English

not defined

no text concepts found

CHAPTER 6 ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.1 The system is the specific part of the universe that is of interest to us. The surroundings are the rest of the universe outside the system. An open system can exchange mass and energy, usually in the form of heat with its surroundings. A closed system allows the transfer of energy (heat) but not mass. An isolated system does not allow the transfer of either mass or energy. Thermal energy is the energy associated with the random motion of atoms and molecules. Chemical energy is stored within the structural units of chemical substances. Potential energy is energy available by virtue of an object’s position. Kinetic energy is the energy produced by a moving object. The law of conservation of energy states that the total quantity of energy in the universe is assumed constant. 6.2 Heat is the transfer of thermal energy between two bodies that are at different temperatures. Thermal energy is the energy associated with the random motion of atoms and molecules. 6.3 The units of energy commonly employed in chemistry are the Joule (J) and the kilojoule (kJ). 6.4 No, the kinetic energy of motion is converted to heat through the friction between the brakes, wheels, tires, and the road. 6.5 Turning on a flashlight converts chemical energy to electrical energy to electromagnetic energy, which includes visible light and infrared (heat) radiation. 6.6 (a) (b) (c) (d) 6.7 Thermochemistry is the study of heat change in chemical reactions. An exothermic process is any process that gives off heatthat is, transfers thermal energy to the surroundings. In an endothermic process, heat has to be supplied to the system by the surroundings. 6.8 The law of conservation of energy. 6.9 The combustion of methane and the freezing of water are exothermic processes. The decomposition of limestone (CaCO3) and the vaporization of water are endothermic processes. 6.10 To decompose a substance, bonds must be broken (an endothermic process). In combination reactions, bonds are formed (an exothermic process). 6.11 The first law of thermodynamics is based on the law of conservation of energy. The sign conventions for q and w are as follows: q is positive for an endothermic process and negative for an exothermic process, and w is positive for work done on the system by the surroundings and negative for work done by the system on the surroundings. Mechanical energy to potential energy to kinetic energy. Chemical energy to electrical energy to electromagnetic energy. Mechanical energy to potential energy to kinetic energy. Chemical energy to thermal energy 130 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.12 State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Energy and temperature are state functions. Work and heat are not state functions. 6.13 (a) Yes. (b) q > 0. Heat absorbed by the gas from the surroundings enables the gas to expand at constant temperature. (c) U = 0. Note that the internal energy of an ideal gas depends only on temperature. Because there is no change in temperature, there is no change in internal energy of the gas. 6.14 In (a) and (c), work is done by the system on the surroundings. In (b), work is done by the surroundings on the system. In (d), no work is done. 6.15 Recall that the work in gas expansion is equal to the product of the external, opposing pressure and the change in volume. (a) w PV w (0)(5.4 1.6)L 0 (b) w PV w (0.80 atm)(5.4 1.6)L 3.0 Latm To convert the answer to joules, we write w 3.0 L atm (c) 101.3 J 3.0 102 J 1 L atm w PV w (3.7 atm)(5.4 1.6)L 14 Latm To convert the answer to joules, we write w 14 L atm 6.16 (a) 101.3 J 1.4 103 J 1 L atm Because the external pressure is zero, no work is done in the expansion. w PV (0)(89.3 26.7)mL w 0 (b) The external, opposing pressure is 1.5 atm, so w PV (1.5 atm)(89.3 26.7)mL w 94 mL atm 0.001 L 0.094 L atm 1 mL To convert the answer to joules, we write: w 0.094 L atm 101.3 J 9.5 J 1 L atm CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS (c) 131 The external, opposing pressure is 2.8 atm, so w PV (2.8 atm)(89.3 26.7)mL w (1.8 102 mL atm) 0.001 L 0.18 L atm 1 mL To convert the answer to joules, we write: w 0.18 L atm 101.3 J 18 J 1 L atm 6.17 An expansion implies an increase in volume, therefore w must be 325 J (see the defining equation for pressure-volume work.) If the system absorbs heat, q must be 127 J. The change in energy (internal energy) is: U q w 127 J 325 J 198 J 6.18 Strategy: Compression is work done on the gas, so what is the sign for w? Heat is released by the gas to the surroundings. Is this an endothermic or exothermic process? What is the sign for q? Solution: To calculate the energy change of the gas (U), we need Equation (6.1) of the text. Work of compression is positive and because heat is given off by the gas, q is negative. Therefore, we have: U q w 26 J 74 J 48 J As a result, the energy of the gas increases by 48 J. 6.19 We first find the number of moles of hydrogen gas formed in the reaction: 50.0 g Sn 1 mol H 2 1 mol Sn 0.421 mol H 2 118.7 g Sn 1 mol Sn The next step is to find the volume occupied by the hydrogen gas under the given conditions. This is the change in volume. V nRT (0.421 mol)(0.0821 L atm / K mol)(298 K) 10.3 L H 2 P 1.00 atm The pressure-volume work done is then: w PV (1.00 atm)(10.3 L) 10.3 L atm 6.20 101.3 J 1.04 103 J 1 L atm Strategy: The work done in gas expansion is equal to the product of the external, opposing pressure and the change in volume. w PV We assume that the volume of liquid water is zero compared to that of steam. How do we calculate the volume of the steam? What is the conversion factor between Latm and J? 132 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS Solution: First, we need to calculate the volume that the water vapor will occupy (Vf). Using the ideal gas equation: VH 2O nH 2O RT P L atm (1 mol) 0.0821 (373 K) mol K = 31 L (1.0 atm) It is given that the volume occupied by liquid water is negligible. Therefore, V Vf Vi 31 L 0 L 31 L Now, we substitute P and V into Equation (6.3) of the text to solve for w. w PV (1.0 atm)(31 L) 31 Latm The problems asks for the work done in units of joules. The following conversion factor can be obtained from Appendix 1 of the text. 1 Latm 101.3 J Thus, we can write: w 31 L atm 101.3 J 3.1 103 J 1 L atm Check: Because this is gas expansion (work is done by the system on the surroundings), the work done has a negative sign. 6.21 Enthalpy is a thermodynamic quantity used to describe heat changes taking place at constant pressure. The enthalpy of reaction, H, is the difference between the enthalpies of the products and the enthalpies of the reactants. At constant pressure, the heat of reaction is equal to the enthalpy change of the same reaction. 6.22 The amount of heat absorbed or released in a reaction will depend on the physical states of each substance. 6.23 This is an exothermic process. When four moles of NH3(g) react with 5 moles of O2(g) to produce four moles of NO(g) and six moles of H2O(g), 904 kJ of thermal energy is released. 6.24 (a) 2905.6 kJ 6.25 The equation as written shows that 805.6 kJ of heat is released when two moles of CuS react. We want to calculate the amount of heat released when 1 g of CuS reacts. (c) 1276.8 kJ (b) +1452.8 kJ The heat evolved per gram of CuS roasted is: heat evolved = 6.26 805.6 kJ 1 mol CuS 4.213 kJ / g CuS 2 mol CuS 95.62 g CuS Strategy: The thermochemical equation shows that for every 2 moles of NO2 produced, 114.6 kJ of heat are given off (note the negative sign). We can write a conversion factor from this information. 114.6 kJ 2 mol NO2 4 How many moles of NO2 are in 1.26 10 g of NO2? What conversion factor is needed to convert between grams and moles? CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 133 4 Solution: We need to first calculate the number of moles of NO2 in 1.26 10 g of the compound. Then, we can convert to the number of kilojoules produced from the exothermic reaction. The sequence of conversions is: grams of NO2 moles of NO2 kilojoules of heat generated Therefore, the heat change is: (1.26 104 g NO2 ) 1 mol NO2 114.6 kJ 1.57 104 kJ 46.01 g NO 2 2 mol NO2 4 This is an exothermic reaction. The amount of heat given off is 1.57 × 10 kJ. 6.27 We can calculate U using Equation (6.10) of the text. U H − RTn We initially have 2.0 moles of gas. Since our products are 2.0 moles of H2 and 1.0 mole of O2, there is a net gain of 1 mole of gas (2 reactant 3 product). Thus, n 1. Looking at the equation given in the problem, it requires 483.6 kJ to decompose 2.0 moles of water (H 483.6 kJ). Substituting into the above equation: 3 U 483.6 10 J/mol − (8.314 J/molK)(398 K)(1) 5 2 U 4.80 10 J/mol 4.80 10 kJ/mol 6.28 We initially have 6 moles of gas (3 moles of chlorine and 3 moles of hydrogen). Since our product is 6 moles of hydrogen chloride, there is no change in the number of moles of gas. Therefore there is no volume change; V 0. w PV (1 atm)(0 L) 0 U H PV PV 0, so U H H 3H rxn 3(184.6 kJ/mol) 553.8 kJ/mol We need to multiply H rxn by three, because the question involves the formation of 6 moles of HCl; whereas, the equation as written only produces 2 moles of HCl. U H 553.8 kJ/mol 6.29 The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius. Units of specific heat are J/gC, and units of heat capacity are J/C. Specific heat is an intensive property; whereas, heat capacity is an extensive property. 6.30 Calorimetry is the measurement of heat changes. Constant-volume and constant-pressure are two types of calorimetry. We need to know the amount of heat absorbed per degree Celsius change (heat capacity) of the calorimeter so that the heat change of the reaction can be determined. The heat capacity of the calorimeter is determined by calibrating it by burning a substance with an accurately know heat of combustion. 134 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.31 Choice (d) will take place when the two metals are brought into contact. Heat will flow from Cu to Al because Cu is at a higher temperature. The definition of heat is the transfer of thermal energy between two bodies that are at different temperatures. 6.32 Specific heat of a substance is the amount of heat required to raise the temperature of one gram of the substance one degree Celsius. Comparing two substances of equal mass, the substance with a larger specific heat will require that more heat be applied to raise its temperature a given amount. Therefore, under the same heating conditions, it will take longer for metal A to reach a temperature of 21°C. 6.33 Specific heat 6.34 q mCusCut (6.22 10 g)(0.385 J/gC)(324.3C 20.5C) 7.28 10 J 728 kJ 6.35 See Table 6.2 of the text for the specific heat of Hg. C 85.7 J/ C 0.237 J / g C m 362 g 3 5 3 q mst (366 g)(0.139 J/g·C)(12.0 77.0)C 3.31 10 J 3.31 kJ 6.36 Strategy: We know the masses of gold and iron as well as the initial temperatures of each. We can look up the specific heats of gold and iron in Table 6.2 of the text. Assuming no heat is lost to the surroundings, we can equate the heat lost by the iron sheet to the heat gained by the gold sheet. With this information, we can solve for the final temperature of the combined metals. Solution: Treating the calorimeter as an isolated system (no heat lost to the surroundings), we can write: or qAu qFe 0 qAu qFe The heat gained by the gold sheet is given by: qAu mAusAut (10.0 g)(0.129 J/gC)(tf 18.0)C where m and s are the mass and specific heat, and t tfinal tinitial. The heat lost by the iron sheet is given by: qFe mFesFet (20.0 g)(0.444 J/gC)(tf 55.6)C Substituting into the equation derived above, we can solve for tf. qAu qFe (10.0 g)(0.129 J/gC)(tf 18.0)C (20.0 g)(0.444 J/gC)(tf 55.6)C 1.29 tf 23.2 8.88 tf 494 10.2 tf 517 tf 50.7C Check: Must the final temperature be between the two starting values? CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.37 135 The heat gained by the calorimeter is: q Cpt q (3024 J/C)(1.126C) 3.405 103 J The amount of heat given off by burning Mg in kJ/g is: (3.405 103 J) 1 kJ 1 24.76 kJ/g Mg 1000 J 0.1375 g Mg The amount of heat given off by burning Mg in kJ/mol is: 24.76 kJ 24.31 g Mg 601.9 kJ/mol Mg 1 g Mg 1 mol Mg If the reaction were endothermic, what would happen to the temperature of the calorimeter and the water? 6.38 Strategy: The neutralization reaction is exothermic. 56.2 kJ of heat are released when 1 mole of H reacts with 1 mole of OH . Assuming no heat is lost to the surroundings, we can equate the heat lost by the reaction to the heat gained by the combined solution. How do we calculate the heat released during the reaction? Are we reacting 1 mole of H with 1 mole of OH ? How do we calculate the heat absorbed by the combined solution? Solution: Assuming no heat is lost to the surroundings, we can write: or qsoln qrxn 0 qsoln qrxn First, let's set up how we would calculate the heat gained by the solution, qsoln msolnssolnt where m and s are the mass and specific heat of the solution and t tf ti. We assume that the specific heat of the solution is the same as the specific heat of water, and we assume that the density of the solution is the same as the density of water (1.00 g/mL). Since the density is 1.00 g/mL, the mass of 400 mL of solution (200 mL 200 mL) is 400 g. Substituting into the equation above, the heat gained by the solution can be represented as: 2 qsoln (4.00 10 g)(4.184 J/gC)(tf 20.48C) Next, let's calculate qrxn, the heat released when 200 mL of 0.862 M HCl are mixed with 200 mL of 0.431 M Ba(OH)2. The equation for the neutralization is: 2HCl(aq) Ba(OH)2(aq) 2H2O(l) BaCl2(aq) There is exactly enough Ba(OH)2 to neutralize all the HCl. Note that 2 mole HCl 1 mole Ba(OH)2, and that the concentration of HCl is double the concentration of Ba(OH)2. The number of moles of HCl is: (2.00 102 mL) 0.862 mol HCl 0.172 mol HCl 1000 mL 136 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS The amount of heat released when 1 mole of H is reacted is given in the problem (56.2 kJ/mol). The amount of heat liberated when 0.172 mole of H is reacted is: qrxn 0.172 mol 56.2 103 J 9.67 103 J 1 mol Finally, knowing that the heat lost by the reaction equals the heat gained by the solution, we can solve for the final temperature of the mixed solution. qsoln qrxn 2 3 (4.00 10 g)(4.184 J/gC)(tf 20.48C) (9.67 10 J) 3 4 3 (1.67 10 )tf (3.43 10 ) 9.67 10 J tf 26.3C 6.39 Substances are said to be in the standard state at 1 atmosphere of pressure. 6.40 By convention, the standard enthalpy of formation of any element in its most stable form is zero. The standard enthalpy of formation of a compound is the heat change when 1 mole of the compound is formed from its elements at a pressure of 1 atm. 6.41 The standard enthalpy of reaction, H rxn , is defined as the enthalpy of a reaction carried out at 1 atmosphere pressure. 6.42 H rxn = nHf(products) mHf(reactants), where m and n denote the stoichiometric coefficients (in moles) for the reactants and products, Hf is the standard enthalpy of formation, and (sigma) means “the sum of.” 6.43 Hess’s law can be stated as follows: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. The law enables us to determine the standard enthalpy of formation of a compound from its elements by an indirect route when direct combination of the elements is not feasible. Using Hess’s law, the standard enthalpy change for the reaction of interest can be easily calculated. 6.44 See the example for CH4 given in Section 6.6 of the text. To determine the standard enthalpy of formation of methane, the heat of combustion of methane along with the heats of combustion of carbon and hydrogen are used. 6.45 CH4(g) and H(g). All the other choices are elements in their most stable form ( H f 0 ). The most stable form of hydrogen is H2(g). 6.46 The standard enthalpy of formation of any element in its most stable form is zero. Therefore, since H f (O2 ) 0, O2 is the more stable form of the element oxygen at this temperature. CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.47 H2O(l) H2O(g) 137 Endothermic H rxn H f [H 2 O( g )] H f [H 2 O(l )] 0 H f [H 2O( l )] is more negative since H rxn 0. You could also solve the problem by realizing that H2O(l) is the stable form of water at 25C, and therefore will have the more negative H f value. 6.48 (a) Br2(l) is the most stable form of bromine at 25C; therefore, H f [Br2 (l )] 0. Since Br2(g) is less stable than Br2(l), H f [Br2 ( g )] 0. (b) I2(s) is the most stable form of iodine at 25C; therefore, H f [I2 ( s )] 0. Since I2(g) is less stable than I2(s), H f [I2 ( g )] 0. 6.49 2H2O2(l) 2H2O(l) O2(g) H2O2(l) has a tendency to decompose because H2O(l) has a more negative H f than H2O2(l). 6.50 Strategy: What is the reaction for the formation of Ag2O from its elements? What is the Hf value for an element in its standard state? Solution: The balanced equation showing the formation of Ag2O(s) from its elements is: 2Ag(s) 1 2 O2(g) Ag2O(s) Knowing that the standard enthalpy of formation of any element in its most stable form is zero, and using Equation (6.18) of the text, we write: H rxn nH f (products) mH f (reactants) H rxn [H f (Ag 2 O)] [2H f (Ag) 1 H (O )] f 2 2 H rxn [H f (Ag 2 O)] [0 0] H f (Ag 2O) H rxn In a similar manner, you should be able to show that H f (CaCl 2 ) H rxn for the reaction Ca(s) Cl2(g) CaCl2(s) 6.51 H [H f (CaO) H f (CO2 )] H f (CaCO3 ) H [(1)(635.6 kJ/mol) (1)(393.5 kJ/mol)] (1)(1206.9 kJ/mol) 177.8 kJ/mol 138 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.52 Strategy: The enthalpy of a reaction is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. The enthalpy of each species (reactant or product) is given by the product of the stoichiometric coefficient and the standard enthalpy of formation, Hf , of the species. Solution: We use the Hf values in Appendix 2 and Equation (6.18) of the text. H rxn nH f (products) mH f (reactants) (a) HCl(g) H (aq) Cl (aq) H rxn H f (H ) H f (Cl ) H f (HCl) 74.9 kJ/mol 0 H f (Cl ) (1)(92.3 kJ/mol) H f (Cl ) 167.2 kJ/mol (b) The neutralization reaction is: and, H (aq) OH (aq) H2O(l) H rxn H f [H 2 O(l )] [H f (H ) H f (OH )] H f [H 2 O(l )] 285.8 kJ/mol (See Appendix 2 of the text.) H rxn (1)(285.8 kJ/mol) [(1)(0 kJ/mol) (1)(229.6 kJ/mol)] 56.2 kJ/mol 6.53 (a) H 2H f (H 2 O) 2H f (H 2 ) H f (O 2 ) H (2)(285.8 kJ/mol) (2)(0) (1)(0) 571.6 kJ/mol (b) H 4H f (CO 2 ) 2H f (H 2 O) 2H f (C2 H 2 ) 5H f (O 2 ) H (4)(393.5 kJ/mol) (2)(285.8 kJ/mol) (2)(226.6 kJ/mol) (5)(0) 2599 kJ/mol 6.54 (a) H = [2H f (CO2 ) 2H f (H 2 O)] [H f (C2 H 4 ) 3H f (O2 )] ] H [(2)(393.5 kJ/mol) (2)(285.8 kJ/mol)] [(1)(52.3 kJ/mol) (3)(0)] H 1410.9 kJ/mol (b) H = [2H f (H 2 O) 2H f (SO2 )] [2H f (H 2S) 3H f (O 2 )] H [(2)(285.8 kJ/mol) (2)(296.1 kJ/mol)] [(2)(20.15 kJ/mol) (3)(0)] H 1123.5 kJ/mol 6.55 The given enthalpies are in units of kJ/g. We must convert them to units of kJ/mol. (a) 22.6 kJ 32.04 g 724 kJ/mol 1g 1 mol (b) 29.7 kJ 46.07 g 1.37 103 kJ/mol 1g 1 mol CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS (c) 6.56 33.4 kJ 60.09 g 2.01 103 kJ/mol 1g 1 mol H rxn nH f (products) mH f (reactants) The reaction is: H2(g) H(g) H(g) and, H rxn [H f (H) H f (H)] H f (H 2 ) H f (H 2 ) 0 H rxn 436.4 kJ/mol 2H f (H) (1)(0) H f (H) 6.57 436.4 kJ/mol 218.2 kJ/mol 2 H 6H f (CO 2 ) 6H f (H 2 O) [H f (C6 H12 ) 9H f (O 2 )] H (6)(393.5 kJ/mol) (6)(285.8 kJ/mol) (1)(151.9 kJ/mol) (l)(0) 3924 kJ/mol Why is the standard heat of formation of oxygen zero? 6.58 The equation as written shows that 879 kJ of heat is released when two moles of ZnS react. We want to calculate the amount of heat released when 1 g of ZnS reacts. Let H be the heat change per gram of ZnS roasted. We write: H 879 kJ 1 mol ZnS 4.51 kJ/g ZnS 2 mol ZnS 97.46 g ZnS This is an exothermic reaction. The amount of heat evolved per gram of ZnS roasted is 4.51 kJ/g ZnS. 6.59 4 This is an exothermic reaction. The amount of heat given off when 1.26 × 10 g of NH3 are produced is: (1.26 104 g NH3 ) 6.60 1 mol NH3 92.6 kJ 3.43 104 kJ 17.03 g NH3 2 mol NH3 H rxn nH f (products) mH f (reactants) The balanced equation for the reaction is: CaCO3(s) CaO(s) CO2(g) H rxn [H f (CaO) H f (CO 2 )] H f (CaCO3 ) H rxn [(1)(635.6 kJ/mol) (1)(393.5 kJ/mol)] (1)(1206.9 kJ/mol) 177.8 kJ/mol 139 140 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS The enthalpy change calculated above is the enthalpy change if 1 mole of CO2 is produced. The problem asks for the enthalpy change if 66.8 g of CO2 are produced. We need to use the molar mass of CO2 as a conversion factor. H 66.8 g CO 2 6.61 1 mol CO 2 177.8 kJ 2.70 102 kJ 44.01 g CO 2 1 mol CO 2 H (kJ/mol) Reaction S(rhombic) O2(g) SO2(g) SO2(g) S(monoclinic) O2(g) S(rhombic) S(monoclinic) 296.06 296.36 H rxn 0.30 kJ/mol Which is the more stable allotropic form of sulfur? 6.62 Strategy: Our goal is to calculate the enthalpy change for the formation of C2H6 from is elements C and H2. This reaction does not occur directly, however, so we must use an indirect route using the information given in the three equations, which we will call equations (a), (b), and (c). Solution: Here is the equation for the formation of C2H6 from its elements. H rxn ? C2H6(g) 2C(graphite) 3H2(g) Looking at this reaction, we need two moles of graphite as a reactant. So, we multiply Equation (a) by two to obtain: (d) H rxn 2(393.5 kJ/mol) 787.0 kJ/mol 2C(graphite) 2O2(g) 2CO2(g) Next, we need three moles of H2 as a reactant. So, we multiply Equation (b) by three to obtain: (e) 3H2(g) 3 2 H rxn 3(285.8 kJ/mol) 857.4 kJ/mol O2(g) 3H2O(l) Last, we need one mole of C2H6 as a product. Equation (c) has two moles of C2H6 as a reactant, so we need to reverse the equation and divide it by 2. (f) 2CO2(g) 3H2O(l) C2H6(g) 7 2 O2(g) H rxn 1 (3119.6 2 kJ/mol) 1559.8 kJ/mol Adding Equations (d), (e), and (f) together, we have: H (kJ/mol) Reaction (d) 2C(graphite) 2O2(g) 2CO2(g) (e) 3H2(g) (f) C2H6(g) 2CO2(g) 3H2O(l) 3 2 787.0 O2(g) 3H2O(l) 857.4 7 2 2C(graphite) 3H2(g) C2H6(g) O2(g) 1559.8 H 84.6 kJ/mol CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.63 141 H (kJ/mol) Reaction CO2(g) 2H2O(l) CH3OH(l) 3 2 O2(g) 726.4 C(graphite) O2(g) CO2(g) 2H2(g) O2(g) 2H2O(l) C(graphite) 2H2(g) 1 2 393.5 2(285.8) H rxn 238.7 kJ/mol O2(g) CH3OH(l) . In this case, the We have just calculated an enthalpy at standard conditions, which we abbreviate H rxn reaction in question was for the formation of one mole of CH3OH from its elements in their standard state. that we calculated is also, by definition, the standard heat of formation H f of Therefore, the H rxn CH3OH (238.7 kJ/mol). 6.64 The second and third equations can be combined to give the first equation. 2Al(s) 3 2 O2(g ) Al2O3(s) 3 2 2Fe(s) Fe2O3(s) H 1669.8 kJ/mol H 822.2 kJ/mol O2(g) 2Al(s) Fe2O3(s) 2Fe(s) Al2O3(s) H 847.6 kJ/mol 6.65 In a chemical reaction the same elements and the same numbers of atoms are always on both sides of the equation. This provides a consistent reference which allows the energy change in the reaction to be interpreted in terms of the chemical or physical changes that have occurred. In a nuclear reaction the same elements are not always on both sides of the equation and no common reference point exists. 6.66 Rearrange the equations as necessary so they can be added to yield the desired equation. 6.67 2B A H1 C A H2 2B C H H2 H1 The reaction corresponding to standard enthalpy of formation, H f , of AgNO2(s) is: Ag(s) 1 2 N2(g) O2(g) AgNO2(s) Rather than measuring the enthalpy directly, we can use the enthalpy of formation of AgNO3(s) and the H rxn provided. AgNO3(s) AgNO2(s) H rxn H f (AgNO2 ) 1 2 O2(g) 1 H (O ) f 2 2 H f (AgNO3 ) 78.67 kJ/mol H f (AgNO2) 0 (123.02 kJ/mol) H f (AgNO 2 ) 44.35 kJ/mol 142 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.68 (a) H rxn nH f (products) mH f (reactants) H rxn [4H f (NH3 ) H f (N 2 )] 3H f (N 2 H 4 ) H rxn [(4)(46.3 kJ/mol) (0)] (3)(50.42 kJ/mol) 336.5 kJ/mol (b) The balanced equations are: (1) N2H4(l) O2(g) N2(g) 2H2O(l) (2) 4NH3(g) 3O2(g) 2N2(g) 6H2O(l) The standard enthalpy change for equation (1) is: H rxn H f (N 2 ) 2H f [H 2 O(l )] {H f [N 2 H 4 (l )] H f (O2 )} ΔH rxn [(1)(0) (2)(285.8 kJ/mol)] [(1)(50.42 kJ/mol) (1)(0)] 622.0 kJ/mol The standard enthalpy change for equation (2) is: H rxn [2H f (N 2 ) 6H f (H 2 O)] [4H f (NH3 ) 3H f (O2 )] ΔH rxn [(2)(0) (6)(285.8 kJ/mol)] [(4)(46.3 kJ/mol) (3)(0)] 1529.6 kJ/mol We can now calculate the enthalpy change per kilogram of each substance. H rxn above is in units of kJ/mol. We need to convert to kJ/kg. N 2 H 4 (l ): H rxn 1 mol N 2 H 4 622.0 kJ 1000 g 1.941 104 kJ / kg N 2 H 4 1 mol N 2 H 4 32.05 g N 2 H 4 1 kg NH3 ( g ): H rxn 1 mol NH3 1529.6 kJ 1000 g 2.245 104 kJ / kg NH3 4 mol NH3 17.03 g NH3 1 kg Since ammonia, NH3, releases more energy per kilogram of substance, it would be a better fuel. 6.69 We initially have 8 moles of gas (2 of nitrogen and 6 of hydrogen). Since our product is 4 moles of ammonia, there is a net loss of 4 moles of gas (8 reactant 4 product). The corresponding volume loss is V nRT (4.0 mol)(0.0821 L atm / K mol)(298 K) 98 L P 1 atm w PV (1 atm)(98 L) 98 L atm H U PV or 101.3 J 9.9 103 J 9.9 kJ 1 L atm U H PV Using H as 185.2 kJ (2 92.6 kJ), (because the question involves the formation of 4 moles of ammonia, not 2 moles of ammonia for which the standard enthalpy is given in the question), and PV as 9.9 kJ (for which we just solved): U 185.2 kJ 9.9 kJ 175.3 kJ CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.70 143 The reaction is, 2Na(s) Cl2(g) 2NaCl(s). First, let's calculate H for this reaction using H f values in Appendix 2. H rxn 2H f (NaCl) [2H f (Na) H f (Cl2 )] H rxn 2(411.0 kJ/mol) [2(0) 0] 822.0 kJ/mol This is the amount of heat released when 1 mole of Cl2 reacts (see balanced equation). We are not reacting 1 mole of Cl2, however. From the volume and density of Cl2, we can calculate grams of Cl2. Then, using the molar mass of Cl2 as a conversion factor, we can calculate moles of Cl2. Combining these two calculations into one step, we find moles of Cl2 to be: 2.00 L Cl2 1.88 g Cl2 1 mol Cl2 0.0530 mol Cl2 1 L Cl2 70.90 g Cl2 Finally, we can use the H rxn calculated above to find the heat change when 0.0530 mole of Cl2 reacts. 0.0530 mol Cl2 822.0 kJ 43.6 kJ 1 mol Cl2 This reaction is exothermic. The amount of heat released is 43.6 kJ. 6.71 (a) Although we cannot measure H rxn for this reaction, the reverse process, is the combustion of glucose. We could easily measure H rxn for this combustion in a bomb calorimeter. C6H12O6(s) 6O2(g) 6CO2(g) 6H2O(l) (b) We can calculate H rxn using standard enthalpies of formation. H rxn H f [C6 H12 O6 ( s )] 6H f [O 2 ( g )] {6H f [CO 2 ( g )] 6H f [H 2 O(l )]} H rxn [(1)(1274.5 kJ/mol) 0] [(6)(393.5 kJ/mol) (6)(285.8 kJ/mol)] 2801.3 kJ/mol H rxn has units of kJ/1 mol glucose. We want the H change for 7.0 10 calculate how many moles of glucose are in 7.0 10 following strategy to solve the problem. 14 14 kg glucose. We need to kg glucose. You should come up with the kg glucose g glucose mol glucose kJ (H) H (7.0 1014 kg) 1 mol C6 H12 O6 1000 g 2801.3 kJ 1.1 1019 kJ 1 kg 180.16 g C6 H12 O6 1 mol C6 H12 O6 6.72 The initial and final states of this system are identical. Since enthalpy is a state function, its value depends only upon the state of the system. The enthalpy change is zero. 6.73 From the balanced equation we see that there is a 1:2 mole ratio between hydrogen and sodium. The number of moles of hydrogen produced is: 0.34 g Na 1 mol H 2 1 mol Na 7.4 103 mol H 2 22.99 g Na 2 mol Na 144 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS Using the ideal gas equation, we write: V nRT (7.4 103 mol)(0.0821 L atm / K mol)(273 K) 0.17 L H 2 P (1 atm) V 0.17 L w PV (1.0 atm)(0.17 L) 0.17 L atm 6.74 H(g) Br(g) HBr(g) 101.3 J 17 J 1 L atm H rxn ? Rearrange the equations as necessary so they can be added to yield the desired equation. H(g) Br(g) 1 2 H2(g) 1 2 1 2 1 2 H2(g) Br2(g) Br2(g) HBr(g) H(g) Br(g) HBr(g) 6.75 H rxn 1 ( 436.4 2 kJ/mol) 218.2 kJ/mol H rxn 1 (192.5 2 kJ/mol) 96.25 kJ/mol H rxn 1 (72.4 2 kJ/mol) 36.2 kJ/mol H 350.7 kJ/mol Using the balanced equation, we can write: H rxn [2H f (CO 2 ) 4H f (H 2 O)] [2H f (CH3OH) 3H f (O 2 )] 1452.8 kJ/mol (2)(393.5 kJ/mol) (4)(285.8 kJ/mol) (2) H f (CH3OH) (3)(0 kJ/mol) 477.4 kJ/mol (2) H f (CH3OH) H f (CH3OH) 238.7 kJ/mol 6.76 qsystem 0 qmetal qwater qcalorimeter qmetal qwater qcalorimeter 0 mmetalsmetal(tfinal tinitial) mwaterswater(tfinal tinitial) Ccalorimeter(tfinal tinitial) 0 All the needed values are given in the problem. All you need to do is plug in the values and solve for smetal. (44.0 g)(smetal)(28.4 99.0)C (80.0 g)(4.184 J/gC)(28.4 24.0)C (12.4 J/C)(28.4 24.0)C 0 3 3 (3.11 10 )smetal (gC) 1.53 10 J smetal 0.492 J/gC 6.77 The original volume of ammonia is: (a) V nRT (1.00 mol)(0.0821 L atm / K mol)(298 K) 1.75 L NH3 P 14.0 atm T2 P2V2T1 (1 atm)(23.5 L)(298 K) 286 K (14.0 atm)(1.75 L) PV 1 1 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS (b) 145 t (286 298)C 12C q mst (17.03 g)(0.0258 J/gC)(12C) 5.27 J w PV (1 atm)(23.5 1.75)L 21.75 L atm 3 101.3 J 2.20 103 J 1 L atm 3 U q w 5.27 J (2.20 10 J) 2.21 10 J 2.21 kJ 6.78 A good starting point would be to calculate the standard enthalpy for both reactions. Calculate the standard enthalpy for the reaction: C(s) 1 2 O2(g) CO(g) This reaction corresponds to the standard enthalpy of formation of CO, so we use the value of 110.5 kJ/mol (see Appendix 2 of the text). Calculate the standard enthalpy for the reaction: CO(g) H2(g) C(s) H2O(g) H rxn [H f (CO) H f (H 2 )] [H f (C) H f (H 2 O)] H rxn [(1)(110.5 kJ/mol) (1)(0)] [(1)(0) (1)(241.8 kJ/mol)] 131.3 kJ/mol The first reaction, which is exothermic, can be used to promote the second reaction, which is endothermic. Thus, the two gases are produced alternately. 6.79 As energy consumers, we are interested in the availability of usable energy. 6.80 First, calculate the energy produced by 1 mole of octane, C8H18. C8H18(l) 25 2 O2(g) 8CO2(g) 9H2O(l) H rxn 8H f (CO2 ) 9H f [H 2 O(l )] [H f (C8 H18 ) 25 H (O )] f 2 2 H rxn [(8)(393.5 kJ/mol) (9)(285.8 kJ/mol)] [(1)(249.9 kJ/mol) ( 25 )(0)] 2 5470 kJ/mol The problem asks for the energy produced by the combustion of 1 gallon of octane. H rxn above has units of kJ/mol octane. We need to convert from kJ/mol octane to kJ/gallon octane. The heat of combustion for 1 gallon of octane is: H 5470 kJ 1 mol octane 2660 g 1.274 105 kJ / gal 1 mol octane 114.22 g octane 1 gal The combustion of hydrogen corresponds to the standard heat of formation of water: H2(g) 1 2 O2(g) H2O(l) Thus, H rxn is the same as H f for H2O(l), which has a value of 285.8 kJ/mol. The number of moles of 5 hydrogen required to produce 1.274 10 kJ of heat is: nH 2 (1.274 105 kJ) 1 mol H 2 445.8 mol H 2 285.8 kJ 146 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS Finally, use the ideal gas law to calculate the volume of gas corresponding to 445.8 moles of H2 at 25C and 1 atm. L atm (445.8 mol) 0.0821 (298 K) nH 2 RT mol K 1.09 104 L = VH 2 (1 atm) P That is, the volume of hydrogen that is energy-equivalent to 1 gallon of gasoline is over 10,000 liters at 1 atm and 25C! 6.81 C2H6(l) The combustion reaction is: 7 2 O2(g) 2CO2(g) 3H2O(l) The heat released during the combustion of 1 mole of ethane is: H rxn [2H f (CO2 ) 3H f (H 2 O)] [H f (C2 H6 ) 7 H (O )] f 2 2 H rxn [(2)(393.5 kJ/mol) (3)(285.8 kJ/mol)] [(1)(84.7 kJ/mol ( 72 )(0)] 1559.7 kJ/mol The heat required to raise the temperature of the water to 98C is: 5 q mH2O sH 2O t (855 g)(4.184 J/gC)(98.0 25.0)C 2.61 10 J 261 kJ The combustion of 1 mole of ethane produces 1559.7 kJ; the number of moles required to produce 261 kJ is: 261 kJ 1 mol ethane 0.167 mol ethane 1559.7 kJ The volume of ethane is: nRT = Vethane P 6.82 L atm (0.167 mol) 0.0821 (296 K) mol K 4.10 L 1 atm 752 mmHg 760 mmHg The heat gained by the liquid nitrogen must be equal to the heat lost by the water. qN 2 qH 2O If we can calculate the heat lost by the water, we can calculate the heat gained by 60.0 g of the nitrogen. Heat lost by the water qH2O = mH 2O sH2O t 2 4 q H 2O (2.00 10 g)(4.184 J/gC)(41.0 55.3)C 1.20 10 J The heat gained by 60.0 g nitrogen is the opposite sign of the heat lost by the water. qN 2 qH 2O qN 2 1.20 104 J The problem asks for the molar heat of vaporization of liquid nitrogen. Above, we calculated the amount of heat necessary to vaporize 60.0 g of liquid nitrogen. We need to convert from J/60.0 g N2 to J/mol N2. H vap 1.20 104 J 28.02 g N 2 5.60 103 J/mol 5.60 kJ/mol 60.0 g N 2 1 mol N 2 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.83 147 The reaction is: 2CO 2NO 2CO2 N2 The limiting reagent is CO (NO is in excess). H [2H f (CO2 ) H f (N 2 )] [2H f (CO) 2H f (NO)] H [(2)(393.5 kJ/mol) (1)(0)] [(2)(110.5 kJ/mol) (2)(90.4 kJ/mol)] 747 kJ/mol 6.84 Recall that the standard enthalpy of formation ( Hf ) is defined as the heat change that results when 1 mole of a compound is formed from its elements at a pressure of 1 atm. Only in choice (a) does H rxn H f . In choice (b), C(diamond) is not the most stable form of elemental carbon under standard conditions; C(graphite) is the most stable form. 6.85 (a) No work is done by a gas expanding in a vacuum, because the pressure exerted on the gas is zero. (b) w PV w (0.20 atm)(0.50 0.050)L 0.090 Latm Converting to units of joules: w 0.090 L atm (c) 101.3 J 9.1 J L atm The gas will expand until the pressure is the same as the applied pressure of 0.20 atm. We can calculate its final volume using the ideal gas equation. nRT V P L atm (0.020 mol) 0.0821 (273 20)K K mol 2.4 L 0.20 atm The amount of work done is: w PV (0.20 atm)(2.4 0.050)L 0.47 Latm Converting to units of joules: w 0.47 L atm 6.86 (a) 101.3 J 48 J L atm The more closely packed, the greater the mass of food. Heat capacity depends on both the mass and specific heat. C ms The heat capacity of the food is greater than the heat capacity of air; hence, the cold in the freezer will be retained longer. (b) Tea and coffee are mostly water; whereas, soup might contain vegetables and meat. Water has a higher heat capacity than the other ingredients in soup; therefore, coffee and tea retain heat longer than soup. 148 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.87 The balanced equation is: C6H12O6(s) 2C2H5OH(l) 2CO2(g) H rxn [2H f (C2 H5 OH) 2H f (CO 2 )] H f (C6 H12 O6 ) H rxn [(2)(276.98 kJ/mol) (2)(393.5 kJ/mol)] (1)(1274.5 kJ/mol) 66.5 kJ/mol 6.88 4Fe(s) 3O2(g) 2Fe2O3(s). This equation represents twice the standard enthalpy of formation of Fe2O3. From Appendix 2, the standard enthalpy of formation of Fe2O3 822.2 kJ/mol. So, H for the given reaction is: H rxn (2)(822.2 kJ/mol) 1644 kJ/mol Looking at the balanced equation, this is the amount of heat released when four moles of Fe react. But, we are reacting 250 g of Fe, not 4 moles. We can convert from grams of Fe to moles of Fe, then use H as a conversion factor to convert to kJ. 250 g Fe 1 mol Fe 1644 kJ 1.84 103 kJ 55.85 g Fe 4 mol Fe 3 The amount of heat produced by this reaction is 1.84 × 10 kJ. 6.89 One conversion factor needed to solve this problem is the molar mass of water. The other conversion factor is given in the problem. It takes 44.0 kJ of energy to vaporize 1 mole of water. 1 mol H 2 O 44.0 kJ You should come up with the following strategy to solve the problem. 4000 kJ mol H2O g H2O ? g H 2O 4000 kJ 6.90 1 mol H 2 O 18.02 g H 2 O = 1.64 103 g H 2O 44.0 kJ 1 mol H 2 O The heat required to raise the temperature of 1 liter of water by 1C is: 4.184 J 1g 1000 mL 1C 4184 J/L g C 1 mL 1L Next, convert the volume of the Pacific Ocean to liters. 3 3 1000 m 100 cm 1L (7.2 10 km ) 7.2 1020 L 3 1 km 1 m 1000 cm 8 3 The amount of heat needed to raise the temperature of 7.2 10 (7.2 1020 L) 4184 J 3.0 1024 J 1L 20 L of water is: CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 149 Finally, we can calculate the number of atomic bombs needed to produce this much heat. (3.0 1024 J) 6.91 1 atomic bomb 1.0 1015 J 3.0 109 atomic bombs 3.0 billion atomic bombs First calculate the final volume of CO2 gas: 1 mol 19.2 g (0.0821 L atm / K mol)(295 K) 44.01 g nRT V = = = 10.6 L 0.995 atm P w PV (0.995 atm)(10.6 L) 10.5 Latm w 10.5 L atm 101.3 J 1.06 103 J 1.06 kJ 1 L atm The expansion work done is 1.06 kJ. 6.92 Strategy: The heat released during the reaction is absorbed by both the water and the calorimeter. How do we calculate the heat absorbed by the water? How do we calculate the heat absorbed by the calorimeter? How much heat is released when 1.9862 g of benzoic acid are reacted? The problem gives the amount of heat that is released when 1 mole of benzoic acid is reacted (3226.7 kJ/mol). Solution: The heat of the reaction (combustion) is absorbed by both the water and the calorimeter. qrxn (qwater qcal) If we can calculate both qwater and qrxn, then we can calculate qcal. First, let's calculate the heat absorbed by the water. qwater mwaterswatert 4 qwater (2000 g)(4.184 J/gC)(25.67 21.84)C 3.20 10 J 32.0 kJ Next, let's calculate the heat released (qrxn) when 1.9862 g of benzoic acid are burned. Hrxn is given in units of kJ/mol. Let’s convert to qrxn in kJ. 1 mol benzoic acid 3226.7 kJ qrxn 1.9862 g benzoic acid 52.48 kJ 122.12 g benzoic acid 1 mol benzoic acid And, qcal qrxn qwater qcal 52.48 kJ 32.0 kJ 20.5 kJ To calculate the heat capacity of the bomb calorimeter, we can use the following equation: qcal Ccalt Ccal 6.93 (a) qcal 20.5 kJ 5.35 kJ/ C t (25.67 21.84)C We carry an extra significant figure throughout this calculation to avoid rounding errors. The number of moles of water present in 500 g of water is: moles of H 2 O 500 g H 2 O 1 mol H 2 O 27.75 mol H 2 O 18.02 g H 2 O 150 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS From the equation for the production of Ca(OH)2, we have 1 mol H2O 1 mol CaO 1 mol Ca(OH)2. Therefore, the heat generated by the reaction is: 27.75 mol Ca(OH)2 65.2 kJ 1.809 103 kJ 1 mol Ca(OH)2 Knowing the specific heat and the number of moles of Ca(OH)2 produced, we can calculate the temperature rise using Equation (6.12) of the text. First, we need to find the mass of Ca(OH)2 in 27.75 moles. 27.75 mol Ca(OH) 2 74.10 g Ca(OH)2 2.056 103 g Ca(OH)2 1 mol Ca(OH)2 From Equation (6.12) of the text, we write: q mst Rearranging, we get t t q ms 1.809 106 J (2.056 103 g)(1.20 J/g C) 733C and the final temperature is t tfinal tinitial tfinal 733C 25C 758C A temperature of 758C is high enough to ignite wood. (b) The reaction is: CaO(s) H2O(l) Ca(OH)2(s) H rxn H f [Ca(OH) 2 ] [H f (CaO) H f (H 2 O)] H rxn is given in the problem (65.2 kJ/mol). Also, the H f values of CaO and H2O are given. Thus, we can solve for H f of Ca(OH)2. 65.2 kJ/mol H f [Ca(OH)2 ] [(1)(635.6 kJ/mol (1)(285.8 kJ/mol)] H f [Ca(OH)2 ] 986.6 kJ/mol 6.94 First, let’s calculate the standard enthalpy of reaction. H rxn 2H f (CaSO4 ) [2H f (CaO) 2H f (SO2 ) H f (O2 )] (2)(1432.69 kJ/mol) [(2)(635.6 kJ/mol) (2)(296.1 kJ/mol) 0] 1002 kJ/mol CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS This is the enthalpy change for every 2 moles of SO2 that are removed. The problem asks to calculate the 5 enthalpy change for this process if 6.6 10 g of SO2 are removed. (6.6 105 g SO2 ) 6.95 1 mol SO2 1002 kJ 5.2 106 kJ 64.07 g SO2 2 mol SO2 First, we need to calculate the volume of the balloon. V (a) 4 3 4 1000 L r (8 m)3 (2.1 103 m3 ) 2.1 106 L 3 3 3 1m We can calculate the mass of He in the balloon using the ideal gas equation. 1 atm 98.7 kPa (2.1 106 L) 2 1.01325 10 kPa PV nHe 8.6 104 mol He L atm RT 0.0821 mol K (273 18)K mass He (8.6 104 mol He) (b) 4.003 g He 3.4 105 g He 1 mol He Work done PV 1 atm 98.7 kPa (2.1 106 L) 2 1.01325 10 kPa (2.0 106 L atm) 101.3 J 1 L atm 8 Work done 2.0 10 J 6.96 (a) The heat needed to raise the temperature of the water from 3C to 37C can be calculated using the equation: q mst First, we need to calculate the mass of the water. 4 glasses of water 2.5 102 mL 1 g water 1.0 103 g water 1 glass 1 mL water 3 The heat needed to raise the temperature of 1.0 10 g of water is: 3 5 2 q mst (1.0 10 g)(4.184 J/gC)(37 3)C 1.4 10 J 1.4 10 kJ (b) We need to calculate both the heat needed to melt the snow and also the heat needed to heat liquid water form 0C to 37C (normal body temperature). The heat needed to melt the snow is: (8.0 102 g) 1 mol 6.01 kJ 2.7 102 kJ 18.02 g 1 mol 151 152 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS The heat needed to raise the temperature of the water from 0C to 37C is: 2 5 2 q mst (8.0 10 g)(4.184 J/gC)(37 0)C 1.2 10 J 1.2 10 kJ The total heat lost by your body is: 2 2 2 (2.7 10 kJ) (1.2 10 kJ) 3.9 10 kJ 6.97 The reaction we are interested in is the formation of ethanol from its elements. 2C(graphite) 1 2 O2(g) 3H2(g) C2H5OH(l) Along with the reaction for the combustion of ethanol, we can add other reactions together to end up with the above reaction. Reversing the reaction representing the combustion of ethanol gives: 2CO2(g) 3H2O(l) C2H5OH(l) 3O2 (g) H 1367.4 kJ/mol We need to add equations to add C (graphite) and remove CO2 and H2O from the reactants side of the equation. We write: C2H5OH(l) 3O2(g) 2CO2(g) 3H2O(l) H 1367.4 kJ/mol 2CO2(g) 2C(graphite) 2O2(g) H 2(393.5 kJ/mol) 3H2(g) 3 2 O2(g) 3H2O(l) 2C(graphite) 6.98 1 2 H 3(285.8 kJ/mol) H f 277.0 kJ/mol O2(g) 3H2(g) C2H5OH(l) Heat gained by ice Heat lost by the soft drink mice 334 J/g msdssdt mice 334 J/g (361 g)(4.184 J/gC)(0 23)C mice 104 g 6.99 The heat required to heat 200 g of water (assume d 1 g/mL) from 20C to 100C is: q mst 4 q (200 g)(4.184 J/gC)(100 20)C 6.7 10 J Since 50% of the heat from the combustion of methane is lost to the surroundings, twice the amount of heat 4 5 2 needed must be produced during the combustion: 2(6.7 10 J) 1.3 10 J 1.3 10 kJ. Use standard enthalpies of formation (see Appendix 2) to calculate the heat of combustion of methane. CH4(g) 2O2(g) CO2(g) 2H2O(l) H 890.3 kJ/mol 2 The number of moles of methane needed to produce 1.3 10 kJ of heat is: (1.3 102 kJ) 1 mol CH 4 0.15 mol CH 4 890.3 kJ CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 153 The volume of 0.15 mole CH4 at 1 atm and 20C is: V nRT (0.15 mol)(0.0821 L atm / K mol)(293 K) 3.6 L P 1.0 atm 3 Since we have the volume of methane needed in units of liters, let's convert the cost of natural gas per 15 ft to the cost per liter. 3 3 1 ft 1 in 1000 cm3 $3.1 103 1L 1 L CH 4 15 ft 3 12 in 2.54 cm $1.30 The cost for 3.6 L of methane is: 3.6 L CH 4 6.100 $3.1 103 $0.011 or about 1.1¢ 1 L CH 4 From Chapter 5, we saw that the kinetic energy (or internal energy) of 1 mole of a gas is 1 mole of an ideal gas, PV RT. We can write: internal energy 3 RT . For 2 3 3 RT PV 2 2 3 5 3 3 (1.2 10 Pa)(5.5 10 m ) 2 8 3 9.9 10 Pam 3 1 Pam 1 N m 2 3 m 1 Nm 1 J 8 Therefore, the internal energy is 9.9 10 J. 6 The final temperature of the copper metal can be calculated. (10 tons 9.072 10 g) q mCusCut 8 6 9.9 10 J (9.072 10 g)(0.385 J/gC)(tf 21C) 6 (3.49 10 )tf 1.06 10 9 tf 304C 6.101 Energy must be supplied to break a chemical bond. By the same token, energy is released when a bond is formed. 6.102 (a) CaC2(s) 2H2O(l) Ca(OH)2(s) C2H2(g) (b) The reaction for the combustion of acetylene is: 2C2H2(g) 5O2(g) 4CO2(g) 2H2O(l) 154 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS We can calculate the enthalpy change for this reaction from standard enthalpy of formation values given in Appendix 2 of the text. H rxn [4 H f (CO 2 ) 2H f (H 2 O)] [2H f (C2 H 2 ) 5H f (O2 )] H rxn [(4)(393.5 kJ/mol) (2)(285.8 kJ/mol)] [(2)(226.6 kJ/mol) (5)(0)] H rxn 2599 kJ/mol Looking at the balanced equation, this is the amount of heat released when two moles of C2H2 are reacted. The problem asks for the amount of heat that can be obtained starting with 74.6 g of CaC2. From this amount of CaC2, we can calculate the moles of C2H2 produced. 74.6 g CaC2 1 mol CaC2 1 mol C2 H 2 1.16 mol C2 H 2 64.10 g CaC2 1 mol CaC2 Now, we can use the H rxn calculated above as a conversion factor to determine the amount of heat obtained when 1.16 moles of C2H2 are burned. 1.16 mol C2 H 2 6.103 2599 kJ 1.51 103 kJ 2 mol C2 H 2 When 1.034 g of naphthalene are burned, 41.56 kJ of heat are evolved. Let's convert this to the amount of heat evolved on a molar basis. The molar mass of naphthalene is 128.2 g/mol. q 128.2 g C10 H8 41.56 kJ 5153 kJ/mol 1.034 g C10 H8 1 mol C10 H8 q has a negative sign because this is an exothermic reaction. This reaction is run at constant volume (V 0); therefore, no work will result from the change. w PV 0 From Equation (6.4) of the text, it follows that the change in energy is equal to the heat change. U q w qv 5153 kJ/mol To calculate H, we rearrange Equation (6.10) of the text. U H RTn H U RTn To calculate H, n must be determined, which is the difference in moles of gas products and moles of gas reactants. Looking at the balanced equation for the combustion of naphthalene: C10H8(s) 12O2(g) 10CO2(g) 4H2O(l) n 10 12 2 H U RTn H 5153 kJ/mol (8.314 J/mol K)(298 K)(2) H 5158 kJ/mol Is H equal to qp in this case? 1 kJ 1000 J CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.104 (a) 155 If no heat exchange occurs between the system and surroundings, then q 0 (called an adiabatic process). U q w w Because the system does work on the surroundings, w is a negative quantity, and there is a decrease in the system's energy. In Section 5.6 of the text, we saw that the average kinetic energy of a gas is directly proportional to the absolute temperature [Equation (5.15) of the text]. It follows that as the energy of the system decreases, the temperature also decreases. It is this cooling effect (or the decrease in the kinetic energy of the water molecules) that is responsible for the formation of snow. (b) This process is approximately adiabatic, so that q 0 and U w. Because work is done on the gas (system), w is positive, and there is an increase in the system's energy. As discussed in part (a) above, the kinetic energy of a gas is directly proportional to the absolute temperature. Thus, the temperature of the system increases, which we notice as a warming effect at the valve stem. (c) We assume that when the car is stopped, its kinetic energy is completely converted into heat (friction of the brakes and friction between the tires and the road). Thus, q 1 mu 2 2 Thus the amount of heat generated must be proportional to the braking distance, d: dq du 2 2 2 Therefore, as u increases to 2u, d increases to (2u) 4u which is proportional to 4d. 6.105 Water has a larger specific heat than air. Thus cold, damp air can extract more heat from the body than cold, dry air. By the same token, hot, humid air can deliver more heat to the body. 6.106 Since the humidity is very low in deserts, there is little water vapor in the air to trap and hold the heat radiated back from the ground during the day. Once the sun goes down, the temperature drops dramatically. 40F temperature drops between day and night are common in desert climates. Coastal regions have much higher humidity levels compared to deserts. The water vapor in the air retains heat, which keeps the temperature at a more constant level during the night. In addition, sand and rocks in the desert have small specific heats compared with water in the ocean. The water absorbs much more heat during the day compared to sand and rocks, which keeps the temperature warmer at night. 6.107 Let's write balanced equations for the reactions between Mg and CO2, and Mg and H2O. Then, we can for each reaction from H f values. calculate H rxn (1) 2Mg(s) CO2(g) 2MgO(s) C(s) (2) Mg(s) 2H2O(l) Mg(OH)2(s) H2(g) is: For reaction (1), H rxn H rxn 2H f [MgO( s )] H f [C( s )] {2H f [Mg( s )] H f [CO2 ( g )]} H rxn (2)(601.8 kJ/mol) (1)(0) [(2)(0) (1)(393.5 kJ/mol)] 8.10 102 kJ/mol 156 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS For reaction (2), H rxn is: H rxn H f [Mg(OH)2 ( s )] H f [H 2 ( g )] {H f [Mg( s )] 2H f [H 2 O(l )]} H rxn (1)(924.66 kJ/mol) (1)(0) [(1)(0) (2)( 285.8 kJ/mol)] 353.1 kJ/mol Both of these reactions are highly exothermic, which will promote the fire rather than extinguishing it. 6.108 First, we calculate H for the combustion of 1 mole of glucose using data in Appendix 2 of the text. We can then calculate the heat produced in the calorimeter. Using the heat produced along with H for the combustion of 1 mole of glucose will allow us to calculate the mass of glucose in the sample. Finally, the mass % of glucose in the sample can be calculated. C6H12O6(s) 6O2(g) 6CO2(g) 6H2O(l) H rxn (6)(393.5 kJ/mol) (6)(285.8 kJ/mol) (1)(1274.5 kJ/mol) 2801 kJ/mol The heat produced in the calorimeter is: (3.134C)(19.65 kJ/C) 61.58 kJ Let x equal the mass of glucose in the sample: x g glucose 1 mol glucose 2801 kJ 61.58 kJ 180.16 g glucose 1 mol glucose x 3.961 g % glucose 6.109 (a) (b) (c) (d) (e) (f) q 3.961 g 100% 96.21% 4.117 g w 0 U 0 H 0 In (b), the internal energy of an ideal gas depends only on temperature. Since temperature is held constant, U 0. Also, H 0 because H U (PV) U (nRT) 0. 6.110 (a) From the mass of CO2 produced, we can calculate the moles of carbon in the compound. From the mass of H2O produced, we can calculate the moles of hydrogen in the compound. 1.419 g CO 2 1 mol CO 2 1 mol C 0.03224 mol C 44.01 g CO 2 1 mol CO 2 0.290 g H 2 O 1 mol H 2 O 2 mol H 0.03219 mol H 18.02 g H 2 O 1 mol H 2 O The mole ratio between C and H is 1:1, so the empirical formula is CH. CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS (b) 157 The empirical molar mass of CH is 13.02 g/mol. molar mass 76 g 5.8 6 empirical molar mass 13.02 g Therefore, the molecular formula is C6H6, and the hydrocarbon is benzene. The combustion reaction is: 2C6H6(l) 15O2(g) 12CO2(g) 6H2O(l) 17.55 kJ of heat is released when 0.4196 g of the hydrocarbon undergoes combustion. We can now ) for the above reaction in units of kJ/mol. Then, from the calculate the enthalpy of combustion ( H rxn enthalpy of combustion, we can calculate the enthalpy of formation of C6H6. 78.11 g C6 H6 17.55 kJ 2 mol C6 H6 6534 kJ/mol 0.4196 g C6 H6 1 mol C6 H6 H rxn (12)H f (CO2 ) (6)H f (H 2 O) (2)H f (C6 H 6 ) 6534 kJ/mol (12)(393.5 kJ/mol) (6)(285.8 kJ/mol) (2)H f (C6 H6 ) H f (C6 H6 ) 49 kJ/mol 6.111 If the body absorbs all the heat released and is an isolated system, the temperature rise, t, is: q mst t q 1.0 107 J 48C (50, 000 g)(4.184 J/g C) ms If the body temperature is to remain constant, the heat released by metabolic activity must be used for the evaporation of water as perspiration, that is, 1 g H2O (1.0 104 kJ) 4.1 103 g H 2O 2.41 kJ Assuming that the density of perspiration is 1 g/mL, this mass corresponds to a volume of 4.1 L. The actual amount of perspiration is less than this because part of the body heat is lost to the surroundings by convection and radiation. 6.112 AB BC CD DA w 0, because V 0 w PV (2 atm)(2 1)L 2 Latm w 0, because V 0 w PV (1 atm)(1 2)L 1 Latm The total work done (2 Latm) (1 Latm) 1 Latm Converting to units of joules, 1 L atm 101.3 J 101.3 J 1 L atm In a cyclic process, the change in a state function must be zero. We therefore conclude that work is not a state function. Note that the total work done equals the area of the enclosure. 158 CHAPTER 6: ENERGY RELATIONSHIPS IN CHEMICAL REACTIONS 6.113 (a) Heating water at room temperature to its boiling point. (b) Heating water at its boiling point. (c) A chemical reaction taking place in a bomb calorimeter (an isolated system) where there is no heat exchange with the surroundings. 6.114 C (graphite) C (diamond) H U PV ΔH ΔU PΔV ΔH ΔU PΔV The pressure is 50,000 atm. From the given densities, we can calculate the volume in liters occupied by one mole of graphite and one mole of diamond. Taking the difference will give ΔV. We carry additional significant figures throughout the calculations to avoid rounding errors. 1 cm3 1L 12.01 g graphite 0.0053378 L/mol graphite 2.25 g graphite 1000 cm3 1 mol graphite 1 cm3 1L 12.01 g diamond 0.0034119 L/mol diamond 3 3.52 g diamond 1000 cm 1 mol diamond ΔH ΔU PΔV (50,000 atm)(0.0034119 L/mol 0.0053378 L/mol) H U 96.295 L atm 101.3 J 9.75 103 J / mol mol 1 L atm