Chapter 12. (a) n = = (10.0 g) = mol; T = = 2.77 K = 2.77 C. T f = T i + T = = 17.8 C. T = = K.

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1 12.1 A system is isolated if it exchanges neither energy nor matter with its surroundings, closed if it exchanges energy but not matter, and open if it exchanges both: (a) Open, energy and matter are constantly entering and leaving the system; (b) (almost) isolated, because the thermos minimizes energy transfer to the coffee; (c) open, energy and matter can flow in and out; and (d) closed, the balloon isolates the helium from the outside A system is isolated if it exchanges neither energy nor matter with its surroundings, closed if it exchanges energy but not matter, and open if it exchanges both: (a) Closed, because it can exchange energy but not matter; (b) (almost) isolated, because the chest minimizes energy transfer; (c) open, because energy and matter flow in and out; and (d) closed, because the satellite is self-contained but exchanges energy with its surroundings. 12. State variables are properties that are independent of the past history of the system. (a) not a state variable; (b) state variable; (c) state variable (mass is conserved); and (d) not a state variable State variables are properties that are independent of the past history of the system. (a) state variable; (b) not a state variable; (c) not a state variable; and (d) state variable A system is any part of the universe that can be defined with specified boundaries. Among possible systems are the automobile and driver, the driver, the automobile, the cooling system, the engine, a cylinder, the battery, etc A system is any part of the universe that can be defined with specified boundaries. Among possible systems are the entire body, heart, lungs, digestive system, liver, kidneys, etc Calculate the temperature change resulting from an energy input using Equation 12-1 and C values from Table 12-1 of your textbook: q = nc T, which can be rearranged to T = nc q m (a) n = = (10.0 g) = mol; MM g 25.0 J T = = 2.77 K = 2.77 C. (0.706 mol)(24.5 J mol K ) T f = T i + T = = 17.8 C. m (b) n = = (25.0 g) = mol; MM g 25.0 J T = = K. ( mol)(24.5 J mol K ) 84

2 T f = T i + T = = 296 K. m (c) n = = (25.0 g) = mol; MM g 25.0 J T = = K. (0.217 mol)(25.51 J mol K ) T f = T i + T = = 299 K. m (d) n = = (25.0 g) = 1.87 mol; MM g 25.0 J T = = K = C. (1.87 mol)( J mol K ) T f = T i + T = = 22.2 C Calculate the temperature change resulting from an energy input using Equation 12-1 and C values from Table 12-1 of your textbook: q = nc T, which can be rearranged to T = nc q m (a) n = = (5.0 g) = mol; MM g 65.0 J T = == K = C. (1.297 mol)(24.5 J mol K ) T f = T i + T = ( 2.058) = 6.0 C. m (b) n = = (50.0 g) = 1.85 mol; MM g 65.0 J T = = K = C. (1.85 mol)(24.5 J mol K ) T f = T i + T = ( 1.441) = 6.6 C. m (c) n = = (50.0 g) = mol; MM g 65.0 J T = = 5.5 K = 5.5 C (0.464 mol)(25.51 J mol K ) T f = T i + T = ( 5.5) = 59.5 C m (d) n = = (50.0 g) = mol; MM g 85

3 65.0 J T = = 0.11 K. (2.775 mol)(75.291j mol K ) T f = T i + T = ( 0.11) = 24.7 K Calculate an energy change accompanying a temperature change using Equation 12-1 and C values from Table 12-1 of your textbook: q = nc T. T = = 72.0 C = 72.0 K; 1000 g n kettle = 1.5 kg = 24.2 mol Fe 1kg g q kettle = (24.2 mol)(25.10 J mol -1 K -1 )(72.0 K) = 4.7 x 10 4 J; 1000 g n water = 2.75 kg = 15 mol water 1kg g q water = (15 mol)( J mol -1 K -1 )(72.0 K) = 8.29 x 10 5 J Calculate an energy change accompanying a temperature change using Equation 12-1 and C values from Table 12-1 of your textbook: q = nc T. T = = 1.6 C = 1.6 K; m nsilver = = 15.0 g = mol g Ag MM q silver = (0.190 mol)(25.51 J mol -1 K -1 )(1.6 K) = 47.9 J; m nwater = = 25.0 g = 1.87 mol g water MM q water = (1.87 mol)( J mol -1 K -1 )(1.6 K) = 1.42 x 10 J When two objects at different temperatures are placed in contact, energy flows from the warmer to the cooler object until the two objects are at the same temperature. Let that temperature be T. The energy lost by the warmer object equals the energy gained by the cooler object: q cool = q warm, and q = nc T. Water: n = 7.5 g = mol water; g q = (2.081 mol)(75.51 J mol -1 K -1 )(T 20.5 C) = (156.8 J K -1 )(T 20.5 C). Coin: n = (27.4 g) = mol Ag; g q = (0.259 mol)(25.51 J mol -1 K -1 )(T C) = (6.47 J K -1 )(T C). Substitute and solve for T: (156.8 J K -1 )(T 20.5 C) = (6.47 J K -1 )(T C) T = 6.47 T ; 16.2 T = 858.2, and 86

4 T = 2.6 C When two objects at different temperatures are placed in contact, energy flows from the warmer to the cooler object until the two objects are at the same temperature. Let that temperature be T. The energy lost by the warmer object equals the energy gained by the cooler object: q cool = q warm, and q = nc T g Coffee: n = 85.0 ml = mol; 1mL g J q coffee = (4.717 mol) (T 84.0 C). K Spoon: n = 24.7 g = mol; g J q spoon = (0.442 mol) (T 18.5 C). K Substitute and solve for T: J J (0.442 mol) (T 18.5 C) = (4.717 mol) (T 84.0 C). K K T = T + 29,82; T = 0,07, and T = 82.0 C Calculate the energy change for a temperature change using Equation 12-1 and C values from Table 12-1 of your textbook: q = nc T, and T = = 66.1 C = 66.1 K; m 1.00 g n water = = 475 ml = 26.6 mol 1mL g ; MM J q water = (26.6 mol) (66.1 K) = 1.1 x 10 K 5 J = 11 kj Calculate an energy change accompanying a temperature change using Equation 12-1 and C values from Table 12-1 of your textbook: q = nc T. T = = 48.5 C = 48.5 K; m n water = = 145 g = mol g MM J q water = (8.047 mol) ( 48.5 K) = 2.94 x 10 K 4 J = 29.4 kj Expansion work can be calculated using Equation 12-, w sys = P ext V sys. The external pressure opposing inflation of a balloon is atmospheric pressure, 1.00 atm: V f = 2.5 L, V i = 0.0 L 87

5 V= 2.5 L 0.0 L = 2.5 L w sys = P ext V sys = (1.00 atm)(2.5 L) = 2.5 L atm. Convert to joules: J w sys = 2.5 L atm = 2.5 x 10 1L atm 2 J Expansion work can be calculated using Equation 12-, wsys = Pext Vsys. The external pressure opposing inflation of a balloon is 755 torr. 1atm Converted to atmospheres: 755 torr = 0.99 atm: 760 torr The final volume of the balloon is 19.5m, converted to liters: 10 L 19.5 m = 19,500 L 1m Assuming the initial volume of the balloon is 0 L then, wsys = Pext Vsys = (0.99 atm)(19,500l) = L atm. Convert to joules: J wsys = L atm = 1.97x 10 1L atm 6 J To work this problem, use data in Chemistry and Life Box. First determine the energy difference( E) between chicken and beef: 6.0 kj Chicken: 250 g = 1500 kj; 1g 16 kj beef: 250 g = 4000 kj; 1g E = 4000 kj 1500 kj = 2500 kj; According to data in Chemistry and Life Box, a 55-kg person walking 6.0 km/hr consumes 1090 kj/hr: 1hr t = 2500 kj = 2. hr kj Therefore, to consume the additional energy a person must walk: 6.0 km distance = 2. hr = 1.4 km. 1hr To work this problem, use data in Chemistry and Life Box. The energy content of sugar is 16 kj/g, so one pound of sugar has an energy content of: 10 g 16 kj kg = 7.28 x 10 1kg 1g kj. A person weighing 85 kg consumes 4245 kj/hr when running at 16 km/hr: 88

6 t = 7.28 x 10 1hr kj = 1.7 hr kj Therefore, to consume the energy of the sugar a person must run: 16 km distance = 2. hr = 27 km. 1hr The heat capacity of a calorimeter can be found from energy and temperature data using Equation 12-5, q calorimeter = C cal T. Here, the heat released by the combustion process is absorbed by the calorimeter so: kj q calorimeter = q glucose = g = kj 1g T = = 1.89 C qcalorimete r kj C cal = = = 14.4 kj/ C. o T 1.89 C The heat capacity of a calorimeter can be found from energy and temperature data using Equation 12-5, q calorimeter = C cal T. Here, q calorimeter = 1150 J T = = 1.80 C qcalorimete r 1150 J C cal = = = 69 J/ C = 69 J/K o T 1.80 C The heat capacity of 125 ml of water is: 1.00 g J C water = 125 ml = 522 J/K 1mL g K Thus, the percentage due to water is: Cwater 522 J/K % = = 100% = 81.7%. C 69 J/K cal (a) In a combustion reaction, the products are CO 2 and H 2 O: C 7 H 6 O 2 + O 2 CO 2 + H 2 O (unbalanced) Follow standard procedures to balance the equation. Give CO 2 a coefficient of 7 to balance C, H 2 O a coefficient of to balance H: C 7 H 6 O 2 + O 2 7 CO 2 + H 2 O 7C + 6H + 4O 7C + 6H + 17O Balance O by giving O 2 a coefficient of 15/2, then multiply by 2 to clear fractions: 2 C 7 H 6 O O 2 14 CO H 2 O (b) Find energy per mole from energy per gram using the molar mass (122.1 g/mol): 89

7 E (kj/mol) = [ E (kj/g)][mm (g/mol)] = 5.61kJ 122.1g =.221 x g kj/mol (c) 15 moles O 2 is consumed for each 2 mol of benzoic acid, so the energy released per mol of O 2 is:.221 x 10 kj 2 mol acid E = = x 10 acid 15 mol O 2 kj/mol (a) In a combustion reaction, the products are CO 2 and H 2 O: C 2 H 2 + O 2 CO 2 + H 2 O (unbalanced) Follow standard procedures to balance the equation. Give CO 2 a coefficient of 2 to balance C: C 2 H 2 + O 2 2CO 2 + H 2 O 2C + 2H + 2O 2C + 2H + 5O Give O 2 a coefficient of 5/2 to balance O, then multiply by 2 to clear fractions: 2 C 2 H O 2 4 CO H 2 O (b) Find energy per mole from energy per gram using the molar mass (26.04 g/mol): 48.2 kj g E (kj/mol) = [ E (kj/g)][mm (g/mol)] = = 1.26 x g kj/mol (c) 5 moles O 2 is consumed for every 2 moles of acetylene, so the energy released per mol of O 2 is: E = 1.26 x 10 kj 2 mol acetylene = 5.04 x 10 acetylene 5 mol O 2 kj/mol Calculate the energy released during combustion from the temperature increase and the total heat capacity of the calorimeter. Then convert to molar energy using molar mass. q = C T = (7.85 kj/k)(02.04 K K) = 4.46 kj E = q = 4.46 kj 4.46 kj g E (kj/mol) = [ E (kj/g)][mm (g/mol)]= = 4.96 x g kj/mol Calculate the heat capacity of the calorimeter from the electrical energy added and the temperature change during electrical heating. Then find the energy released during combustion from the temperature increase and the total heat capacity of the calorimeter. Finally convert to molar energy using molar mass. qcalorimete r kj C = = = 4.68 kj/ C o T 4.22 C 90

8 4.68 kj q combustion = C T combustion = o (8.47 C) = 9.64 kj 1 C - q 9.64 kj E = = = kj/g m 1.75 g The molar heat of combustion for methanol is obtained by multiplying the energy per gram by the molar mass: kj 2.04 g E (kj/mol) = [ E (kj/g)][mm (g/mol)] = = 7.26 x 10 1g 2 kj/mol Standard enthalpy changes are calculated from Equation using standard enthalpies of formation, which can be found in Appendix D of your textbook: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f(reactants) (a) H o reaction= [2 mol( 9.5 kj/mol) + 2 mol( kj/mol)] [1 mol(52.4 kj/mol) + mol(0 kj/mol)] = kj; (b) H o reaction= 1 mol (0 kj/mol) + mol(0 kj/mol) 2 mol( 45.9 kj/mol) = 91.8 kj; (c) H o reaction= [1 mol( kj/mol) + 5 mol(0kj/mol)] [5 mol( kj/mol) + 4 mol(0 kj/mol)] = kj; (d) H o reaction= [1 mol( kj/mol) + 4 mol( 92. kj/mol)] [( kj/mol) + 2 mol( kj/mol)] = 21.2 kj Standard enthalpy changes are calculated from Equation using standard enthalpies of formation, which can be found in Appendix D of your textbook: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) (a) H o reaction= 2 mol( kj/mol) [2 mol(0 kj/mol) + mol(0 kj/mol)] = kj; (b) H o reaction= [2 mol( 1.9 kj/mol) + 1 mol(91. kj/mol)] [ mol(.2 kj/mol) + 1 mol( kj/mol)] = 9.7 kj; (c) H o reaction= [4 mol( 9.5 kj/mol) + 2 mol( kj/mol)] [2 mol(227.4 kj/mol) + 5 mol(0 kj/mol)] = kj Reaction energies are related to reaction enthalpies (see Prob ) through Equation 12-9: H reaction E reaction + (nrt) gases Since temperature does not change, we can rearrange the equation to: E reaction H reaction RT (n) (gases) Begin by calculating the change in moles of gases, then use the rearranged equation to determine the E reaction (a) n gases = 2 ( + 1) = 2 mol; 8.14 J K kj E reaction kj ( 2 mol) ( ) K 1J 91

9 = = kj; (b) n gases = = 2 mol; 8.14 J K E reaction 91.8 kj (2 mol) ( ) K 1J (c) all compounds are solid, n gases = 0 mol; E reaction kj; (d) n gases = 4 mol; E reaction 21.2 kj (4 mol) ( ) K 1J 8.14 J K kj = = 86.8 kj; kj = = 11. kj Reaction energies are related to reaction enthalpies (see Prob ) through Equation 12-9: H reaction E reaction + (nrt) gases Since temperature does not change, we can rearrange the equation to: E reaction H reaction RT (n) (gases) Begin by calculating the change in moles of gases, then use the rearranged equation to determine the E reaction (a) n gases = 0 = mol; 8.14 J K E reaction kj ( mol) ( ) K 1J = kj; (b) n gases = = 0 mol; E reaction 9.7 kj; (c) n gases = 4 (5 + 2) = mol; 8.14 J K E reaction kj ( mol) ( ) K 1J = kj. kj = kj = A formation reaction has elements in their standard states as reactants and 1 mol of a single product: (a) K(s) + P(s) + 2 O 2 (g) K PO 4 (s); (b) 2 C(graphite) + 2 H 2 (g) + O 2 (g) CH CO 2 H(l); (c) C(graphite) + 9/2 H 2 (g) +1/2 N 2 (g) (CH ) N(g); (d) 2 Al(s) + /2 O 2 (g) Al 2 O (s). 92

10 12.0 A formation reaction has elements in their standard states as reactants and 1 mol of a single product: (a) 4 C(graphite) + 5 H 2 (g) +1/2 O 2 (g) C 4 H 9 OH(l); (b) 2 Na(s) + C(graphite) + /2 O 2 (g) Na 2 CO (s); (c) /2 O 2 (g) O (g); (d) Fe(s) + 2 O 2 (g) Fe O 4 (s) Standard enthalpy changes are calculated from Equation using standard enthalpies of formation, which can be found in Appendix D of your textbook: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) (a) H o reaction = [4 mol(91. kj/mol) + 6 mol( kj/mol)] [4 mol( 45.9 kj/mol) + 5 mol(0 kj/mol)] = kj; (b) H o reaction = [2 mol(0 kj/mol) + 6 mol( kj/mol)] [4 mol( 45.9 kj/mol) + mol(0 kj/mol)] = kj Standard enthalpy changes are calculated from Equation using standard enthalpies of formation, which can be found in Appendix D of your textbook: H o reaction= Σ coeff p H o f (products) Σ coeff r H o f (reactants) (a) H o reaction= 2 mol( 82.0 kj/mol) [1 mol( kj/mol) + mol( kj/mol)] = 5.7 kj; (b) H o reaction= 2 mol( 1094.) [( kj/mol) + mol( kj/mol)] = 57.6 kj. 12. Calculate the energy released during the dissolving process from the temperature increase and the total heat capacity of the calorimeter. Then convert to molar energy using molar mass. Assume that the heat capacity of the calorimeter is the heat capacity of its water contents: J C cal = g = J/K = J/ C; g K q = C cal T = ( J/ C)(29.7 C 22.0 C) =.54 x 10 J; remember that the heat absorbed by the calorimeter is the heat lost by the solution process: - q.54 x 10 J H = = = 745 J/g m 4.75 g H (J/mol) = [ H (J/g)][MM (g/mol)] = 745 J g = 8. x 10 1g 4 J/mol Calculate the energy absorbed during the dissolving process from the temperature decrease and the total heat capacity of the calorimeter. Then convert to molar energy using MM. Assume that the heat capacity of the calorimeter is the heat capacity of its water contents: 9

11 75.291J C cal = 50.0 g = J/K; g K q = C cal T = (208.9 J/K)(296.6 K K) = 42.6 J; - q ( 42.6 J) H = = = 42.6 J/g m 1.00 g 42.6 J g H (J/mol) = [ H (J/g)][MM (g/mol)] = = 4.20 x 10 1g 4 J/mol (a) The positive value for H signals that reaction to form solid MgSO 4 is endothermic. Dissolving, the reverse reaction, is exothermic, releasing energy, which is absorbed by the water. (b) Use the molar enthalpy and number of moles dissolving to calculate the energy released: m n = = 2.55 g = x 10 MM g -2 mol MgSO 4 ; q water = q salt = n H = (2.119 x kj mol) = 1.9 kj. (c) Use Equation 12-1, q = nc T, to determine the temperature change: T = nc q ; m n = = (5.00 x g ml) MM = mol; 1mL g 1.9 x 10 J T = = K -1-1 ( mol)(75.291j mol K ) 12.6 (a) The negative value for H signals that reaction to form solid NH 4 NO is exothermic. Dissolving, the reverse reaction, is endothermic, absorbing energy from the water. (b) Use the molar enthalpy and number of moles dissolving to calculate the energy released: m n = = 25.0 g = 0.12 mol; MM g 21.1kJ q water = q salt = n H = 0.12 mol = 6.59 kj. (c) Use Equation 12-1, q = nc T, to determine the temperature change: T = nc q ; m n = = 2.50 x g ml MM = 1.87 mol; 1mL g 6.59 x 10 J T = = 6.1K -1-1 (1.87 mol)(75.291j mol K ) 94

12 12.7 The formation of a solid from ions in solution involves two opposing energy effects. Energy is released because of the coulombic attraction between cations and anions, but energy is absorbed to overcome the ion-dipole attractions between ions and water molecules. Solid formation is endothermic when the sum of all ion-dipole attractions in solution is greater than the ion-ion interactions in the solid The formation of a solid from ions in solution involves two opposing energy effects. Energy is released because of the coulombic attraction between cations and anions, but energy is absorbed to overcome the ion-dipole attractions between ions and water molecules. Solid formation is exothermic when the ion-ion interactions in the solid are greater than the sum of all ion-dipole attractions in solution The heats of phase changes indicate the strengths of intermolecular forces; the larger the intermolecular forces, the larger the heat of the phase change. (a) Methane has a lower heat of vaporization than ethane because, being a smaller molecule, it has smaller dispersion forces. (b) Ethanol has a significantly higher heat of vaporization than diethyl ether because of strong hydrogen bonding. (c) Argon (18 electrons) has a higher heat of fusion than methane (10 electrons) because it has a higher polarizability due to its larger number of electrons The heats of phase changes indicate the strengths of intermolecular forces; the larger the intermolecular forces, the larger the heat of the phase change. (a) Water has a significantly higher heat of vaporization than methane because of strong hydrogen bonding. (b) Benzene has a higher heat of fusion than ethane because, being a larger molecule, it has larger dispersion forces. (c) Oxygen (12 electrons, two atoms rather than one) has a higher heat of vaporization than argon (18 electrons, a single atom) because it has a higher polarizability due to its larger size (two atoms rather than one) Because E is a state function, the energy released ( E) when one gram of gasoline burns is the same regardless of the conditions. Work (w) is done when an automobile accelerates but no work is done when an automobile idles. Thus, q idle = E and q accel = E w accel, so more heat must be removed under idling conditions (this is part of the reason why automobiles tend to overheat in traffic jams) When a liquid droplet vaporizes in a vacuum, the opposing pressure is zero, so no work must be done against an external pressure and E = q. Thus, the heat absorbed under these conditions is E for the phase change To work a problem involving heat transfers, it is useful to set up a block diagram illustrating the process. In this problem, a copper block transfers energy to ice: 95

13 Thus, q ice = q Cu, q Cu, = n Cu C T, and q ice = n ice H fus. Substituting gives: n ice H f = n Cu C T Here are the data needed for the calculation: m n Cu = = 12.7 g = mol; MM g C = J/mol K, T = (0.0 C C) = 200 C = 200 K; H f = 6.01 kj/mol = 6.01 x 10 J/mol; Substitute and solve for the amount of ice that melts: -1-1 ncuc T ( mol)(24.45 J mol K )(-200K) n = = -1 H f 6.01 x 10 J mol Finally, convert to mass: g m ice = n MM = mol = 2.9 g of ice melts. ice = The heat that is lost by the coin will be absorbed by the ice. Thus, q ice = q Au, q Au, = n Au C T, and q ice = n ice H fus. Substituting gives: n ice H f = n Au C T Here are the data needed for the calculation: n Au = 7.65 g = mol, g T = ( )K - ( )K = 100 K, C Au = 25.4 J/mol K = kj/mol K mol Finally, convert to mass: 25.4 J q Au = mol ( 100 K) = 98.6 J K q Au = q ice, n ice = kj = mol; 6.01kJ Now use mass-mole calculations to determine the mass of ice that melts: 96

14 18.02 g m ice = mol = g ice melted (a) In a combustion reaction, a substance reacts with molecular oxygen to form CO 2 and H 2 O: C 6 H 12 O 6 + O 2 CO 2 + H 2 O Balance the equation using standard procedures. Give CO 2 a coefficient of 6 and H 2 O a coefficient of 6 to balance C and H; C 6 H 12 O 6 + O 2 6 CO H 2 O 6C + 12H + 8 O 6C + 12H + 18O then give O 2 a coefficient of 6 to balance O: C 6 H 12 O O 2 (g) 6 CO 2 (g) + 6 H 2 O(l) (b) To determine the molar heat of combustion, multiply the heat released (negative, indicating an exothermic reaction) in burning one gram by the molar mass: kj g H molar = = 2.8 x 10 1g kj/mol; (c) Use the molar heat of combustion along with Equation and data from Appendix D to determine the heat of formation of glucose: o H reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) 2.8 x 10 kj/mol = [6 mol( 9.5 kj/mol) + 6 mol( kj/mol)] [6 mol(0 kj/mol) + H f o (glucose)] H o f (glucose) = (2.8 x x x 10 ) kj/mol = 1.25 x 10 kj/mol (a) In this reaction, urea reacts with molecular oxygen to form CO 2, H 2 O, and N 2 : (NH 2 ) 2 CO + O 2 CO 2 + H 2 O + N 2 Give CO 2 a coefficient of 1, H 2 O a coefficient of 2, and N 2 a coefficient of 1 to balance C, H, and O. (NH 2 ) 2 CO + O 2 CO H 2 O + N 2 2N + 4H + C + O 2N + 4H + C + 4O Then give O 2 a coefficient of /2 to balance O (retain the fractional coefficient because a combustion reaction refers to one mole of substance burning): (NH 2 ) 2 CO(s) + /2 O 2 (g) CO 2 (g) + 2 H 2 O(l) + N 2 (g) 97

15 (b) A heat of combustion refers to one mole of substance burning. When 1 mol of urea burns, 2 mol of H 2 O form. Thus, the heat generated per mole of H 2 O formed is: 62.2 kj urea q H2 O = = 16.1 kj. urea 2 mol H2O (c) Use the molar heat of combustion along with Equation and data from Appendix D to determine the heat of formation of urea: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) 62.2 kj/mol = [1 mol( 9.5 kj/mol) + 2 mol( kj/mol) + 1 mol(0 kj/mol)] [1.5 mol(0 kj/mol) + H o f (urea)] H o f (urea) = ( ) kj/mol =.0 kj/mol Living organisms require both matter and energy to carry out their life processes. An isolated system receives neither matter nor energy from its surroundings, so a living organism that is isolated will quickly die When a moving automobile brakes and slows, its kinetic energy of motion is reduced. This energy is transferred to the brakes and tires as heat, causing the brakes and tires to heat up. The net transformation is conversion of kinetic energy of the automobile into thermal energy of the brakes and tires A molar heat capacity can be calculated from a temperature change using q = n C T: T = K K = 19.6 K; n = 52.5 g = mol; g q C = n T = (100.0 J) = 20.1 J/mol K. (0.254 mol)(19.6 K) To determine the molar heat capacity, C m, of the rhodium metal we will need to use the equation: q q = n C m T, or C m = n T. T = T f T i = 27 K 7 K = 100 K In order to solve the equation for C m we need to find q. To find q we must notice that the heat lost by the rhodium metal is equal to the heat gained by the ice, or q Rh = q ice. The q gained by the ice is only in the form of phase change energy so we must use the formula: 6.01kJ q p = n H fus = 0.16 g = kj g Using this energy and the above eqn we calculate molar heat capacity of Rh: 98

16 n Rh = 4.5 g = 0.42 mol g q kj C m = = = kj mol -1 K -1 n T (0.42 mol)( 100 K) In this process, neither P nor V is constant, so we cannot calculate q directly. Instead, we can calculate w and E. An ideal gas has no intermolecular forces, so its energy does not change when it expands or contracts. Thus, for T = 0, E = 0. Calculate w using w = P ext V: w = (4.00 atm)(20.0 L 0.0 L) = 40.0 L atm; Convert to joules, recalling that 1 L atm = J: J w = (40.0 L atm) = 4.05 x 10 J 1L atm q = E w = x 10 J = 4.05 x 10 J In this process, the pressure can be assumed to be atmospheric pressure, so this is a constant-pressure cooling, for which H = q = nc p T. Once H has been calculated, find E using H = E + (PV). m n He = = g MM = 4.92 x 10-2 mol; 4.00 g C p = 20.8 J/mol K; T = (255 K 25 K) = 70 K H = (4.92 x J mol) ( 70 K) = 71.7 J; K (PV) = (nrt) = nr T = (4.92 x J mol) ( 70 K) = 28.6 J K E = H (PV) = 71.7 J ( 28.6 J) = 4.1 J Standard enthalpy changes are calculated from Equation using standard enthalpies of formation, which can be found in Appendix D of your textbook: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) (a) H o reaction = 2 mol( 95.7 kj/mol) [2 mol( kj/mol) + 1 mol(0 kj/mol)] = kj; (b) H o reaction = 1 mol(11.1 kj/mol) 2 mol(.2 kj/mol) = 55. kj; (c) H o reaction = 1 mol( kj/mol) + 2 mol(0 kj/mol) [1 mol( kj/mol) + 2 mol(0 kj/mol)] = kj Standard enthalpy changes are calculated from Equation using standard enthalpies of formation, which can be found in Appendix D of your textbook: H o reaction = Σ coeffp H o f (products) Σ coeffr H o f (reactants) (a) H o reaction =2 mol(15.1 kj/mol) + 6 mol( kj/mol) 99

17 [2 mol( 45.9 kj/mol) + mol(0.0 kj/mol) +2 mol( 74.6 kj/mol)] = 99.8 kj; (b) H o reaction = [2 mol( kj/mol) + 4 mol( 9.5 kj/mol)] [5 mol(0.0 kj/mol) + 2 mol(227.4 kj/mol)] = kj; (c) H o reaction = [ ( kj/mol) + (0 kj/mol)] [1 mol(52.4 kj/mol) + 1 mol(142.7 kj/mol)] = 61. kj An enthalpy of formation can be calculated from the heat of a reaction provided all other enthalpies of formation are known. For this reaction, formation enthalpies of three of the four reagents appear in Appendix D of your textbook: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) 677 kj = 1 mol( kj/mol) + mol( kj/mol) [ H f o (P4 S ) + 8 mol(0 kj/mol)] H f o (P4 S ) = 677 kj kj kj = 197 kj Standard enthalpy changes are calculated from Equation using standard enthalpies of formation, which can be found in Appendix D of your textbook: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) First balance the overall reaction. Give PH a coefficient of 4 to balance P and H 2 O a coefficient of 6 to balance H. This gives 16 O on the right, requiring a coefficient of 8 for O 2 : 4 PH (g) + 8 O 2 (g) P 4 O 10 (s) + 6 H 2 O(g) H o reaction = 1 mol( kj/mol) + 6 mol( kj/mol) [4 mol(5.4 kj/mol) + 8 mol(0 kj/mol)] = kj To work a problem involving heat transfers, it is useful to set up a block diagram illustrating the process. In this problem, the unknown metal transfers energy to water: Thus, qmetal = qwater = -(nc T)water m n water = = 80.0 g MM = 4.44 mol water; g C = J/mol C, and T = ( ) =.6 C; J q metal = q water = (4.44 mol) (.6 C) = 120 J. o C 400

18 Without knowing the identity of the metal, we cannot determine nmetal, but heat capacity values are provided in units of J/g K, so we can use q = mc T: T = = 71.6 C = 71.6 K; q C = m T = ( 120 J) (44.0 g)(-71.6 K) = 0.82 J / g K This matches the heat capacity of Cu (0.85 J/gK), so the metal is copper To work a problem involving heat transfers, it is useful to set up a block diagram illustrating the process. In this problem, the coin transfers energy to water: Thus, q coin = q water = (nc T) water m n water = = 21.5 g MM = 1.19 mol water; g C = J/mol C, and T = ( ) = 6.0 C; J q coin = -q water = (1.19 mol) (6.0 C) = 59 J. o C Now use q = mc T to calculate C in J/g K: T = = 78.5 C = 78.5 K; q ( 59 J) C = = = 0.44 J/g K m T (15.5 g)(-78.5k) Calculate C for Ni and Ag: 25.51J Ag: C = = 0.25 J/g K K g J Ni: C = = J/g K. K g C matches that of nickel, so the coin is counterfeit The process shown is condensation of a gas to form a solid, the reverse of sublimation. Sublimation always is endothermic, as energy must be provided to overcome intermolecular forces of attraction; thus the depicted process is exothermic. (a) H sys is negative (exothermic process); (b) E surr is positive (surroundings must absorb the energy released in the process); (c) E universe = 0 (total energy always is conserved). 401

19 12.60 The process shown is a constant-volume decomposition of diatomic molecules to atoms. Energy must be provided to the system to break the chemical bonds holding the molecules together, so the depicted process is endothermic. (a) w sys = 0 because there is no volume change; (b) q sys is positive because heat must be provided to break the bonds; (c) E surr is negative because the surroundings provide the energy required to break the bonds The enthalpy of a reaction can be calculated from Equation and standard heats of formation (see Appendix D): H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) For this "reaction," the calculation is simple: H o reaction = 1 mol( kj/mol) 1 mol( kj/mol) = kj Compare this value with H vap = kj/mol. The values differ because H vap refers to vaporization of liquid water at the normal boiling point, 7 K, while o H reaction refers to vaporization of liquid water under standard thermodynamic conditions, 298 K An enthalpy of formation can be calculated from the heat of a reaction provided all other enthalpies of formation are known. For this reaction, formation enthalpies of four of the five reagents appear in Appendix D of your textbook: H o reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) 1196 kj = [1 mol(0 kj/mol) + 6 mol( 27. kj/mol) + 1 mol(0 kj/mol)] [2 mol( 45.9 kj/mol) + 2 H f o (ClF )] H o f (ClF ) = ½ ( )kJ/mol = kj/mol The enthalpy of a reaction can be calculated from Equation and standard heats of formation (see Appendix D): H reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) (a) The question asks for H when 1 mol of hydrazine burns, so the appropriate reaction is: N 2 H 4 (l) + 1/2 N 2 O 4 (g) /2 N 2 (g) + 2 H 2 O(g) H reaction = [1.5 mol(0 kj/mol) + 2 mol( kj/mol)] [1 mol(50.6 kj/mol) mol(11.1 kj/mol)] = 59.8 kj/mol (b) When hydrazine burns in oxygen, the reaction is: N 2 H 4 (l) + O 2 (g) N 2 (g) + 2 H 2 O(g) H reaction = [1 mol(0 kj/mol) + 2 mol( kj/mol)] [1 mol(50.6 kj/mol) + 1 mol(0 kj/mol)] = 54. kj/mol 402

20 The reaction with dinitrogen tetroxide is slightly more exothermic because that compound is slightly less stable than the elements from which it forms. Dinitrogen tetroxide is used in preference to molecular oxygen because the reaction occurs on contact, while oxygen would require an ignition device Convert heats of combustion to a per-gram basis by dividing by the molar mass of the substance: kj (a) H per gram = = kj/g; g kj (b) H per gram = = 22.7 kj/g; 2.04 g kj (c) H per gram = = 48.9 kj/g; g Octane has the highest energy content and methanol the lowest per gram of fuel (octane's high energy content is a main reason why it is widely used as an automobile fuel) The difference between H reaction and E reaction can be estimated using Equation 12-9: H reaction E reaction + (nrt) gases so H reaction E reaction (nrt) (gases) (a) n gases = 0, so H reaction E reaction 0; (b) n gases = 1 mol + 1 mol 0 mol = 2 mol gases; 8.14 J H reaction E reaction (2 mol) (298 K) = 4.96 x 10 J; K (c) The reaction is C 4 H 9 OH(l) + 6 O 2 (g) 4 CO 2 (g) + 5 H 2 O(l); n gases = 4 mol 6 mol = 2 mol; 8.14 J H reaction E reaction (2 mol) (298 K) = 4.96 x 10 J. K The difference between H reaction and E reaction can be estimated using Equation 12-9: H reaction E reaction + (nrt) gases so E reaction H reaction (nrt) (gases) For a vaporization process, T = T vap and n = 1; 8.14 J 10 - kj Argon: E reaction (6. kj/mol) 1 (87 K) K = 5.6 kj/mol; 1J 8.14 J 10 - kj Ethane: E reaction (15.5 kj/mol) 1 (184 K) K = 14.0 kj/mol; 1J 40

21 8.14 J 10 - kj Mercury: E reaction (59.0 kj/mol) 1 (60 K) K = 5.8 kj/mol; 1J % difference = (100%) E reaction E H reaction reaction (5.6 kj / mol - 6. kj / mol) Argon: % difference = (100%) = 1%; (5.6 kj / mol) (14.0 kj/mol kj/mol) Ethane: % difference = (100%) = 11%; (14.0 kj/mol) (5.8 kj / mol kj / mol) Mercury: % difference = (100%) = 9.7%; (5.8 kj / mol) Argon has the highest percentage difference Use the definition of each type of process to identify the reaction associated with each process: (a) I 2 (s) I 2 (g); (b) 1/2 I 2 (s) I(g); (c) 2 C(graphite) + /2 H 2 (g) + 1/2 Cl 2 (g) C 2 H Cl(g); (d) Na 2 SO 4 (s) 2 Na + (aq) + SO 4 2- (aq) Use the definition of each type of process to identify the reaction associated with each process: (a) Br 2 (l) Br 2 (g); (b) 1/2 Br 2 (l) Br(g); (c) C(graphite) + 2H 2 (g) + ½O 2 (g) CH OH(g); (d) MgCl 2 (s) Mg 2+ (aq) + 2Cl - (aq) (a) This is a cooling process, for which q = nc T: m n = = (2.50 g) MM = mol; g T = ( ) C = 62.5 C = 62.5 K; J q = (0.187 mol) ( 62.5 K) = 65 J. o C (b) This is a phase change, for which q = n H phase, followed by the same cooling process as in part (a): kj 10 J q = (0.187 mol) 65 J = 5658 J 65 J = 6.1 x 10 J. 1kJ Steam at C releases nearly ten times as much heat as boiling water. ; 404

22 12.70 Balance the reactions by inspection, starting with N. Then use Equation to calculate heats of reaction: H reaction = Σ coeff p H o f (products) Σ coeff r H o f (reactants) (a) N 2 + O 2 2 NO; H reaction = 2 mol(91. kj/mol) [1 mol(0 kj/mol) + 1 mol(0 kj/mol)] = kj; H mol N = kj = 91. kj/mol; 2 mol N (b) 2 N 2 O + O 2 4 NO; H reaction = 4 mol(91. kj/mol) [2 mol(81.6 kj/mol) + 1 mol(0 kj/mol)] = kj; kj H mol N = = 50.5 kj/mol; 4 mol N (c) 2 NO + O 2 2 NO 2 ; H reaction = 2 mol(.2 kj/mol) [2 mol(91. kj/mol) + 1(0)] = kj; kj H mol N = = 58.1 kj/mol; 2 mol N (d) 4 NO 2 + O 2 2 N 2 O 5 ; H reaction = 2 mol(1. kj/mol) [4 mol(.2 kj/mol) + 1 mol(0 kj/mol)] = kj; kj H mol N = = 26.6 kj/mol 4 mol N Compression work can be calculated using Equation 12-, w sys = P ext V sys. The external pressure is the pressure exerted by the compressor, 15.0 atm. The air is initially at 1.00 atm and occupies a volume that can be calculated using P f V f = P i V i : V i = P f V f (15.0 atm)(0.0 L) = = 450 L P i (1.00 atm) V = V f V i = 0.0 L 450 L = 420 L; w sys = P ext V sys = (15.0 atm)( 420 L) = 6.0 x 10 L atm. Convert to joules: w sys = (6.0 x J L atm) = 6.8 x 10 5 kg m 2 /s 2 = 68 kj. 1L atm Data in Problem indicate that glucose combustion releases 15.7 kj/g of energy. First determine how much glucose would be required at 100% efficiency, then correct for inefficiency and sugar content: 1g glucose 100% 100 g cereal Mass required = 220 kj = 1 g cereal 15.7 kj 0% 5 g glucose 405

23 12.7 (a)the energy required in a heating process is calculated from Equation 12-1: q = T nc : 2 m 10 cm n water = = (155 m 1.00 g ) = 8.60 x 10 6 mol; MM 1m 1cm g C = J/mol K, T = (0 20) = 10 C = 10 K; q = (8.60 x J mol) (10 K) = 6.48 x 10 9 J K (b) If heating is 80% efficient, the heat required is (6.48 x % J) = 8.1 x 10 9 J; 80% Given that the heat of combustion of methane is 80 kj/mol, the amount of methane required is (8.1 x kj J) = 1.0 x 10 4 mol; 1J 80 kj Multiply by MM to convert to grams: (1.0 x g mol) = 1.6 x 10 5 g methane First calculate the energy required to heat the water; then calculate the energy per mole of photons; finally determine how many photons are needed: q = n C T; m 10 ml 1.00 g n water = = 40.0 L MM = 2.22 x 10 mol; 1L 1mL g T = 50 C 25 C = 25 C = 25 K; q = (2.22 x J mol) (25 K) = 4.18 x 10 6 J; K Now determine the energy per mole of photons and compare N Ahc (6.022 x 10 /mol)(6.626 x 10 J s)(2.998 x 10 m/s) E photons = = -7 λ 5.15 x 10 m E photons = 2.2 x 10 5 J/mol; Photons required = 6 100% 4.18 x 10 J 2 mol 5 = 2.2 x 10 J 80% To work a problem involving heat transfers, it is useful to set up a block diagram illustrating the process. In this problem, a silver spoon absorbs energy from coffee (water): 406

24 Thus, q water = q Ag, q Ag = (nc Ag T) Ag, and q water = (nc H2 O T) water For Ag, m n = = 99 g = mol Ag; MM g C Ag = J/mol K, and T = (x 280) K; For water, m 1.00 g n = = 205 ml MM = 11.8 mol H 2 O; 1mL g C H2O = J/mol K, and T = (x 50) K; Substitute and solve for x: 25.51J J (0.918 mol) (x 280) K = (11.8 mol) (x 50) K; K K 2.27 x 6516 = x , x = 0690 x = 48 K. The final temperature of the coffee is 48 K. To determine whether an Al spoon would be more or less effective, compare nc for the two spoons J Al: = 99 g = 89 J/K; g K 25.51J Ag: = 99 g = 2 J/K; g K An Al spoon requires more energy per degree temperature increase, so an Al spoon would be more effective in cooling the coffee To determine the mass of methane required, first determine the heat required to convert the water into steam. The conversion of water at 25.0 C into steam at C takes place in two steps, heating to C and vaporization at that temperature. The total heat required is the sum of the heats for the two steps: q = nc T + n H vap : m 10 g n water = = 2.50 kg = 18.7 mol; MM 1kg g 407

25 75.291J 10 - kj q water = (18.7 mol)[ (75.0 K) K kj/mol] = 6.44 x 10 1J kj; q methane = - q water = x 10 kj The heat of combustion of methane could be calculated from Equation using data from Appendix D. However, the value is given in Problem 12.7: H comb = 80 kj/mol g Mass required = x 10 kj =129 g. -80 kj The energy required in a heating process is calculated from Equation 12-1: q = nc T. To determine q, we must first compute the amounts of N 2 and O 2 in the room, using the PV ideal gas equation: n = (assume no air escapes the room as it is heated): RT 2 10 cm 1L V = (.0 m)(5.0 m)(4.0 m) = 6.0 x 10 1m 10 cm 4 L; p N2 = (1 atm)(0.78) = 0.78 atm; p O2 = (1 atm)(0.22) = 0.22 atm; T = = 288 K; 4 (0.78atm)(6.0 x 10 L) n N 2 = = 1.98 x 10 mol L atm ( )(288 K) mol K 4 (0.22 atm)(6.0 x 10 L) n O 2 = = x 10 mol L atm ( mol K )(288 K) T = = 10 C = 10 K. q N2 = (1.98 x J mol) (10 K) = 5.77 x 10 5 J K q O2 = (0.559 x J mol) (10 K) = 1.64 x 10 5 J K q tot = q N2 + q O2 = 5.77 x 10 5 J+1.64 x 10 5 J = 7.4 x 10 2 kj. (Value only has two sig. figs. because the gas pressures are known to only two sig. figs.) To determine the amount of methane required, we need the heat of combustion of methane, which can be calculated from standard heats of formation using Equation using data from Appendix D. However, the value is given in Problem 12.7: H comb = 80 kj/mol. Divide the energy required by the heat of combustion to determine amount of methane required: (7.4 x 10 2 kj) = 0.92 mol. 80 kj 408

26 16.04 g Multiply by molar mass to convert to grams: 0.92 mol = 15 g CH 4 required The expression that always is correct for q is (e) q = E w, which is derived from the first law of thermodynamics. For the others: (a) E q when volume changes; (b) H q when pressure changes; (c) q v q when volume changes; and (d) q p q when pressure changes The process shows an expansion at constant temperature. As a gas expands, it does work on its surroundings, but if temperature remains constant, the energy of the gas does not change: (a) w sys is negative (system does work on surroundings); (b) E surr = 0 because E sys = 0 and total energy is conserved; (c) q sys is positive because w sys is negative and E sys = (a) Work done on the system compresses it, so the new figure should show a smaller volume. (b) An exothermic reaction increases the temperature of the system, leading to expansion and a larger volume: First calculate the heat required to warm the bear. Then calculate the heat of combustion of arachadonic acid. Finally, determine the amount of arachidonic acid required to do the job: 10 g J q = mc T = 500 kg o (25 C 5 C) = 4.18 x 10 1kg 1g C 7 J; H comb = [20 mol( 9.5 kj/mol) + 16 mol( kj/mol)] [1 mol( 66 kj/mol) + 27 mol(0 kj/mol)] H comb = 1.18 x 10 4 kj Mass required = 7 1kJ g 4.18 x 10 J 1.1 x 10 g 4 =. 10 J 1.18 x 10 kj First compute the quantity of gasoline consumed in traveling 1 km, then determine the energy content of that quantity of gasoline: 409

27 1L 10 ml 0.68 g Mass of gasoline = 1.0 km = 11 g 6.0 km 1L 1mL 48 kj Energy consumed = 11 g = 5.4 x 10 1g kj To work a problem involving heat transfers, it is useful to set up a block diagram illustrating the process. In this problem, a copper block transfers energy to water: Thus, q water = q Cu, q Cu = (nc T) Cu, and q water = (nc T) water m For Cu, n = = 9.50 g = mol Cu; MM g C = J/mol C, and T = (x 200.0) C; m 1.00 g For water, n = = 200 ml = mol water; MM 1mL g C = J/mol C, and T = (x 5.00) C; Substitute and solve for x: J J ( mol) o (x 200.0) C = (11.10 mol) o (x 5.00) C; C C.65 x 70.6 = 85.7 x , 89.4 x = 4910, x = 5.85 C This problem asks us to construct a graph that summarizes the phase changes of water. The first step is to determine the number of moles of water in 75.0 ml: 1.00 g 75.0 ml 4.16 mol 1mL g = Now determine the energy losses for the processes described: kj Energy lost during liquid cooling = 4.16 mol (5K) = 11 kj K 410

28 6.01kJ Energy lost during freezing = 4.16 mol = 25.0 kj kj Energy lost during ice cooling = 4.16 mol (25 K) =.9 kj K Using these values construct the graph: To determine the thermodynamic values first calculate the number of moles of the ammonia and the heat of vaporization: 0.81g moles NH = 275 ml = 1.1 mol 1mL g 2.2 kj q = n H vap = 1.1 mol = 04 kj Since this system is at constant pressure, H = q = 04 kj The only work on the system is expansion work. Thus, w = -P V = -nrt vap 8.14 kj w = -1.1 mol (240 K) = x 10 4 J or kj K Finally E = H - P V = H - nrt vap = 04kJ 26.1 kj = 2.8 x 10 2 kj. (density of NH is known to only two significant figures) A system should be defined in such a way that energy transfers can be conveniently specified. (a) Water is the system, and you are part of the surroundings. The water absorbs heat as it evaporates, and this heat must be provided by the surroundings (your body). (b) If our focus is on the beaker of water, it is the convenient choice for the system. The Bunsen burner and methane then are part of the surroundings. The beaker absorbs heat from the burning methane. (If our focus were on how a Bunsen burner operates, we would choose the methane as the system; then the system would release heat to the surroundings.) 411

29 (c) The acid and base solutions inside the thermos constitute a convenient system. This is a (nearly) isolated system, with minimum heat flow. The energy released by the reacting species remains in the system and causes the temperature rise of the mixed solutions The cooling process takes place at constant pressure, so H can be calculated using H = nc T. To determine E, use H = E + PV, from which E = H (PV): m n = = 1.25 g = 9.40 x 10 MM 12.9 g - mol; T = 20.0 K 50.0 K = 0.0 K; H cooling = (9.40 x J mol) ( 0.0 K) = 22.8 J; K 10 - L J (PV) = P V = (1.00 atm)( ml) = 2.6 J; 1mL 1L atm E cooling = 22.8 J ( 2.6 J) = 20.2 J The energy that is added to a gaseous monatomic substance goes entirely into translational energy of the atoms. The energy that is added to a gaseous diatomic substance goes partly into translational energy and partly into rotational energy of the molecules. Thus, a given amount of added energy increases the total translational energy (and the temperature) of a diatomic substance by a smaller amount, and the heat capacities of diatomic gases are larger than those of monatomic substances The total heat released in the reaction can be calculated from the temperature rise: q = n water C T; 1.00 g n = (200.0 ml) = mol; 1mL g T = = 6.8 C = 6.8 K; J 10 - kj q = ( mol) (6.8 K) K = 5.7 kj; 1J To determine the molar heat, divide the total heat by the number of moles reacting: q H molar = ; n acid 10 - L 1.00 mol n acid = MV = ml = mol; 1mL 1L 5.7 kj H molar = = 57 kj/mol mol This problem asks for the energy released from burning 1 gallon of ethanol. Begin by determining the balanced chemical equation using standard procedures: The unbalanced equation is: C 2 H 5 OH + O 2 CO 2 + H 2 O 412

30 Give CO 2 a coefficient of 2 to balance C, and H 2 O a coefficient of to balance H. C 2 H 5 OH + O 2 2 CO 2 + H 2 O This leaves 7 O atoms on the product side and 1 O atom in C 2 H 5 OH. Give O 2 a coefficient of to balance the O atoms and obtain the balanced equation: C 2 H 5 OH + O 2 2 CO 2 + H 2 O To determine the energy released by burning 1 gallon of ethanol we first need the reaction enthalpy. The reaction enthalpy can be obtained using the average bond energies in Table 9-2 or by using the heats of formation in Appendix D. We ll show both. Using bond energies: Bonds Broken Energy (kj/mol) Bonds Formed Energy kj/mol C-C 45 4 C=O 4(-800) 5 C-H 5(415) 6 H-O 6(-460) C-O 60 O-H 460 O=O (495) Total H reaction = 4725 kj/mol 5960 kj/mol = 125 kj/mol Using heats of formation: H reaction = [ mol( kj/mol) + 2 mol(-9.5 kj/mol)] [ mol(0 kj/mol) + 1 mol( kj/mol)] = 125 kj/mol Note that both methods yield the same heat of reaction. Now use stoichiometric calculations to determine the energy for 1 gallon of ethanol..785 L 10 ml g 1.25 x 10 kj 1 gallon = 7.98 x 10 1gallon 1L 1mL g 4 kj 41