# Chapter 16.6: End of Chapter Material

+ H_2O \rightarrow \underset{lactic acid}{4CH_{3}CH\left ( OH \right )CO_{2}H} \)

The combustion of sucrose, however, occurs as follows:

C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
1. Which reaction is thermodynamically more favorable—the anaerobic conversion of sucrose to lactic acid or the aerobic oxidation to CO2 and H2O? The values of ΔHf are as follows: lactic acid, −694.1 kJ/mol; sucrose, −222 kJ/mol; CO2, −393.5 kJ/mol; and H2O, −285.8 kJ/mol.
2. What is ΔE for the combustion of 12.0 g of sucrose at normal body temperature (37°C)?
• Phosphorus exists as several allotropes, the most common being red, black, and white phosphorus. White phosphorus consists of tetrahedral P4 molecules and melts at 44.15°C; it is converted to red phosphorus by heating at 400°C for several hours. The chemical differences between red and white phosphorus are considerable: white phosphorus burns in air, whereas red phosphorus is stable; white phosphorus is soluble in organic compounds, whereas red phosphorus is not; white phosphorus melts at 44.15°C, whereas red phosphorus melts at 597°C. If the enthalpy of fusion of white phosphorus is 0.659 kJ/mol, what is its ΔS? Black phosphorus is even less reactive than red. Based on this information, which allotrope would you predict to have the highest entropy? the lowest? Why?

• ♦ Ruby and sapphire have a common mineral name: corundum (Al2O3). Although they are crystalline versions of the same compound, the nature of the imperfections determines the identity of the gem. Outline a method for measuring and comparing the entropy of a ruby with the entropy of a sapphire. How would you expect the entropies to compare with the entropy of a perfect corundum crystal?

• ♦ Tin has two crystalline forms—α and β—represented in the following equilibrium equation:

$$\underset{gray}{\alpha-tin} + H_2O \overset{18^{o}C}{\rightleftharpoons} \underset{white}{\beta-tin} \overset{232^{o}C}{\rightleftharpoons} Sn\left ( l \right )$$

The earliest known tin artifacts were discovered in Egyptian tombs of the 18th dynasty (1580–1350 BC), although archaeologists are surprised that so few tin objects exist from earlier eras. It has been suggested that many early tin objects were either oxidized to a mixture of stannous and stannic oxides or transformed to powdery, gray tin. Sketch a thermodynamic cycle similar to part (b) in Figure 18.4.3 to show the conversion of liquid tin to gray tin. Then calculate the change in entropy that accompanies the conversion of Sn(l) to α-Sn using the following data: Cp(white) = 26.99, Cp(gray) = 25.77 J/(mol·K), ΔHfus = 7.0 kJ/mol, ΔHβ → α = −2.2 kJ/mol.

• The reaction of SO2 with O2 to produce SO3 has great industrial significance because SO3 is converted to H2SO4 by reaction with water. Unfortunately, the reaction is also environmentally important because SO3 from industrial smokestacks is a primary source of acid rain. ΔH for the reaction of SO2 with O2 to form SO3 is −23.49 kJ/mol, and ΔS is −22.66 J/(mol·K). Does this reaction occur spontaneously at 25°C? Does it occur spontaneously at 800°C assuming no change in ΔH and ΔS? Why is this reaction usually carried out at elevated temperatures?

• Pollutants from industrial societies pose health risks to individuals from exposure to metals such as lead, mercury, and cadmium. The biological effects of a toxic metal may be reduced by removing it from the system using a chelating agent, which binds to the metal and forms a complex that is eliminated from the system without causing more damage. In designing a suitable chelating agent, one must be careful, however, because some chelating agents form metal complexes that are more toxic than the metal itself. Both methylamine (CH3NH2) and ethylenediamine (H2NCH2CH2NH2, abbreviated en) could, in principle, be used to treat heavy metal poisoning. In the case of cadmium, the reactions are as follows:

$$Cd^{2+} + 2CH_{3}NH_{2} \rightarrow Cd\left (CH_{3}NH_{2} \right )_{2}^{2+}$$

$$\Delta H = -7.03\; kcal/mol \;\;\; \Delta S=-1.58 \; cal/\left ( mol\cdot K \right )$$

$$Cd^{2+} + en \rightarrow Cd\left (en \right )_{2}^{2+}$$

$$\Delta H = -7.02\; kcal/mol \;\;\; \Delta S=-3.15 \; cal/\left ( mol\cdot K \right )$$

Based strictly on thermodynamic arguments, which would you choose to administer to a patient suffering from cadmium toxicity? Why? Assume a body temperature of 37°C.

• ♦ Explosive reactions often have a large negative enthalpy change and a large positive entropy change, but the reaction must also be kinetically favorable. For example, the following equation represents the reaction between hydrazine, a rocket propellant, and the oxidizer dinitrogen tetroxide:

2N2H4(l) + N2O4(l) → 4H2O(g) + 3N2(g) ΔH° = −249 kJ/mol, ΔS° = 218 J/(mol·K)
1. How much free energy is produced from this reaction at 25°C?
2. Is the reaction thermodynamically favorable?
3. What is K?
4. This reaction requires thermal ignition. Why?
• ♦ Cesium, a silvery-white metal used in the manufacture of vacuum tubes, is produced industrially by the reaction of CsCl with CaC2:

2CsCl(l) + CaC2(s) → CaCl2(l) + 2C(s) + 2Cs(g)

Compare the free energy produced from this reaction at 25°C and at 1227°C, the temperature at which it is normally run, given these values:

ΔH298 K° = 32.0 kJ/mol, ΔS298 K°= 8.0 J/(mol·K); ΔH1500 K° = −0.6 kJ/mol, ΔS1500 K° = 3.6 J/(mol·K).

1. If you wanted to minimize energy costs in your production facility by reducing the temperature of the reaction, what is the lowest temperature at which products are favored over reactants (assuming the reaction is kinetically favorable at the lower temperature)? Assume ΔH° and ΔS° vary linearly with temperature.
2. What is the ratio K1500 K/K298 K?
• Dessicants (drying agents) can often be regenerated by heating, although it is generally not economically worthwhile to do so. A dessicant that is commonly regenerated is CaSO4·2H2O:

$$CaSO_{4}\cdot 2H_{2}O \left ( s \right ) \rightarrow CaSO_{4} \left ( s \right ) + 2H_{2}O\left ( g \right )$$

$$\Delta H_{298}^{o} = -7.02\; kcal/mol \;\; \Delta S_{298}^{o}=-3.15 \; cal/\left ( mol\cdot K \right )$$

Regeneration is carried out at 250°C.

1. What is ΔG° for this reaction?
2. What is the equilibrium constant at 25°C?
3. What is the ratio K250°C/K25°C?
4. What is the equilibrium constant at 250°C?
5. Is regeneration of CaSO4(s)·2H2O an enthalpy- or entropy-driven process? Explain your answer.
• The nitrogen triiodide complex with ammonia (NI3·NH3) is a simple explosive that can be synthesized from common household products. When detonated, it produces N2 and I2. It can be painted on surfaces when wet, but it is shock sensitive when dry (even touching it with a feather can cause an explosion). Do you expect ΔG for the explosion reaction to be positive or negative? Why doesn’t NI3·NH3 explode spontaneously?

• Adenosine triphosphate (ATP) contains high-energy phosphate bonds that are used in energy metabolism, coupling energy-yielding and energy-requiring processes. Cleaving a phosphate link by hydrolysis (ATP hydrolysis) can be described by the reaction ATP + H2O ⇌ ADP + Pi + H+ where Pi symbolizes phosphate. Glycerol and ATP react to form glycerol-3-phosphate, ADP, and H+, with an overall K = 6.61 × 105 at 37°C. The reaction of glycerol with phosphate to form glycerol-3-phosphate and water has an equilibrium constant of 2.82 × 10−2. What is the equilibrium constant for ATP hydrolysis? How much free energy is released from the hydrolysis of ATP?

• ♦ Consider the biological reduction of molecular nitrogen, for which the following is the minimal reaction stoichiometry under optimal conditions (Pi = phosphate):

8H+ + 8e + N2 + 16ATP → H2 + 2NH3 + 16ADP + 16Pi
1. What is the approximate ratio of Keq for this reaction and for the same reaction in the absence of ATP?
2. Given the fact that at pH 7 both the reaction of protons and electrons to give H2 and the reaction of H2 with N2 to give ammonia are thermodynamically spontaneous (i.e., K >> 1), suggest a reason that nitrogen-fixing bacteria use such a large energy input to drive a reaction that is already spontaneous.

1. −4315 kJ/mol B5H9
2. 1340 kJ
3. −1.55 × 106 kJ per 50 lb of B5H9
1. −2046 kJ/mol

1. aerobic conversion
2. −268 kJ
2. Yes, the reaction is spontaneous at 25°C, but its rate is very slow. The reaction is not spontaneous at 800°C (ΔG = 0.82 kJ/mol), but the reaction rate is much greater.

1. −314 kJ/mol
2. yes
3. 2.10 × 1045
4. Ignition is required to overcome the high activation energy to reaction.
1. 4.4 kJ/mol
2. 0.17
3. 78.3
4. 13
5. entropy-driven; ΔH° > 0, so ΔS° must be positive for the reaction to be spontaneous.
3. 2.34 × 107; −43.7 kJ/mol

## Contributors

• Anonymous

Modified by Joshua Halpern (Howard University), Scott Sinex, and Scott Johnson (PGCC)