1. Topic 15 Energetics AHL Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 19: Chemical Thermodynamics
Topic 15 Energetics AHL
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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2. First Law of Thermodynamics
You will recall from Chapter 5 that energy cannot be created nor destroyed.
Therefore, the total energy of the universe is a constant.
Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.
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3. Spontaneous Processes
Spontaneous processes are those that can proceed without any outside intervention.
The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.
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4. Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
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5. Spontaneous Processes
Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
Above 0 C it is spontaneous for ice to melt.
Below 0 C the reverse process is spontaneous.
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6. Sample Exercise 19.1 Identifying Spontaneous Processes
Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150 °C is added to water at 40 °C, the water gets hotter. (b) Water at room temperature decomposes into H2(g) and O2(g), (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 °C.
Solution
Analyze: We are asked to judge whether each process will proceed spontaneously in the direction indicated, in the reverse direction, or in neither direction.
Plan: We need to think about whether each process is consistent with our experience about the natural direction of events or whether we expect the reverse process to occur.
Solve: (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere between the initial temperatures of the metal and the water.
(b) Experience tells us that this process is not spontaneous—we certainly have never seen hydrogen and oxygen gases spontaneously bubbling up out of water! Rather, the reverse process—the reaction of H2 and O2 to form H2O—is spontaneous. (c) By definition, the normal boiling point is the temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. If the temperature were below 80.1 °C, condensation would be spontaneous.
Under 1 atm pressure CO2(s) sublimes at –78 °C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at –100 °C and 1 atm pressure?
Answer: No, the reverse process is spontaneous at this temperature.
Practice Exercise
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7. Reversible Processes
In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.
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8. Irreversible Processes
Irreversible processes cannot be undone by exactly reversing the change to the system.
Spontaneous processes are irreversible.
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9. Entropy
Entropy (S) is a term coined by Rudolph Clausius in the 19th century.
Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, q T
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10. Entropy
Entropy can be thought of as a measure of the randomness of a system.
It is related to the various modes of motion in molecules.
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11. Entropy
Like total energy, E, and enthalpy, H, entropy is a state function.
Therefore,
S = Sfinal  Sinitial
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12. Entropy
For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:
S =
qrev T
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13. Second Law of Thermodynamics
The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.
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14. Second Law of Thermodynamics
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
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15. Second Law of Thermodynamics
These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.
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16. Sample Exercise 19.2 Calculating ΔS for a Phase Change
The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is
–38.9 °C, and its molar enthalpy of fusion is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?.
Solution
Analyze: We first recognize that freezing is an exothermic process; heat is transferred from the system to the surroundings when a liquid freezes (q < 0). The enthalpy of fusion is ΔH for the melting process. Because freezing is the reverse of melting, the enthalpy change that accompanies the freezing of 1 mol of Hg is –ΔHfusion = –2.29 kJ/mol.
Plan: We can use –ΔHfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg:
We can use this value of q as qrev in Equation We must first, however, convert the temperature to K:
Solve: We can now calculate the value of ΔSsys
Check: The entropy change is negative because heat flows from the system, making qrev negative.
Comment: The procedure we have used here can be used to calculate ΔS for other isothermal phase changes, such as the vaporization of a liquid at its boiling point.
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17. Sample Exercise 19.2 Calculating ΔS for a Phase Change
The normal boiling point of ethanol, C2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is
38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point?
Answer: –163 J/K
Practice Exercise
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18. Entropy on the Molecular Scale
Ludwig Boltzmann described the concept of entropy on the molecular level.
Temperature is a measure of the average kinetic energy of the molecules in a sample.
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19. Entropy on the Molecular Scale
Molecules exhibit several types of motion:
Translational: Movement of the entire molecule from one place to another.
Vibrational: Periodic motion of atoms within a molecule.
Rotational: Rotation of the molecule on about an axis or rotation about  bonds. 
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20. Entropy on the Molecular Scale
Boltzmann envisioned the motions of a sample of molecules at a particular instant in time.
This would be akin to taking a snapshot of all the molecules.
He referred to this sampling as a microstate of the thermodynamic system.
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21. Entropy on the Molecular Scale
Each thermodynamic state has a specific number of microstates, W, associated with it.
Entropy is
S = k lnW
where k is the Boltzmann constant, 1.38  1023 J/K.
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22. Entropy on the Molecular Scale
The change in entropy for a process, then, is
S = k lnWfinal  k lnWinitial
lnWfinal
lnWinitial
S = k ln
Entropy increases with the number of microstates in the system.
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23. Entropy on the Molecular Scale
The number of microstates and, therefore, the entropy tends to increase with increases in
Temperature.
Volume.
The number of independently moving molecules.
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24. Entropy and Physical States
Entropy increases with the freedom of motion of molecules.
Therefore,
S(g) > S(l) > S(s)
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25. Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.
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26. Entropy Changes In general, entropy increases when
Gases are formed from liquids and solids;
Liquids or solutions are formed from solids;
The number of gas molecules increases;
The number of moles increases.
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27. Sample Exercise 19.3 Predicting the Sign of ΔS
Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature:
Solution
Analyze: We are given four equations and asked to predict the sign of ΔS for each chemical reaction.
Plan: The sign of ΔS will be positive if there is an increase in temperature, an increase in the volume in which the molecules move, or an increase in the number of gas particles in the reaction. The question states that the temperature is constant. Thus, we need to evaluate each equation with the other two factors in mind.
Solve:
(a) The evaporation of a liquid is accompanied by a large increase in volume. One mole of water (18 g) occupies about 18 mL as a liquid and if it could exist as a gas at STP it would occupy 22.4 L. Because the molecules are distributed throughout a much larger volume in the gaseous state than in the liquid state, an increase in motional freedom accompanies vaporization. Therefore, ΔS is positive.
(b) In this process the ions, which are free to move throughout the volume of the solution, form a solid in which they are confined to a smaller volume and restricted to more highly constrained positions. Thus, ΔS is negative.
(c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer microstates) than do the molecules of a gas. Because O2 gas is converted into part of the solid product Fe2O3, ΔS is negative.
(d) The number of moles of gases is the same on both sides of the equation, and so the entropy change will be small. The sign of ΔS is impossible to predict based on our discussions thus far, but we can predict that ΔS will be close to zero.
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28. Sample Exercise 19.3 Predicting the Sign of ΔS
Indicate whether each of the following processes produces an increase or decrease in the entropy of the system:
Answer: (a) increase, (b) decrease, (c) decrease, (d) decrease
Practice Exercise
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29. Sample Exercise 19.4 Predicting Which Sample of Matter Has the Higher Entropy
Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.
Solution
Analyze: We need to select the system in each pair that has the greater entropy.
Plan: To do this, we examine the state of the system and the complexity of the molecules it contains.
Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the number of microstates and twice the entropy when they are at the same pressure. (c) The HCl sample has the higher entropy because the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms cannot.
Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or
1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(s) at 0 °C or 1 mol of H2O(l) at 25 °C,
(c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.
Answers: (a) 1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(l) at 25 °C, (c) 1 mol of SO2(g) at STP, (d) 2 mol of NO2(g) at STP
Practice Exercise
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30. Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0.
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31. Standard Entropies
These are molar entropy values of substances in their standard states.
Standard entropies tend to increase with increasing molar mass.
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32. Standard Entropies
Larger and more complex molecules have greater entropies.
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33. S = nS(products) — mS(reactants)
Entropy Changes
Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:
S = nS(products) — mS(reactants)
where n and m are the coefficients in the balanced chemical equation.
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34. Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies
Calculate ΔSº for the synthesis of ammonia from N2(g) and H2(g) at 298 K:
N2(g) + 3 H2(g) → 2 NH3(g)
Solution
Analyze: We are asked to calculate the entropy change for the synthesis of NH3(g) from its constituent elements.
Plan: We can make this calculation using Equation 19.8 and the standard molar entropy values for the reactants and the products that are given in Table 19.2 and in Appendix C.
Solve: Using Equation 19.8, we have
Substituting the appropriate S° values from Table 19.2 yields
ΔS° = 2S°(NH3) - [S°(N2) + 3S°(H2)]
ΔS° = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K)
+ (3 mol)(130.6 J/mol-K)] = J/K
Check: The value for ΔS° is negative, in agreement with our qualitative prediction based on the decrease in the number of molecules of gas during the reaction.
Using the standard entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K:
Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g)
Answers: J/K
Practice Exercise
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35. Entropy Changes in Surroundings
Heat that flows into or out of the system changes the entropy of the surroundings.
For an isothermal process:
Ssurr =
qsys T
At constant pressure, qsys is simply H for the system.
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36. Entropy Change in the Universe
The universe is composed of the system and the surroundings.
Therefore,
Suniverse = Ssystem + Ssurroundings
For spontaneous processes
Suniverse > 0
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37. Entropy Change in the Universe
qsystem T
Since Ssurroundings =
and qsystem = Hsystem
This becomes:
Suniverse = Ssystem +
Multiplying both sides by T, we get
TSuniverse = Hsystem  TSsystem
Hsystem T
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38. Gibbs Free Energy
TDSuniverse is defined as the Gibbs free energy, G.
When Suniverse is positive, G is negative.
Therefore, when G is negative, a process is spontaneous.
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39. Gibbs Free Energy
If DG is negative, the forward reaction is spontaneous.
If DG is 0, the system is at equilibrium.
If G is positive, the reaction is spontaneous in the reverse direction.
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40. Standard Free Energy Changes
Analogous to standard enthalpies of formation are standard free energies of formation, G. f
DG = SnDG (products)  SmG (reactants) f
where n and m are the stoichiometric coefficients.
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41. Free Energy Changes At temperatures other than 25°C, DG° = DH  TS
How does G change with temperature?
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42. Free Energy and Temperature
There are two parts to the free energy equation:
H— the enthalpy term
TS — the entropy term
The temperature dependence of free energy, then comes from the entropy term.
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43. Free Energy and Temperature
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44. Free Energy and Equilibrium
Under any conditions, standard or nonstandard, the free energy change can be found this way:
G = G + RT lnQ
(Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)
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45. Free Energy and Equilibrium
At equilibrium, Q = K, and G = 0.
The equation becomes
0 = G + RT lnK
Rearranging, this becomes
G = RT lnK
or,
K = e
-G RT
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