CHAPTER 6 THERMOCHEMISTRY: ENERGY FLOW AND CHEMICAL CHANGE

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1 CHAPTER 6 THERMOCHEMISTRY: ENERGY FLOW AND CHEMICAL CHANGE CHEMICAL CONNECTIONS BOXED READING PROBLEMS B6.1 Plan: Convert the given mass in kg to g, divide by the molar mass to obtain moles, and convert moles to kj of energy. Sodium sulfate decahydrate will transfer 54 kj/mol. 10 g 1 mol Na2SO 4 10H2O 54 kj Heat (kj) = kg Na2SO 4 10H 2O 1 kg g Na 2 SO 4 10H 2 O 1 mol Na 2 SO 4 10H 2 O = 5.49x10 5 = 5.49x10 5 kj B6.2 Plan: Three reactions are given. Equation 1) must be multiplied by 2, and then the reactions can be added, canceling substances that appear on both sides of the arrow. Add the H values for the three reactions to get the Hrxn for the overall gasification reaction of 2 moles of coal. Use the relationship Hrxn = m Hf (products) n Hf (reactants) to find the heat of combustion of 1 mole of methane. Then find the Hrxn for the gasification of 1.00 kg of coal and Hrxn for the combustion of the methane produced from 1.00 kg of coal and sum these values. a) 1) 2C(coal) + 2H 2 O(g) 2CO(g) + 2H 2 (g) Hrxn = 2(129.7 kj) 2) CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) Hrxn = 41 kj ) CO(g) + H 2 (g) CH 4 (g) + H 2 O(g) Hrxn = 206 kj 2C(coal) + 2H 2 O(g) CH 4 (g) + CO 2 (g) b) The total may be determined by doubling the value for equation 1) and adding to the other two values. H rxn = 2(129.7 kj) + ( 41 kj) + ( 206 kj) = 12.4 = 12 kj c) Calculating the heat of combustion of CH 4 : CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H rxn = m Hf (products) n Hf (reactants) H rxn = [(1 mol CO 2 )( Hf of CO 2 ) + (2 mol H 2 O)( Hf of H 2 O)] [(1 mol CH 4 )( Hf of CH 4 ) + (2 mol O 2 )( Hf of O 2 )] H rxn = [(1 mol)( 95.5 kj/mol) + (2 mol)( kj/mol)] [(1 mol)( 74.87kJ/mol) + (2 mol )(0.0 kj/mol)] H rxn = kj/mol CH 4 Total heat for gasification of 1.00 kg coal: 10 g 1 mol coal 12.4 kj H = 1.00 kg coal = kj 1 kg g coal 2 mol coal Total heat from burning the methane formed from 1.00 kg of coal: 10 g 1 mol coal 1molCH kj H = 1.00 kg coal = kj 1 kg g coal 2 mol coal 1 mol CH4 Total heat = kj + ( kj) = =.0x10 4 kj rxn 6-1

2 END OF CHAPTER PROBLEMS 6.1 The sign of the energy transfer is defined from the perspective of the system. Entering the system is positive, and leaving the system is negative. 6.2 No, an increase in temperature means that heat has been transferred to the surroundings, which makes q negative. 6. E = q + w = w, since q = 0. Thus, the change in work equals the change in internal energy. 6.4 Plan: Remember that an increase in internal energy is a result of the system (body) gaining heat or having work done on it and a decrease in internal energy is a result of the system (body) losing heat or doing work. The internal energy of the body is the sum of the cellular and molecular activities occurring from skin level inward. The body s internal energy can be increased by adding food, which adds energy to the body through the breaking of bonds in the food. The body s internal energy can also be increased through addition of work and heat, like the rubbing of another person s warm hands on the body s cold hands. The body can lose energy if it performs work, like pushing a lawnmower, and can lose energy by losing heat to a cold room. 6.5 a) electric heater b) sound amplifier c) light bulb d) automobile alternator e) battery (voltaic cell) 6.6 Plan: Use the law of conservation of energy. The amount of the change in internal energy in the two cases is the same. By the law of energy conservation, the change in energy of the universe is zero. This requires that the change in energy of the system (heater or air conditioner) equals an opposite change in energy of the surroundings (room air). Since both systems consume the same amount of electrical energy, the change in energy of the heater equals that of the air conditioner. 6.7 Heat energy; sound energy (impact) Kinetic energy (falling text) Potential energy (raised text) Mechanical energy (raising of text) Chemical energy (biological process to move muscles) 6.8 Plan: The change in a system s energy is E = q + w. If the system receives heat, then its q final is greater than q initial so q is positive. Since the system performs work, its w final < w initial so w is negative. E = q + w E = (+425 J) + ( 425 J) = 0 J 6.9 q + w = 255 cal + ( 428 cal) = 68 cal 6.10 Plan: The change in a system s energy is E = q + w. A system that releases thermal energy has a negative value for q and a system that has work done on it has a positive value for work. Convert work in calories to work in joules J Work (J) = 50 cal 1 cal = J E = q + w = 675 J J = = 1.54x10 J 6-2

3 6.11 E = q + w = kj 10 J 1kJ kcal 10 cal J 1 kcal 1 cal = = 1.65x10 J 6.12 Plan: Convert 6.6x10 10 J to the other units using conversion factors. C(s) + O 2 (g) CO 2 (g) + 6.6x10 10 J (2.0 tons) a) E (kj) = (6.6 x kJ J) = 6.6x10 10 J 7 kj b) E (kcal) = (6.6 x cal 1 kcal J) J = 1.577x10 7 = 1.6x10 7 kcal 10 cal c) E (Btu) = (6.6 x Btu J) = 6.256x10 7 = 6.x10 7 Btu 1055 J 6.1 CaCO (s) + 9.0x10 6 kj CaO(s) + CO 2 (g) (5.0 tons) a) E (J) = (9.0x J kj) = 9.0x10 1kJ 9 J b) E (cal) = (9.0x J 1 cal kj) = x10 1 kj 9 = 2.2x10 9 cal J c) E (Btu) = (9.0x J 1 Btu kj) = 8.508x10 1kJ 6 = 8.5x10 6 Btu 1055J 6.14 E (J) = (4.1x10 10 cal J Calorie) = x10 1 Calorie 7 = 1.7x10 7 J 1 cal E (kj) = (4.1x10 10 cal J 1 kj Calorie) 1 Calorie 1 cal = x10 4 = 1.7x10 4 kj 10 J 6.15 Plan: 1.0 lb of body fat is equivalent to about 4.1x10 Calories. Convert Calories to kj with the appropriate conversion factors. 4.1x10 Cal 10 cal J 1 kj h Time = 1.0lb = = 8.8 h 1.0 lb 1 Cal 1 cal 10 J 1950 kj 6.16 The system does work and thus its internal energy is decreased. This means the sign will be negative Since many reactions are performed in an open flask, the reaction proceeds at constant pressure. The determination of H (constant pressure conditions) requires a measurement of heat only, whereas E requires measurement of heat and PV work The hot pack is releasing (producing) heat, thus H is negative, and the process is exothermic. 6-

4 6.19 Plan: An exothermic process releases heat and an endothermic process absorbs heat. a) Exothermic, the system (water) is releasing heat in changing from liquid to solid. b) Endothermic, the system (water) is absorbing heat in changing from liquid to gas. c) Exothermic, the process of digestion breaks down food and releases energy. d) Exothermic, heat is released as a person runs and muscles perform work. e) Endothermic, heat is absorbed as food calories are converted to body tissue. f) Endothermic, the wood being chopped absorbs heat (and work). g) Exothermic, the furnace releases heat from fuel combustion. Alternatively, if the system is defined as the air in the house, the change is endothermic since the air s temperature is increasing by the input of heat energy from the furnace The internal energy of a substance is the sum of kinetic (E K ) and potential (E P ) terms. E K (total) = E K (translational) + E K (rotational) + E K (vibrational) E P = E P (atom) + E P (bonds) E P (atom) has nuclear, electronic, positional, magnetic, electrical, etc., components H = E + PV (constant P) a) H < E, PV is negative. b) H = E, a fixed volume means PV = 0. c) H > E, PV is positive for the transformation of solid to gas Plan: An exothermic reaction releases heat, so the reactants have greater H (H initial ) than the products (H final ). H = H final H initial < 0. Reactants Increasing, H Products H = (), (exothermic) 6.2 Increasing, H Products Reactants H = (+), (endothermic) 6.24 Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield carbon dioxide gas, water vapor, and heat. Combustion reactions are exothermic. The freezing of liquid water is an exothermic process as heat is removed from the water in the conversion from liquid to solid. An exothermic reaction or process releases heat, so the reactants have greater H (H initial ) than the products (H final ). 6-4

5 a) Combustion of ethane: 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) + heat 2C 2 H 6 + 7O 2 (initial) Increasing, H 4CO 2 + 6H 2 O (final) H = (), (exothermic) b) Freezing of water: H 2 O(l) H 2 O(s) + heat H 2 O(l) (initial) H = (), (exothermic) Increasing, H H 2 O(s) (final) 6.25 a) Na(s) + 1/2Cl 2 (g) NaCl(s) + heat Na(s) + 1/2Cl 2 (g) Increasing, H NaCl(s) exothermic b) C 6 H 6 (l) + heat C 6 H 6 (g) C 6 H 6 (g) Increasing, H C 6 H 6 (l) H = (+), (endothermic) 6.26 Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield carbon dioxide gas, water vapor, and heat. Combustion reactions are exothermic. An exothermic reaction releases heat, so the reactants have greater H (H initial ) than the products (H final ). If heat is absorbed, the reaction is endothermic and the products have greater H (H final ) than the reactants (H initial ). a) 2CH OH(l) + O 2 (g) 2CO 2 (g) + 4H 2 O(g) + heat 2CH OH + O 2 (initial) Increasing, H 2CO 2 + 4H 2 O (final) H = (), (exothermic) 6-5

6 b) Nitrogen dioxide, NO 2, forms from N 2 and O 2. 1/2N 2 (g) + O 2 (g) + heat NO 2 (g) NO 2 (final) Increasing, H 1/2N 2 + O 2 (initial) H = (+), (endothermic) 6.27 a) CO 2 (s) + heat CO 2 (g) CO 2 (g) Increasing, H H = (+), (endothermic) CO 2 (s) b) SO 2 (g) + 1/2O 2 (g) SO (g) + heat SO 2 (g) + 1/2O 2 (g) Increasing, H SO (g) H = (), (exothermic) 6.28 Plan: Recall that q sys is positive if heat is absorbed by the system (endothermic) and negative if heat is released by the system (exothermic). Since E = q + w, the work must be considered in addition to q sys to find ΔE sys. a) This is a phase change from the solid phase to the gas phase. Heat is absorbed by the system so q sys is positive (+). b) The system is expanding in volume as more moles of gas exist after the phase change than were present before the phase change. So the system has done work of expansion and w is negative. ΔE sys = q + w. Since q is positive and w is negative, the sign of ΔE sys cannot be predicted. It will be positive if q > w and negative if q < w. c) ΔE univ = 0. If the system loses energy, the surroundings gain an equal amount of energy. The sum of the energy of the system and the energy of the surroundings remains constant a) There is a volume decrease; V final < V initial so ΔV is negative. Since w sys = PΔV, w is positive, +. b) H sys is as heat has been removed from the system to liquefy the gas. c) E sys = q + w. Since q is negative and w is positive, the sign of ΔE sys and ΔE surr cannot be predicted. ΔE sys will be positive and ΔE surr will be negative if w > q and ΔE sys will be negative and ΔE surr will be positive if w < q. 6.0 The molar heat capacity of a substance is larger than its specific heat capacity. The specific heat capacity of a substance is the quantity of heat required to change the temperature of 1 g of a substance by 1 K while the molar heat capacity is the quantity of heat required to change the temperature of 1 mole of a substance by 1 K. The specific heat capacity of a substance is multiplied by its molar mass to obtain the molar heat capacity. 6.1 To determine the specific heat capacity of a substance, you need its mass, the heat added (or lost), and the change in temperature. 6-6

7 6.2 Specific heat capacity is an intensive property; it is defined on a per gram basis. The specific heat capacity of a particular substance has the same value, regardless of the amount of substance present. 6. Specific heat capacity is the quantity of heat required to raise 1g of a substance by 1 K. Molar heat capacity is the quantity of heat required to raise 1 mole of substance by 1 K. Heat capacity is also the quantity of heat required for a 1 K temperature change, but it applies to an object instead of a specified amount of a substance. Thus, specific heat capacity and molar heat capacity are used when talking about an element or compound while heat capacity is used for a calorimeter or other object. 6.4 In a coffee-cup calorimeter, reactions occur at constant pressure. q p = H. In a bomb calorimeter, reactions occur at constant volume. q v = E. 6.5 Plan: The heat required to raise the temperature of water is found by using the equation q = c x mass x T. The specific heat capacity, c water, is found in Table 6.2. Because the Celsius degree is the same size as the Kelvin degree, T = 100 C 25 C = 75 C = 75 K. J q (J) = c x mass x T = g75 K = = 6.9x10 J gk J g K gk 6.6 q (J) = c x mass x T = = = 18 J 6.7 Plan: Use the relationship q = c x mass x T. We know the heat (change kj to J), the specific heat capacity, and the mass, so T can be calculated. Once T is known, that value is added to the initial temperature to find the final temperature. q (J) = c x mass x T T initial = 1.00 C T final =? mass = 295 g c = J/g K 10 J q = 75.0 kj = 7.50x10 1 kj 4 J 7.50x10 4 J = (0.900 J/g K)(295 g)(t) 7.50x10 J T = J 295 g gk T = K = C 4 T = T final T initial T final T + T initial T final = C C = = 295 C 6.8 q (J) = c x mass x T 688 J = (2.42 J/g K)(27.7 g)(t) 688 J (T) = = K = C 2.42 J 27.7 g gk T = T final T initial T initial T final T T initial = 2.5 C ( C) = = 42.8 C (Because the Celsius degree is the same size as the Kelvin degree, T is the same in either temperature unit.) 6-7

8 6.9 Plan: Since the bolts have the same mass and same specific heat capacity, and one must cool as the other heats (the heat lost by the hot bolt equals the heat gained by the cold bolt), the final temperature is an average of the two initial temperatures. T1 + T C + 55C = = 77.5 C q lost = q gained 2(mass)(c Cu )(T final 105) C = (mass)(c Cu )(T final 45) C 2(T final 105) C = (T final 45) C 2(105 C) 2T final = T final 45 C 210 C + 45 C = T final + 2T final = T final (255 C)/ = T final = 85.0 C 6.41 Plan: The heat lost by the water originally at 85 C is gained by the water that is originally at 26 C. Therefore q lost = q gained. Both volumes are converted to mass using the density. Mass (g) of 75 ml = 75 ml 1.00 g 1.00 g 1 ml = 75 g Mass (g) of 155 ml = 155 ml 1 ml = 155 g q lost = q gained c x mass x T (85 C water)= c x mass x T (26 C water) (4.184 J/g C)(75 g)(t final 85) C =(4.184 J/g C)(155 g)(t final 26) C (75 g)(t final 85) C = (155 g) (T final 26) C T final = 155T final = 155T final + 75T final = 20.T final T final = (10405/20.) = = 45 C 6.42 q lost = q gained [24.4 ml(1.00 g/ml)](4.184 J/g C)( ) C = (mass)(4.184 J/g C)( ) C (24.4)( ) = (mass)( ) (24.4)( 11.5) = (mass)(5.) = (mass)(5.) g = mass 1 ml Volume (ml) = g = = 5 ml 1.00 g 6.4 Plan: Heat gained by water and the container equals the heat lost by the copper tubing so q water + q calorimeter = q copper. T = T final T initial Specific heat capacity in units of J/g K has the same value in units of J/g C since the Celsius and Kelvin unit are the same size. q lost = q gained = q water + q calorimeter (455 g Cu)(0.87 J/g C)(T final 89.5) C = (159 g H 2 O)(4.184 J/g C)(T final 22.8) C + (10.0 J/ C)(T final 22.8) C ( )(T final 89.5) = ( )(T final 22.8) + (10.0)(T final 22.8) T final = T final T final = T final T final T final = T final T final = /(851.41) = = 6.6 C 6-8

9 6.44 q lost = q gained = q water + q calorimeter (0.5 g alloy)(c alloy )( ) C = (50.0 g H 2 O)(4.184 J/g C)( ) C + (9.2 J/ C)( ) C (0.5 g)(c alloy )( 61.9 C) = (50.0 g)(4.184 J/g C)(9.1 C) + (9.2 J/ C)(9.1 C) (c alloy ) = = c alloy = / = = 1.1 J/g C 6.45 Benzoic acid is C 6 H 5 COOH, and will be symbolized as HBz. q reaction = q water + q calorimeter 1 mol HBz 227 kj 10 J q reaction = g HBz g HBz 1 mol HBz 1 kj q water = c x mass x T = J/g C x 1200 g x T q calorimeter = C x T = 165 J/ C x T q reaction = q water + q calorimeter x10 4 J = J/g C x 1200 g x T J/ C x T x10 4 J = (T) + 165(T) x10 4 J = 685.8(T) T = x10 4 /685.8 = = 5.05 C = x10 4 J 6.46 a) Energy will flow from Cu (at C) to Fe (at 0.0 C). b) To determine the final temperature, the heat capacity of the calorimeter must be known. c) q Cu = q Fe + q calorimeter assume q calorimeter = 0. q Cu = q Fe + 0 (20.0 g Cu)(0.87 J/g C)(T final 100.0) C = (0.0 g Fe)(0.450 J/g C)(T final 0.0) C J (20.0 g)(0.87 J/g C)(T final C) = (0.0 g)(0.450 J/g C)(T final 0.0 C) (7.74)(T final 100.0) = (1.5)(T final 0.0) T final = 1.5T final 774 = ( ) T final = 21.24T final T final = 774/21.24 = = 6.4 C 6.47 q hydrocarbon = q water + q calorimeter q hydrocarbon = (2.550 L H 2 O)(1mL /10 L)(1.00g/mL)(4.184 J/g C)( ) C + (40 J/ C)( ) C q hydrocarbon = (2550. g)(4.184 J/g C)(.55 C) + (40 J/ C)(.55 C) q hydrocarbon = ( J) + ( J) = J q hydrocarbon =.9061x10 4 J q hydrocarbon /g = (.9061x10 4 J)/1.520 g = x10 4 = 2.59x10 4 J/g 6.48 The reaction is: 2KOH(aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + 2H 2 O(l) q (kj) = ( ) ml(1.00 g/ml)(4.184 J/g C)( ) C(1 kj/10 J) = kj (The temperature increased so the heat of reaction is exothermic.) Amount (moles) of H 2 SO 4 = (25.0 ml)(0.500 mol H 2 SO 4 /L)(10 L/1 ml) = mol H 2 SO 4 Amount (moles) of KOH = (25.0 ml)(1.00 mol KOH/L)(10 L/1 ml) = mol KOH The moles show that both H 2 SO 4 and KOH are limiting. The enthalpy change could be calculated in any of the following ways: H = kj/ mol H 2 SO 4 = = 112 kj/mol H 2 SO 4 H = kj/ mol KOH = = 55.8 kj/mol KOH (Per mole of K 2 SO 4 gives the same value as per mole of H 2 SO 4, and per mole of H 2 O gives the same value as per mole of KOH.) 6.49 Reactants Products + Energy H rxn = ( ) Thus, energy is a product. 6-9

10 6.50 Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an exothermic reaction in which heat is released. The reaction has a positive H rxn, because this reaction requires the input of energy to break the oxygen-oxygen bond in O 2 : O 2 (g) + energy 2O(g) 6.51 Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an exothermic reaction in which heat is released. As a substance changes from the gaseous state to the liquid state, energy is released so H would be negative for the condensation of 1 mol of water. The value of H for the vaporization of 2 mol of water would be twice the value of H for the condensation of 1 mol of water vapor but would have an opposite sign (+H). H 2 O(g) H 2 O(l) + Energy 2H 2 O(l) + Energy 2H 2 O(g) H condensation = ( ) H vaporization = (+)2[H condensation ] The enthalpy for 1 mole of water condensing would be opposite in sign to and one-half the value for the conversion of 2 moles of liquid H 2 O to H 2 O vapor Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an exothermic reaction in which heat is released. The H rxn is specific for the reaction as written, meaning that 20.2 kj is released when one-eighth of a mole of sulfur reacts. Use the ratio between moles of sulfur and H to convert between amount of sulfur and heat released. a) This reaction is exothermic because H is negative. b) Because H is a state function, the total energy required for the reverse reaction, regardless of how the change occurs, is the same magnitude but different sign of the forward reaction. Therefore, H = kj kj c) H rxn = 2.6 mol S8 = = 4.2x10 1/8 mol S 2 kj 8 d) The mass of S 8 requires conversion to moles and then a calculation identical to part c) can be performed. 1molS kj H rxn = 25.0 g S8 = = 15.7 kj g S 8 1/ 8 mol S MgCO (s) MgO(s) + CO 2 (g) H rxn = 117. kj a) Absorbed b) H rxn (reverse) = 117. kj 117. kj c) H rxn = 5.5 mol CO2 = = 628 kj 1molCO2 2 d) H rxn = 5.5 g CO 2 1molCO 117. kj = = 94.6 kj g CO 1 mol CO Plan: A thermochemical equation is a balanced equation that includes the heat of reaction. Since heat is absorbed in this reaction, H will be positive. Convert the mass of NO to moles and use the ratio between NO and H to find the heat involved for this amount of NO. a) 1/2N 2 (g) + 1/2O 2 (g) (g) H = kj 1 mol NO kj b) H rxn =.50 g NO = = 10.5 kj 0.01 g NO 1 mol NO 6-10

11 6.55 a) KBr(s) K(s) + 1/2Br 2 (l) H rxn = 94 kj 10 g 1 mol KBr 94 kj b) H rxn = 10.0 kg KBr 1 kg g KBr 1 mol KBr =.109x10 4 =.1x10 4 kj 6.56 Plan: For the reaction written, 2 moles of H 2 O 2 release kj of energy upon decomposition. Use this ratio to convert between the given amount of reactant and the amount of heat released. The amount of H 2 O 2 must be converted from kg to g to moles. 2H 2 O 2 (l) 2H 2 O(l) + O 2 (g) H rxn = kj 10 g 1molH2O kj Heat (kj) = q = 652 kg H2O2 = x10 1 kg 6 = 1.88x10 6 kj 4.02 g H2O2 2 mol H2O For the reaction written, 1 mole of B 2 H 6 releases kj of energy upon reaction. B 2 H 6 (g) + 6Cl 2 (g) 2BCl (g) + 6HCl(g) H rxn = kj 10 g 1molB2H kj Heat (kj) = q = 1 kg = 2.700x10 1 kg 4 = 2.70x10 4 kj/kg g B2H6 1 mol B2H Fe(s) + O 2 (g) 2Fe 2 O (s) H rxn = 1.65x10 kj 10 g 1 mol Fe 1.65x10 kj a) Heat (kj) = q = kg Fe = = 1850 kj 1 kg g Fe 4 mol Fe 2molFeO gFeO 2 b) Mass (g) of Fe 2 O = 4.85x10 kj = = 99 g Fe 2 O 1.65x10 kj 1molFe2O HgO(s) 2Hg(l) + O 2 (g) H rxn = kj 1 mol HgO kj a) Heat (kj) = q = 555 g HgO = = 2 kj g HgO 2 mol Hg b) Mass (g) of Hg = 275 kj 2 mol Hg g Hg = = 608 g Hg kj 1 mol Hg 6.60 Plan: A thermochemical equation is a balanced equation that includes the heat of reaction. Heat is released in this reaction so H is negative. Use the ratio between H and moles of C 2 H 4 to find the amount of C 2 H 4 that must react to produce the given quantity of heat. a) C 2 H 4 (g) + O 2 (g) 2CO 2 (g) + 2H 2 O(g) H rxn = 1411 kj 1molCH gCH 2 4 b) Mass (g) of C 2 H 4 = 70.0 kj = = 1.9 g C 2 H kj 1 mol C2H a) C 12 H 22 O 11 (s) + 12O 2 (g) 12CO 2 (g) + 11H 2 O(g) H rxn = 5.64x10 kj b) Heat (kj) = q = 1 g C H O mol C12H22O x10 kj 42.0 g C H O 1 mol C H O = = 16.5 kj/g 6.62 Hess s law: H rxn is independent of the number of steps or the path of the reaction. 6.6 Hess s law provides a useful way of calculating energy changes for reactions which are difficult or impossible to measure directly. 6-11

12 6.64 Plan: Two chemical equations can be written based on the description given: C(s) + O 2 (g) CO 2 (g) H 1 (1) CO(g) + 1/2O 2 (g) CO 2 (g) H 2 (2) The second reaction can be reversed and its H sign changed. In this case, no change in the coefficients is necessary since the CO 2 cancels. Add the two H values together to obtain the H of the desired reaction. C(s) + O 2 (g) CO 2 (g) H 1 CO 2 (g) CO(g) + 1/2O 2 (g) H 2 (reaction is reversed) Total C(s) + 1/2O 2 (g) CO(g) H rxn = H 1 + (H 2 ) How are the H values for each reaction determined? The H 1 can be found by using the heats of formation in Appendix B: H 1 = [H f (CO 2 )] [H f (C) + H f (O 2 )] = [ 9.5 kj/mol] [0 + 0] = 9.5 kj/mol. The H 2 can be found by using the heats of formation in Appendix B: H 2 = [H f (CO 2 )] [H f (CO) + 1/2H f (O 2 )] = [ 9.5] [ kj/mol + 0)] = 28 kj/mol. H rxn = H 1 + (H 2 ) = 9.5 kj + ( 28.0 kj) = kj 6.65 Plan: To obtain the overall reaction, add the first reaction to the reverse of the second. When the second reaction is reversed, the sign of its enthalpy change is reversed from positive to negative. Ca(s) + 1/2O 2 (g) CaO(s) H = 65.1 kj CaO(s) + CO 2 (g) CaCO (s) H = 178. kj (reaction is reversed) Ca(s) + 1/2O 2 (g) + CO 2 (g) CaCO (s) H = 81.4 kj NOCl(g) 2NO(g) + Cl 2 (g) H = 2( 8.6 kj) 2NO(g) N 2 (g) + O 2 (g) H = 2(90. kj) 2NOCl(g) N 2 (g) + O 2 (g) + Cl 2 (g) H = 77.2 kj + ( kj) = 10.4 kj 6.67 Plan: Add the two equations, canceling substances that appear on both sides of the arrow. When matching the equations with the arrows in the Figure, remember that a positive H corresponds to an arrow pointing up while a negative H corresponds to an arrow pointing down. 1) N 2 (g) + O 2 (g) 2NO(g) H = kj 2) 2NO(g) + O 2 (g) 2NO 2 (g) H = kj ) N 2 (g) + 2O 2 (g) 2NO 2 (g) H rxn = kj In Figure P6.67, A represents reaction 1 with a larger amount of energy absorbed, B represents reaction 2 with a smaller amount of energy released, and C represents reaction as the sum of A and B ) P 4 (s) + 6Cl 2 (g) 4PCl (g) H 1 = 1148 kj 2) 4PCl (g) + 4Cl 2 (g) 4PCl 5 (g) H 2 = 460 kj ) P 4 (s) + 10Cl 2 (g) 4PCl 5 (g) H overall = 1608 kj Equation 1) = B, equation 2) = C, equation ) = A 6.69 Plan: Vaporization is the change in state from a liquid to a gas: H 2 O(l) H 2 O(g). The two equations describing the chemical reactions for the formation of gaseous and liquid water can be combined to yield the equation for vaporization. 1) Formation of H 2 O(g): H 2 (g) + 1/2O 2 (g) H 2 O(g) H = kj 2) Formation of H 2 O(l): H 2 (g) + 1/2O 2 (g) H 2 O(l) H = kj 6-12

13 Reverse reaction 2 (change the sign of H) and add the two reactions: H 2 (g) + 1/2O 2 (g) H 2 O(g) H = kj H 2 O(l) H 2 (g) + 1/2O 2 (g) H = kj H 2 O(l) H 2 O(g) H vap = 44.0 kj 6.70 C(s) + 1/4S 8 (s) CS 2 (l) H = kj CS 2 (l) CS 2 (g) H = kj C(s) + 1/4S 8 (s) CS 2 (g) H = kj 6.71 C (diamond) + O 2 (g) CO 2 (g) H = 95.4 kj CO 2 (g) C(graphite) + O 2 (g) H = ( 9.5 kj) C(diamond) C(graphite) H = 1.9 kj 6.72 The standard heat of reaction, Hrxn, is the enthalpy change for any reaction where all substances are in their standard states. The standard heat of formation, Hf, is the enthalpy change that accompanies the formation of one mole of a compound in its standard state from elements in their standard states. Standard state is 1 atm for gases, 1 M for solutes, and the most stable form for liquids and solids. Standard state does not include a specific temperature, but a temperature must be specified in a table of standard values. 6.7 The standard heat of reaction is the sum of the standard heats of formation of the products minus the sum of the standard heats of formation of the reactants multiplied by their respective stoichiometric coefficients. H rxn = m Hf (products) n Hf (reactants) 6.74 Plan: Hf is for the reaction that shows the formation of one mole of compound from its elements in their standard states. a) 1/2Cl 2 (g) + Na(s) NaCl(s) The element chlorine occurs as Cl 2, not Cl. b) H 2 (g) + 1/2O 2 (g) H 2 O(g) The element hydrogen exists as H 2, not H, and the formation of water is written with water as the product. c) No changes 6.75 Plan: Formation equations show the formation of one mole of compound from its elements. The elements must be in their most stable states ( = 0). Hf a) Ca(s) + Cl 2 (g) CaCl 2 (s) b) Na(s) + 1/2H 2 (g) + C(graphite) + /2O 2 (g) NaHCO (s) c) C(graphite) + 2Cl 2 (g) CCl 4 (l) d) 1/2H 2 (g) + 1/2N 2 (g) + /2O 2 (g) HNO (l) 6.76 a) 1/2H 2 (g) + 1/2I 2 (s) HI(g) b) Si(s) + 2F 2 (g) SiF 4 (g) c) /2O 2 (g) O (g) d) Ca(s) + 1/2P 4 (s) + 4O 2 (g) Ca (PO 4 ) 2 (s) 6.77 Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the Hf values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. 6-1

14 H rxn = m Hf (products) n Hf (reactants) a) Hrxn = {2 Hf [SO 2 (g)] + 2 Hf [H 2 O(g)]} {2 Hf [H 2 S(g)] + Hf [O 2 (g)]} = [(2 mol)( kj/mol) + (2 mol)( kj/mol)] [(2 mol)( 20.2 kj/mol) + ( mol)(0.0 kj/mol)] = kj b) The balanced equation is CH 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4HCl(g) H rxn = {1 Hf [CCl 4 (l)] + 4 Hf [HCl(g)]} {1 Hf [CH 4 (g)] + 4 Hf [Cl 2 (g)]} H rxn = [(1 mol)( 19 kj/mol) + (4 mol)( 92.1 kj/mol)] [(1 mol)( kj/mol) + (4 mol)(0 kj/mol)] = 4 kj 6.78 Hrxn = m Hf (products) n Hf (reactants) a) Hrxn = {1 Hf [SiF 4 (g)] + 2 Hf [H 2 O(l)]} {1 Hf [SiO 2 (s)] + 4 Hf [HF(g)]} = [(1 mol)( kj/mol) + (2 mol)( kj/mol)] [(1 mol)( kj/mol) + (4 mol)( 27 kj/mol)] = 184 kj b) 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) H rxn = {4 Hf [CO 2 (g)] + 6 Hf [H 2 O(g)]} {2 Hf [C 2 H 6 (g)] + 7 Hf [ O 2 (g)]} = [(4 mol)( 9.5 kj/mol) + (6 mol)( kj/mol)] [(2 mol)( kj/mol) + (7 mol)(0 kj/mol)] = kj (or kj for reaction of 1 mol of C 2 H 6 ) 6.79 Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the H values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. In this case, H f of CuO must be calculated. H rxn = m Hf (products) n Hf (reactants) Cu 2 O(s) + 1/2O 2 (g) 2CuO(s) Hrxn = kj H rxn = {2 Hf [CuO(s)]} {1 Hf [Cu 2 O(s)] + 1/2 Hf [O 2 (g)]} kj = {(2 mol) Hf [CuO(s)]} {(1 mol)( kj/mol) + (1/2 mol)(0 kj/mol)} kj = 2 mol Hf [CuO(s)] kj H 14.6 kj f [CuO(s)] = = 157. kj/mol 2 mol 6.80 Hrxn = m Hf (products) n Hf (reactants) C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O(g) Hrxn = kj H rxn = {2 Hf [CO 2 (g)] + 1 Hf [H 2 O(g)]} {1 Hf [C 2 H 2 (g)] + 5/2 Hf [O 2 (g)]} kj = {(2 mol)( 9.5 kj/mol) + (1 mol)( kj/mol)} {(1 mol) Hf [C 2 H 2 (g)] + (5/2 mol)(0.0 kj/mol)} kj = kj kj (1 mol) Hf [C 2 H 2 (g)] H f [C 2 H 2 (g)] = kj = kj/mol 1 mol f Hrxn is known and 6-14

15 6.81 Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the Hf values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. Hess s law can also be used to calculate the enthalpy of reaction. In part b), rearrange equations 1) and 2) to give the equation wanted. Reverse the first equation (changing the sign of Hrxn ) and multiply the coefficients (and Hrxn ) of the second reaction by 2. 2PbSO 4 (s) + 2H 2 O(l) Pb(s) + PbO 2 (s) + 2H 2 SO 4 (l) H rxn = m Hf (products) n Hf (reactants) a) Hrxn = {1 Hf [Pb(s)] + 1 Hf [PbO 2 (s)] + 2 Hf [H 2 SO 4 (l)]} {2 Hf [PbSO 4 (s)] + 2 Hf [H 2 O(l)]} = [(1 mol)(0 kj/mol) + (1 mol)( kjmol) + (2 mol)( kj/mol)] [(2 mol)( kj/mol) + (2 mol)( kj/mol)] = 50.9 kj b) Use Hess s law: PbSO 4 (s) Pb(s) + PbO 2 (s) + 2SO (g) Hrxn = ( 768 kj) Equation has been reversed. 2SO (g) + 2H 2 O (l) 2H 2 SO 4 (l) Hrxn = 2( 12 kj) 2PbSO 4 (s) + 2H 2 O(l) Pb(s) + PbO 2 (s) + 2H 2 SO 4 (l) Hrxn = 504 kj 6.82 Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the Hf values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. Convert the mass of stearic acid to moles and use the ratio between stearic acid and Hrxn to find the heat involved for this amount of acid. For part d), use the kcal/g of fat relationship calculated in part c) to convert 11.0 g of fat to total kcal and compare to the 100. Cal amount. a) C 18 H 6 O 2 (s) + 26O 2 (g) 18CO 2 (g) + 18H 2 O(g) b) Hrxn = m Hf (products) n Hf (reactants) H rxn = {18 Hf [CO 2 (g)] + 18 Hf [H 2 O(g)]} {1 Hf [C 18 H 6 O 2 (s)] + 26 Hf [O 2 (g)]} = [(18 mol)( 9.5 kj/mol) + (18 mol)( kj/mol)] [(1 mol)( 948 kj/mol) + (26 mol)(0 kj/mol)] = 10, = 10,488 kj 1molC c) q (kj) = 18H6O2 10, kj 1.00 g C18H6O2 = = 6.9 kj C18H6O2 1 mol C18H6O2 1 kcal q (kcal) = kj = = 8.81 kcal kj kcal d) q (kcal) = 11.0 g fat = = 96.9 kcal 1.0 g fat Since 1 kcal = 1 Cal, 96.9 kcal = 96.9 Cal. The calculated calorie content is consistent with the package information. 6.8 a) H 2 SO 4 (l) H 2 SO 4 (aq) H rxn = {1 Hf [H 2 SO 4 (aq)]} {1 Hf [H 2 SO 4 (l)]} = [(1 mol)( kj/mol)] [(1 mol)( kj/mol)] = 9.52 kj 6-15

16 b) q (J) = c x mass x T 9.52 kj x J J/kJ = x g ml g C 1 ml x T final 25.0 C 9.52x J J = x g x Tfinal 25.0 C g C 9.52 x 10 4 J = (T final )710 J/ºC x10 4 J T final = = 50.2 C c) Adding the acid to a large amount of water releases the heat to a large mass of solution and thus, the potential temperature rise is minimized due to the large heat capacity of the larger volume Plan: Use the ideal gas law, PV = nrt, to calculate the volume of one mole of helium at each temperature. Then use the given equation for ΔE to find the change in internal energy. The equation for work, w = PΔV, is needed for part c), and q P = ΔE + PΔV is used for part d). For part e), recall that ΔH = q P. a) PV = nrt or V = nrt P T = = 288 K and T = = 0 K L atm Initial volume (L) = V = nrt K mol K = = 2.6 L/mol 1.00 atm Final volume (L) = V = nrt P = L atm K P = mol K 1.00 atm = = 24.9 L/mol b) Internal energy is the sum of the potential and kinetic energies of each He atom in the system (the balloon). The energy of one mole of helium atoms can be described as a function of temperature, E = /2nRT, where n = 1 mole. Therefore, the internal energy at 15 C and 0 C can be calculated. The inside back cover lists values of R with different units. E = /2nRT = (/2)(1.00 mol) (8.14 J/mol K)(0 288)K = = 187 J c) When the balloon expands as temperature rises, the balloon performs PV work. However, the problem specifies that pressure remains constant, so work done on the surroundings by the balloon is defined by the equation: w = PV. When pressure and volume are multiplied together, the unit is L atm, so a conversion factor is needed to convert work in units of L atm to joules J w = PV = 1.00 atm ( ) L 1 L atm = = 1.2x102 J d) q P = E + PV = ( J) + ( J) = =.1x10 2 J e) H = q P = 10 J. f) When a process occurs at constant pressure, the change in heat energy of the system can be described by a state function called enthalpy. The change in enthalpy equals the heat (q) lost at constant pressure: H = E + PV = E w = (q + w) w = q P 6.85 a) Respiration: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(g) H rxn = m Hf (products) n Hf (reactants) = {6 Hf [CO 2 (g)] + 6 Hf [H 2 O(g)]} {1 Hf [C 6 H 12 O 6 (s)] + 6 Hf [O 2 (g)]} = [(6 mol)( 9.5 kj/mol) + (6 mol)( kj/mol)] [(1 mol)( 127. kj/mol) + (6 mol)(0.0 kj/mol)] = = kj 6-16

17 Fermentation: C 6 H 12 O 6 (s) 2CO 2 (g) + 2CH CH 2 OH(l) H rxn = {2 Hf [CO 2 (g)] + 2 Hf [CH CH 2 OH(l)]} [1 Hf [C 6 H 12 O 6 (s)]} = [(2 mol)( 9.5 kj/mol) + (2 mol)( kj/mol)] [(1 mol)( 127. kj/mol)] = = 69.0 kj b) Combustion of ethanol: CH CH 2 OH(l) + O 2 (g) 2CO 2 (g) + H 2 O(g) H rxn = {2 Hf [CO 2 (g)] + Hf [H 2 O(g)]} {1 Hf [CH CH 2 OH(l)] + Hf [O 2 (g)]} H rxn = [(2 mol)( 9.5 kj/mol) + ( mol)( kj/mol)] [(1 mol)( kj/mol) + ( mol)(0.0 kj/mol)] = = kj Heats of combustion/mol C: kj 1 mol C6H12O6 Sugar: 1 mol C6H12O 6 6 mol C = = kj/mol C kj 1 mol CHCH2OH Ethanol: 1 mol CHCH2OH 2 mol C = = kj/mol C Ethanol has a higher value a) Reactions: 1) C 21 H 44 (s) + 2O 2 (g) 21CO 2 (g) + 22H 2 O(g) 2) C 21 H 44 (s) + 4/2O 2 (g) 21CO(g) + 22H 2 O(g) ) C 21 H 44 (s) + 11O 2 (g) 21C(s) + 22H 2 O(g) Heats of combustion: 1) Hrxn = {21 Hf [CO 2 (g)] + 22 Hf [H 2 O(g)]} {[1 Hf [C 21 H 44 (s)] + 2 Hf [O 2 (g)]} = [(21 mol)( 9.5 kj/mol) + (22 mol)( kj/mol)] [(1 mol)( 476 kj/mol) + (2 mol)(0.0 kj/mol)] = 1, = 1,108 kj 2) Hrxn = {21 Hf [CO(g)] + 22 Hf [H 2 O(g)]} {1 Hf [C 21 H 44 (s)] + 4/2 Hf [O 2 (g)]} = [(21 mol)( kj/mol) + (22 mol)( kj/mol)] [(1 mol)( 476 kj/mol) + (4/2 mol)(0.0 kj/mol)] = = 7165 kj ) Hrxn = {21 Hf [C(s)] + 22 Hf [H 2 O(g)]} {1 Hf [C 21 H 44 (s)] + 11 Hf [O 2 (g)]} = [(21 mol)(0.0 kj/mol) +(22 mol)( kj/mol)] [(1 mol)( 476 kj/mol) + (11 mol)(0.0 kj/mol)] = = 4844 kj 1molC21H kj b) q (kj) = 254 g C21H44 = x10 4 = 1.12x10 4 kj g C H 1 mol C H c) The moles of C 21 H 44 need to be calculated one time for multiple usage. It must be assumed that the remaining 87.00% of the candle undergoes complete combustion. Moles C 21 H 44 = (254 g C 21 H 44 )(1 mol C 21 H 44 / g C 21 H 44 ) = mol q = (0.87)( mol)( kj/mol) + (0.0800)( mol)( kj/mol) + (0.0500)( mol)( kj/mol)] = x10 4 = 1.05x10 4 kj 6.87 a) EO(l) EO(g) Hvap = J/g(44.05 g/mol)(1 kj/1000 J) = kj/mol H vap = {1 Hf [EO(g)]} {1 Hf [EO(l)]} kj/mol = { Hf [EO(g)]} [(1 mol)( 77.4 kj/mol)] H f [EO(g)] = 52.2 kj/mol EO(g) CH 4 (g) + CO(g) H rxn = {1 Hf [CH 4 (g)] + 1 Hf [CO(g)]} {1 Hf [EO(g)]} H rxn = [(1 mol)( kj/mol) + (1 mol)( kj/mol)] [(1 mol)( 52.2 kj/mol)] 6-17

18 H rxn = 1.0 kj/mol b) Assume that you have 1.00 mole of EO(g) mole of EO(g) produces 1.00 mole or g of CH 4 (g) and 1.00 mole or g of CO(g). There is a total product mass of g g = g. q = c x mass x T 1000 J 1.0 kj q 1 kj ΔT = = c x mass 2.5 J/gC g ΔT = ºC ΔT = T final T initial ºC = T final 9ºC T final = = 101 C 6.88 a) N 2 O 5 (g) + NO(g) 9NO 2 (g) H rxn = {9 Hf [NO 2 (g)]} { Hf [N 2 O 5 (g)] + Hf [NO(g)]} = [(9 mol)(.2 kj/mol)] [( mol)(11 kj/mol) + ( mol)(90.29 kj/mol)] = 5.07 = 5 kj x10 mol 5.07 kj 10 J b) 9 molecules product = = 76.0 J 1 molecule product 9 moles product 1 kj 4 qt 1 L 1 ml g 1molC8H x10 kj 1 gal qt 10 L ml g 1 mol C8H 18 = x10 6 = 2.54x10 6 kj b) Distance (km) = 6 1 h 65 mi 1 km x10 kj 4 5.5x10 kj 1 h 0.62 mi = x10 = 4.8x10 km c) Only a small percentage of the chemical energy in the fuel is converted to work to move the car; most of the chemical energy is lost as waste heat flowing into the surroundings a) Heat (kj) = 20.4 gal 6.90 q = c x mass x T In this situation, all of the samples have the same mass, 50. g, so mass is not a variable. All also have the same q value, 450. J. So, 450. J α (c x ΔT). c, specific heat capacity, and ΔT are inversely proportional. The higher the ΔT, the lower the value of specific heat capacity: ΔT: B > D > C > A Specific heat capacity: B < D < C < A 6.91 ClF(g) + 1/2O 2 (g) 1/2Cl 2 O(g) + 1/2OF 2 (g) Hrxn = 1/2(167.5 kj) = 8.75 kj F 2 (g) + 1/2O 2 (g) OF 2 (g) Hrxn = 1/2( 4.5 kj) = kj 1/2Cl 2 O(g) + /2OF 2 (g) ClF (l) + O 2 (g) Hrxn = 1/2(94.1 kj) = kj ClF(g) + F 2 (g) ClF (l) Hrxn = 15.1 kj 6.92 a) AgNO (aq) + NaI(aq) AgI(s) + NaNO (aq) 10 L 5.0 g AgNO 1 mol AgNO Moles of AgNO = 50.0 ml 1 ml 1 L g AgNO = x10 mol AgNO 10 L 5.0 g NaI 1 mol NaI Moles of NaI = 50.0 ml 1mL 1L 149.9g NaI = x10 mol NaI 6-18

19 The AgNO is limiting, and will be used to finish the problem: 1 mol AgI 24.8 g AgI Mass (g) of AgI = x10 mol AgNO 1 mol AgNO 1 mol AgI = = 0.5 g AgI b) Ag + (aq) + I (aq) AgI(s) H rxn = {1 Hf [AgI(s)]} {1 Hf [Ag + (aq)] + 1 Hf [I (aq)]} = [(1 mol)( 62.8 kj/mol)] [(1 mol)(105.9 kj/mol) + (1 mol)( kj/mol)] = 112. kj c) Hrxn = q = c x mass x T 112. kj 1 mol AgI x10 mol AgNO T = Hrxn mol AgI 1 mol AgNO 10 J /c x mass = J 1.00 g 1kJ mL g K ml = = 0.9 K 6.9 Plan: Use conversion factors to solve parts a) and b). For part c), first find the heat of reaction for the combustion of methane by using the heats of formation of the reactants and products. The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the o Hf values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to reflect the higher number of moles. For part e), convert the amount of water in gal to mass in g and use the relationship q = c x mass x T to find the heat needed; then use the conversion factors between joules and therms and the cost per therm to determine the total cost of heating the water. 1 cal J 45.6 g 1.0C 1.00 lbf a) = = 1.1x10 J/Btu g C 1 cal 1 lb 1.8F 1 Btu 100,000 Btu J b) E = 1.00 therm 1therm Btu = x10 8 = 1.1x10 8 J c) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H rxn = {1 Hf [CO 2 (g)] + 2 Hf [H 2 O(g)]} {1 Hf [CH 4 (g)] + 2 Hf [O 2 (g)]} = [(1 mol)( 9.5 kj/mol) + (2 mol)( kj/mol)] [(1 mol)( kj/mol) + (2 mol)(0.0 kj/mol)] = = 802. kj/mol CH x10 J 1 kj 1 mol CH4 Moles of CH 4 = 1.00 therm 1 therm 10 J kj = = 1.x10 2 mol CH 4 $ therm d) Cost = therm = = $0.0050/mol mol 4qt 1 L 1mL 1.0g 1 gal qt 10 L ml = x10 6 g e) Mass (g) of = 18 gal J 6 q = c x mass x T = x10 g C= x10 8 J gc 8 1therm $0.66 Cost = x10 J 8 = = $ x10 J 1therm 6-19

20 6.94 a) Mass (kg) of = 5600 EJ b) Years = 5600 EJ J 1 kj 1 mol CH g CH4 1 kg 1 EJ 10 J 802 kj 1 mol CH4 10 g = 1.12x10 14 kg CH 4 1yr = 14 yr 4.0x10 EJ 1 L 1 ml 1.00 g J 1 kj 1molCH4 c) Moles of CH 4 = 1.00 qt C qt 10 L ml g C 10 J 802 kj = mol CH 4 Volume (ft g CH4 1L 10 m 5.ft ) of CH 4 = mol CH4 1 mol CH g CH4 1 L 1m = = 0.29 ft d) Volume = 2x10 J 1 1 kj 1 mol CH g CH4 1 L 10 m 5. ft 10 J 802 kj 1 mol CH g CH 4 1 L 1 m = x10 7 = 2x10 7 ft 6.95 The reaction is exothermic. The argon atoms in the chamber after the reaction are moving with greater kinetic energy, indicating an increase in temperature H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) 2H + (aq) + 2OH (aq) 2H 2 O(l) H rxn = {2 Hf [H 2 O(l)]} {2 Hf [H + (aq)] + 2 Hf [OH (aq)]} = [(2 mol)( kj/mol)] [(2 mol)(0 kj/mol) + (2 mol)( kj/mol)] = kj 1 mole of H 2 SO 4 reacts with 2 moles of NaOH L 1.00 ml 1.00 g Mass (g) of H 2 SO 4 solution = 1 mol H2SO mol H2SO 4 10 L 1.00 ml = 2060 g H 2 SO 4 solution g NaOH 100 g solution Mass (g) of NaOH solution = 2 mol NaOH 1 mol NaOH = 200. g NaOH solution 40 g NaOH q = c x mass x T 10 J kj q 1 kj ΔT = = = 11.82ºC c x mass J/g C g 1ºC ºC = = 4ºC This temperature is above the temperature at which a flammable vapor could be formed so the temperature increase could cause the vapor to explode a) 2C 12 H 26 (l) + 7O 2 (g) 24CO 2 (g) + 26H 2 O(g) b) Hrxn = {24 Hf [CO 2 (g)] + 26 Hf [H 2 O(g)]} {2 Hf [C 12 H 26 (g)] + 7 Hf [O 2 (g)]} 1.50x10 4 kj = [(24 mol)( 9.5 kj/mol) + (26 mol)( kj/mol)] [(2 mol) Hf [C 12 H 26 (g)] + (7 mol)(0.0 kj/mol)] 1.50x10 4 kj = kj kj [(2 mol) Hf [C 12 H 26 (g)] kj] 1.50x10 4 kj = 15, kj (2 mol) Hf [C 12 H 26 (g)] 1.50x10 4 kj + 15, kj = (2 mol) H [C 12 H 26 (g)] f 6-20

21 kj = (2 mol) Hf [C 12 H 26 (g)] H f [C 12 H 26 (g)] = =.66x10 2 kj 4 4 qt 1 L 1 ml g C12H26 1 mol C12H x10 kj c) Heat (kj) = 0.50 gal 1 gal qt 10 L ml 170. g C12H 26 2 mol C12H 26 = 6.240x10 4 = 6.2x10 4 kj kj 0.50 gal d) Volume (gal) = Btu 4 = = 1.1x10 2 gal 1Btu 6.240x10 kj 6.98 Plan: Heat of reaction is calculated using the relationship Hrxn = m Hf (products) n Hf (reactants). The heats of formation for all of the species, except SiCl 4, are found in Appendix B. Use reaction, with its given, to find the heat of formation of SiCl 4 (g). Once the heat of formation of SiCl 4 is known, the heat of Hrxn reaction of the other two reactions can be calculated. When reactions 2 and are added to obtain a fourth reaction, the heats of reaction of reactions 2 and are also added to obtain the heat of reaction for the fourth reaction. a) () SiCl 4 (g) + 2H 2 O(g) SiO 2 (s) + 4HCl(g) H rxn = {1 Hf [SiO 2 (s)] + 4 Hf [HCl(g)]} {1 Hf [SiCl 4 (g)] + 2 Hf [H 2 O(g)]} 19.5 kj = [(1 mol)( kj/mol) + (4 mol)( 92.1 kj/mol)] [ Hf [SiCl 4 (g)] + (2 mol)( kj/mol)] 19.5 kj = [ Hf [SiCl 4 (g)] + ( kj)] kj = Hf [SiCl 4 (g)] kj H f [SiCl 4 (g)] = kj/mol The heats of reaction for the first two steps can now be calculated. 1) Si(s) + 2Cl 2 (g) SiCl 4 (g) H rxn = {1 Hf [SiCl 4 (g)]} {1 Hf [Si(s)] + 2 Hf [Cl 2 (g)]} = [(1 mol)( kj/mol)] [(1 mol)(0 kj/mol) + (2 mol)(0 kj/mol)] = = kj 2) SiO 2 (s) + 2C(graphite) + 2Cl 2 (g) SiCl 4 (g) + 2CO(g) H rxn = {1 Hf [SiCl 4 (g)] + 2 Hf [CO(g)]} {1 Hf [SiO 2 (g)] + 2 Hf [C(graphite)] +2 Hf [Cl 2 (g)]} = [(1 mol)( kj/mol) + (2 mol)( kj/mol)] [(1 mol)( kj/mol) + (2 mol)(0 kj/mol) + (2 mol)(0 kj/mol)] = = 2.9 kj b) Adding reactions 2 and yields: (2) SiO 2 (s) + 2C(graphite) + 2Cl 2 (g) SiCl 4 (g) + 2CO(g) Hrxn = kj () SiCl 4 (g) + 2H 2 O(g) SiO 2 (s) + 4HCl(g) Hrxn = 19.5 kj 2C(graphite) + 2Cl 2 (g) + 2H 2 O(g) 2CO(g) + 4HCl(g) Hrxn = kj = kj o Confirm this result by calculating H rxn using Appendix B values. 2C(graphite) + 2Cl 2 (g) + 2H 2 O(g) 2CO(g) + 4HCl(g) H rxn = {2 Hf [CO(g)] + 4 Hf [HCl(g)]} {2 Hf [C(graphite)] + 2 Hf [Cl 2 (g)] + 2 Hf [H 2 O(g)]} = [(2 mol)( kj/mol) + (4 mol)( 92.1 kj) [(2 mol)(0 kj/mol) + (2 mol)(0 kj/mol) + (2 mol)( kj/mol)] = = kj 6-21

22 6.99 Plan: Use PV = nrt to find the initial volume of nitrogen gas at 0 C and then the final volume at 819 C. Then the relationship w = PΔV can be used to calculate the work of expansion. a) PV = nrt L atm Initial volume at 0 C + 27 = 27 K = V = nrt 1 mol K mol K = L 1.00 atm Final volume at 819 C + 27 = 1092 K = V = nrt P = L atm 1 mol K P = mol K 1.00 atm ΔV = V final V initial = L L = L w = PV = (1 atm) x L = atm L 1 J w (J) = atm L 9.87 x 10 atm L = = 6.81x10 J b) q = c x mass x T g Mass (g) of N 2 = 1 mol N2 = g 1 mol N2 q T = ( c )(mass) x10 J = g (1.00 J/g K) = = 24 K = 24 C = L Plan: Note the numbers of moles of the reactants and products in the target equation and manipulate equations 1-5 and their Hrxn values so that these equations sum to give the target equation. Then the manipulated o H rxn values will add to give the Hrxn value of the target equation. Only reaction contains N 2 O 4 (g), and only reaction 1 contains N 2 O (g), so we can use those reactions as a starting point. N 2 O 5 appears in both reactions 2 and 5, but note the physical states present: solid and gas. As a rough start, adding reactions 1,, and 5 yields the desired reactants and products, with some undesired intermediates: Reverse (1) N 2 O (g) NO(g) + NO 2 (g) Hrxn = ( 9.8 kj) = 9.8 kj Multiply () by 2 4NO 2 (g) 2N 2 O 4 (g) Hrxn = 2( 57.2 kj) = kj (5) N 2 O 5 (s) N 2 O 5 (g) Hrxn = (54.1 kj) = 54.1 kj N 2 O (g) + 4NO 2 (g) + N 2 O 5 (s) NO(g) + NO 2 (g) + 2N 2 O 4 (g) + N 2 O 5 (g) To cancel out the N 2 O 5 (g) intermediate, reverse equation 2. This also cancels out some of the undesired NO 2 (g) but adds NO(g) and O 2 (g). Finally, add equation 4 to remove those intermediates: Reverse (1) N 2 O (g) NO(g) + NO 2 (g) Hrxn = ( 9.8 kj) = 9.8 kj Multiply () by 2 4NO 2 (g) 2N 2 O 4 (g) Hrxn = 2( 57.2 kj) = kj (5) N 2 O 5 (s) N 2 O 5 (g) Hrxn = 54.1 kj Reverse (2) N 2 O 5 (g) NO(g) + NO 2 (g) + O 2 (g) Hrxn = ( kj) = (4) 2NO(g) + O 2 (g) 2NO 2 (g) Hrxn = kj Total: N 2 O (g) + N 2 O 5 (s) 2N 2 O 4 (g) Hrxn = 22.2 kj 6-22

23 6.101 Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the H values (Appendix B) are reported as energy per one mole, use f the appropriate stoichiometric coefficient to reflect the higher number of moles. In this case, Hrxn of the second reaction is known and Hf of N 2 H 4 (aq) must be calculated. For part b), calculate Hrxn for the reaction between N 2 H 4 (aq) and O 2, using the value of H for N 2 H 4 (aq) found in part a); then determine the moles of O 2 present f by multiplying volume and molarity and multiply by the Hrxn for the reaction. a) 2NH (aq) + NaOCl(aq) N 2 H 4 (aq) + NaCl(aq) + H 2 O(l) H rxn = {1 Hf [N 2 H 4 (aq)] + 1 Hf [NaCl(aq)] + 1 Hf [H 2 O(l)]} {2 Hf [NH (aq)] + 1 Hf [NaOCl(aq)]} Note that the Appendix B value for N 2 H 4 is for N 2 H 4 (l), not for N 2 H 4 (aq), so this term must be calculated. In addition, Appendix B does not list a value for NaCl(aq), so this term must be broken down into H f [Na + (aq)] and Hf [Cl (aq)]. 151 kj = [ Hf [N 2 H 4 (aq)] + (1 mol)( kj/mol) + (1 mol)( kj/mol) + (1 mol)( kj/mol)] [(2 mol)( 80.8 kj/mol) + (1 mol)( 46 kj/mol)] 151 kj = [ Hf [N 2 H 4 (aq)] + ( kj)] [ kj] 151 kj = [ Hf [N 2 H 4 (aq)] + ( 185. kj) H f [N 2 H 4 (aq)] = 4. = 4 kj x10 mol b) Moles of O 2 = 5.00x10 L = 1.25 mol O 1 L 2 N 2 H 4 (aq) + O 2 (g) N 2 (g) + 2H 2 O(l) H rxn = {1 Hf [N 2 (g)] + 2 Hf [H 2 O(l)]} {1 Hf [N 2 H 4 (aq)] + 1 Hf [O 2 (g)]} = [(1 mol)(0 kj/mol) + (2 mol)( kj/mol)] [(1 mol)(4. kj/mol) + (1 mol)(0 kj/mol] = kj kj Heat (kj) = 1.25 mol O2 = = 757 kj 1 mol O CO(g) + 2H 2 (g) CH OH(l) H rxn = {1 Hf [CH OH(l)]} {1 Hf [CO(g)] + 2 Hf [H 2 (g)]} = [(1 mol)( 28.6 kj/mol)] [(1 mol)( kj mol) + (2 mol)(0.0 kj/mol)] = kj Find the limiting reactant: Moles of CO = PV RT = 112 kpa 15.0 L 1atm = mol CO L atm kpa K mol K 1 mol CH OH mol CO 1 mol CO = mol CH OH Moles of CH OH from CO = Moles of H 2 = PV 744 torr 18.5 L 1atm = mol H torr K mol K RT = L atm Moles of CH OH from H 2 = mol H H 2 is limiting. 2 1 mol CH OH = mol CH OH 2 mol H2 6-2