Chapter 13: Thermodynamics

### 13.1 Spontaneity [Go to section 13.1]

- What is a spontaneous reaction?
- What is a nonspontaneous reaction?
- Indicate whether the following processes are spontaneous or nonspontaneous.
- Liquid water freezing at a temperature below its freezing point
- Liquid water freezing at a temperature above its freezing point
- The combustion of gasoline
- A ball thrown into the air
- A raindrop falling to the ground
- Iron rusting in a moist atmosphere

- Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.

### 13.2 Entropy [Go to section 13.2]

- In Figure 13.2.2., all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, for the system when it is converted from distribution (b) to distribution (d).
- A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
- Consider a system similar to the one in Figure 13.2.2., except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to [latex]\frac{1}{8}[/latex]. What does this comparison tell us about even larger systems?
- In Figure 13.2.2 all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).
- Consider the system shown in Figure 13.2.4. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
- How does the process described in the previous item relate to the system shown in Figure 13.2.3.?
- At room temperature, the entropy of the halogens increases from I
_{2}to Br_{2}to Cl_{2}. Explain. - Consider the system shown in Figure 13.2.4. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?
- Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
- C
_{2}H_{5}OH(l) or C_{3}H_{7}OH(l) - C
_{2}H_{5}OH(l) or C_{2}H_{5}OH(g) - 2H(g) or H(g)

- C
- Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.
- H
_{2}(g), HBrO_{4}(g), HBr(g) - H
_{2}O(l), H_{2}O(g), H_{2}O(s) - He(g), Cl
_{2}(g), P_{4}(g)

- H
- Write the balanced chemical equation for the combustion of benzene, C
_{6}H_{6}(l), to give carbon dioxide and water vapor. Would you expect ΔS to be positive or negative in this process? - Write the balanced chemical equation for the combustion of methane, CH
_{4}(g), to give carbon dioxide and water vapor. Explain why it is difficult to predict whether ΔS is positive or negative for this chemical reaction. - Compare the magnitude of ΔS for the fusion and vaporization of of N
_{2}. Which will be greater? Why?

## Show Selected Solutions

- ∆S = 0
- 1/32
- 1.91 × 10
^{-23}J/K - The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I
_{2}is a solid, Br_{2}is a liquid, and Cl_{2}is a gas. - The answers are as follows:
- C
_{3}H_{7}OH(l) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature. - C
_{2}H_{5}OH(g) as it is in the gaseous state. ( - 2H(g), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).

- C
- ∆S is positive
- ΔS increases in both transitions, but the magnitude of ΔS is greater in the liquid to gas transition because there is a larger change in the number of microstates for that transition.

### 13.3 The Second and Third Laws of Thermodynamics [Go to section 13.3]

- What is the difference between ΔS and ΔS° for a chemical change?
- Calculate ΔS°
_{298}for the following changes.- [latex]\ce{SnCl4}[/latex]
*(I)*[latex]\longrightarrow[/latex] [latex]\ce{SnCl4}[/latex](*g*) - [latex]\ce{CS2}[/latex](
*g*) [latex]\longrightarrow[/latex] [latex]\ce{CS2}[/latex](*I*) - [latex]\ce{Cu}[/latex]
*(**s**)*[latex]\longrightarrow[/latex] [latex]\ce{Cu}[/latex](*g*) - [latex]\ce{H2O}[/latex]
*(I)*[latex]\longrightarrow[/latex] [latex]\ce{H2O}[/latex](*g*) - 2[latex]\ce{H2}[/latex](
*g)*+ [latex]\ce{O2}[/latex](*g*) [latex]\longrightarrow[/latex] 2[latex]\ce{H2O}[/latex](*I*) - 2[latex]\ce{HCl}[/latex](
*g*) + Pb(*s*) [latex]\longrightarrow[/latex] [latex]\ce{PbCl2}[/latex](*s*) + [latex]\ce{H2}[/latex](*g*) - [latex]\ce{Zn}[/latex](
*s*) + [latex]\ce{CuSO4}[/latex](*s*) [latex]\longrightarrow[/latex] [latex]\ce{Cu}[/latex](*s*) + [latex]\ce{ZnSO4}[/latex](*s*)

- [latex]\ce{SnCl4}[/latex]
- Determine the entropy change for the combustion of liquid ethanol, [latex]\ce{C2H5OH}[/latex] under the standard conditions to give gaseous carbon dioxide and liquid water.
- Determine the entropy change for the combustion of gaseous propane,[latex]\ce{C3H8}[/latex] under the standard conditions to give gaseous carbon dioxide and water.
- “Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is [latex]\ce{Fe2O3}[/latex](
*s*) + 2[latex]\ce{Al}[/latex](*s*) [latex]\longrightarrow[/latex] [latex]\ce{Al2O3}[/latex](*s*) + 2[latex]\ce{Fe}[/latex](*s*). Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat. - Using the relevant S°
_{298}values listed in Standard Thermodynamic Properties for Selected Substances, calculate ΔS°_{298 }for the following changes:- [latex]\ce{N2}[/latex](
*g*) + 3[latex]\ce{H2}[/latex](*g*) [latex]\longrightarrow[/latex] 2[latex]\ce{NH3}[/latex](*g*) - [latex]\ce{N2}[/latex](
*g*) + [latex]\frac {5} {2}[/latex][latex]\ce{O2}[/latex](*g*) [latex]\longrightarrow[/latex] [latex]\ce{N2O5}[/latex](*g*)

- [latex]\ce{N2}[/latex](
- From the following information, determine ΔS°
_{298}for the following:

[latex]\begin{array}{lrrr} \ce{N}(g) + \ce{O}(g) \longrightarrow \ce{NO}(g) & \Delta S°_{298} & = & ? \\ \ce{N2}(g) + \ce{O2}(g) \longrightarrow \ce{2NO}(g) & \Delta S°_{298} & = & 24.8 \text{ J/K} \\ \ce{N2}(g) \longrightarrow \ce{2N}(g) & \Delta S°_{298} & = & 115.0 \text{ J/K} \\ \ce{O2}(g) \longrightarrow \ce{2O}(g) & \Delta S°_{298} & = & 117.0 \text{ J/K} \\ \end{array}[/latex]

### 13.4 Free Energy [Go to section 13.4]

- Use the standard free energy of formation data in Standard Thermodynamic Properties for Selected Substances to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
- [latex]\ce{MnO2[/latex](
*s*) [latex]\longrightarrow[/latex] Mn(*s*) + [latex]\ce{O2}[/latex](*g*) - [latex]\ce{H2}[/latex](
*g*) + [latex]\ce{Br2}[/latex](*I*) [latex]\longrightarrow[/latex] 2HBr(*g*) - Cu(
*s*) + S(*s*) [latex]\longrightarrow[/latex] CuS(*s*) - 2LiOH(
*s*) + [latex]\ce{CO2}[/latex](*g*) [latex]\longrightarrow[/latex] [latex]\ce{Li2Co3}[/latex](*s*) + [latex]\ce{H2O}[/latex](*g*) - [latex]\ce{CH4}[/latex](
*g*) + [latex]\ce{O2}[/latex](*g*) [latex]\longrightarrow[/latex] [latex]\ce{C}[/latex](*s*, graphite) + 2[latex]\ce{H2O}[/latex](*g*) - [latex]\ce{CS2}[/latex](
*g*) + 3[latex]\ce{Cl2}[/latex](*g*) [latex]\longrightarrow[/latex] [latex]\ce{CCl4}[/latex](*g*) + [latex]\ce{S2Cl2}[/latex](*g*)

- [latex]\ce{MnO2[/latex](
- What is the difference between ΔG and ΔG° for a chemical change?
- Given:

[latex]\begin{array}{lrr} \ce{P4}(s) + \ce{5O2}(g) \longrightarrow \ce{P4O10}(s) & \Delta G°_{298} & = & -2697.0 \text{ kJ/mol} \\ \ce{2H2}(g) + \ce{O2}(g) \longrightarrow \ce{2H2O}(g) & \Delta G°_{298} & = & -457.18 \text{ kJ/mol} \\ \ce{6H2O}(g) + \ce{P4O10}(s) \longrightarrow \ce{4H3PO4}(I) & \Delta G°_{298} & = & -428.66 \text{ kJ/mol} \\ \end{array}[/latex]- Determine the standard free energy of formation, ΔG°
_{f}, for phosphoric acid. - How does your calculated result compare to the value in Appendix G? Explain.

- Determine the standard free energy of formation, ΔG°
- Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.
- Consider the decomposition of red mercury(II) oxide under standard state conditions. 2HgO(
*s*, red) [latex]\longrightarrow[/latex] 2Hg(*I*) + [latex]\ce{O2}[/latex](*g*)- Is the decomposition spontaneous under standard state conditions?
- Above what temperature does the reaction become spontaneous?

- Use the standard free energy data in Standard Thermodynamic Properties for Selected Substances to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
- C(
*s*, graphite) + [latex]\ce{O2}[/latex](*g*) [latex]\longrightarrow[/latex] [latex]\ce{CO2}[/latex](*g*) - [latex]\ce{O2}[/latex](
*g*) + [latex]\ce{N2}[/latex](*g*) [latex]\longrightarrow[/latex] 2NO(*g*) - 2Cu(
*s*) + S(*g*) [latex]\longrightarrow[/latex] [latex]\ce{Cu2S}[/latex](*s*) - CaO(
*s*) + [latex]\ce{H2O}[/latex](*I*) [latex]\longrightarrow[/latex] [latex]\ce{Ca(OH)2}[/latex](*s*) - [latex]\ce{Fe2O3}[/latex](
*s*) + 3CO(*g*) [latex]\longrightarrow[/latex] 2Fe(*s*) + 3[latex]\ce{CO2}[/latex](*g*) - [latex]\ce{CaSO4 2H2O}[/latex](
*s*) [latex]\longrightarrow[/latex] [latex]\ce{CaSO4}[/latex](*s*) + 2[latex]\ce{H2O}[/latex](*g*)

- C(
- Under what conditions is the evaporation of fluorine spontaneous?
- Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
- Ammonia: 2[latex]\ce{NH3}[/latex](
*g*) [latex]\longrightarrow[/latex] [latex]\ce{N2}[/latex](*g*) + 3[latex]\ce{H2}[/latex])*g*) - Diborane: [latex]\ce{B2H6}[/latex](
*g*) [latex]\longrightarrow[/latex] 2[latex]\ce{B}[/latex](*g*) + 3[latex]\ce{H2}[/latex])*g*) - Hydrazine: [latex]\ce{N2H4}[/latex](
*g*) [latex]\longrightarrow[/latex] [latex]\ce{N2}[/latex](*g*) + 2[latex]\ce{H2}[/latex])*g*) - Hydrogen peroxide: [latex]\ce{H2O2}[/latex](
*I*) [latex]\longrightarrow[/latex] [latex]\ce{H2O}[/latex](*g*) + [latex]\frac {1}{2}[/latex][latex]\ce{O2}[/latex])*g*)

- Ammonia: 2[latex]\ce{NH3}[/latex](

- Above what temperature is [latex]{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\longrightarrow \text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)[/latex] spontaneous?
- Determine ΔG° for the following reactions.
- Antimony pentachloride decomposes at 448 °C. The reaction is: [latex]\ce{SbCl5[/latex](
*g*) [latex]\longrightarrow[/latex] [latex]\ce{SbCl3}[/latex](*g*) + [latex]\ce{Cl2}[/latex](*g*) An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl_{5}, 9.14 g of SbCl_{3}, and 2.84 g of Cl_{2}. - Chlorine molecules dissociate according to this reaction: [latex]\ce{Cl2}[/latex](
*g*) [latex]/longrightarrow[\latex] 2Cl(*g*). 1.00% of Cl_{2}molecules dissociate at 975 K and a pressure of 1.00 atm.

- Antimony pentachloride decomposes at 448 °C. The reaction is: [latex]\ce{SbCl5[/latex](
- Given that the ΔG°
_{f }for Pb^{2+}(*aq*) and Cl^{–}(*aq*) is –24.3 kJ/mole and –131.2 kJ/mole respectively, determine the solubility product, K, for PbCl_{sp}_{2}(*s*). - Determine the standard free energy change, ΔG°
_{f }, for the formation of S^{2–}(*aq*) given that the for Ag^{+}(*aq*) and Ag_{2}S(*s*) are 77.1 kJ/mole and –39.5 kJ/mole respectively, and the solubility product for Ag_{2}S(*s*) is 8 × 10^{–51}. - Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
- When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.
- In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation: Glu + ATP [latex]\longrightarrow[/latex] G6P + ADP ΔG°
_{298}= -17 kJ. In this process, ATP becomes ADP summarized by the following equation: ATP [latex]\longrightarrow[/latex] ADP ΔG°_{298}= -30 kJ. Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process: Glu [latex]\longrightarrow][/latex] G6P ΔG°_{298}= ? - An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur described by the following equation: [latex]\ce{Cu2S][/latex](
*s*) [latex]\longrightarrow[/latex] Cu(*s*) + S(*s*)- Determine ΔG°
_{298 }for the decomposition of Cu_{2}S(*s*). - The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine ΔG°
_{298 }for the process. - The production of copper from chalcocite is performed by roasting the Cu
_{2}S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.

- Determine ΔG°

## Show Selected Solutions

- The answers are as follows:
- 465.1 kJ; nonspontaneous
- -106.86 kJ; spontaneous
- -291.9 kJ; spontaneous
- -83.4 kJ; spontaneous
- -406.7 kJ; spontaneous
- -30.0 kJ; spontaneous

- The answers are as follows:
- -4497.2 kJ/mol
- Dividing this result by four gives the equation of interest. The standard free energy of formation is –1124.3kJ/mol. (b) The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.

- The answers are as follows:
- nonspontaneous
- 566 °C

- High temperatures
- T>287 K
- K
= 1.6 × 10_{sp}^{-5} - The conversion is spontaneous, ΔG°
_{298}< 0, and is both enthalpy driven (exothermic, ΔH°_{298}< 0) and entropy driven, ΔS°_{298}> 0. However, the conversion is so slow that diamonds are effectively very stable. - 13 kJ; As this is a nonspontaneous process, it needs to be coupled to the hydrolysis of ATP to become spontaneous.