered than gas
reactions that have larger numbers of product molecules than reactant molecules
increase in temperature
solids dissociating into ions upon dissolving
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20. Increases in Entropy
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21. DS
For a process where the final condition is more random than the initial condition, DSsystem is positive and the entropy change is favorable for the process to be spontaneous
For a process where the final condition is more orderly than the initial condition, DSsystem is negative and the entropy change is unfavorable for the process to be spontaneous
DSsystem = DSreaction = Sn(S°products) − Sn(S°reactants)
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22. Entropy Change in State Change
When materials change state, the number of macrostates it can have changes as well
the more degrees of freedom the molecules have, the more macrostates are possible
solids have fewer macrostates than liquids, which have fewer macrostates than gases
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23. Entropy Change and State Change
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24. Practice – Predict whether DSsystem is + or − for each of the following
A hot beaker burning your fingers
Water vapor condensing
Separation of oil and vinegar salad dressing
Dissolving sugar in tea
2 PbO2(s)  2 PbO(s) + O2(g)
2 NH3(g)  N2(g) + 3 H2(g)
Ag+(aq) + Cl−(aq)  AgCl(s)
DS is +
DS is −
DS is −
DS is +
DS is +
DS is +
DS is −
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25. The 2nd Law of Thermodynamics
The 2nd Law of Thermodynamics says that 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 must be positive and big for a spontaneous process
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26. Heat Flow, Entropy, and the 2nd Law
When ice is placed in water, heat flows from the water into the ice
According to the 2nd Law, heat must flow from water to ice because it results in more dispersal of heat. The entropy of the universe increases.
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27. Heat Transfer and Changes in Entropy of the Surroundings
The 2nd Law demands that the entropy of the universe increase for a spontaneous process
Yet processes like water vapor condensing are spontaneous, even though the water vapor is more random than the liquid water
If a process is spontaneous, yet the entropy change of the process is unfavorable, there must have been a large increase in the entropy of the surroundings
The entropy increase must come from heat released by the system – the process must be exothermic!
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28. Entropy Change in the System and Surroundings
When the entropy change in system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive), and large to allow the process to be spontaneous
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29. Heat Exchange and 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 its original temperature
the higher the original temperature, the less effect addition or removal of heat has
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30. Temperature Dependence of DSsurroundings
When heat is added to surroundings that are cool it has more of an effect on the entropy than it would have if the surroundings were already hot
Water freezes spontaneously below 0 °C because the heat released on freezing increases the entropy of the surroundings enough to make DS positive
above 0 °C the increase in entropy of the surroundings is insufficient to make DS positive
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31. Quantifying Entropy Changes in Surroundings
The entropy change in the surroundings is proportional to the amount of heat gained or lost
qsurroundings = −qsystem
The entropy change in the surroundings is also inversely proportional to its temperature
At constant pressure and temperature, the overall relationship is
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32. Example 17.2a: Calculate the entropy change of the surroundings at 25ºC for the reaction below C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DHrxn = −2044 kJ
Given:
Find:
DHsystem = −2044 kJ, T = 25 ºC = 298 K
DSsurroundings, J/K
Conceptual Plan:
Relationships: DS
T, DH
Solution:
Check:
combustion is largely exothermic, so the entropy of the surroundings should increase significantly
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33. Practice – The reaction below has DHrxn = +66. 4 kJ at 25 °C
Practice – The reaction below has DHrxn = kJ at 25 °C. (a) Determine the Dssurroundings, (b) the sign of DSsystem, and (c) whether the process is spontaneous 2 O2(g) + N2(g)  2 NO2(g)
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34. DSsurr, J/K, DSreact + or −, DSuniverse + or −
Practice – The reaction 2 O2(g) + N2(g)  2 NO2(g) has DHrxn = kJ at 25 °C. Calculate the entropy change of the surroundings. Determine the sign of the entropy change in the system and whether the reaction is spontaneous.
Given:
Find:
DHsys = kJ, T = 25 ºC = 298 K
DSsurr, J/K, DSreact + or −, DSuniverse + or −
Conceptual Plan:
Relationships: DS
T, DH
Solution:
the major difference is that there are fewer product molecules than reactant molecules, so the DSreaction is unfavorable and (−)
both DSsurroundings and DSreaction are (−), DSuniverse is (−) and the process is nonspontaneous
because DSsurroundings is (−) it is unfavorable
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35. Gibbs Free Energy and Spontaneity
It can be shown that −TDSuniv = DHsys−TDSsys
The Gibbs Free Energy, G, is the maximum amount of work energy that can be released to the surroundings by a system
for a constant temperature and pressure system
the Gibbs Free Energy is often called the Chemical Potential because it is analogous to the storing of energy in a mechanical system
DGsys = DHsys−TDSsys
Because DSuniv determines if a process is spontaneous, DG also determines spontaneity
DSuniv is + when spontaneous, so DG is −
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36. Gibbs Free Energy, DG
A 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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37. DG, DH, and DS
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38. Free Energy Change and Spontaneity
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39. Example 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
Conceptual Plan:
Relationships: DG
T, DH, DS
Solution:
Because 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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40. Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K at 25 °C. Calculate DG and determine if it is spontaneous.
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41. Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K at 25 °C. Calculate DG and determine if it is spontaneous.
Given:
Find:
DH = −847.6 kJ and DS = −41.3 J/K, T = 298 K
DG, kJ
Conceptual Plan:
Relationships: DG
T, DH, DS
Solution:
Because DG is −, the reaction is spontaneous at this temperature. To make it nonspontaneous, we need to increase the temperature.
Answer:
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42. Example 17.3b: 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
Conceptual Plan:
Relationships: T
DG, DH, DS
Solution:
the temperature must be higher than 673K for the reaction to be spontaneous
Answer:
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43. Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K. Calculate the maximum temperature it will be spontaneous.
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44. DH = −847.6 kJ and DS = −41.3 J/K, DG > 0 T, C
Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K at 25 °C. Determine the maximum temperature it is spontaneous.
Given:
Find:
DH = −847.6 kJ and DS = −41.3 J/K, DG > 0
T, C
Conceptual Plan:
Relationships: T
DG, DH, DS
Solution:
2048 − 273 = 1775 C
Answer:
any temperature above 1775 C will make the reaction nonspontaneous
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45. Standard Conditions
The standard state is the state of a material at a defined set of conditions
Gas = pure gas at exactly 1 atm pressure
Solid or Liquid = pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest
usually 25 °C
Solution = substance in a solution with concentration 1 M
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46. 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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47. Standard Absolute Entropies
Extensive
Entropies for 1 mole of a substance 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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48. Standard Absolute Entropies
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49. 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 (l)
70.0
H2O (g)
188.8
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50. 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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51. Relative Standard Entropies: Allotropes
The less constrained the structure of an allotrope is, the larger its entropy
The fact that the layers in graphite are not bonded together makes it less constrained
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52. 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)
CO (g)
28.01
197.7
C2H4 (g)
28.05
219.3
Substance
Molar
Mass
S°,
(J/mol∙K)
Ar (g)
39.948
154.8
NO (g)
30.006
210.8
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53. 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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54. The Standard Entropy Change, DS
The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions
DSºreaction = (∑npSºproducts) − (∑nrSºreactants)
remember: though the standard enthalpy of formation, DHf°, of an element is 0 kJ/mol, the absolute entropy at 25 °C, S°, is always positive
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55. Substance
S, J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
188.8
Example 17.4: Calculate DS for the reaction 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Given:
Find:
standard entropies from Appendix IIB
DS, J/K
Conceptual Plan:
Relationships: DS
SNH3, SO2, SNO, SH2O
Sol’n:
Check:
DS is +, as you would expect for a reaction with more gas product molecules than reactant molecules
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56. Practice – Calculate the DS for the reaction 2 H2(g) + O2(g)  2 H2O(g)
Substance
S, J/molK
H2(g)
130.7
O2(g)
205.2
H2O(g)
188.8
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57. Substance
S, J/molK
H2(g)
130.6
O2(g)
205.2
H2O(g)
188.8
Example 17.4: Calculate DS for the reaction 2 H2(g) + O2(g)  2 H2O(g)
Given:
Find:
standard entropies from Appendix IIB
DS, J/K
Conceptual Plan:
Relationships: DS
SH2, SO2, SH2O
Solution:
Check:
DS is −, as you would expect for a reaction with more gas reactant molecules than product molecules
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58. Calculating DG
At 25 C
DGoreaction = SnDGof(products) - SnDGof(reactants)
At temperatures other than 25 C
assuming the change in DHoreaction and DSoreaction is negligible
DGreaction = DHreaction – TDSreaction or
DGtotal = DGreaction 1 + DGreaction
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59. Example 17. 6: The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = −98
Example 17.6: The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = −98.9 kJ and DS = −94.0 J/K at 25 °C. Calculate DG at 125 C and determine if it is more or less spontaneous than at 25 °C with DG° = −70.9 kJ/mol SO3.
Given:
Find:
DH = −98.9 kJ, DS = −94.0 J/K, T = 398 K
DG, kJ
Conceptual Plan:
Relationships:
DG
T, DH, DS
Solution:
because DG is −, the reaction is spontaneous at this temperature, though less so than at 25 C
Answer:
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60. Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g)
Substance
DH, kJ/mol
S, J/mol
H2O2(g)
−136.3
232.7
O2(g)
205.2
H2O(g)
−241.8
188.8
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61. Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g)
Given:
Find:
standard energies from Appendix IIB, T = 298 K
DG, kJ
DH = kJ, DS = −117.4 J/K, T = 298 K
DG, kJ
Conceptual Plan:
Relationships:
DG
DHf° Sº of prod & react
DH, DSº
Solution:
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62. Standard Free Energies of Formation
The free energy of formation (DGf°) is the change in free energy when 1 mol of a compound forms from its constituent elements in their standard states
The free energy of formation of pure elements in their standard states is zero
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63. Substance
DGf°, kJ/mol
CH4(g)
−50.5
O2(g)
0.0
CO2(g)
−394.4
H2O(g)
−228.6
O3(g)
+163.2
Example 17.7: 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
Conceptual Plan:
Relationships:
DG
DGf° of prod & react
Solution:
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64. Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g)
Substance
DG, kJ/mol
H2O2(g)
−105.6
O2(g)
H2O(g)
−228.6
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65. Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g)
Given:
Find:
standard free energies of formation from Appendix IIB
DG, kJ
Conceptual Plan:
Relationships:
DG
DGf° of prod & react
Solution:
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66. 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 amount 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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67. Example 17.8: Find DGºrxn for the following reaction using the given equations 3 C(s) + 4 H2(g)  C3H8(g)
Given:
Find:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DGº = −2074 kJ
C(s) + O2(g)  CO2(g) DGº = −394.4 kJ
2 H2(g) + O2(g)  2 H2O(g) DGº = −457.1 kJ
DGº of 3 C(s) + 4 H2(g)  C3H8(g)
Rel:
Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the DGº’s is the DGº of the reaction you want to find.
Solution:
3 CO2(g) + 4 H2O(g)  C3H8(g) + 5 O2(g) DGº = kJ
3 C(s) + 3 O2(g)  3 CO2(g) DGº = − kJ
4 H2(g) + 2 O2(g)  4 H2O(g) DGº = −914.2 kJ
3 C(s) + 4 H2(g)  C3H8(g) DGº = −23 kJ
Given:
Find:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DGº = −2074 kJ
C(s) + O2(g)  CO2(g) DGº = −394.4 kJ
2 H2(g) + O2(g)  2 H2O(g) DGº = −457.1 kJ
DGº of 3 C(s) + 4 H2(g)  C3H8(g)
Rel:
Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the DGº’s is the DGº of the reaction you want to find.
Solution:
3 CO2(g) + 4 H2O(g)  C3H8(g) + 5 O2(g) DGº = kJ
3 x [C(s) + O2(g)  CO2(g) ] DGº = 3(−394.4 kJ)
2 x [2 H2(g) + O2(g)  2 H2O(g)] DGº = 2(−457.1 kJ)
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68. Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g)
Substance
DG, kJ/mol
H2 (g) + O2(g)  H2O2(g)
−105.6
2 H2 (g) + O2(g)  2 H2O(g)
−457.2
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69. Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g)
Given:
Find:
H2(g) + O2(g)  H2O2(g) DGº = −105.6 kJ
2 H2(g) + O2(g)  2 H2O(g) DGº = −457.2 kJ
DGº of 2 H2O(g) + O2(g)  2 H2O2(g)
Rel:
Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the DGº’s is the DGº of the reaction you want to find.
Solution:
2 H2(s) + 2 O2(g)  2 H2O2(g) DGº = −211.2 kJ
2 H2O(g)  2 H2(g) + O2(g) DGº = kJ
2 H2O(g) + O2(g)  2 H2O2(g) DGº = kJ
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70. What’s “Free” About Free Energy?
The free energy is the maximum amount of energy released from a system that is available to do work on the surroundings
For many exothermic reactions, some of the heat released due to the enthalpy change goes into increasing the entropy of the surroundings, so it is not available to do work
And even some of this free energy is generally lost to heating up the surroundings
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71. Free Energy of an Exothermic Reaction
C(s, graphite) + 2 H2(g) → CH4(g)
DH°rxn = −74.6 kJ = exothermic
DS°rxn = −80.8 J/K = unfavorable
DG°rxn = −50.5 kJ = spontaneous
DG° is less than DH° because some of the released heat energy is lost to increase the entropy of the surroundings
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72. 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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73. 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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74. DG under Nonstandard Conditions
DG = DG only when the reactants and products are in their standard states
their 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
Tro: Chemistry: A Molecular Approach, 2/e

75. Tro: Chemistry: A Molecular Approach, 2/e

76. non-standard conditions, DGº, T DG, kJ
Example 17.9: Calculate DG at 298 K for the reaction under the given conditions 2 NO(g) + O2(g)  2 NO2(g) DGº = −71.2 kJ
Substance
P, atm
NO(g)
0.100
O2(g)
NO2(g)
2.00
Given:
Find:
non-standard conditions, DGº, T
DG, kJ
Conceptual Plan:
Relationships: DG
DG, PNO, PO2, PNO2
Solution:
Tro: Chemistry: A Molecular Approach, 2/e

77. Practice – Calculate DGrxn for the given reaction at 700 K under the given conditions N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ
Substance
P, atm
N2(g) 33
H2(g) 99
NH3(g)
2.0
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78. non-standard conditions, DGº, T DG, kJ
Substance
P, atm
N2(g) 33
H2(g) 99
NH3(g)
2.0
Practice – Calculate DGrxn for the given reaction at 700 K under the given conditions N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ
Given:
Find:
non-standard conditions, DGº, T
DG, kJ
Conceptual Plan:
Relationships: DG
DG, PNO, PO2, PNO2
Solution:
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79. DGº and K Because DGrxn = 0 at equilibrium, then DGº = −RTln(K)
When K < 1, DGº is + and the reaction is spontaneous in the reverse direction under standard conditions
nothing will happen if there are no products yet!
When K > 1, DGº is − and the reaction is spontaneous in the forward direction under standard conditions
When K = 1, DGº is 0 and the reaction is at equilibrium under standard conditions
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81. Substance DGf°, kJ/mol N2O4(g) +99.8 NO2(g) +51.3
Example 17.10: Calculate K at 298 K for the reaction N2O4(g)  2 NO2(g)
Given:
Find:
standard free energies of formation from Appendix IIB K
Conceptual Plan:
Relationships:
DG
DGf of prod & react K
DGº = −RTln(K)
Solution:
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82. Practice – Estimate the equilibrium constant for the given reaction at 700 K N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ
Tro: Chemistry: A Molecular Approach, 2/e

83. Practice – Estimate the Equilibrium Constant for the given reaction at 700 K N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ
Given:
Find:
DGº K
Conceptual Plan:
Relationships:
DG K
DGº = −RTln(K)
Solution:
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84. Why Is the Equilibrium Constant Temperature Dependent?
Combining these two equations
DG° = DH° − TDS°
DG° = −RTln(K)
It can be shown that
This equation is in the form y = mx + b
The graph of ln(K) versus inverse T is a straight line with slope and y-intercept
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