Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 12: Thermodynamics

36 problems · 4 figures

Example 12.1 §12.1 source ↗

Describe how energy is redistributed when a spoon at room temperature is placed in a cup of hot coffee.

Answer

Heat will spontaneously flow from the hotter object (coffee) to the colder object (spoon), resulting in a more uniform distribution of thermal energy as the spoon warms and the coffee cools.

Example 12.2 §12.2 source ↗

Consider the system shown in Figure 12.9. What is the change in entropy for the process where all the energy is transferred from the hot object (AB) to the cold object (CD)?
Figure 12.9 in the book ↗
This shows a microstate model describing the flow of heat from a hot object to a cold object. (a) Before the heat flow occurs, the object comprised of particles A and B contains both units of energy and as represented by a distribution of three microstates. (b) If the heat flow results in an even dispersal of energy (one energy unit transferred), a distribution of four microstates results. (c) If both energy units are transferred, the resulting distribution has three microstates.

Answer

0 J/K

Example 12.3 §12.2 source ↗

Predict the sign of the entropy change for the following processes. Give a reason for your prediction.

(a) NaNO3(s)⟶ Na+(aq)+ NO3−(aq)

(b) the freezing of liquid water

(c) CO2(s)⟶ CO2(g)

(d) CaCO3(s)⟶ CaO(s)+ CO2(g)

Answer

(a) Positive; The solid dissolves to give an increase of mobile ions in solution. (b) Negative; The liquid becomes a more ordered solid. (c) Positive; The relatively ordered solid becomes a gas. (d) Positive; There is a net increase in the amount of gaseous species.

Example 12.4 §12.3 source ↗

Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of Suniv?

Answer

Entropy is a state function, so ΔSfreezing = −ΔSmelting = −22.1 J/K and qsurr = +6.00 kJ. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.

Example 12.5 §12.3 source ↗

Calculate the standard entropy change for the following process: H2(g)+C2H4(g)⟶C2H6(g)

Answer

−120.6 J K–1 mol–1

Example 12.6 §12.3 source ↗

Calculate the standard entropy change for the following reaction: Ca(OH)2(s)⟶CaO(s)+H2O(l)

Answer

24.7 J/K

Example 12.7 §12.4 source ↗

Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process? C2H6(g)⟶H2(g)+C2H4(g)

Answer

ΔG°= 102.0 kJ/mol; the reaction is nonspontaneous (not spontaneous) at 25 °C.

Example 12.8 §12.4 source ↗

Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C? C2H4(g)⟶H2(g)+C2H2(g)

Answer

(a) 140.8 kJ/mol, nonspontaneous
(b) 141.5 kJ/mol, nonspontaneous

Example 12.9 §12.4 source ↗

What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions? FeS(s)+ O 2 (g)→Fe(s)+ SO 2 (g)

Answer

–199.7 kJ; spontaneous

12-1 §12.1 source ↗

What is a spontaneous reaction?

Answer

A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source.

12-3 §12.1 source ↗

Indicate whether the following processes are spontaneous or nonspontaneous.

(a) Liquid water freezing at a temperature below its freezing point

(b) Liquid water freezing at a temperature above its freezing point

(c) The combustion of gasoline

(d) A ball thrown into the air

(e) A raindrop falling to the ground

(f) Iron rusting in a moist atmosphere

Answer

(a) spontaneous; (b) nonspontaneous; (c) spontaneous; (d) nonspontaneous; (e) spontaneous; (f) spontaneous

12-5 §12.1 source ↗

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.

Answer

Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.

12-7 §12.2 source ↗

In Figure 12.8 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).
Figure 12.8 in the book ↗
The sixteen microstates associated with placing four particles in two boxes are shown. The microstates are collected into five distributions—(a), (b), (c), (d), and (e)—based on the numbers of particles in each box.

Answer

There are four initial microstates and four final microstates.
ΔS= klnWfWi= 1.38×10−23J/K×ln44= 0

12-9 §12.2 source ↗

Consider a system similar to the one in Figure 12.8, 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 18. What does this comparison tell us about even larger systems?
Figure 12.8 in the book ↗
The sixteen microstates associated with placing four particles in two boxes are shown. The microstates are collected into five distributions—(a), (b), (c), (d), and (e)—based on the numbers of particles in each box.

Answer

The probability for all the particles to be on one side is 132. This probability is noticeably lower than the 18 result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.

12-11 §12.2 source ↗

Consider the system shown in Figure 12.9. 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)?
Figure 12.9 in the book ↗
This shows a microstate model describing the flow of heat from a hot object to a cold object. (a) Before the heat flow occurs, the object comprised of particles A and B contains both units of energy and as represented by a distribution of three microstates. (b) If the heat flow results in an even dispersal of energy (one energy unit transferred), a distribution of four microstates results. (c) If both energy units are transferred, the resulting distribution has three microstates.

Answer

There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states.
ΔS= kln(WfWi)= 1.38×10−23J/K×ln(41)= 1.91×10−23J/K

12-13 §12.2 source ↗

At room temperature, the entropy of the halogens increases from I2 to Br2 to Cl2. Explain.

Answer

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, I2 is a solid, Br2 is a liquid, and Cl2 is a gas.

12-15 §12.2 source ↗

Indicate which substance in the given pairs has the higher entropy value. Explain your choices.

(a) C2H5OH(l) or C3H7OH(l)

(b) C2H5OH(l) or C2H5OH(g)

(c) 2H(g) or H(g)

Answer

(a) C3H7OH(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. (b) C2H5OH(g) as it is in the gaseous state. (c) 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).

12-17 §12.2 source ↗

Predict the sign of the entropy change for the following processes. Give a reason for your prediction.

(a) Na+(aq)+ Cl−(aq)⟶ NaCl(s)

(b) 2Fe(s)+ 32O2(g)⟶ Fe2O2(s)

(c) 2C6H14(l)+ 19O2(g)⟶ 14H2O(g)+ 12CO2(g)

Answer

(a) Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution. (b) Negative. There is a net loss of three moles of gas from reactants to products. (c) Positive. There is a net increase of seven moles of gas from reactants to products.

12-19 §12.2 source ↗

Write the balanced chemical equation for the combustion of benzene, C6H6(l), to give carbon dioxide and water vapor. Would you expect ΔS to be positive or negative in this process?

Answer

C6H6(l)+ 7.5O2(g)⟶ 3H2O(g)+ 6CO2(g)
There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and ΔS is positive.

12-21 §12.3 source ↗

Calculate ΔS° for the following changes.

(a) SnCl4(l)⟶ SnCl4(g)

(b) CS2(g)⟶ CS2(l)

(c) Cu(s)⟶Cu(g)

(d) H2O(l)⟶ H2O(g)

(e) 2H2(g)+ O2(g)⟶ 2H2O(l)

(f) 2HCl(g)+ Pb(s)⟶ PbCl2(s)+ H2(g)

(g) Zn(s)+ CuSO4(s)⟶ Cu(s)+ ZnSO4(s)

Answer

(a) 107 J/K; (b) −86.7 J/K; (c) 133.2 J/K; (d) 118.8 J/K; (e) −326.6 J/K; (f) −171.9 J/K; (g) −7.2 J/K

12-23 §12.3 source ↗

Determine the entropy change for the combustion of gaseous propane, C3H8, under the standard conditions to give gaseous carbon dioxide and water.

Answer

100.3 J/K

12-25 §12.3 source ↗

Using the relevant S° values listed in Appendix G, calculate ΔS°298 for the following changes:

(a) N2(g)+ 3H2(g)⟶ 2NH3(g)

(b) N2(g)+ 52O2(g)⟶ N2O5(g)

Answer

(a) −198.1 J/K; (b) −348.9 J/K

12-27 §12.3 source ↗

By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
SNaCl(s)°= 72.11Jmol·K SNaCl(l)°= 95.06Jmol·K ΔHfusion°= 27.95 kJ/mol

What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?

Answer

As ΔSuniv < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.

12-29 §12.3 source ↗

Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.

(a) C(s, graphite)+ O2(g)⟶ CO2(g)

(b) O2(g)+ N2(g)⟶ 2NO(g)

(c) 2Cu(s)+ S(g)⟶ Cu2S(s)

(d) CaO(s)+ H2O(l)⟶ Ca(OH)2(s)

(e) Fe2O3(s)+ 3CO(g)⟶ 2Fe(s)+ 3CO2(g)

(f) CaSO4·2H2O(s)⟶ CaSO4(s)+ 2H2O(g)

Answer

(a) 2.86 J/K; (b) 24.8 J/K; (c) −113.2 J/K; (d) −24.7 J/K; (e) 15.5 J/K; (f) 290.0 J/K

12-31 §12.4 source ↗

A reaction has ΔH° = 100 kJ/mol and ΔS°= 250 J/mol·K. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?

Answer

The reaction is nonspontaneous at room temperature.
Above 400 K, ΔG will become negative, and the reaction will become spontaneous.

12-33 §12.4 source ↗

Use the standard free energy of formation data in Appendix G 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.

(a) MnO2(s)⟶ Mn(s)+ O2(g)

(b) H2(g)+ Br2(l)⟶ 2HBr(g)

(c) Cu(s)+ S(g)⟶ CuS(s)

(d) 2LiOH(s)+ CO2(g)⟶ Li2CO3(s)+ H2O(g)

(e) CH4(g)+ O2(g)⟶ C(s,graphite)+ 2H2O(g)

(f) CS2(g)+ 3Cl2(g)⟶ CCl4(g)+ S2Cl2(g)

Answer

(a) 465.1 kJ nonspontaneous; (b) −106.86 kJ spontaneous; (c) −291.9 kJ spontaneous; (d) −83.4 kJ spontaneous; (e) −406.7 kJ spontaneous; (f) −154.3 kJ spontaneous

12-35 §12.4 source ↗

Given:
P4(s)+5O2(g)⟶P4O10(s)ΔG°=−2697.0 kJ/mol 2H2(g)+O2(g)⟶2H2O(g)ΔG°=−457.18 kJ/mol 6H2O(g)+P4O10(s)⟶4H3PO4(l)ΔG°=−428.66 kJ/mol

(a) Determine the standard free energy of formation, ΔGf°, for phosphoric acid.

(b) How does your calculated result compare to the value in Appendix G? Explain.

Answer

(a) The standard free energy of formation is –1124.3 kJ/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.

12-37 §12.4 source ↗

Consider the decomposition of red mercury(II) oxide under standard state conditions.
2HgO(s,red)⟶ 2Hg(l)+ O2(g)

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer

(a) The reaction is nonspontaneous; (b) Above 566 °C the process is spontaneous.

12-39 §12.4 source ↗

Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.

(a) N2(g)+ O2(g)⟶ 2NO(g) T= 2000°C Kp= 4.1×10−4

(b) H2(g)+ I2(g)⟶ 2HI(g) T= 400°C Kp= 50.0

(c) CO2(g)+ H2(g)⟶ CO(g)+ H2O(g) T= 980°C Kp= 1.67

(d) CaCO3(s)⟶ CaO(s)+ CO2(g) T= 900°C Kp= 1.04

(e) HF(aq)+ H2O(l)⟶ H3O+(aq)+ F−(aq) T= 25°C Ka= 7.2×10−4

(f) AgBr(s)⟶ Ag+(aq)+ Br−(aq) T= 25°C Ksp= 3.3×10−13

Answer

(a) 1.5 × 102 kJ; (b) −21.9 kJ; (c) −5.34 kJ; (d) −0.383 kJ; (e) 18 kJ; (f) 71 kJ

12-41 §12.4 source ↗

Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.

(a)O2(g)+2F2(g)⟶2OF2(g)ΔG°=−9.2 kJ (b)I2(s)+Br2(l)⟶2IBr(g)ΔG°=7.3 kJ (c)2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)ΔG°=−79 kJ (d)N2O3(g)⟶NO(g)+NO2(g)ΔG°=−1.6 kJ (e)SnCl4(l)⟶SnCl4(l)ΔG°=8.0 kJ

Answer

(a) K = 41; (b) K = 0.053; (c) K = 6.9 × 1013; (d) K = 1.9; (e) K = 0.04

12-42 §12.4 source ↗

Determine ΔGº for the following reactions.

(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
SbCl5(g)⟶ SbCl3(g)+ Cl2(g)

An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.

(b) Chlorine molecules dissociate according to this reaction:
Cl2(g)⟶ 2Cl(g)

1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.

Answer

(a) 22.1 kJ;
(b) 98.9 kJ/mol

12-44 §12.4 source ↗

Determine the standard free energy change, ΔGf°, for the formation of S2−(aq) given that the ΔGf° for Ag+(aq) and Ag2S(s) are 77.1 kJ/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 × 10−51.

Answer

90 kJ/mol

12-46 §12.4 source ↗

The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
H2O(l)⇌ H2O(g) ΔG°= 8.58 kJ

(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?

(b) Determine the equilibrium constant, KP, for this physical process.

(c) By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, PH2O, is 0.011 atm.

(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of PH2O in the air?

Answer

(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) Kp = 0.031; (c) The evaporation of water is spontaneous; (d) PH2O must always be less than Kp or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.

12-48 §12.4 source ↗

One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
G6P⇌ F6P ΔG°= 1.7 kJ

(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?

(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.

Answer

(a) Nonspontaneous as ΔG°>0; (b) ΔG°= ΔG°+RTlnQ, ΔG= 1.7×103+ (8.314×310×ln28120)= −2.1 kJ. The forward reaction to produce F6P is spontaneous under these conditions.

12-50 §12.4 source ↗

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.

Answer

ΔG is negative as the process is spontaneous. ΔH is positive as with the solution becoming cold, the dissolving must be endothermic. ΔS must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.

12-52 §12.4 source ↗

What happens to ΔG° (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

(a) S(s)+ O2(g)⟶ SO2(g)

(b) 2SO2(g)+ O2(g)⟶ SO3(g)

(c) HgO(s)⟶ Hg(l)+ O2(g)

Answer

(a) Increasing PO2 will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(b) Increasing PO2 will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(c) Increasing PO2 will shift the equilibrium the reactants, which decreases the value of K. ΔG° therefore becomes more positive.

Loading chapter 13…

Problems and answers are adapted from Chemistry: Atoms First 2e by Paul Flowers, Edward J. Neth, William R. Robinson, Klaus Theopold and Richard Langley, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.