Chapter 18

Entropy, Free Energy, and Equilibrium

18.6

Strategy:

According to equation 18.2, the number of ways of arranging N particles in X cells is W = X^N. Use this equation to calculate the number of arrangements and equation 18.1 to calculate the entropy.

Setup:

For the setup in Figure 18.2, X = 2 when the barrier is in place and X = 4 when it is absent. To simplify the calculations, use the power property of natural log to write:

.

Solution:

a.	With barrier: N = 10; W = 2^10 = 1024; S = (N×k)(ln W) = (10)(1.38 × 10^–23 J/K)ln(2) = 9.57 × 10^–23 J/K. Without barrier: N = 10; W = 4^10 = 1.48 × 10^6; S = 1.91 × 10^–22 J/K.
b.	With barrier: N = 50; W = 2^50 = 1.13 × 10^15; S = 4.78 × 10^–22 J/K. Without barrier: N = 10; W = 4^50 = 1.27 × 10^30; S = 9.57 × 10^–22 J/K.
c.	With barrier: N = 100; W = 2^100 = 1.26 × 10^30; S = 9.57 × 10^–22 J/K. Without barrier: N = 100; W = 4^100 = 1.61 × 10^60; S = 1.91 × 10^–21 J/K.

18.7

Strategy:

According to equation 18.2, the number of ways of arranging N particles in X cells is W = X^N. Use this equation to calculate the number of arrangements and equation 18.1 to calculate the entropy.

Setup:

For the setup in the figure, X = 4 when the barrier is in place and X = 8 when it is absent.

Solution:

a.

With barrier: N = 2; W = 4^2 = 16; Without barrier: N = 2; W = 8^2 = 64.

b.

From part (a), we know that 16 of the 64 arrangements have both particles in the left side of the container. Similarly, there are 16 ways for the particles to be found on the right-hand side. The number of arrangements with one particle per side is 64 – 16 – 16 = 32. Both particles on one side: S = k ln W^N = (1.38 × 10^–23 J/K)ln(32) = 3.83 × 10^–23 J/K; particles on opposite sides: S = (1.38 × 10^–23 J/K)ln(32) = 4.78 × 10^–23. The most probable state is the one with the larger entropy; that is, the state in which the particles are on opposite sides.

18.12

Strategy:

Equation 18.4 gives the entropy change for the isothermal expansion of an ideal gas. Substitute the given values into the equation and compute DS.

Solution:

a.
b.
c.

18.13

Strategy:

Equation 18.4 gives the entropy change for the isothermal expansion of an ideal gas. Substitute the given values into the equation and compute DS.

Solution:

a.
b.
c.

18.14

a.

Li(l); The liquid form of any substance always has greater entropy.

b.

At first glance there may seem to be no apparent difference between the two substances that might affect the entropy (the molecular formulas are identical). However, the first has the -O-H structural feature which allows it to participate in hydrogen bonding with other molecules. This results in fewer possible arrangements of molecules in the liquid state. The standard entropy of CH₃OCH₃ is larger.

c.

Both are monatomic species. However, the Xe atom has a greater molar mass than Ar. Xenon has the higher standard entropy.

d.

Due to the extra oxygen atom, carbon dioxide molecules have a more complex molecular structure than do carbon monoxide molecules. Specifically, carbon dioxide has vibrational and rotational modes that carbon monoxide does not. So, carbon dioxide gas has the higher standard entropy (see Appendix 2). There is also a mass effect. For gas phase molecules of similar complexity, increasing molar mass tends to increase the molar entropy.

e.

O₃ has a greater molecular complexity than O₂ and thus has the higher standard entropy. Ozone’s molecules are also more massive (see (d) above).

f.

Because of its greater molecular complexity and greater mass (see answer (d)), one mole of N₂O₄ has a larger standard entropy than one mole of NO₂. Compare values in Appendix 2.

Think About It:

Use the data in Appendix 2 to compare the standard entropy of one mole of N2O4 with that of two moles of NO2. In this situation the number of atoms is the same for both. Which is higher and why?

18.15

In order of increasing entropy per mole at 25°C:

(c) < (d) < (e) < (a) < (b)

a.	Ne(g): a monatomic gas of higher molar mass than H2. (For gas phase molecules, increasing molar mass tends to increase the molar entropy. While Ne(g) is less complex structurally than H2(g), its much larger molar mass more than offsets the complexity difference.)
b.	SO2(g): a polyatomic gas of higher complexity and higher molar mass than Ne(g) (see the explanation for (a) above).
c.	Na(s): highly ordered, crystalline material.
d.	NaCl(s): highly ordered crystalline material, but with more particles per mole than Na(s).
e.	H2: a diatomic gas, hence of higher entropy than a solid.

18.16

Using Equation 18.5 of the text to calculate :

a.
b.
c.	DS= 4S°(CO2) + 6S°(H2O(l)) - [2S°(C2H6) + 7S°(O2)] DS= (4)(213.6 J/K×mol) + (6)(69.9 J/K×mol) - [(2)(229.5 J/K×mol) + (7)(205.0 J/K×mol)] = -620.2 J/K×mol

18.17

Strategy:

To calculate the standard entropy change of a reaction, we look up the standard entropies of reactants and products in Appendix 2 of the text and apply Equation 18.7. As in the calculation of enthalpy of reaction, the stoichiometric coefficients have no units, so is expressed in units of J/K×mol.

Solution:

The standard entropy change for a reaction can be calculated using the following equation.

a.	= (1)(33.3 J/K×mol) + (1)(188.7 J/K×mol) - [(1)(131.0 J/K×mol) + (1)(43.5 J/K×mol)] = 47.5 J/K×mol
b.	= (1)(50.99 J/K×mol) + (3)(41.6 J/K×mol) - [(2)(28.3 J/K×mol) + (3)(43.9 J/K×mol)] = -12.5 J/K×mol
c.	= (1)(213.6 J/K×mol) + (2)(69.9 J/K×mol) - [(1)(186.2 J/K×mol) + (2)(205.0 J/K×mol)] = -242.8 J/K×mol

Why was the entropy value for water different in parts (a) and (c)?

18.20

Strategy:

Assume all reactants and products are in their standard states. The entropy change in the surroundings is related to the enthalpy change of the system by Equation 18.7. For each reaction in Exercise 18.16, use Appendix 2 to calculate DH_sys ( = DH_rxn) (Section 5.3) and then use Equation 18.7 to calculate DS_surr. Finally, use Equation 18.8 to compute DS_univ. If DS_univ is positive, then according to the second law of thermodynamics, the reaction is spontaneous. The values for DS_sys are found in the solution for problem 18.16.

Solution:

a.	spontaneous
b.	not spontaneous
c.	spontaneous

18.21

Strategy:

Assume all reactants and products are in their standard states. The entropy change in the surroundings is related to the enthalpy change of the system by Equation 18.7. For each reaction in Exercise 18.17, use Appendix 2 to calculate DH_sys ( = DH_rxn) (Section 5.3) and then use Equation 18.7 to calculate DS_surr. Finally, use Equation 18.8 to compute DS_univ. If DS_univ is positive, then according to the second law of thermodynamics, the reaction is spontaneous. The values for DS_sys are found in the solution for problem 18.16.

Solution:

a.	spontaneous
b.	spontaneous
c.	spontaneous

18.22

Strategy:

Assume all reactants and products are in their standard states. According to Equations 18.7, the entropy change in the surroundings for an isothermal process is :

.

Also, Equation 18.8 states that the entropy change of the universe is:

.

Use Appendix 2 to calculate DS_sys ( = DS_rxn) and DH_sys ( = DH_rxn). Substitute these results into the above equations and determine the sign of DS_univ. If DS_univ is positive, then according to the second law of thermodynamics, the reaction is spontaneous.

Solution:

a.	not spontaneous
b.	spontaneous
c.	spontaneous
d.	To calculate DHrxn for N2 ® 2N, use the bond enthalpy from Table 8.6. Calculate DSsurr and DSuniv: not spontaneous

Think About It:

We could have assumed that the reaction in (d) occurred at constant volume instead of constant pressure. Would this have changed the conclusion about the spontaneity of the reaction?

18.23

Strategy:

Assume all reactants and products are in their standard states. According to Equations 18.7, the entropy change in the surroundings for an isothermal process is :

.

Also, Equation 18.8 states that the entropy change of the universe is:

.

Use Appendix 2 to calculate DS_sys ( = DS_rxn) and DH_sys ( = DH_rxn). Substitute these results into the above equations and determine the sign of DS_univ. If DS_univ is positive, then according to the second law of thermodynamics, the reaction is spontaneous.

Solution:

a.	spontaneous
b.	not spontaneous
c.	To calculate DHrxn for H2 ® 2H, use the bond enthalpy from Table 8.6. Calculate DSsurr and DSuniv: not spontaneous
d.	spontaneous

Think About It:

We could have assumed that the reaction in (c) occurred at constant volume instead of constant pressure. Would this have changed the conclusion about the spontaneity of the reaction?

18.24

Strategy:	We assume the denaturation is a single-step equilibrium process. Therefore, because DG is zero at equilibrium, we can use Equation 18.10, substituting 0 for DG and solving for DS, to determine the entropy change associated with the process.
Setup:	The melting temperature of the protein is 63 + 273.15 = 336.2 K.
Solution:	The unfolding of the protein increases the volume it occupies and increases its entropy. Therefore, we estimate the entropy of denaturation to be positive. Calculating the entropy gives, DSdenaturation = 1.52 kJ/K×mol or 1.52 ´ 10^3 J/K×mol 16 J/K×mol per amino acid

18.29

Strategy:	Use Equation 18.10 to solve for DG.
Setup:	From problem 18.24, DS = 1.52 kJ/K×mol and DH = 510 kJ/mol. T = (20 + 273.15) = 293.2 K.
Solution:	DG = DH - TDS = 510 kJ/mol - (293.2 K)(1.52 kJ/K×mol) = 64 kJ/mol

18.30

Using Equation 18.12 of the text to solve for the change in standard free energy,

a.
b.
c.	= (4)(-394.4 kJ/mol) + (2)(-237.2 kJ/mol) - (2)(209.2 kJ/mol) - (5)(0) = -2470 kJ/mol

18.31

Strategy:

To calculate the standard free-energy change of a reaction, we look up the standard free energies of formation of reactants and products in Appendix 2 of the text and apply Equation 18.13. Note that all the stoichiometric coefficients have no units so is expressed in units of kJ/mol. The standard free energy of formation of any element in its stable allotropic form at 1 atm and 25°C is zero.

Solution:

The standard free energy change for a reaction can be calculated using the following equation.

a.
b.
c.

18.32

Reaction A: First apply Equation 18.10 of the text to compute the free energy change at 25°C (298 K)

DG = DH - TDS = 10,500 J/mol - (298 K)(30 J/K×mol) = 1560 J/mol

The +1560 J/mol shows the reaction is not spontaneous at 298 K. The DG will change sign (i.e., the reaction will become spontaneous) above the temperature at which DG = 0.

0 = DH - TDS

Solving for T gives

Reaction B: Calculate DG.

DG = DH - TDS = 1800 J/mol - (298 K)(-113 J/K×mol) = 35,500 J/mol

The free energy change is positive, which shows that the reaction is not spontaneous at 298 K. Since both terms are positive, there is no temperature at which their sum is negative. The reaction is not spontaneous at any temperature.

18.33

a.

Calculate DG from DH and DS. DG = DH - TDS = -126,000 J/mol - (298 K)(84 J/K×mol) = -151,000 J/mol The free energy change is negative so the reaction is spontaneous at 298 K. Since DH is negative and DS is positive, the reaction is spontaneous at all temperatures.

b.

Calculate DG. DG = DH - TDS = -11,700 J/mol - (298 K)(-105 J/K×mol) = +19,600 J The free energy change is positive at 298 K which means the reaction is not spontaneous at that temperature. The positive sign of DG results from the large negative value of DS. At lower temperatures, the -TDS term will be smaller thus allowing the free energy change to be negative. DG will equal zero when DH = TDS. Rearranging, At temperatures below 111 K, DG will be negative and the reaction will be spontaneous.

18.34

DS_fus = = 48.8 J/K×mol

DS_vap = = 74.0 J/K×mol

18.35

Strategy:	Equation 18.7 from the text relates the entropy change of the surroundings to the enthalpy change of the system and the temperature at which a phase change occurs. DSsurr = We know that DSsys = -DSsurr, so we can rewrite Equation 18.7 as DSsys =
Solution:	DSfus = = 99.9 J/K×mol DSvap = = 93.6 J/K×mol
Think About It:	Remember that Celsius temperatures must be converted to Kelvin.

18.36

Using Equation 18.12 from the text,

DG = 2DG(C₂H₅OH) + 2DG(CO₂) - DG( C₆H₁₂O₆)

DG = (2)(-174.18 kJ/mol) + (2)(-394.4 kJ/mol) - (-910.56 kJ/mol) = -226.6 kJ/mol

18.37

Using Equation 18.2 from the text,

DG = DG(NO) - [1/2 DG(O₂) + DG( NO)]

= (-110.5 kJ/mol) - [(0 kJ/mol) + (-34.6 kJ/mol) = -75.9 kJ/mol

75.9 kJ of Gibbs free energy are released.

18.41

Strategy:	According to Equation 18.14 of the text, the equilibrium constant for the reaction is related to the standard free energy change; that is, DG° = -RT ln K. Since we are given DG° in the problem, we can solve for the equilibrium constant. What temperature unit should be used?
Solution:	Solving Equation 18.14 for K gives

18.42

Since we are given the equilibrium constant in the problem, we can solve for DG° using Equation 18.14.

DG° = -RTln K

Substitute K_w, R, and T into the above equation to calculate the standard free energy change, DG°. The temperature at which K_w = 1.0 ´ 10^-14 is 25°C = 298 K.

DG° = -RTln K_w

DG° = -(8.314 J/mol×K)(298 K) ln (1.0 ´ 10^-14) = 8.0 ´ 10⁴ J/mol = 8.0 ´ 10¹ kJ/mol

18.43

K_sp = [Fe²⁺][OH^-]² = 1.6 ´ 10^-14

DG° = -RTln K_sp = -(8.314 J/K×mol)(298 K)ln (1.6 ´ 10^-14) = 7.9 ´ 10⁴ J/mol = 79 kJ/mol

18.44

Use standard free energies of formation from Appendix 2 to find the standard free energy difference.

We can calculate K_P using the following equation. We carry additional significant figures in the calculation to minimize rounding errors when calculating K_P.

DG° = -RTln K_P

4.572 ´ 10⁵ J/mol = -(8.314 J/mol×K)(298 K) ln K_P

-184.54 = ln K_P

Taking the anti-ln of both sides,

e^-184.54 = K_P

K_P = 7.2 ´ 10^-81

18.45

a.

We first find the standard free energy change of the reaction.

= (1)(-269.6 kJ/mol) + (1)(0) - (1)(-305.0 kJ/mol) = 35.4 kJ/mol

We can calculate K_P by rearranging Equation 18.14 of the text.

b.

We are finding the free energy difference between the reactants and the products at their nonequilibrium values. The result tells us the direction of and the potential for further chemical change. We use the given nonequilibrium pressures to compute Q_P.

The value of DG (notice that this is not the standard free energy difference) can be found using Equation 18.13 of the text and the result from part (a).

DG = DG° + RTln Q = (35.4 ´ 10³ J/mol) + (8.314 J/K×mol)(298 K)ln (37) = 44.6 kJ/mol

Think About It:

Which way is the direction of spontaneous change for this system? What would be the value of DG if the given data were equilibrium pressures? What would be the value of QP in that case?

18.46

a.

The equilibrium constant is related to the standard free energy change by the following equation. DG° = -RTln K Substitute KP, R, and T into the above equation to the standard free energy change, DG°. DG° = -RTln KP DG° = -(8.314 J/mol×K)(2000 K) ln (4.40) = -2.464 ´ 10^4 J/mol = -24.6 kJ/mol

b.

Under non-standard-state conditions, DG is related to the reaction quotient Q by the following equation. DG = DG° + RTln QP We are using QP in the equation because this is a gas-phase reaction. Step 1: DG° was calculated in part (a). We must calculate QP. We carry additional significant figures in this calculation to minimize rounding errors. Step 2: Substitute DG° = -2.46 ´ 10^4 J/mol and QP into the following equation to calculate DG. DG = DG° + RTln QP DG = -2.464 ´ 10^4 J/mol + (8.314 J/mol×K)(2000 K) ln (4.062) DG = (-2.464 ´ 10^4 J/mol) + (2.331 ´ 10^4 J/mol) DG = -1.33 ´ 10^3 J/mol = -1.33 kJ/mol

18.47

The expression of K_P is:

Thus you can predict the equilibrium pressure directly from the value of the equilibrium constant. The only task at hand is computing the values of K_P using Equations 18.10 and 18.14 of the text.

a.

At 25°C, DG° = DH° - TDS° = (177.8 ´ 10³ J/mol) - (298 K)(160.5 J/K×mol)

= 130.0 ´ 10³ J/mol

b.

At 800°C, DG° = DH° - TDS° = (177.8 ´ 10³ J/mol) - (1073 K)(160.5 J/K×mol)

= 5.58 ´ 10³ J/mol

Think About It:

What assumptions are made in the second calculation?

18.48

We use the given K_P to find the standard free energy change.

DG° = -RTln K

DG° = -(8.314 J/K×mol)(298 K) ln (5.62 ´ 10³⁵) = –2.04 ´ 10⁵ J/mol = -204 kJ/mol

The standard free energy of formation of one mole of COCl₂ can now be found using the standard free energy of reaction calculated above and the standard free energies of formation of CO(g) and Cl₂(g).

18.49

The equilibrium constant expression is:

We are actually finding the equilibrium vapor pressure of water (compare to Problem 18.47). We use Equation 18.14 of the text.

or 23.6 mmHg

Think About It:

The positive value of DG° indicates that reactants are favored at equilibrium at 25°C. Is that what you would expect?

18.50

The standard free energy change is given by:

You can look up the standard free energy of formation values in Appendix 2 of the text.

Thus, the formation of graphite from diamond is favored under standard-state conditions at 25°C. However, the rate of the diamond to graphite conversion is very slow (due to a high activation energy) so that it will take millions of years before the process is complete.

18.53

C₆H₁₂O₆ + 6O₂ ® 6CO₂ + 6H₂O DG° = -2880 kJ/mol

ADP + H₃PO₄ ® ATP + H₂O DG° = +31 kJ/mol

Maximum number of ATP molecules synthesized:

18.54

The equation for the coupled reaction is:

glucose + ATP ® glucose 6-phosphate + ADP

DG° = 13.4 kJ/mol - 31 kJ/mol = -18 kJ/mol

As an estimate:

K = 1 ´ 10³

18.55

Strategy:

Melting is an endothermic process, DH_fus > 0. Also, a liquid generally has a higher entropy than the solid at the same temperature, so DS_fus > 0. To determine the sign of DG_fus, use the fact that melting is spontaneous (DG_fus < 0) for temperatures above the freezing point and is not spontaneous (DG_fus > 0) for temperatures below the freezing point.

Solution:

a.	The temperature is above the freezing point, so melting is spontaneous and DGfus < 0.
b.	The temperature is at the freezing point, so the solid and the liquid are in equilibrium and DGfus = 0.
c.	The temperature is below the freezing point, so melting is not spontaneous and DGfus > 0.

18.56

In each part of this problem we can use the following equation to calculate DG.

DG = DG° + RTln Q

or

DG = DG° + RTln( [H⁺][OH^-])

a.	In this case, the given concentrations are equilibrium concentrations at 25°C. Since the reaction is at equilibrium, DG = 0. This is advantageous, because it allows us to calculate DG°. Also recall that at equilibrium, Q = K. We can write: DG° = -RTln Kw DG° = -(8.314 J/K×mol)(298 K) ln (1.0 ´ 10^-14) = 8.0 ´ 10^4 J/mol
b.	DG = DG° + RTln Q = DG° + RTln ([H^+][OH^-]) DG = (8.0 ´ 10^4 J/mol) + (8.314 J/K×mol)(298 K) ln [(1.0 ´ 10^-3)(1.0 ´ 10^-4)] = 4.0 ´ 10^4 J/mol
c.	DG = DG° + RTln Q = DG° + RTln ([H^+][OH^-]) DG = (8.0 ´ 10^4 J/mol) + (8.314 J/K×mol)(298 K) ln [(1.0 ´ 10^-12)(2.0 ´ 10^-8)] = -3.2 ´ 10^4 J/mol
d.	DG = DG° + RTln Q = DG° + RTln [H^+][OH^-] DG = (8.0 ´ 10^4 J/mol) + (8.314 J/K×mol)(298 K) ln [(3.5)(4.8 ´ 10^-4)] = 6.4 ´ 10^4 J/mol

18.57

Only U and H are associated with the first law alone.

18.58

One possible explanation is simply that no reaction is possible. In other words, there is an unfavorable free energy difference between products and reactants (DG > 0).

A second possibility is that the potential for spontaneous change exists (DG < 0), but that the reaction is extremely slow (very large activation energy).

A remote third choice is that the student accidentally prepared a mixture in which the components were already at their equilibrium concentrations.

Think About It:

Which of the above situations would be altered by the addition of a catalyst?

18.59

Setting DG equal to zero and solving Equation 18.10for T gives

T = 42°C

18.60

For a reaction to be spontaneous, DG must be negative. If DS is negative, as it is in this case, then the reaction must be exothermic. When water freezes, it gives off heat (exothermic). Consequently, the entropy of the surroundings increases and DS_univ > 0.

18.61

If the process is spontaneous as well as endothermic, the signs of DG and DH must be negative and positive, respectively. Since DG = DH - TDS, the sign of DS must be positive (DS > 0) for DG to be negative.

18.62

The equation is: BaCO₃(s) BaO(s) + CO₂(g)

DG° = (1)(-528.4 kJ/mol) + (1)(-394.4 kJ/mol) - (1)(-1138.9 kJ/mol) = 216.1 kJ/mol

DG° = -RTln K_P

18.63

a.	Using the relationship: benzene DSvap = 87.8 J/K×mol hexane DSvap = 90.1 J/K×mol mercury DSvap = 93.7 J/K×mol toluene DSvap = 91.8 J/K×mol Trouton’s rule is a statement about DS. In most substances, the molecules are in constant and random motion in both the liquid and gas phases, so DS = 90 J/K×mol.
b.	Using the data in Table 12.6 of the text, we find: ethanol DSvap = 111.9 J/K×mol water DSvap = 109.4 J/K×mol In ethanol and water, there are fewer possible arrangements of the molecules due to the network of H-bonds, so DS is greater.

18.64

Evidence shows that HF, which is strongly hydrogen-bonded in the liquid phase, is still considerably hydrogen-bonded in the vapor state such that its DS_vap is smaller than most other substances.

18.65

a.	2CO + 2NO ® 2CO2 + N2
b.	The oxidizing agent is NO; the reducing agent is CO.
c.	DG° = (2)(-394.4 kJ/mol) + (0) - (2)(-137.3 kJ/mol) - (2)(86.7 kJ/mol) = -687.6 kJ/mol DG° = -RTln KP KP = 3 ´ 10^120
d.	QP = 1.2 ´ 10^14 Since QP << KP, the reaction will proceed to the right.
e.	DH° = (2)(-393.5 kJ/mol) + (0) - (2)(-110.5 kJ/mol) - (2)(90.4 kJ/mol) = -746.8 kJ/mol Since DH° is negative, raising the temperature will decrease KP (Le Chatelier’s principle), thereby increasing the amount of reactants and decreasing the amount of products. No, the formation of N2 and CO2 is not favored by raising the temperature.

18.66

a.

At two different temperatures T₁ and T₂,

(1)

(2)

Rearranging Equations (1) and (2),

(3)

(4)

Subtracting equation (3) from equation (4) gives,

b.

Using the equation that we just derived, we can calculate the equilibrium constant at 65°C.

K₁ = 4.63 ´ 10^-3 T₁ = 298 K

K₂ = ? T₂ = 338 K

Taking the anti-ln of both sides of the equation,

K₂ = 0.074

Think About It:

K2 > K1, as we would predict for a positive DH°. Recall that an increase in temperature will shift the equilibrium towards the endothermic reaction; that is, the decomposition of N2O4.

18.67

The equilibrium reaction is:

AgCl(s) Ag⁺(aq) + Cl^-(aq)

K_sp = [Ag⁺][Cl^-] = 1.6 ´ 10^-10

We can calculate the standard enthalpy of reaction from the standard enthalpies of formation in Appendix 2 of the text.

DH° = (1)(105.9 kJ/mol) + (1)(-167.2 kJ/mol) - (1)(-127.0 kJ/mol) = 65.7 kJ/mol

From Problem 18.66(a):

K₁ = 1.6 ´ 10^-10 T₁ = 298 K

K₂ = ? T₂ = 333 K

K₂ = 2.6 ´ 10^-9

The increase in K indicates that the solubility increases with temperature.

18.68

At absolute zero. A substance can never have a negative entropy.

18.69

Assuming that both DH° and DS° are temperature independent, we can calculate both DH° and DS°.

DH° = (1)(-110.5 kJ/mol) + (1)(0)] - [(1)(-241.8 kJ/mol) + (1)(0)]

DH° = 131.3 kJ/mol

DS° = S°(CO) + S°(H₂) - [S°(H₂O) + S°(C)]

DS° = [(1)(197.9 J/K×mol) + (1)(131.0 J/K×mol)] - [(1)(188.7 J/K×mol) + (1)(5.69 J/K×mol)]

DS° = 134.5 J/K×mol

It is obvious from the given conditions that the reaction must take place at a fairly high temperature (in order to have red-hot coke). Setting DG° = 0

0 = DH° - TDS°

The temperature must be greater than 703°C for the reaction to be spontaneous.

18.70

a.	We know that HCl is a strong acid and HF is a weak acid. Thus, the equilibrium constant will be less than 1 (K < 1).
b.	The number of particles on each side of the equation is the same, so DS° » 0. Therefore DH° will dominate.
c.	HCl is a weaker bond than HF (see Table 9.4 of the text), therefore DH° > 0.

18.71

Begin by writing the equation for the process described in the problem. Hydrogen ions must be transported from a region where their concentration is 10^-7.4 (4 ´ 10^-8 M) to a region where their concentration is 10^-1 (0.1 M).

H⁺(aq, 4 ´ 10^-8 M) H⁺(aq, 0.1 M)

DG° for this process is 0. We use Equation 18.13 to calculate DG. (T = 37°C + 273 = 310 K.)

DG = DG° + RTlnQ = (8.314 ´ 10^-3 kJ/mol)(310 K)= 38 kJ/mol

Therefore, the Gibbs free energy required for the secretion of 1 mole of H⁺ ions from the blood plasma to the stomach is 38 kJ.

18.72

For a reaction to be spontaneous at constant temperature and pressure, DG must be negative. The process of crystallization results in fewer possible arrangements of ions, so DS < 0. We also know that

DG = DH - TDS

Since DG must be negative, and since the entropy term will be positive (-TDS, where DS is negative), then DH must be negative (DH < 0). The reaction will be exothermic.

18.73

For the reaction: CaCO₃(s) CaO(s) + CO₂(g)

Using the equation from Problem 18.66:

Substituting,

Solving,

DH° = 1.74 ´ 10⁵ J/mol = 174 kJ/mol

18.74

For the reaction to be spontaneous, DG must be negative.

DG = DH - TDS

Given that DH = 19 kJ/mol = 19,000 J/mol, then

DG = 19,000 J/mol - (273 K + 72 K)(DS)

Solving the equation with the value of DG = 0

0 = 19,000 J/mol - (273 K + 72 K)(DS)

DS = 55 J/K×mol

This value of DS which we solved for is the value needed to produce a DG value of zero. Any value of DS that is larger than 55 L/mol×K will result in a spontaneous reaction.

18.75

a.	DS is positive

b.	DS is negative

c.	DS is positive

d.	DS is positive

18.76

The second law states that the entropy of the universe must increase in a spontaneous process. But the entropy of the universe is the sum of two terms: the entropy of the system plus the entropy of the surroundings. One of the entropies can decrease, but not both. In this case, the decrease in system entropy is offset by an increase in the entropy of the surroundings. The reaction in question is exothermic, and the heat released increases the entropy of the surroundings.

18.77

At the temperature of the normal boiling point the free energy difference between the liquid and gaseous forms of mercury (or any other substances) is zero, i.e. the two phases are in equilibrium. We can therefore use Equation 18.10 of the text to find this temperature. For the equilibrium,

Hg(l) Hg(g)

DG = DH - TDS = 0

DS = S°[Hg(g)] - S°[Hg(l)] = 174.7 J/K×mol - 77.4 J/K×mol = 97.3 J/K×mol

T_bp = 625 K

Think About It:

What assumptions are made? Notice that the given enthalpies and entropies are at standard conditions, namely 25°C and 1.00 atm pressure. In performing this calculation we assume that DH° and DS° do not depend on temperature. The actual normal boiling point of mercury is 356.58°C. Is the assumption of the temperature independence of these quantities reasonable?

18.78

Setting DG equal to zero, because the system is at equilibrium, and rearranging Equation 18.10, we write

The problem asks for the change in entropy for the vaporization of 0.50 moles of ethanol. The DS calculated above is for 1 mole of ethanol.

DS for 0.50 mol = (112 J/mol×K)(0.50 mol) = 56 J/K

18.79

No. A negative DG° tells us that a reaction has the potential to happen, but gives no indication of the rate.

18.80

For the given reaction we can calculate the standard free energy change from the standard free energies of formation. Then, we can calculate the equilibrium constant, K_P, from the standard free energy change.

DG° = (1)(-587.4 kJ/mol) - [(4)(-137.3 kJ/mol) + (1)(0)] = -38.2 kJ/mol = -3.82 ´ 10⁴ J/mol

Substitute DG°, R, and T (in K) into the following equation to solve for K_P.

DG° = -RTln K_P

K_P = 4.5 ´ 10⁵

18.81

a.	DG° = -106.4 kJ/mol KP = 4 ´ 10^18
b.	DG° = -53.2 kJ/mol KP = 2 ´ 10^9 The KP in (a) is the square of the KP in (b). Both DG° and KP depend on the number of moles of reactants and products specified in the balanced equation.

18.82

We carry additional significant figures throughout this calculation to minimize rounding errors. The equilibrium constant is related to the standard free energy change by the following equation:

DG° = -RTln K_P

2.12 ´ 10⁵ J/mol = -(8.314 J/mol×K)(298 K) ln K_P

K_P = 6.894 ´ 10^-38

We can write the equilibrium constant expression for the reaction.

This pressure is far too small to measure.

18.83

Strategy:	We assume the denaturation is a single-step equilibrium process. Therefore, because DG is zero at equilibrium, we can use Equation 18.10, substituting 0 for DG and solving for T, to determine the temperature at which the denaturation becomes spontaneous.
Setup:	DH = 512 kJ/mol and DS = 1.60 kJ/K ∙ mol.
Solution:	DH and DS are both positive, indicating this is an endothermic process with high entropy. 320 K = 47°C

18.84

Both (a) and (b) apply to a reaction with a negative DG° value. Statement (c) is not always true. An endothermic reaction that has a positive DS° (increase in entropy) will have a negative DG° value at high temperatures.

18.85

a.	If DG° for the reaction is 173.4 kJ/mol, then,
b.	DG° = -RTln KP 173.4 ´ 10^3 J/mol = -(8.314 J/K×mol)(298 K)ln KP KP = 4 ´ 10^-31
c.	DH° for the reaction is 2 ´ (NO) = (2)(90.4 kJ/mol) = 180.8 kJ/mol Using the equation in Problem 18.66: K2 = 3 ´ 10^-6
d.	Lightning supplies the energy necessary to drive this reaction, converting the two most abundant gases in the atmosphere into NO(g). The NO gas dissolves in the rain, which carries it into the soil where it is converted into nitrate and nitrite by bacterial action. This “fixed” nitrogen is a necessary nutrient for plants.

18.86

We write the two equations as follows. The standard free energy change for the overall reaction will be the sum of the two steps.

CuO(s) Cu(s) + O₂(g) DG° = 127.2 kJ/mol

C(graphite) + O₂(g) CO(g) DG° = -137.3 kJ/mol

CuO + C(graphite) Cu(s) + CO(g) DG° = -10.1 kJ/mol

We can now calculate the equilibrium constant from the standard free energy change, DG°.

ln K = 1.81

K = 6.1

18.87

As discussed in Chapter 18 of the text for the decomposition of calcium carbonate, a reaction favors the formation of products at equilibrium when

DG° = DH° - TDS° < 0

If we can calculate DH° and DS°, we can solve for the temperature at which decomposition begins to favor products. We use data in Appendix 2 of the text to solve for DH° and DS°.

DH° = -601.8 kJ/mol + (-393.5 kJ/mol) - (-1112.9 kJ/mol) = 117.6 kJ/mol

DS° = S°[MgO(s)] + S°[CO₂(g)] - S°[MgCO₃(s)]

DS° = 26.78 J/K×mol + 213.6 J/K×mol - 65.69 J/K×mol = 174.7 J/K×mol

For the reaction to begin to favor products,

DH° - TDS° < 0

or

T > 673.2 K

18.88

According to Table 18.4, a reaction for which DH and DS are both negative is spontaneous at low temperatures, (d).

18.89

a.	DG° = (1)(0) + (1)(-84.9 kJ/mol) - (1)(0) - (2)(0) DG° = -84.9 kJ/mol DG° = -RTln K -84.9 ´ 10^3 J/mol = -(8.314 J/mol×K)(298 K) ln K K = 7.6 ´ 10^14
b.	DG° = 64.98 kJ/mol DG° = -RTln K 64.98 ´ 10^3 J/mol = -(8.314 J/mol×K)(298 K) ln K K = 4.1 ´ 10^-12 The activity series is correct. The very large value of K for reaction (a) indicates that products are highly favored; whereas, the very small value of K for reaction (b) indicates that reactants are highly favored.

18.90

2NO + O₂ 2NO₂

DG° = (2)(51.8 kJ/mol) - (2)(86.7 kJ/mol) - 0 = -69.8 kJ/mol

DG° = -RTln K

-69.8 ´ 10³ J/mol = -(8.314 J/mol×K)(298 K)ln K

K = 1.7 ´ 10¹² M^-1

k_r = 4.2 ´ 10^-3 M^-1s^-1

18.91

a.	It is the reverse of a disproportionation redox reaction.
b.	DG° = (2)(-228.6 kJ/mol) - (2)(-33.0 kJ/mol) - (1)(-300.4 kJ/mol) DG° = -90.8 kJ/mol -90.8 ´ 10^3 J/mol = -(8.314 J/mol×K)(298 K) ln K K = 8.2 ´ 10^15 Because of the large value of K, this method is feasible for removing SO2.
c.	DH° = (2)(-241.8 kJ/mol) + (3)(0) - (2)(-20.15 kJ/mol) - (1)(-296.4 kJ/mol) DH° = -146.9 kJ/mol DS° = (2)(188.7 J/K×mol) + (3)(31.88 J/K×mol) - (2)(205.64 J/K×mol) - (1)(248.5 J/K×mol) DS° = -186.7 J/K×mol DG° = DH° - TDS° Due to the negative entropy change, DS°, the free energy change, DG°, will become positive at higher temperatures. Therefore, the reaction will be less effective at high temperatures.

18.92

(1) Measure K and use DG° = -RT

(2) Measure DH° and DS° and use DG° = DH° - TDS°

18.93

2O₃ 3O₂

DG° = -326.8 kJ/mol

-326.8 ´ 10³ J/mol = -(8.314 J/mol×K)(243 K) ln K_P

K_P = 1.8 ´ 10⁷⁰

Due to the large magnitude of K, you would expect this reaction to be spontaneous in the forward direction. However, this reaction has a large activation energy, so the rate of reaction is extremely slow.

18.94

DS_denaturation = 1.2 kJ/K×mol or 1.2 ´ 10³ J/K×mol

9.8 J/K×mol per amino acid

18.95

First convert to moles of ice.

For a phase transition:

DS_sys =91.1 J/K

DS_surr =-91.1 J/K

DS_univ = DS_sys + DS_surr = 0

The system is at equilibrium.

Think About It:

We ignored the freezing point depression of the salt water. How would inclusion of this effect change our conclusion?

18.96

Heating the ore alone is not a feasible process. Looking at the coupled process:

Cu₂S ® 2Cu + S DG° = 86.1 kJ/mol

S + O₂ ® SO₂ DG° = -300.4 kJ/mol

Cu₂S + O₂ ® 2Cu + SO₂ DG° = -214.3 kJ/mol

Since DG° is a large negative quantity, the coupled reaction is feasible for extracting sulfur.

18.97

Since we are dealing with the same ion (K⁺), Equation 18.13 of the text can be written as:

DG = DG° + RTln Q

DG = 8.5 ´ 10³ J/mol = 8.5 kJ/mol

18.98

First, we need to calculate DH° and DS° for the reaction in order to calculate DG°.

DH° = -41.2 kJ/mol DS° = -42.0 J/K×mol

Next, we calculate DG° at 300°C or 573 K, assuming that DH° and DS° are temperature independent.

DG° = DH° - TDS°

DG° = -41.2 ´ 10³ J/mol - (573 K)(-42.0 J/K×mol)

DG° = -1.71 ´ 10⁴ J/mol

Having solved for DG°, we can calculate K_P.

DG° = -RTln K_P

-1.71 ´ 10⁴ J/mol = -(8.314 J/K×mol)(573 K) ln K_P

ln K_P = 3.59

K_P = 36

Due to the negative entropy change calculated above, we expect that DG° will become positive at some temperature higher than 300°C. We need to find the temperature at which DG° becomes zero. This is the temperature at which reactants and products are equally favored (K_P = 1).

DG° = DH° - TDS°

0 = DH° - TDS°

T = 981 K = 708°C

This calculation shows that at 708°C, DG° = 0 and the equilibrium constant KP= 1. Above 708°C, DG° is positive and K_P will be smaller than 1, meaning that reactants will be favored over products. Note that the temperature 708°C is only an estimate, as we have assumed that both DH° and DS° are independent of temperature.

Using a more efficient catalyst will not increase K_P at a given temperature, because the catalyst will speed up both the forward and reverse reactions. The value of K_P will stay the same.

18.99

a.	DG° for CH3COOH: DG° = -(8.314 J/mol×K)(298 K) ln (1.8 ´ 10^-5) DG° = 2.7 ´ 10^4 J/mol = 27 kJ/mol DG° for CH2ClCOOH: DG° = -(8.314 J/mol×K)(298 K) ln (1.4 ´ 10^-3) DG° = 1.6 ´ 10^4 J/mol = 16 kJ/mol
b.	The TDS° term determines the value of DG°. The system’s entropy change dominates.
c.	The breaking and making of specific O-H bonds. Other contributions include solvent separation and ion solvation.
d.	The CH3COO^- ion, which is smaller than CH2ClCOO^-, can participate in hydration to a greater extent, leading to solutions with fewer possible arrangements.

18.100

butane ® isobutane

DG° = (1)(-18.0 kJ/mol) - (1)(-15.9 kJ/mol)

DG° = -2.1 kJ/mol

For a mixture at equilibrium at 25°C:

DG° = -RTln K_P

-2.1 ´ 10³ J/mol = -(8.314 J/mol×K)(298 K) ln K_P

K_P = 2.3

This shows that there are 2.3 times as many moles of isobutane as moles of butane. Or, we can say for every one mole of butane, there are 2.3 moles of isobutane.

By difference, the mole % of butane is 30%.

Yes, this result supports the notion that straight-chain hydrocarbons like butane are less stable than branched-chain hydrocarbons like isobutane.

18.101

We can calculate K_P from DG°.

DG° = (1)(-394.4 kJ/mol) + (0) - (1)(-137.3 kJ/mol) - (1)(-255.2 kJ/mol)

DG° = -1.9 kJ/mol

-1.9 ´ 10³ J/mol = -(8.314 J/mol×K)(1173 K) ln KP

K_P = 1.2

Now, from K_P, we can calculate the mole fractions of CO and CO2.

We assumed that DG° calculated from values was temperature independent. The values in Appendix 2 of the text are measured at 25°C, but the temperature of the reaction is 900°C.

18.102

DG° = -RTln K

and,

DG = DG° + RTln Q

Substituting,

DG = -RTln K + RTln Q

DG = RT(ln Q - ln K)

If Q > K, DG > 0, and the net reaction will proceed from right to left (see Section 15.4 of the text).

If Q < K, DG < 0, and the net reaction will proceed from left to right.

If Q = K, DG = 0. The system is at equilibrium.

18.103

For a phase transition, DG = 0. We write:

DG = DH - TDS

0 = DH - TDS

Substituting DH and the temperature, (-78° + 273°)K = 195 K, gives

This value of DS_sub is for the sublimation of 1 mole of CO₂. We convert to the DS value for the sublimation of 84.8 g of CO₂.

18.104

Equation 18.10 represents the standard free-energy change for a reaction, and not for a particular compound like CO₂. The correct form is:

DG° = DH° - TDS°

For a given reaction, DG° and DH° would need to be calculated from standard formation values (graphite, oxygen, and carbon dioxide) first, before plugging into the equation. Also, DS° would need to be calculated from standard entropy values.

C(graphite) + O₂(g) ® CO₂(g)

18.105

We can calculate DS_sys from standard entropy values in Appendix 2 of the text. We can calculate DS_surr from the DH_sys value given in the problem. Finally, we can calculate DS_univ from the DS_sys and DS_surr values.

DS_sys = (2)(69.9 J/K×mol) - [(2)(131.0 J/K×mol) + (1)(205.0 J/K×mol)] = -327 J/K×mol

DS_univ = DS_sys + DS_surr = (-327 + 1918) J/K×mol = 1591 J/K×mol

18.106

The equation representing the process described in the problem can be written as

glucose(12.3 mM) glucose(0.12 mM)

The temperature is (37°C + 273) = 310 K. DG° for this process is zero. We use Equation 18.13 to calculate the free energy change.

DG = DG° + RTlnQ = (8.314 ´ 10^-3 kJ/K×mol)(310 K)= -11.9 kJ/mol

The overall free energy change for the transport of three moles of glucose into the cell is 3 mol ´ -11.9 kJ/mol = -35.8 kJ.

18.107

q, and w are not state functions. Recall that state functions represent properties that are determined by the state of the system, regardless of how that condition is achieved. Heat and work are not state functions because they are not properties of the system. They manifest themselves only during a process (during a change). Thus their values depend on the path of the process and vary accordingly.

18.108

(d) will not lead to an increase in entropy of the system. The gas is returned to its original state. The entropy of the system does not change.

18.109

Since the adsorption is spontaneous, DG must be negative. When hydrogen bonds to the surface of the catalyst, there is a decrease in the number of possible arrangements of atoms in the system, DS must be negative. Since there is a decrease in entropy, the adsorption must be exothermic for the process to be spontaneous, DH must be negative.

18.110

a.	An ice cube melting in a glass of water at 20°C. The value of DG for this process is negative so it must be spontaneous.
b.	A "perpetual motion" machine. In one version, a model has a flywheel which, once started up, drives a generator which drives a motor which keeps the flywheel running at a constant speed and also lifts a weight.
c.	A perfect air conditioner; it extracts heat energy from the room and warms the outside air without using any energy to do so. (Note: this process does not violate the first law of thermodynamics.)
d.	Same example as (a).
e.	The conversion of NO2(g) to N2O4(g) in a closed flask at constant temperature containing NO2(g) and N2O4(g) at equilibrium. (Note that the conversion of N2O4(g) to NO2(g) is also an equilibrium process in this example.)

18.111

a.	Each CO molecule has two possible orientations in the crystal, CO or OC If there is no preferred orientation, then for one molecule there are two, or 2^1, choices of orientation. Two molecules have four or 2^2 choices, and for 1 mole of CO there are choices. From Equation 18.1 of the text: S = 5.76 J/K×mol
b.	The fact that the actual residual entropy is 4.2 J/K×mol means that the orientation is not totally random.

18.112

We can calculate DG° at 872 K from the equilibrium constant, K₁.

DG° = 6.25 × 10⁴ J/mol = 62.5 kJ/mol

We use the equation derived in Problem 18.66 to calculate DH°.

DH° = 157.8 kJ/mol

Now that both DG° and DH° are known, we can calculate DS° at 872 K.

DG° = DH° - TDS°

62.5 × 10³ J/mol = (157.8 × 10³ J/mol) - (872 K)DS°

DS° = 109 J/K×mol

18.113

We use data in Appendix 2 of the text to calculate DH° and DS°.

DH° = 82.93 kJ/mol - 49.04 kJ/mol = 33.89 kJ/mol

DS° = S°[C₆H₆(g)] - S°[C₆H₆(l)]

DS° = 269.2 J/K×mol - 172.8 J/K×mol = 96.4 J/K×mol

We can now calculate DG° at 298 K.

DG° = DH° - TDS°

DG° = 5.2 kJ/mol

DH° is positive because this is an endothermic process. We also expect DS° to be positive because this is a
liquid ® vapor phase change. DG° is positive because we are at a temperature that is below the boiling point of benzene (80.1°C).

18.114

The reaction mixtures are at equilibrium if K = Q. Use Equation 18.14 to calculate K.

DG° = -RTln K

−3400 J/mol = -(8.314 J/mol×K)(298 K) ln K

1.372 = ln K

Taking the anti-ln of both sides,

e^-1.372 = K

K = 3.9 ≈ 4

To find Q of each reaction mixture, take the concentration of products over concentration of reactants raised to the appropriate power.

For the reaction, A₂(g) + B₂(g) 2AB(g)

Q =

The partial pressure is equal to the number of molecules times 0.10 atm. Therefore, the partial pressure for 3 molecules of AB is 3 × 0.10 atm = 0.30 atm.

Q_i =

Q_ii =

Q_iii =

a.	Since K = Q for reaction (iii), then reaction (iii) is at equilibrium.
b.	Use Equation 18.13 to calculate DG. Reaction i has a negative DG value.
c.	Reactions ii and iii have positive DG values.

18.115

Strategy:	Use Equation 18.14 to solve for the equilibrium constant, K.
Setup:	R = 8.314 × 10^−3 kJ/K ∙ mol and T = (20 + 273) = 293 K. From problem 18.25, DG = 64 kJ/mol.
Solution:	Solving Equation 18.14 for K gives