0 1 NaN Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Application Problems Problems marked with a ♦ involve multiple concepts. Electric utilities have been exploring thermal energy storage as a potentially attractive energy-storage solution for peak use. Thermal energy is extracted as steam and stored in rock, oil, or water for later conversion to electricity via heat exchangers. Which steps involve heat transfer? Which involve work done? ♦ During World War II, German scientists developed the first rocket-powered airplane to be flown in combat, the Messerschmitt 163 Komet . The Komet was powered by the reaction of liquid hydrogen peroxide (H 2 O 2 ) and hydrazine (N 2 H 4 ) as follows: 2H 2 O 2 (l) + N 2 H 4 (l) → N 2 (g) + 4H 2 O(g) Determine the standard molar enthalpy of reaction. What amount of work can be done by the reaction of 1.00 kg of hydrazine with a stoichiometric amount of hydrogen peroxide at 25°C at a constant pressure of 1.00 atm? What is the change in internal energy under these conditions? ♦ During the 1950s, pentaborane-9 was tested as a potential rocket fuel. However, the idea was abandoned when it was discovered that B 2 O 3 , the product of the reaction of pentaborane-9 with O 2 , was an abrasive that destroyed rocket nozzles. The reaction is represented by the equation 2B 5 H 9 (l) + 12O 2 (g) → 5B 2 O 3 (s) + 9H 2 O(g) Determine the standard molar enthalpy of reaction. How much work is done against a pressure of 1.0 atm when 50 lb of fuel are allowed to react with sufficient oxygen to cause the reaction to go to completion at 25°C? What is Δ E for the reaction? ♦ Polar explorers must be particularly careful to keep their clothes from becoming damp because the resulting heat loss could be fatal. If a polar explorer’s clothes absorbed 1.0 kg of water and the clothes dried from the polar wind, what would be the heat loss [Δ H vap (H 2 O) = 44 kJ/mol]? How much glucose must be consumed to make up for this heat loss to prevent death [Δ H comb (glucose) = −802 kJ/mol]? Propane gas is generally preferred to kerosene as a fuel for stoves in the boating industry because kerosene stoves require more maintenance. Propane, however, is much more flammable than kerosene and imposes an added risk because it is denser than air and can collect in the bottom of a boat and ignite. The complete combustion of propane produces CO 2 gas and H 2 O vapor and has a value of Δ H = −2044 kJ/mol at 25°C. What is Δ E ? The propane in Problem 5 can be produced from the hydrogenation of propene according to the following reaction: C 3 H 6 (g) + H 2 (g) → C 3 H 8 (g); Δ H = −124 kJ/mol. Given that the reaction H 2 (g) + 1/2 O 2 (g) → H 2 O(g) has Δ H = −241.8 kJ/mol, what is the standard enthalpy of combustion of propene? value of Δ E for the combustion of propene? ♦ The anaerobic conversion of sucrose, a sweetening agent, to lactic acid, which is associated with sour milk, can be described as follows: \( \underset{sucrose}{C_{12}H_{22}O_{11}} + H_2O \rightarrow \underset{lactic acid}{4CH_{3}CH\left ( OH \right )CO_{2}H} \) The combustion of sucrose, however, occurs as follows: C 12 H 22 O 11 (s) + 12O 2 (g) → 12CO 2 (g) + 11H 2 O(l) Which reaction is thermodynamically more favorable—the anaerobic conversion of sucrose to lactic acid or the aerobic oxidation to CO 2 and H 2 O? The values of Δ H f are as follows: lactic acid, −694.1 kJ/mol; sucrose, −222 kJ/mol; CO 2 , −393.5 kJ/mol; and H 2 O, −285.8 kJ/mol. What is Δ E 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 P 4 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 Δ S ? 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 (Al 2 O 3 ). 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: C p (white) = 26.99, C p (gray) = 25.77 J/(mol·K), Δ H fus = 7.0 kJ/mol, Δ H β → α = −2.2 kJ/mol. The reaction of SO 2 with O 2 to produce SO 3 has great industrial significance because SO 3 is converted to H 2 SO 4 by reaction with water. Unfortunately, the reaction is also environmentally important because SO 3 from industrial smokestacks is a primary source of acid rain. Δ H for the reaction of SO 2 with O 2 to form SO 3 is −23.49 kJ/mol, and Δ S 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 Δ H and Δ S ? 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 (CH 3 NH 2 ) and ethylenediamine (H 2 NCH 2 CH 2 NH 2 , 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: 2N 2 H 4 (l) + N 2 O 4 (l) → 4H 2 O(g) + 3N 2 (g) Δ H ° = −249 kJ/mol, Δ S ° = 218 J/(mol·K) How much free energy is produced from this reaction at 25°C? Is the reaction thermodynamically favorable? What is K ? 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 CaC 2 : 2CsCl(l) + CaC 2 (s) → CaCl 2 (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: ΔH 298 K ° = 32.0 kJ/mol, ΔS 298 K ° = 8.0 J/(mol·K); ΔH 1500 K ° = −0.6 kJ/mol, ΔS 1500 K ° = 3.6 J/(mol·K). 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 Δ H ° and Δ S ° vary linearly with temperature. What is the ratio K 1500 K / K 298 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 CaSO 4 ·2H 2 O: \( CaSO_{4}\cdot 2H_{2}O \left ( s \right ) \rightarrow CaSO_{4} \left ( s \right ) + 2H_{2}O\left ( g \right ) \) \( \Delta H_{298}^{o} = -7.02\; kcal/mol \;\; \Delta S_{298}^{o}=-3.15 \; cal/\left ( mol\cdot K \right )\) Regeneration is carried out at 250°C. What is Δ G ° for this reaction? What is the equilibrium constant at 25°C? What is the ratio K 250°C / K 25°C ? What is the equilibrium constant at 250°C? Is regeneration of CaSO 4 (s)·2H 2 O an enthalpy- or entropy-driven process? Explain your answer. The nitrogen triiodide complex with ammonia (NI 3 ·NH 3 ) is a simple explosive that can be synthesized from common household products. When detonated, it produces N 2 and I 2 . 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 Δ G for the explosion reaction to be positive or negative? Why doesn’t NI 3 ·NH 3 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 + H 2 O ⇌ ADP + Pi + H + where P i symbolizes phosphate. Glycerol and ATP react to form glycerol-3-phosphate, ADP, and H + , with an overall K = 6.61 × 10 5 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 (P i = phosphate): 8H + + 8e − + N 2 + 16ATP → H 2 + 2NH 3 + 16ADP + 16P i What is the approximate ratio of K eq for this reaction and for the same reaction in the absence of ATP? Given the fact that at pH 7 both the reaction of protons and electrons to give H 2 and the reaction of H 2 with N 2 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. Answers −4315 kJ/mol B 5 H 9 1340 kJ −1.55 × 10 6 kJ per 50 lb of B 5 H 9 −2046 kJ/mol aerobic conversion −268 kJ Yes, the reaction is spontaneous at 25°C, but its rate is very slow. The reaction is not spontaneous at 800°C (Δ G = 0.82 kJ/mol), but the reaction rate is much greater. −314 kJ/mol yes 2.10 × 10 45 Ignition is required to overcome the high activation energy to reaction. 4.4 kJ/mol 0.17 78.3 13 entropy-driven; Δ H ° > 0, so Δ S ° must be positive for the reaction to be spontaneous. 2.34 × 10 7 ; −43.7 kJ/mol Contributors Anonymous Modified by Joshua B. Halpern