1. Free Energy and Thermodynamics
Chapter 17
Free Energy and Thermodynamics

2. 17.1 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

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 is a state function
internal energy change independent of how done

4. The Energy Tax You can’t break even!
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

5. Note: Fewer steps generally results in a lower total heat tax

6. 17.2 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 of the system before the reaction with the free energy of the system after the reaction.
if the system after reaction has less chemical potential than before the reaction, the reaction is thermodynamically favorable.
Spontaneity ≠ fast or slow

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

8. Reversibility of Process
Any spontaneous process is irreversible because there is a net release of chemical potential when it proceeds in that direction
it will proceed in only one direction
If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction

9. Thermodynamics vs. Kinetics
Thermodynamics deals with the relative chemical potentials of the reactants and products. It enables us to predict if a reaction is spontaneous and find out how much work it can do.
Kinetics deals with the chemical potential of intermediate states. It enables us to determine why a reaction is SLOW or FAST.

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)

11. Spontaneous Processes: Enthalpy and Entropy
Spontaneous processes occur because they release energy (chemical potential) from the system.
Most spontaneous processes proceed from a system of higher enthalpy to a system at lower enthalpy.
Exothermic
But there are some spontaneous processes that proceed from a system of lower enthalpy to a system at higher enthalpy
Endothermic
How can something absorb energy (enthalpy), yet have a net release of chemical potential (free energy)? – disorder/entropy

12. 17.3 Entropy and the Second Law of Thermodynamics
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.

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.
DH = DE + P DV
The entropy change, DS, is the difference in randomness of the reactants compared to the products.

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

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.

16. 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

17. 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)

18. 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.

19. Entropy Change and State Change

20. Water vapor condensing Separation of oil and vinegar salad dressing
Practice – Predict whether DSsystem is + or − for each of the following
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 −

21. EXAMPLE 17.1 Predicting the Sign of Entropy Change
Predict the sign of ∆S for each process:
SOLUTION
(a) Since a gas has a greater entropy than a liquid, the entropy decreases and S is negative.
(b) Since a solid has a lower entropy than a gas, the entropy increases and S is positive.
(c) Since the number of moles of gas increases, the entropy increases and S is positive.
FOR PRACTICE 17.1
© 2011 Pearson Education, Inc.

22. 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

23. 17.4 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!

24. 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.

25. 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

26. 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 DSuniverse positive.
above 0 °C the increase in entropy of the surroundings is insufficient to make DSuniverse positive

27. 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

28. EXAMPLE 17.2 Calculating Entropy Changes in the Surroundings
Consider the combustion of propane gas:
(a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25 C .
(b) Determine the sign of the entropy change for the system.
(c) Determine the sign of the entropy change for the universe. Will the reaction be spontaneous?
(a) The entropy change of the surroundings is given by Equation Substitute the value of Hrxn and the temperature in kelvins and calculate Ssurr.
SOLUTION
(b) Determine the number of moles of gas on each side of the reaction. An increase in the number of moles of gas implies a positive Ssys.
(c) The change in entropy of the universe is the sum of the entropy changes of the system and the surroundings. If the entropy changes of the system and surroundings are both the same sign, the entropy change for the universe will also have the same sign.
Therefore, Suniv is positive and the reaction is spontaneous.
© 2011 Pearson Education, Inc.

29. Given: Find: Conceptual Plan: Relationships: Solution:
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 29

30. 17.5 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 −

31. Free Energy Change and Spontaneity
Gibbs Free Energy is often called the Chemical Potential because it determines the direction of spontaneous change for chemical systems.

32. Gibbs Free Energy, DG
A process will be spontaneous when DG is negative
DGsys = DHsys−TDSsys
DG will be negative when
DHsys is negative and DSsys is positive
exothermic and more random
DHsys is negative and large and DSsys is negative but small
DHsys is positive but small and DSsys is positive and large
DG will be positive when DHsys is + and DSsys is −
never spontaneous at any temperature
When DG = 0 the reaction is at equilibrium

33. DG, DH, DS, and T on Spontaneity
When DHsys & DSsys are of the same signs, temperature comes into play by judging the DSsurr and how it affects DSsys + DSsurr = DSuniverse
When DHsys & DSsys are positive, reaction is endothermic. System is receiving heat from surrounding.
When the temperature is low, the loss of heat from surrounding is significant, and DSsurr is a HIGH NEGATIVE VALUE. Since DSsys + DSsurr = DSuniverse, DSuniverse would end up being negative, and reaction is nonspontaneous.
When the temperature is high, the loss of heat from surrounding is insignificant, and DSsurr is a LOW NEGATIVE VALUE. Since DSsys + DSsurr = DSuniverse, DSuniverse would end up being positive, and reaction is spontaneous.

34. Continued…
When DHsys & DSsys are negative, reaction is exothermic. Surrounding is receiving heat from the system.
When the temperature is low, the gain of heat in the surrounding is significant, and DSsurr is a HIGH POSITIVE VALUE.
Since DSsys + DSsurr = DSuniverse, DSuniverse would end up being positive, and reaction is spontaneous.
When the temperature is high, the gain of heat in the surrounding is insignificant, and DSsurr is a LOW POSITIVE VALUE.
Since DSsys + DSsurr = DSuniverse, DSuniverse would end up being negative, and reaction is nonspontaneous.

35. 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:
Tro: Chemistry: A Molecular Approach, 2/e

36. 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:
Tro: Chemistry: A Molecular Approach, 2/e

37. 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.
Tro: Chemistry: A Molecular Approach, 2/e

38. 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:
Tro: Chemistry: A Molecular Approach, 2/e 38

39. 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.
Tro: Chemistry: A Molecular Approach, 2/e

40. 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
Tro: Chemistry: A Molecular Approach, 2/e 40

41. 17.6 Calculating Standard Entropy Changes in Chemical Reactions (DS°rxn) 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

42. 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 +

43. Standard Absolute Entropies
Extensive property (dependent on the amount of matter)
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

44. Standard Absolute Entropies

45. 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

46. Relative Standard Entropies: Molar Mass
The larger the molar mass, the larger the entropy.

47. 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.

48. 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

49. 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

50. DS, J/K DS SNH3, SO2, SNO, SH2O
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
Tro: Chemistry: A Molecular Approach, 2/e

51. 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
Tro: Chemistry: A Molecular Approach, 2/e

52. 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:
Relationship s: 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
Tro: Chemistry: A Molecular Approach, 2/e

53. 17.7 Free Energy Changes in Chemical Reactions : ∆G°rxn
Three ways to calculate ∆G°rxn.
Calculate the ∆H°rxn and ∆S°rxn, based on the standard formation values and using ∆G°rxn = ∆H°rxn - T ∆S°rxn.
Calculate ∆G°f from ∆G°rxn = Total products ∆G°f - Total reactants ∆G°f
DGoreaction = SnDGof(products) - SnDGof(reactants)
Calculate ∆G°rxn based on the ∆G°rxn of multiple stepwise reactions (same as Hess’ Law for ∆H°rxn )

54. EXAMPLE 17.5 Calculating the Standard Change in Free Energy for a Reaction using Grxn = Hrxn – T Srxn
One of the possible initial steps in the formation of acid rain is the oxidation of the pollutant SO2 to SO3 by the reaction:
Calculate Grxn at 25 C and determine whether the reaction is spontaneous.
Begin by looking up (in Appendix IIB) the standard enthalpy of formation and the standard entropy for each reactant and product.
SOLUTION
Calculate Hrxn using Equation 6.15.
Calculate Srxn using Equation
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55. Calculate Grxn using the calculated values of Hrxn and Srxn and Equation The temperature must be converted to kelvins.
The reaction is spontaneous at this temperature, because Grxn is negative.
© 2011 Pearson Education, Inc.

56. EXAMPLE 17.6 Estimating the Standard Change in Free Energy for a Reaction at a Temperature Other than 25 C Using Grxn = Hrxn – T Srxn
For the reaction in Example 17.5, estimate the value of Grxn at 125 C . Does the reaction become more or less spontaneous at this elevated temperature; that is, does the value of Grxn become more negative (more spontaneous) or more positive (less spontaneous)?
Estimate Grxn at the new temperature using the calculated values of Hrxn and Srxn from Example For T, convert the given temperature to kelvins. Make sure to use the same units for Hrxn and Srxn (usually joules).
SOLUTION
Since the value of Grxn at this elevated temperature is less negative (or more positive) than the value of Grxn at 25 C (which is –70.9 kJ), the reaction is less spontaneous.
© 2011 Pearson Education, Inc.

57. 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. (same rule as for enthalpy, H)

58. EXAMPLE 17.7 Calculating Grxn from Standard Free Energies of Formation
Ozone in the lower atmosphere is a pollutant that can be formed by the following reaction involving the oxidation of unburned hydrocarbons:
Use the standard free energies of formation to determine Grxn for this reaction at 25 C.
Begin by looking up (in Appendix IIB) the standard free energies of formation for each reactant and product. Remember that the standard free energy of formation of a pure element in its standard state is zero.
SOLUTION
Calculate Grxn by substituting into Equation
© 2011 Pearson Education, Inc.

60. EXAMPLE 17.8 Calculating Grxn for a Stepwise Reaction
Find Grxn for the following reaction:
Use the following reactions with known ∆G’s:
SOLUTION
To work this problem, manipulate the reactions with known Grxn’s in such a way as to get the reactants of interest on the left, the products of interest on the right, and other species to cancel.
Since the first reaction has C3H8 as a reactant, and the reaction of interest has C3H8 as a product, reverse the first reaction and change the sign of Grxn.
The second reaction has C as a reactant and CO2 as a product, just as required in the reaction of interest. However, the coefficient for C is 1, and in the reaction of interest, the coefficient for C is 3. Therefore, multiply this equation and its Grxn by 3.
continued…
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61. The third reaction has H2(g) as a reactant, as required
The third reaction has H2(g) as a reactant, as required. However, the coefficient for H2 is 2, and in the reaction of interest, the coefficient for H2 is 4. Multiply this reaction and its Grxn by 2.
Lastly, rewrite the three reactions after multiplying through by the indicated factors and show how they sum to the reaction of interest. Grxn for the reaction of interest is then the sum of the G’s for the steps.
© 2011 Pearson Education, Inc.

62. 17.8 & 17.9 Free Energy Changes at NonStandard States & Relating Free Energy To The Equilibrium Constant
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

63. EXAMPLE 17.9 Calculating Grxn under Nonstandard Conditions
Consider the following reaction at 298 K:
Calulate Grxn under the following conditions:
Is the reaction more or less spontaneous under these conditions than under standard conditions?
Use the law of mass action to calculate Q.
SOLUTION
Substitute Q, T, and Grxn into Equation to calculate Grxn. (Since the units of R include joules, write Grxn in joules.)
The reaction is spontaneous under these conditions, but less spontaneous than it was under standard conditions (because Grxn is less negative than Grxn).
© 2011 Pearson Education, Inc.

64. 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

65. EXAMPLE 17.10 The Equilibrium Constant and Gorxn
Use tabulated free energies of formation to calculate the equilibrium constant for the following reaction at 298 K:
Begin by looking up (in Appendix IIB) the standard free energies of formation for each reactant and product.
SOLUTION
Calculate Grxn by substituting into Equation
Calculate K from Grxn by solving Equation for K and substituting the values of Grxn and temperature.
© 2011 Pearson Education, Inc.