74 Free Energy

What is the difference between ΔG and ΔG° for a chemical change?

A reaction has \(\text{Δ}H\text{°}\) = 100 kJ/mol and \(\text{Δ}S\text{°}=\text{250 J/mol·K.}\) Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?

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

Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.

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) \({\text{MnO}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Mn}\left(s\right)+{\text{O}}_{2}\left(g\right)\)

(b) \({\text{H}}_{2}\left(g\right)+{\text{Br}}_{2}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2HBr}\left(g\right)\)

(c) \(\text{Cu}\left(s\right)+\text{S}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CuS}\left(s\right)\)

(d) \(\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\)

(e) \({\text{CH}}_{4}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+{\text{2H}}_{2}\text{O}\left(g\right)\)

(f) \({\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)\)

(a) 465.1 kJ nonspontaneous; (b) −106.86 kJ spontaneous; (c) −53.6 kJ spontaneous; (d) −83.4 kJ spontaneous; (e) −406.7 kJ spontaneous; (f) −30.0 kJ spontaneous

Use the standard free energy 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) \(\text{C}\left(s\text{, graphite}\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\)

(b) \({\text{O}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2NO}\left(g\right)\)

(c) \(\text{2Cu}\left(s\right)+\text{S}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cu}}_{2}\text{S}\left(s\right)\)

(d) \(\text{CaO}\left(s\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ca}{\left(\text{OH}\right)}_{2}\left(s\right)\)

(e) \({\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)+\text{3CO}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Fe}\left(s\right)+{\text{3CO}}_{2}\left(g\right)\)

(f) \({\text{CaSO}}_{4}\text{·}{\text{2H}}_{2}\text{O}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CaSO}}_{4}\left(s\right)+{\text{2H}}_{2}\text{O}\left(g\right)\)

Given:
\(\begin{array}{cccc}{\text{P}}_{4}\left(s\right)+{\text{5O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{P}}_{4}{\text{O}}_{10}\left(s\right)\hfill & & & \text{Δ}G\text{°}=\text{−2697.0 kJ/mol}\hfill \\ {\text{2H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}_{2}\text{O}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−457.18 kJ/mol}\hfill \\ {\text{6H}}_{2}\text{O}\left(g\right)+{\text{P}}_{4}{\text{O}}_{10}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{4H}}_{3}{\text{PO}}_{4}\left(l\right)\hfill & & & \text{Δ}G\text{°}=\text{−428.66 kJ/mol}\hfill \end{array}\)

(a) Determine the standard free energy of formation, \(\text{Δ}{G}_{\text{f}}^{°},\) for phosphoric acid.

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

(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.

Is the formation of ozone (O₃(g)) from oxygen (O₂(g)) spontaneous at room temperature under standard state conditions?

Consider the decomposition of red mercury(II) oxide under standard state conditions.
\(\text{2HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{red}\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)\)

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

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

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

(a) Ammonia: \({\text{2NH}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)\)

(b) Diborane: \({\text{B}}_{2}{\text{H}}_{6}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2B}\left(g\right)+{\text{3H}}_{2}\left(g\right)\)

(c) Hydrazine: \({\text{N}}_{2}{\text{H}}_{4}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\left(g\right)+{\text{2H}}_{2}\left(g\right)\)

(d) Hydrogen peroxide: \({\text{H}}_{2}{\text{O}}_{2}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\)

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

\(\begin{array}{ccccccc}\text{(a)}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2NO}\left(g\right)\hfill & & & \text{T}=2000\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\hfill \\ \text{(b)}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2HI}\left(g\right)\hfill & & & \text{T}=400\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=50.0\hfill \\ \text{(c)}\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \text{T}=980\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=1.67\hfill \\ \text{(d)}\phantom{\rule{0.2em}{0ex}}{\text{CaCO}}_{3}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)\hfill & & & \text{T}=900\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=1.04\hfill \\ \text{(e)}\phantom{\rule{0.2em}{0ex}}\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=7.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\hfill \\ \text{(f)}\phantom{\rule{0.2em}{0ex}}\text{AgBr}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=3.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−13}}\hfill \end{array}\)

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

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

\(\begin{array}{ccccccc}\text{(a)}\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2BrCl}\left(g\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−2}}\hfill \\ \text{(b)}\phantom{\rule{0.2em}{0ex}}{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{2SO}}_{3}\left(g\right)\hfill & & & \text{T}=500\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=48.2\hfill \\ \text{(c)}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \text{T}=60\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=\text{0.196 atm}\hfill \\ \text{(d)}\phantom{\rule{0.2em}{0ex}}\text{CoO}\left(s\right)+\text{CO}\left(g\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}\text{Co}\left(s\right)+{\text{CO}}_{2}\left(g\right)\hfill & & & \text{T}=550\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.90\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\hfill \\ \text{(e)}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\hfill \\ \text{(f)}\phantom{\rule{0.2em}{0ex}}{\text{PbI}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Pb}}^{2+}\left(aq\right)+{\text{2I}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=8.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−9}}\hfill \end{array}\)

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

\(\begin{array}{cccc}\text{(a)}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}\left(g\right)+{\text{2F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2OF}}_{2}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−9.2 kJ}\hfill \\ \text{(b)}\phantom{\rule{0.2em}{0ex}}{\text{I}}_{2}\left(s\right)+{\text{Br}}_{2}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2IBr}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{7.3 kJ}\hfill \\ \text{(c)}\phantom{\rule{0.2em}{0ex}}\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−79 kJ}\hfill \\ \text{(d)}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−1.6 kJ}\hfill \\ \text{(e)}\phantom{\rule{0.2em}{0ex}}{\text{SnCl}}_{4}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SnCl}}_{4}\left(l\right)\hfill & & & \text{Δ}G\text{°}=\text{8.0 kJ}\hfill \end{array}\)

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

Determine ΔGº for the following reactions.

(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
\({\text{SbCl}}_{5}\left(g\right)⟶{\text{SbCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\)

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:
\({\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{Cl}\left(g\right)\)

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

(a) 22.1 kJ;
(b) 61.6 kJ/mol

Given that the \(\text{Δ}{G}_{\text{f}}^{°}\) for Pb²⁺(aq) and Cl⁻(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, K_sp, for PbCl₂(s).

Determine the standard free energy change, \(\text{Δ}{G}_{\text{f}}^{°},\) for the formation of S²⁻(aq) given that the \(\text{Δ}{G}_{\text{f}}^{°}\) for Ag⁺(aq) and Ag₂S(s) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag₂S(s) is 8 \(×\) 10⁻⁵¹.

90 kJ/mol

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.

The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
\({\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{8.58 kJ}\)

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

(b) Determine the equilibrium constant, K_P, 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, \({P}_{{\text{H}}_{2}\text{O}},\) 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 \({P}_{{\text{H}}_{2}\text{O}}\) in the air?

(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.

In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
\(\text{Glu}+\text{ATP}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{G6P}+\text{ADP}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{−17 kJ}\)

In this process, ATP becomes ADP summarized by the following equation:
\(\text{ATP}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{ADP}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{−30 kJ}\)

Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
\(\text{Glu}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{G6P}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=?\)

One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
\(\text{G6P}\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}\text{F6P}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{1.7 kJ}\)

(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.

(a) Nonspontaneous as \(\text{Δ}G\text{°}>0;\) (b) \(\text{Δ}G\text{°}=\text{Δ}{G}^{°}\text{+}RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}Q,\) \(\text{Δ}G=1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}+\phantom{\rule{0.2em}{0ex}}\left(8.314\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}335\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.4em}{0ex}}\frac{28}{128}\right)\phantom{\rule{0.2em}{0ex}}=\text{−2.5 kJ}.\) The forward reaction to produce F6P is spontaneous under these conditions.

Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.

(a) \({\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2NH}}_{3}\left(g\right)\)

(b) \(\text{HCl}\left(g\right)+{\text{NH}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}\text{Cl}\left(s\right)\)

(c) \({\left({\text{NH}}_{4}\right)}_{2}{\text{Cr}}_{2}{\text{O}}_{7}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cr}}_{2}{\text{O}}_{3}\left(s\right)+{\text{4H}}_{2}\text{O}\left(g\right)+{\text{N}}_{2}\left(g\right)\)

(d) \(\text{2Fe}\left(s\right)+{\text{3O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)\)

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.

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.

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:
\({\text{Cu}}_{2}\text{S}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cu}\left(s\right)+\text{S}\left(s\right)\)

(a) Determine \(\text{Δ}G\text{°}\) for the decomposition of Cu₂S(s).

(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine \(\text{Δ}G\text{°}\) for the process.

(c) The production of copper from chalcocite is performed by roasting the Cu₂S 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.

What happens to \(\text{Δ}G\text{°}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

(a) \(\text{S}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{2}\left(g\right)\)

(b) \({\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{3}\left(g\right)\)

(c) \(\text{HgO}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)\)

(a) Increasing \({P}_{{\text{O}}_{\text{2}}}\) will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(b) Increasing \({P}_{{\text{O}}_{\text{2}}}\) will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(c) Increasing \({P}_{{\text{O}}_{\text{2}}}\) will shift the equilibrium the reactants, which decreases the value of K. ΔG° therefore becomes more positive.