Chapter 16

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

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

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.

There are four initial microstates and four final microstates.
$\text{Δ}S=k\text{ln}\frac{W_{\text{f}}}{W_{\text{i}}}=1.38\times {\text{10}}^{\mathrm{-23}}\text{J/K}\times \text{ln}\frac{4}{4}=0$

The probability for all the particles to be on one side is $\frac{1}{32}.$ This probability is noticeably lower than the $\frac{1}{8}$ 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.

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.
$\text{Δ}S=k\text{ln}\left(\frac{W_{\text{f}}}{W_{\text{i}}}\right)=1.38\times {10}^{-23}\text{J/K}\times \text{ln}\left(\frac{4}{1}\right)=1.91\times {10}^{-23}\text{J/K}$

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

(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).

(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.

${\text{C}}_6{\text{H}}_6(l)+7.5{\text{O}}_2(g)⟶{\text{3H}}_2\text{O(}g)+{\text{6CO}}_2(g)$
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.

(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

100.3 J/K

(a) −198.1 J/K; (b) −348.9 J/K

As ΔS_univ < 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.

(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

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

(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

(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.

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

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

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

In each of the following, the value of ΔG is not given at the temperature of the reaction. Therefore, we must calculate ΔG from the values ΔH° and ΔS and then calculate ΔG from the relation ΔG = ΔH° − TΔS°. (a) K = 1.07 × 10⁻¹³; (b) K = 2.51 $\times$ 10⁻³; (c) K = 2.73 $\times$ 10⁴; (d) K = 0.219; (e) K = 16.1

The standard free energy change is $\text{Δ}{G}^{°}=\text{−}RT\text{ln}K=\text{4.84 kJ/mol}.$ When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and $\text{Δ}G$ becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.

The reaction will be spontaneous at temperatures greater than 287 K.

K = 5.35 $\times$ 10¹⁵; The process is exothermic.

1.0 $\times$ 10⁻⁸ atm. This is the maximum pressure of the gases under the stated conditions.

$x=1.29\times {10}^{-5}\text{atm}=P_{{\text{O}}_2}$

−0.16 kJ

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

90 kJ/mol

(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) K_p = 0.031; (c) The evaporation of water is spontaneous; (d) $P_{{\text{H}}_2\text{O}}$ must always be less than K_p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.

(a) Nonspontaneous as $\text{Δ}{G}^{°}>0;$ (b) $\text{Δ}G=\text{Δ}{G}^{\text{°}}+RT\text{ln}Q,$ $\text{Δ}G=1.7\times {10}^3+\left(8.314\times 310\times \text{ln}\frac{28}{120}\right)=\text{−2.1 kJ}.$ The forward reaction to produce F6P is spontaneous under these conditions.

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.

(a) Increasing the oxygen partial pressure will yield a decrease in Q and $\text{Δ}G$ thus becomes more negative. (b) Increasing the oxygen partial pressure will yield a decrease in Q and $\text{Δ}G$ thus becomes more negative. (c) Increasing the oxygen partial pressure will yield an increase in Q and $\text{Δ}G$ thus becomes more positive.