CHAPTER 6 THERMOCHEMISTRY

Transcription

1 CHAPTER 6 THERMOCHEMISTRY Problem Categories Biological: 6.81, 6.10, 6.141, Conceptual: 6.1, 6.47, 6.48, 6.49, 6.50, 6.7, 6.95, 6.97, 6.10, 6.11, 6.117, 6.119, 6.17, 6.18, 6.1, 6.19, 6.14, Descriptive: 6.71, 6.76, 6.11, 6.1. Environmental: 6.9, 6.110, 6.111, 6.15, 6.15, 6.154, 6.155, Industrial: 6.5, 6.57, Organic: 6.55, 6.87, 6.108, 6.15, 6.1. Difficulty Level Easy: 6.15, 6.16, 6.17, 6.18, 6.5, 6.6, 6., 6., 6.4, 6.45, 6.46, 6.48, 6.51, 6.5, 6.54, 6.55, 6.57, 6.59, 6.61, 6.64, 6.71, 6.8, 6.87, 6.97, 6.98, 6.99, 6.101, 6.10, 6.114, 6.10, Medium: 6.19, 6.0, 6.7, 6.8, 6.1, 6.5, 6.6, 6.7,6.8, 6.47, 6.49, 6.50, 6.5, 6.56, 6.58, 6.60, 6.6, 6.6, 6.7, 6.7, 6.74, 6.75, 6.76, 6.77, 6.78, 6.79, 6.80, 6.81, 6.8, 6.84, 6.86, 6.88, 6.90, 6.95, 6.96, 6.100, 6.10, 6.104, 6.105, 6.107, 6.108, 6.109, 6.110, 6.11, 6.11, 6.115, 6.117, 6.118, 6.119, 6.11, 6.1, 6.1, 6.16, 6.17, 6.18, 6.10, 6.1, 6.1, 6.16, 6.141, 6.14, 6.146, Difficult: 6.85, 6.89, 6.91, 6.9, 6.9, 6.94, 6.106, 6.111, 6.116, 6.14, 6.15, 6.19, 6.14, 6.15, 6.17, 6.18, 6.19, 6.140, 6.14, 6.144, Recall that the work in gas expansion is equal to the product of the external, opposing pressure and the change in volume. (a) (b) w PV w (0)( )L 0 w PV w (0.80 atm)( )L.0 Latm To convert the answer to joules, we write.0 L atm 101. J w 1 L atm.0 10 J (c) w PV w (.7 atm)( )L 14 Latm To convert the answer to joules, we write w 14 L atm 101. J 1 L atm J 6.16 (a) Because the external pressure is zero, no work is done in the expansion. w PV (0)( )mL w 0

2 190 CHAPTER 6: THERMOCHEMISTRY (b) The external, opposing pressure is 1.5 atm, so w PV (1.5 atm)( )mL L w 94 ml atm Latm 1 ml To convert the answer to joules, we write: w L atm 101. J 1 L atm 9.5 J (c) The external, opposing pressure is.8 atm, so w PV (.8 atm)( )mL L w ( ml atm) 0.18 Latm 1 ml To convert the answer to joules, we write: 0.18 L atm 101. J w 1 L atm 18 J 6.17 An expansion implies an increase in volume, therefore w must be 5 J (see the defining equation for pressure-volume work.) If the system absorbs heat, q must be 17 J. The change in energy (internal energy) is: U q w 17 J 5 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 6 J 74 J 48 J As a result, the energy of the gas increases by 48 J We first find the number of moles of hydrogen gas formed in the reaction: 1 mol Sn 1 mol H 50.0 g Sn 0.41 mol H 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. nrt (0.41 mol)(0.081 Latm/ K mol)(98 K) V 10. L H P 1.00 atm The pressure-volume work done is then: (1.00 atm)(10. L) 10. L atm 101. J w PV J 1 L atm

3 CHAPTER 6: THERMOCHEMISTRY 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? Solution: First, we need to calculate the volume that the water vapor will occupy (V f ). Using the ideal gas equation: V HO n HO P RT Latm (1 mol) (7 K) mol K = 1 L (1.0 atm) It is given that the volume occupied by liquid water is negligible. Therefore, V V f V i 1 L 0 L 1 L Now, we substitute P and V into Equation (6.) of the text to solve for w. w PV (1.0 atm)(1 L) 1 Latm The problems asks for the work done in units of joules. The following conversion factor can be obtained from Appendix of the text. Thus, we can write: 1 Latm 101. J w 1 L atm 101. J 1 L atm.1 10 J Check: Because this is gas expansion (work is done by the system on the surroundings), the work done has a negative sign. 6.5 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. The heat evolved per gram of ZnS roasted is: heat evolved 879 kj 1 mol ZnS = mol ZnS g ZnS 4.51 kj/g ZnS 6.6 Strategy: The thermochemical equation shows that for every moles of NO produced, kj of heat are given off (note the negative sign of ΔH). We can write a conversion factor from this information kj mol NO How many moles of NO are in g of NO? What conversion factor is needed to convert between grams and moles?

4 19 CHAPTER 6: THERMOCHEMISTRY Solution: We need to first calculate the number of moles of NO in 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 NO moles of NO kilojoules of heat generated Therefore, the heat given off is: 1 mol NO kj ( g NO ) g NO mol NO kj 6.7 We can calculate U using Equation (6.10) of the text. U H RTn We initially have.0 moles of gas. Since our products are.0 moles of H and 1.0 mole of O, there is a net gain of 1 mole of gas ( reactant product). Thus, n 1. Looking at the equation given in the problem, it requires 48.6 kj to decompose.0 moles of water (H 48.6 kj). Substituting into the above equation: U J (8.14 J/molK)(98 K)(1 mol) U J kj 6.8 We initially have 6 moles of gas ( moles of chlorine and 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 PV 0, so U H PV U H H H rxn (184.6 kj/mol) 55.8 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 moles of HCl. U H 55.8 kj/mol 6.1 (d). The definition of heat is the transfer of thermal energy between two bodies that are at different temperatures. 6. C 85.7 J/ C Specific heat 0.7 J/g C m 6 g 6. q m Cu s Cu t (6. 10 g)(0.85 J/gC)(4. 0.5)C J 78 kj 6.4 See Table 6. of the text for the specific heat of Hg. q mst (66 g)(0.19 J/g C)( )C.1 10 J.1 kj The amount of heat liberated is.1 kj.

5 CHAPTER 6: THERMOCHEMISTRY 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. 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 q Au q Fe 0 q Au q Fe The heat gained by the gold sheet is given by: q Au m Au s Au t (10.0 g)(0.19 J/gC)(t f 18.0)C where m and s are the mass and specific heat, and t t final t initial. The heat lost by the iron sheet is given by: q Fe m Fe s Fe t (0.0 g)(0.444 J/gC)(t f 55.6)C Substituting into the equation derived above, we can solve for t f. q Au q Fe (10.0 g)(0.19 J/gC)(t f 18.0)C (0.0 g)(0.444 J/gC)(t f 55.6)C 1.9 t f t f t f 517 t f 50.7C Check: Must the final temperature be between the two starting values? 6.6 Strategy: We know the mass of aluminum and the initial and final temperatures of water and aluminum. We can look up the specific heats of water and aluminum in Table 6. of the text. Assuming no heat is lost to the surroundings, we can equate the heat lost by the aluminum to the heat gained by the water. With this information, we can solve for the mass of the water. Solution: Treating the calorimeter as an isolated system (no heat lost to the surroundings), we can write: or q q H O Al 0 q q H O Al The heat gained by water is given by: q m s t m (4.184 J/g C)(4.9.4) C H O H O H O H O where m and s are the mass and specific heat, and t t final t initial. The heat lost by the aluminum is given by: qal mal sal t (1.1 g)(0.900 J/g C)( ) C

6 194 CHAPTER 6: THERMOCHEMISTRY Substituting into the equation derived above, we can solve for q q m H O HO Al m HO. (4.184 J/g C)(4.9.4) C (1.1 g)(0.900 J/g C)( ) C (6.8)( m ) 619 m HO HO 98.6 g 6.7 The heat gained by the calorimeter is: q C p t q (04 J/C)(1.16C) J The amount of heat given off by burning Mg in kj/g is: 1 kj 1 ( J) 4.76 kj/g Mg 1000 J g Mg The amount of heat given off by burning Mg in kj/mol is: 4.76 kj 4.1 g Mg 1 g Mg 1 mol Mg kj/mol Mg If the reaction were endothermic, what would happen to the temperature of the calorimeter and the water? 6.8 The heat released by the reaction heats the solution and the calorimeter: q rxn +(q soln q cal ) The heat released by the reaction is: mol HCl 56. kj 85.0 ml 4.0 kj 1000 ml soln 1 mol We assume that the density of the solution is 1.00 g/ml, and its specific heat is J/g ºC. q rxn +(q soln q cal ) J = mst + Ct J = (170. g)(4.184 J/g ºC)(t f 18.4ºC) + (5 J/ºC)(t f 18.4ºC) = 711.8t f t f t f = t f =.9ºC 6.45 CH 4 (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 H (g) The standard enthalpy of formation of any element in its most stable form is zero. Therefore, since H f(o ) 0, O is the more stable form of the element oxygen at this temperature.

7 CHAPTER 6: THERMOCHEMISTRY H O(l) H O(g) Endothermic H H [H O( g)] H [H O( l )] 0 rxn f f Hf[HO( l )] is more negative since H rxn 0. You could also solve the problem by realizing that H O(l) is the stable form of water at 5C, and therefore will have the more negative H f value (a) Br (l) is the most stable form of bromine at 5C; therefore, Hf[Br ( l )] 0. Since Br (g) is less stable than Br (l), Hf[Br ( g )] 0. (b) I (s) is the most stable form of iodine at 5C; therefore, Hf[I ( s )] 0. Since I (g) is less stable than I (s), Hf[I ( g )] H O (l) H O(l) O (g) Because H O(l) has a more negative H f than H O (l) Strategy: What is the reaction for the formation of Ag O from its elements? What is the H f value for an element in its standard state? Solution: The balanced equation showing the formation of Ag O(s) from its elements is: Ag(s) 1 O (g) Ag O(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 nh (products) mh (reactants) rxn f f 1 rxn f f f H [ H (Ag O)] [ H (Ag) H (O )] H [ H (Ag O)] [0 0] rxn f H (Ag O) H f rxn In a similar manner, you should be able to show that H (CaCl ) H for the reaction f rxn Ca(s) Cl (g) CaCl (s) H [ H (CaO) H (CO )] H (CaCO ) 6.51 f f f H [(1)(65.6 kj/mol) (1)(9.5 kj/mol)] (1)(106.9 kj/mol) kj/mol 6.5 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

8 196 CHAPTER 6: THERMOCHEMISTRY product of the stoichiometric coefficient and the standard enthalpy of formation, H f, of the species.

9 CHAPTER 6: THERMOCHEMISTRY 197 Solution: We use the H f values in Appendix and Equation (6.18) of the text. H nh (products) mh (reactants) rxn f f (a) HCl(g) H (aq) Cl (aq) rxn f f f H H (H ) H (Cl ) H (HCl) 74.9 kj/mol 0 H f (Cl ) (1)(9. kj/mol) H f (Cl ) 167. kj/mol (b) The neutralization reaction is: and, H (aq) OH (aq) H O(l) rxn f f f H H [H O( l)] [ H (H ) H (OH )] Hf[HO( l )] 85.8 kj/mol (See Appendix of the text.) H rxn (1)(85.8 kj/mol) [(1)(0 kj/mol) (1)(9.6 kj/mol)] 56. kj/mol 6.5 (a) H Hf (HO) Hf (H ) H f (O ) H ()(85.8 kj/mol) ()(0) (1)(0) kj/mol H 4 H (CO ) H (H O) H (C H ) 5 H (O ) (b) f f f f H (4)(9.5 kj/mol) ()(85.8 kj/mol) ()(6.6 kj/mol) (5)(0) 599 kj/mol 6.54 (a) H = [ Hf (CO ) Hf (HO)] [ Hf (CH 4) H f (O )] ] H [()(9.5 kj/mol) ()(85.8 kj/mol)] [(1)(5. kj/mol) ()(0)] H 1411 kj/mol H = [ H (H O) H (SO )] [ H (H S) H (O )] (b) f f f f H [()(85.8 kj/mol) ()(96.1 kj/mol)] [()(0.15 kj/mol) ()(0)] H 114 kj/mol 6.55 The given enthalpies are in units of kj/g. We must convert them to units of kj/mol. (a) (b) (c).6 kj.04 g 74 kj/mol 1 g 1 mol 9.7 kj g 1 g 1 mol.4 kj g 1 g 1 mol kj/mol kj/mol

10 198 CHAPTER 6: THERMOCHEMISTRY H nh (products) mh (reactants) 6.56 rxn f f The reaction is: and, H (H ) 0 f H (g) H(g) H(g) H [ H (H) H (H)] H (H ) rxn f f f H 46.4 kj/mol H (H) (1)(0) rxn H f (H) 46.4 kj/mol f 18. kj/mol H 6 H (CO ) 6 H (H O) [ H (C H ) 9 H (O )] 6.57 f f f 6 1 f H (6)(9.5 kj/mol) (6)(85.8 kj/mol) (1)(151.9 kj/mol) (l)(0) 94 kj/mol Why is the standard heat of formation of oxygen zero? 6.58 Using the H f values in Appendix and Equation (6.18) of the text, we write H [5 H (B O ) 9 H (H O)] [ H (B H ) 1 H (O )] rxn f f f 5 9 f H [(5)(16.6 kj/mol) (9)(85.8 kj/mol)] [()(7. kj/mol) (l)(0 kj/mol)] H kj/mol Looking at the balanced equation, this is the amount of heat released for every moles of B 5 H 9 reacted. We can use the following ratio kj mol B H 5 9 to convert to kj/g B 5 H 9. The molar mass of B 5 H 9 is 6.1 g, so heat kj 1 mol B H released per gram B 5 9 5H kj/g B 5 H 9 mol B5H9 6.1 g B5H The amount of heat given off is: 1 mol NH 9.6 kj ( g NH ) g NH mol NH kj H nh (products) mh (reactants) 6.60 rxn f f The balanced equation for the reaction is: CaCO (s) CaO(s) CO (g) H [ H (CaO) H (CO )] H (CaCO ) rxn f f f

11 CHAPTER 6: THERMOCHEMISTRY 199 Hrxn [(1)( 65.6 kj/mol) (1)( 9.5 kj/mol)] (1)( kj/mol) kj/mol

12 00 CHAPTER 6: THERMOCHEMISTRY The enthalpy change calculated above is the enthalpy change if 1 mole of CO is produced. The problem asks for the enthalpy change if 66.8 g of CO are produced. We need to use the molar mass of CO as a conversion factor. 1 mol CO kj H 66.8 g CO kj g CO 1 mol CO 6.61 Reaction H (kj/mol) S(rhombic) O (g) SO (g) SO (g) S(monoclinic) O (g) 96.6 S(rhombic) S(monoclinic) H Which is the more stable allotropic form of sulfur? rxn 0.0 kj/mol 6.6 Strategy: Our goal is to calculate the enthalpy change for the formation of C H 6 from is elements C and H. 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 C H 6 from its elements. C(graphite) H (g) C H 6 (g) H rxn? Looking at this reaction, we need two moles of graphite as a reactant. So, we multiply Equation (a) by two to obtain: (d) C(graphite) O (g) CO (g) H rxn ( 9.5 kj/mol) kj/mol Next, we need three moles of H as a reactant. So, we multiply Equation (b) by three to obtain: (e) H (g) O (g) H O(l) H rxn ( 85.8 kj/mol) kj/mol Last, we need one mole of C H 6 as a product. Equation (c) has two moles of C H 6 as a reactant, so we need to reverse the equation and divide it by. (f) CO (g) H O(l) C H 6 (g) 7 O 1 (g) H (119.6 kj/mol) kj/mol rxn Adding Equations (d), (e), and (f) together, we have: Reaction (d) (e) H (kj/mol) C(graphite) O (g) CO (g) H (g) O (g) H O(l) (f) CO (g) H O(l) C H 6 (g) 7 O (g) C(graphite) H (g) C H 6 (g) H 84.6 kj/mol

13 CHAPTER 6: THERMOCHEMISTRY Reaction H (kj/mol) CO (g) H O(l) CH OH(l) O (g) 76.4 C(graphite) O (g) CO (g) 9.5 H (g) O (g) H O(l) (85.8) C(graphite) H (g) 1 O (g) CH OH(l) Hrxn 8.7 kj/mol We have just calculated an enthalpy at standard conditions, which we abbreviate H rxn. In this case, the reaction in question was for the formation of one mole of CH OH from its elements in their standard state. Therefore, the Hrxn that we calculated is also, by definition, the standard heat of formation H f of CH OH (8.7 kj/mol) The second and third equations can be combined to give the first equation. Al(s) O (g ) Al O (s) H kj/mol Fe O (s) Fe(s) O (g) H 8. kj/mol Al(s) Fe O (s) Fe(s) Al O (s) H kj/mol 6.71 Cl(g). The standard enthalpy of formation of an element in its most stable form is zero. The most stable form of chlorine is Cl (g). 6.7 The gas will expand until the pressure is the same as the external pressure of 1.0 atm. We can calculate its final volume using the ideal gas equation. V nrt (.70 mol)(0.081 L atm/k mol)(7 78.)K = = = L P 1.0 atm We assume that the initial volume of the liquid is negligible compared to the volume of the vapor. w PV (1.0 atm)( L) Latm 101. J 4 w L atm J 11 kj 1 Latm The expansion work done is 11 kj. 6.7 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 Reaction H (kj/mol) BrF(g) 1 Br (l) + 1 F (g) ½(188) 1 Br (l) + F (g) BrF (g) ½( 768) BrF(g) + F (g) BrF (g) Hrxn.9010 kj/mol

14 0 CHAPTER 6: THERMOCHEMISTRY 6.75 The reaction corresponding to standard enthalpy of formation, H f, of AgNO (s) is: Ag(s) 1 N (g) O (g) AgNO (s) Rather than measuring the enthalpy directly, we can use the enthalpy of formation of AgNO (s) and the H rxn provided. AgNO (s) AgNO (s) 1 O (g) 1 rxn f f f H H (AgNO ) H (O ) H (AgNO ) kj/mol H f (AgNO ) 0 (1.0 kj/mol) H (AgNO ) f 44.5 kj/mol 6.76 (a) Hrxn nhf (products) mh f (reactants) H [4 H (NH ) H (N )] H (N H ) H rxn f f f 4 rxn [(4)( 46. kj/mol) (0)] ()(50.4 kj/mol) 6.5 kj/mol (b) The balanced equations are: (1) N H 4 (l) O (g) N (g) H O(l) () 4NH (g) O (g) N (g) 6H O(l) The standard enthalpy change for equation (1) is: H H (N ) H [H O( l)] { H [N H ( l)] H (O )} rxn f f f 4 f H rxn [(1)(0) ()(85.8 kj/mol)] [(1)(50.4 kj/mol) (1)(0)] 6.0 kj/mol The standard enthalpy change for equation () is: H [ H (N ) 6 H (H O)] [4 H (NH ) H (O )] rxn f f f f H rxn [()(0) (6)(85.8 kj/mol)] [(4)(46. kj/mol) ()(0)] kj/mol We can now calculate the enthalpy change per kilogram of each substance. H rxn kj/mol. We need to convert to kj/kg. 6.0 kj 1 mol N H 1000 g N H ( l): H 4 above is in units of 4 4 rxn kj/kg NH4 1 mol NH4.05 g NH4 1 kg kj 1 mol NH 1000 g NH ( g): H 4 rxn kj/kg NH 4 mol NH 17.0 g NH 1 kg Since ammonia, NH, releases more energy per kilogram of substance, it would be a better fuel.

15 CHAPTER 6: THERMOCHEMISTRY The heat produced by the reaction heats the solution: q rxn +q soln The heat produced by the reaction is: 0.86 mol HCl 56. kj ml 9.69 kj 1000 ml soln 1 mol We assume that the density of the solution is 1.00 g/ml, and its specific heat is J/g ºC. q rxn +q soln J = mst J = (400. g)(4.184 J/g ºC)(t f 0.48ºC) J = ( J/ C)t f J t f = 6.ºC 6.78 The heat lost by the water was absorbed by the dissolution of NH 4 NO. The heat lost by the water is: q = mst = (80.0 g)(4.184 J/g ºC)( ) ºC q = J Thus, J of heat is absorbed during the dissolution process. H soln is expressed in kj/mol. H J g NH4NO 1 kj soln g NH4NO 1 mol NH4NO 1000 J kj/mol 6.79 We initially have 8 moles of gas ( 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 nrt (4.0 mol)(0.081 Latm/ Kmol)(98 K) V 98 L P 1 atm (1 atm)( 98 L) 98 L atm 101. J w PV J 9.9 kj 1 L atm H U PV or U H PV Using H as 185. kj ( 9.6 kj), (because the question involves the formation of 4 moles of ammonia, not 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. kj 9.9 kj 175. kj 6.80 The reaction is, Na(s) Cl (g) NaCl(s). First, let's calculate H for this reaction using H f values in Appendix. H H (NaCl) [ H (Na) H (Cl )] rxn f f f Hrxn ( kj/mol) [(0) 0] 8.0 kj/mol

16 04 CHAPTER 6: THERMOCHEMISTRY This is the amount of heat released when 1 mole of Cl reacts (see balanced equation). We are not reacting 1 mole of Cl, however. From the volume and density of Cl, we can calculate grams of Cl. Then, using the

17 CHAPTER 6: THERMOCHEMISTRY 05 molar mass of Cl as a conversion factor, we can calculate moles of Cl. Combining these two calculations into one step, we find moles of Cl to be: 1.88 g Cl 1 mol Cl.00 L Cl mol Cl 1 L Cl g Cl Finally, we can use the H rxn calculated above to find the amount of heat released when mole of Cl reacts. 8.0 kj mol Cl 4.6 kj 1 mol Cl The amount of heat released is 4.6 kj (a) Although we cannot measure H rxn for this reaction, the reverse process, is the combustion of glucose. We could easily measure C 6 H 1 O 6 (s) 6O (g) 6CO (g) 6H O(l) H rxn for this combustion in a bomb calorimeter. (b) We can calculate H rxn using standard enthalpies of formation. H H [C H O ( s)] 6 H [O ( g)] {6 H [CO ( g)] 6 H [H O( l )]} rxn f f f f H rxn [(1)(174.5 kj/mol) 0] [(6)(9.5 kj/mol) (6)(85.8 kj/mol)] 801. kj/mol Hrxn has units of kj/1 mol glucose. We want the H change for kg glucose. We need to calculate how many moles of glucose are in kg glucose. You should come up with the following strategy to solve the problem. kg glucose g glucose mol glucose kj (H) g 1 mol C6H1O kj 19 H ( kg) kj 1 kg 180. g C H O 1 mol C H O 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. 6.8 From the balanced equation we see that there is a 1: mole ratio between hydrogen and sodium. The number of moles of hydrogen produced is: 1 mol Na 1 mol H 0.4 g Na mol H.99 g Na mol Na Using the ideal gas equation, we write: nrt ( mol)(0.081 L atm/ K mol)(7 K) V 0.17 L H P (1 atm) V 0.17 L (1.0 atm)(0.17 L) 0.17 L atm 101. J w PV 17 J 1 L atm

18 06 CHAPTER 6: THERMOCHEMISTRY 6.84 H(g) Br(g) HBr(g) H rxn? Rearrange the equations as necessary so they can be added to yield the desired equation The reactions are: H(g) 1 H 1 (g) H ( 46.4 kj/mol) 18. kj/mol rxn Br(g) 1 Br 1 (g) H ( 19.5 kj/mol) 96.5 kj/mol rxn 1 H (g) 1 Br (g) HBr(g) H 1 ( 7.4 kj/mol) 6. kj/mol rxn H(g) Br(g) HBr(g) H 50.7 kj/mol H (g) + O (g) H O(l) (1) CH 4 (g) + O (g) CO (g) + H O(l) () We use Equation (6.18) of the text to calculate the heat released in each reaction. From Equation (1): H () H [H O( l )] f H ()(85.8 kj/mol) kj/mol From Equation (): H H (CO ) H [H O( l)] H (CH ) f f f 4 H (9.5 kj/mol) + ()(85.8 kj/mol) ( kj/mol)] 890. kj/mol Next, we use the ideal gas equation to calculate the total moles of gas in the mixture. n PV (. atm)(15.6 L) = = = 1.45 mol RT (0.081 L atm/k mol)(7 19.4)K 8.4 mole percent is hydrogen and 71.6 mole percent is methane. mol H = (0.84)(1.45 mol) = 0.41 mol mol CH 4 = (0.716)(1.45 mol) = 1.04 mol The heat released during the combustion is: kj 890. kj heat released 0.41 mol H 1.04 mol CH kj mol H 1 mol CH The standard enthalpy of formation is the heat change that results when 1 mole of a compound is formed from its elements at a pressure of 1 atm. The formation reaction for BaO(s) is: Ba(s) + ½O (g) BaO(s) In the problem, we are given the heat change per.740 g of Ba. We convert to units of kj/mol. o kj 17. g Ba Hf 558. kj/mol.740 g Ba 1 mol Ba

19 CHAPTER 6: THERMOCHEMISTRY Using the balanced equation, we can write: H [ H (CO ) 4 H (H O)] [ H (CH OH) H (O )] rxn f f f f kj/mol ()(9.5 kj/mol) (4)(85.8 kj/mol) () H f (CH OH) ()(0 kj/mol) kj/mol () H f (CH OH) H f (CH OH) 8.7 kj/mol 6.88 q system 0 q metal q water q calorimeter q metal q water q calorimeter 0 m metal s metal (t final t initial ) m water s water (t final t initial ) C calorimeter (t final t initial ) 0 All the needed values are given in the problem. All you need to do is plug in the values and solve for s metal. (44.0 g)(s metal )( )C (80.0 g)(4.184 J/gC)( )C (1.4 J/C)( )C 0 ( )s metal (gc) J s metal 0.49 J/gC 6.89 The reaction is: CO NO CO N The limiting reagent is CO (NO is in excess). H [ H (CO ) H (N )] [ H (CO) H (NO)] f f f f H [()(9.5 kj/mol) (1)(0)] [()(110.5 kj/mol) ()(90.4 kj/mol)] kj/mol 6.90 A good starting point would be to calculate the standard enthalpy for both reactions. Calculate the standard enthalpy for the reaction: C(s) 1 O (g) CO(g) This reaction corresponds to the standard enthalpy of formation of CO, so we use the value of kj/mol (see Appendix of the text). Calculate the standard enthalpy for the reaction: C(s) H O(g) CO(g) H (g) H [ H (CO) H (H )] [ H (C) H (H O)] rxn f f f f H rxn [(1)(110.5 kj/mol) (1)(0)] [(1)(0) (1)(41.8 kj/mol)] 11. 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 Let's start with the combustion of methane: CH 4 (g) O (g) CO (g) H O(l) H H (CO ) H (H O) H (CH ) rxn f f f 4 H (1)( 9.5 kj/mol) ()( 85.8 kj/mol) (1)( kj/mol) 890. kj/ mol rxn

20 08 CHAPTER 6: THERMOCHEMISTRY Now, let's calculate the heat produced by the combustion of water gas. We will consider the combustion of H and CO separately. We can look up the H f of H O(l) in Appendix. H (g) 1 O (g) H O(l) H rxn 85.8 kj/mol For the combustion of CO(g), we use CO(g) 1 O (g) CO (g) H rxn? 1 rxn f f f H H (CO ) H (CO) H (O ) rxn H f values from Appendix to calculate the H rxn. H (1)( 9.5 kj/mol) (1)( kj/mol) 8.0 kj/mol The H rxn values calculated above are for the combustion of 1 mole of H and 1 mole of CO, which equals moles of water gas. The total heat produced during the combustion of 1 mole of water gas is: ( )kJ/mol 84.4 kj/ mol of water gas which is less heat than that produced by the combustion of 1 mole of methane. Additionally, CO is very toxic. Natural gas (methane) is easier to obtain compared to carrying out the high temperature process of producing water gas. 6.9 First, calculate the energy produced by 1 mole of octane, C 8 H 18. C 8 H 18 (l) 5 O (g) 8CO (g) 9H O(l) 5 rxn f f f 8 18 f H 8 H (CO ) 9 H [H O( l)] [ H (C H ) H (O )] H rxn [(8)(9.5 kj/mol) (9)(85.8 kj/mol)] [(1)(49.9 kj/mol) ( 5 )(0)] 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: 5470 kj 1 mol octane 660 g 5 H kj / gal 1 mol octane 114. g octane 1 gal The combustion of hydrogen corresponds to the standard heat of formation of water: H (g) 1 O (g) H O(l) Thus, H rxn is the same as H f for H O(l), which has a value of 85.8 kj/mol. The number of moles of hydrogen required to produce kj of heat is: n 5 H ( kj) mol H 1 mol H 85.8 kj

21 CHAPTER 6: THERMOCHEMISTRY 09 Finally, use the ideal gas law to calculate the volume of gas corresponding to moles of H at 5C and 1 atm. Latm n (445.8 mol) (98 K) H RT mol K = 4 V H L P (1 atm) That is, the volume of hydrogen that is energy-equivalent to 1 gallon of gasoline is over 10,000 liters at 1 atm and 5C! 6.9 The reaction for the combustion of octane is: C 8 H 18 (l) 5 O (g) 8CO (g) 9H O(l) Hrxn for this reaction was calculated in problem 6.86 from standard enthalpies of formation. H rxn 5470 kj/mol Heat/gal of octane 5470 kj 1 mol C8H g 785 ml 1 mol C H 114. g 1 ml 1 gal 8 18 Heat/gal of octane kj/gal gasoline The reaction for the combustion of ethanol is: C H 5 OH(l) O (g) CO (g) H O(l) H H (CO ) H (H O) H (C H OH) H (O ) rxn f f f 5 f H rxn ()(9.5 kj/mol) ()(85.8 kj/mol) (1)(77.0 kj/mol) 167 kj/mol Heat/gal of ethanol 167 kj 1 mol CH5OH g 785 ml 1 mol C H OH g 1 ml 1 gal 5 Heat/gal of ethanol kj/gal ethanol For ethanol, what would the cost have to be to supply the same amount of heat per dollar of gasoline? For gasoline, it cost $4.50 to provide kj of heat. $ kj kj 1 gal ethanol 4 $.14 / gal ethanol 6.94 The combustion reaction is: C H 6 (l) 7 O (g) CO (g) H O(l) The heat released during the combustion of 1 mole of ethane is: 7 rxn f f f 6 f H [ H (CO ) H (H O)] [ H (C H ) H (O )] H rxn [()(9.5 kj/mol) ()(85.8 kj/mol)] [(1)(84.7 kj/mol ( 7 )(0)] 1560 kj/mol The heat required to raise the temperature of the water to 98C is: q mh OsH Ot (855 g)(4.184 J/gC)( )C J 61 kj

22 10 CHAPTER 6: THERMOCHEMISTRY The combustion of 1 mole of ethane produces 1560 kj; the number of moles required to produce 61 kj is: 1 mol ethane 61 kj mol ethane 1560 kj The volume of ethane is: V ethane nrt P = Latm (0.167 mol) (96 K) mol K 1 atm 75 mmhg 760 mmhg 4.10 L 6.95 As energy consumers, we are interested in the availability of usable energy The heat gained by the liquid nitrogen must be equal to the heat lost by the water. q q N H O 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 qh O = mh OsH Ot q H O ( g)(4.184 J/gC)( )C J The heat gained by 60.0 g nitrogen is the opposite sign of the heat lost by the water. q q N H O 4 q N 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 N to J/mol N J 8.0 g N H vap J/mol 5.60 kj/mol 60.0 g N 1 mol N 6.97 The evaporation of ethanol is an endothermic process with a fairly high H value. When the liquid evaporates, it absorbs heat from your body, hence the cooling effect Recall that the standard enthalpy of formation ( H f ) 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 w PV (1.0 atm)( )L Latm Using the conversion factor 1 Latm 101. J: 101. J w = ( Latm) = Latm 0.16 J 0.16 J of work are done by water as it expands on freezing.

23 CHAPTER 6: THERMOCHEMISTRY (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 (0.0 atm)( )L Latm Converting to units of joules: w L atm 101. J L atm 9.1 J (c) The gas will expand until the pressure is the same as the applied pressure of 0.0 atm. We can calculate its final volume using the ideal gas equation. Latm (0.00 mol) (7 0)K nrt mol K V.4 L P 0.0 atm The amount of work done is: w PV (0.0 atm)( )L 0.47 Latm Converting to units of joules: w 0.47 L atm 101. J L atm 48 J The equation corresponding to the standard enthalpy of formation of diamond is: Adding the equations: C(graphite) C(diamond) C(graphite) O (g) CO (g) CO (g) C(diamond) O (g) C(graphite) C(diamond) H 9.5 kj/mol H 95.4 kj/mol H 1.9 kj/mol Since the reverse reaction of changing diamond to graphite is exothermic, need you worry about any diamonds that you might have changing to graphite? 6.10 (a) 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 The balanced equation is: C 6 H 1 O 6 (s) C H 5 OH(l) CO (g) H [ H (C H OH) H (CO )] H (C H O ) rxn f 5 f f 6 1 6

24 1 CHAPTER 6: THERMOCHEMISTRY H rxn [()(76.98 kj/mol) ()(9.5 kj/mol)] (1)(174.5 kj/mol) 66.5 kj/mol

25 CHAPTER 6: THERMOCHEMISTRY Fe(s) O (g) Fe O (s). This equation represents twice the standard enthalpy of formation of Fe O. From Appendix, the standard enthalpy of formation of Fe O 8. kj/mol. So, H for the given reaction is: H rxn ()( 8. 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 50 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. 1 mol Fe 1644 kj 50 g Fe kj g Fe 4 mol Fe The amount of heat produced by this reaction is kj 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 HO 44.0 kj You should come up with the following strategy to solve the problem kj mol H O g H O 1 mol H O 18.0 g H O? g H O g H O 4000 kj = 44.0 kj 1 mol HO The heat required to raise the temperature of 1 liter of water by 1C is: J 1 g 1000 ml C 4184 J/L gc 1 ml 1 L Next, convert the volume of the Pacific Ocean to liters m 100 cm 1 L 0 (7. 10 km ) L 1 km 1 m 1000 cm The amount of heat needed to raise the temperature of L of water is: J 4 (7. 10 L).0 10 J 1L Finally, we can calculate the number of atomic bombs needed to produce this much heat. 4 1 atomic bomb 9 (.0 10 J).0 10 atomic bombs.0 billion atomic bombs J First calculate the final volume of CO gas: V 1 mol 19. g (0.081 Latm/ K mol)(95 K) nrt g = = = 10.6 L P atm

26 14 CHAPTER 6: THERMOCHEMISTRY w PV (0.995 atm)(10.6 L) 10.5 Latm 10.5 L atm 101. J w J 1.06 kj 1 L atm The expansion work done is 1.06 kj 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 g of benzoic acid are reacted? The problem gives the amount of heat that is released when 1 mole of benzoic acid is reacted (6.7 kj/mol). Solution: The heat of the reaction (combustion) is absorbed by both the water and the calorimeter. q rxn (q water q cal ) If we can calculate both q water and q rxn, then we can calculate q cal. First, let's calculate the heat absorbed by the water. q water m water s water t q water (000 g)(4.184 J/gC)( )C J.0 kj Next, let's calculate the heat released (q rxn ) when g of benzoic acid are burned. H rxn is given in units of kj/mol. Let s convert to q rxn in kj. 1 mol benzoic acid 6.7 kj q rxn g benzoic acid 5.49 kj 1.1 g benzoic acid 1 mol benzoic acid And, q cal q rxn q water q cal 5.49 kj.0 kj 0.5 kj To calculate the heat capacity of the bomb calorimeter, we can use the following equation: q cal C cal t C cal qcal 0.5 kj t ( ) C 5.5 kj/ C From thermodynamic data in Appendix of the text, we can calculate the amount of heat released per mole of H and CH 4. H (g) ½ O (g) H O(l) CH 4 (g) O (g) CO (g) H O(l) H 85.8 kj/mol H 890. kj/mol We know that 54 kj of heat is released from the combustion of 5.0 g of the gaseous mixture of H and CH 4. and mh m CH 5.0 g CH 4 m 5.0 g m H 4 We know the amount of heat released per mole of each substance. We need to convert from grams to moles of each substance, and then we can convert to kj of heat released.

27 CHAPTER 6: THERMOCHEMISTRY 15 1 mol H 85.8 kj 1 mol CH kj mh (5.0 m H ) 54 kj.016 g H 1 mol H g CH4 1 mol CH4 H m ( ) m m 964 m H H H 11. g m CH 5.0 g 11. g 1.8 g First, let s calculate the standard enthalpy of reaction. H H (CaSO ) [ H (CaO) H (SO ) H (O )] rxn f 4 f f f ()(14.7 kj/mol) [()(65.6 kj/mol) ()(96.1 kj/mol) 0] 100 kj/mol This is the enthalpy change for every moles of SO that are removed. The problem asks to calculate the enthalpy change for this process if g of SO are removed. 1 mol SO 100 kj ( g SO ) g SO mol SO kj L 5 Volume of room (.80 m 10.6 m 17. m) L 1m PV nrt n air PV (1.0 atm)( L) mol air RT (0.081 L atm / Kmol)( 7)K g air 5 mass air ( mol air) g air 1 mol air Heat to be removed from air: q m air s air t q ( g)(1. J/gC)(8.C) q J kj Conservation of energy: q air q salt 0 q air nh fus 0 q air m salt Hfus M salt 0

28 16 CHAPTER 6: THERMOCHEMISTRY m m salt salt q airm H fus salt ( kj)(. g/mol) g 74.4 kj/mol m salt 5 kg 6.11 First, we need to calculate the volume of the balloon L 6 V r (8 m) (.1 10 m ).1 10 L 1m (a) We can calculate the mass of He in the balloon using the ideal gas equation. n He 1 atm kpa (.1 10 L) PV kpa mol He RT Latm (7 18)K mol K g He 5 mass He ( mol He).4 10 g He 1 mol He (b) Work done PV 1 atm kpa (.1 10 L) kpa J (.0 10 Latm) 1 Latm Work done J 6.11 First, we calculate Hrxn for three reactions. Using the (1) C H (g) + H (g) C H 4 (g) () C H 4 (g) + H (g) C H 6 (g) () C H (g) + H (g) C H 6 (g) o H f values, we write: (1) () () o H 1 5. kj/mol 6.6 kj/mol 174. kj/mol o H 84.7 kj/mol 5. kj/mol 17.0 kj/mol o H 84.7 kj/mol 6.6 kj/mol 11. kj/mol

29 CHAPTER 6: THERMOCHEMISTRY 17 The following diagram summarizes the results. C H + H H 11. kj/mol H kj/mol C H 4 + H C H 6 H 17.0 kj/mol We use Equation (6.18) of the text. f f f H H [Fe(OH) ] { H (Fe ) ()[ H (OH )]} H (84.5 kj/mol) [(47.7 kj/mol) + ()( 9.94 kj/mol)] 86.7 kj/mol The heat produced by the reaction heats the solution and the calorimeter: q rxn (q soln q cal ) q soln mst (50.0 g)(4.184 J/gC)( )C 611 J q cal Ct (98.6 J/C)( )C 88 J q rxn (q soln q cal ) (611 88)J 899 J The 899 J produced was for 50.0 ml of a M AgNO solution mol Ag 50.0 ml mol Ag 1000 ml soln On a molar basis the heat produced was: 899 J mol Ag J/mol Ag 180 kj/mol Ag The balanced equation involves moles of Ag, so the heat produced is mol 180 kj/mol 60 kj Since the reaction produces heat (or by noting the sign convention above), then: q rxn 60 kj/mol Zn (or 60 kj/ mol Ag ) (a) The heat needed to raise the temperature of the water from C to 7C can be calculated using the equation: q mst First, we need to calculate the mass of the water ml 1 g water 4 glasses of water g water 1 glass 1 ml water The heat needed to raise the temperature of g of water is: q mst ( g)(4.184 J/gC)(7 )C J kj

30 18 CHAPTER 6: THERMOCHEMISTRY (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 7C (normal body temperature). The heat needed to melt the snow is: 1 mol 6.01 kj ( g).7 10 kj 18.0 g 1 mol The heat needed to raise the temperature of the water from 0C to 7C is: q mst ( g)(4.184 J/gC)(7 0)C J kj The total heat lost by your body is: (.7 10 kj) (1. 10 kj).9 10 kj 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 Thus the amount of heat generated must be proportional to the braking distance, d: d q d u Therefore, as u increases to u, d increases to (u) 4u which is proportional to 4d (a) H Hf (F ) Hf (HO) [ Hf (HF) H f (OH )] H [(1)(9.1 kj/mol) (1)(85.8 kj/mol)] [(1)(0.1 kj/mol) (1)(9.6 kj/mol) H 65. kj/mol (b) We can add the equation given in part (a) to that given in part (b) to end up with the equation we are interested in. HF(aq) OH (aq) F (aq) H O(l) H 65. kj/mol H O(l) H (aq) OH (aq) H 56. kj/mol HF(aq) H (aq) F (aq) H 9.0 kj/mol 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 The equation we are interested in is the formation of CO from its elements. C(graphite) 1 O (g) CO(g) H? Try to add the given equations together to end up with the equation above. C(graphite) O (g) CO (g) H 9.5 kj/mol CO (g) CO(g) 1 O (g) H 8.0 kj/mol

31 C(graphite) 1 O (g) CO(g) H kj/mol CHAPTER 6: THERMOCHEMISTRY 19

32 0 CHAPTER 6: THERMOCHEMISTRY We cannot obtain H f for CO directly, because burning graphite in oxygen will form both CO and CO Energy intake for mechanical work: 000 J g.6 10 J 1g 1 J 1 kgm s J mgh 5 kg m.6 10 (46 kg)(9.8 m/s ) h s h m 6.1 (a) mass kg Potential energy mgh ( kg)(9.8 m/s )(51 m) Potential energy 0.50 J (b) Kinetic energy 1 mu 0.50 J 1 ( kg) u 0.50 J u m /s u m/s (c) q mst 0.50 J (1.0 g)(4.184 J/gC)t t 0.1C 6.1 For Al: (0.900 J/gC)(6.98 g) 4. J/C This law does not hold for Hg because it is a liquid The reaction we are interested in is the formation of ethanol from its elements. C(graphite) 1 O (g) H (g) C H 5 OH(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: CO (g) H O(l) C H 5 OH(l) O (g) H kj/mol

33 CHAPTER 6: THERMOCHEMISTRY 1 We need to add equations to add C (graphite) and remove H O from the reactants side of the equation. We write: CO (g) H O(l) C H 5 OH(l) O (g) H kj/mol C(graphite) O (g) CO (g) H (9.5 kj/mol) H (g) O (g) H O(l) H (85.8 kj/mol) C(graphite) 1 O (g) H (g) C H 5 OH(l) H f 77.0 kj/mol 6.15 (a) C 6 H 6 (l) 15 O (g) 6CO (g) H O(l) H 67.4 kj/mol (b) C H (g) 5 O (g) CO (g) H O(l) H kj/mol (c) C(graphite) O CO (g) H 9.5 kj/mol (d) H (g) 1 O (g) H O(l) H 85.8 kj/mol Using Hess s Law, we can add the equations in the following manner to calculate the standard enthalpies of formation of C H and C 6 H 6. C H : (b) (c) (d) C(graphite) H (g) C H (g) H 6.6 kj/mol Therefore, H f (CH ) 6.6 kj/mol C 6 H 6 : (a) 6(c) (d) 6C(graphite) H (g) C 6 H 6 (l) H 49.0 kj/mol Therefore, H f (C6H 6) 49.0 kj/mol Finally: C H (g) C 6 H 6 (l) H rxn (1)(49.0 kj/mol) ()(6.6 kj/mol) 60.8 kj/mol 6.16 Heat gained by ice Heat lost by the soft drink m ice 4 J/g m sd s sd t m ice 4 J/g (61 g)(4.184 J/gC)(0 )C m ice 104 g 6.17 Immediately after the candle is blown out, the smoke is rich in paraffin wax vapor, which is highly flammable. The paraffin vapor ignites, and then the flame moves toward the wick which is still soaked in liquid wax, re-lighting the candle.

34 CHAPTER 6: THERMOCHEMISTRY 6.18 The decomposition reaction is: NH 4 Cl(s) NH (g) + HCl(g) Using the H f values in Appendix and Equation (6.18) of the text, we write: H [ H (NH ) H (HCl)] H (NH Cl) rxn f f f 4 H [(46. kj/mol) (9. kj/mol)] ( 15.9 kj/mol) H kj/mol Looking at the balanced equation, this is the amount of heat released for every 1 mole of NH 4 Cl decomposed. We can use the following ratio kj 1 mol NH Cl 4 to convert to the amount of heat required to decompose 89.7 grams of NH 4 Cl. 1 mol NH4Cl kj 89.7 g NH4Cl 96 kj 5.49 g NH Cl 1 mol NH Cl The heat required to heat 00 g of water (assume d 1 g/ml) from 0C to 100C is: q mst q (00 g)(4.184 J/gC)(100 0)C J Since 50% of the heat from the combustion of methane is lost to the surroundings, twice the amount of heat needed must be produced during the combustion: ( J) J kj. Use standard enthalpies of formation (see Appendix ) to calculate the heat of combustion of methane. CH 4 (g) O (g) CO (g) H O(l) H 890. kj/mol The number of moles of methane needed to produce kj of heat is: 1 mol CH (1. 10 kj) mol CH 890. kj 4 The volume of 0.15 mole CH 4 at 1 atm and 0C is: V nrt (0.15 mol)(0.081 L atm/ K mol)(9 K) P 1.0 atm.6 L 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. $1.0 1 ft 1 in 1000cm $ ft 1 in.54 cm 1 L 1 L CH4 The cost for.6 L of methane is: $ L CH4 $0.011 or about 1.1? 1 L CH 4

35 CHAPTER 6: THERMOCHEMISTRY 6.10 From Chapter 5, we saw that the kinetic energy (or internal energy) of 1 mole of a gas is RT. For 1 mole of an ideal gas, PV RT. We can write: 1 Pam 1 internal energy N m m 1 Nm 1 J Therefore, the internal energy is J. RT PV ( Pa)( m ) Pam The final temperature of the copper metal can be calculated. (10 tons g) q m Cu s Cu t J ( g)(0.85 J/gC)(t f 1C) ( )t f t f 04C 6.11 Energy must be supplied to break a chemical bond. By the same token, energy is released when a bond is formed. 6.1 (a) CaC (s) H O(l) Ca(OH) (s) C H (g) (b) The reaction for the combustion of acetylene is: C H (g) 5O (g) 4CO (g) H O(l) We can calculate the enthalpy change for this reaction from standard enthalpy of formation values given in Appendix of the text. H [4 H (CO ) H (H O)] [ H (C H ) 5 H (O )] rxn f f f f H rxn [(4)(9.5 kj/mol) ()(85.8 kj/mol)] [()(6.6 kj/mol) (5)(0)] H rxn 599 kj/mol Looking at the balanced equation, this is the amount of heat released when two moles of C H are reacted. The problem asks for the amount of heat that can be obtained starting with 74.6 g of CaC. From this amount of CaC, we can calculate the moles of C H produced. 1 mol CaC 1 mol C H 74.6 g CaC 1.16 mol C H g CaC 1 mol CaC 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 C H are burned.

36 4 CHAPTER 6: THERMOCHEMISTRY 599 kj 1.16 mol CH mol C H kj

37 CHAPTER 6: THERMOCHEMISTRY 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 When 1.04 g of naphthalene are burned, 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 18. g/mol kj 18. g C10 H q kj/mol 1.04 g C H 1 mol C H 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 q v 515 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: C 10 H 8 (s) 1O (g) 10CO (g) 4H O(l) Is H equal to q p in this case? n 10 1 H U RTn 1 kj H 515 kj/mol (8.14 J/mol K)(98 K)( ) 1000 J H 5158 kj/mol 6.15 Let's write balanced equations for the reactions between Mg and CO, and Mg and H O. Then, we can calculate H rxn for each reaction from H f values. (1) Mg(s) CO (g) MgO(s) C(s) () Mg(s) H O(l) Mg(OH) (s) H (g) For reaction (1), H rxn is: H H [MgO( s)] H [C( s)] { H [Mg( s)] H [CO ( g )]} rxn f f f f H rxn ()( kj/mol) (1)(0) [()(0) (1)( 9.5 kj/mol)] kj/mol

38 6 CHAPTER 6: THERMOCHEMISTRY For reaction (), H rxn is: H H [Mg(OH) ( s)] H [H ( g)] { H [Mg( s)] H [H O( l )]} rxn f f f f H rxn (1)( kj/mol) (1)(0) [(1)(0) ()( 85.8 kj/mol)] 5.1 kj/mol Both of these reactions are highly exothermic, which will promote the fire rather than extinguishing it We know that U q w. H q, and w PV RTn. Using thermodynamic data in Appendix of the text, we can calculate H. H (g) O (g) H O(l), H (85.8 kj/mol) kj/mol Next, we calculate w. The change in moles of gas (n) equals. w PV RTn w (8.14 J/molK)(98 K)() J/mol 7.4 kj/mol U q w U kj/mol 7.4 kj/mol 564. kj/mol Can you explain why U is smaller (in magnitude) than H? 6.17 (a) 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: 1 mol H O moles of H O 500 g H O 7.75 mol H O 18.0 g HO From the equation for the production of Ca(OH), we have 1 mol H O 1 mol CaO 1 mol Ca(OH). Therefore, the heat generated by the reaction is: 65. kj 7.75 mol Ca(OH) kj 1 mol Ca(OH) Knowing the specific heat and the number of moles of Ca(OH) produced, we can calculate the temperature rise using Equation (6.1) of the text. First, we need to find the mass of Ca(OH) in 7.7 moles g Ca(OH) 7.75 mol Ca(OH) g Ca(OH) From Equation (6.1) of the text, we write: q mst Rearranging, we get t q ms 1 mol Ca(OH) J t 7C ( g)(1.0 J/g C) 6

39 CHAPTER 6: THERMOCHEMISTRY 7 and the final temperature is t t final t initial t final 7C 5C 758C A temperature of 758C is high enough to ignite wood. (b) The reaction is: CaO(s) H O(l) Ca(OH) (s) H H [Ca(OH) ] [ H (CaO) H (H O)] rxn f f f H rxn is given in the problem (65. kj/mol). Also, the H f values of CaO and H O are given. Thus, we can solve for H f of Ca(OH). 65. kj/mol H [Ca(OH) ] [(1)( 65.6 kj/mol (1)( 85.8 kj/mol)] f f H [Ca(OH) ] kj/mol 6.18 First, we calculate H for the combustion of 1 mole of glucose using data in Appendix 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. C 6 H 1 O 6 (s) 6O (g) 6CO (g) 6H O(l) H rxn (6)( 9.5 kj/mol) (6)( 85.8 kj/mol) (1)( kj/mol) 801. kj/mol The heat produced in the calorimeter is: (.14C)(19.65 kj/c) kj Let x equal the mass of glucose in the sample: 1 mol glucose 801. kj x g glucose kj 180. g glucose 1 mol glucose x.961 g.961 g % glucose 100% 96.1% g 6.19 q w U H (a) (b) 0 0 (c) (d) (e) 0 (f)