CHEM 101 Homework Chapter 5 Assignment Solution, Fall 2013

Chem 101 Recitation 9 Chapter 6 Chang 9 th 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  PV w  (0)(5.4  1.6)L  0 (b) w  PV w  (0.80 atm)(5.4  1.6)L  3.0 Latm To convert the answer to joules, we write (c) w  PV w  (3.7 atm)(5.4  1.6)L  14 Latm To convert the answer to joules, we write 6.19 We first find the number of moles of hydrogen gas formed in the reaction: The next step is to find the volume occupied by the hydrogen gas under the given conditions. This is the change in volume. The pressure-volume work done is then: 6.25 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: 6.27 We can calculate E using Equation (6.10) of the text. E  H − RTn 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: E  483.6  103 J − (8.314 J/molK)(398 K)(1 mol) E  4.80  105 J  4.80  102 kJ 1 Chem 101 Recitation 9 Chapter 6 Chang 9 th 6.33 6.34 q  mCusCut  (6.22  103 g)(0.385 J/gC)(324.3C  20.5C)  7.28  105 J = 728 kJ 6.35 See Table 6.1 of the text for the specific heat of Hg. q  mst  (366 g)(0.139 J/g·C)(12.0C  77.0C)  3.31  103 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: qAu  qFe  0 or qAu  qFe The heat gained by the gold sheet is given by: qAu  mAusAut  (10.0 g)(0.129 J/gC)(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  mFesFet  (20.0 g)(0.444 J/gC)(tf  55.6)C Substituting into the equation derived above, we can solve for tf. qAu  qFe (10.0 g)(0.129 J/gC)(tf  18.0)C  (20.0 g)(0.444 J/gC)(tf  55.6)C 1.29 tf  23.2  8.88 tf  494 10.2 tf  517 tf  50.7C Check: Must the final temperature be between the two starting values? 6.37 The heat gained by the calorimeter is: q  Cpt q  (3024 J/C)(1.126C)  3.405  103 J The amount of heat given off by burning Mg in kJ/g is: The amount of heat given off by burning Mg in kJ/mol is: 2 Chem 101 Recitation 9 Chapter 6 Chang 9 th If the reaction were endothermic, what would happen to the temperature of the calorimeter and the water? 6.45 CH4(g) and H(g). All the other choices are elements in their most stable form ( ). 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 O2 is the more stable form of the element oxygen at this temperature. 6.50 Strategy: What is the reaction for the formation of Ag2O from its elements? What is the value for an element in its standard state? Solution: The balanced equation showing the formation of Ag2O(s) from its elements is: 2Ag(s)  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: In a similar manner, you should be able to show that for the reaction Ca(s)  Cl2(g) CaCl2(s) 6.51 H  [(1)(635.6 kJ/mol)  (1)(393.5 kJ/mol)]  (1)(1206.9 kJ/mol)  177.8 kJ/mol 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, , of the species. Solution: We use the values in Appendix 3 and Equation (6.18) of the text. (a) HCl(g)  H(aq)  Cl(aq) 3 Chem 101 Recitation 9 Chapter 6 Chang 9 th 74.9 kJ/mol  0  (Cl)  (1)(92.3 kJ/mol) (b) The neutralization reaction is: H(aq)  OH(aq)  H2O(l) and, (See Appendix 3 of the text.)  (1)(285.8 kJ/mol)  [(1)(0 kJ/mol)  (1)(229.6 kJ/mol)]  56.2 kJ/mol 6.57 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? 6.59 The amount of heat given off is: 6.60 The balanced equation for the reaction is: CaCO3(s) CaO(s)  CO2(g) 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. 6.63 Reaction  H  (kJ/mol) CO2(g)  2H2O(l)  CH3OH(l)  O2(g) 726.4 C(graphite)  O2(g)  CO2(g) 393.5 2H2(g)  O2(g)  2H2O(l) 2(285.8) 4 Chem 101 Recitation 9 Chapter 6 Chang 9 th C(graphite)  2H2(g)  O2(g)  CH3OH(l) We have just calculated an enthalpy at standard conditions, which we abbreviate . In this case, the reaction in question was for the formation of one mole of CH3OH from its elements in their standard state. Therefore, the that we calculated is also, by definition, the standard heat of formation of CH3OH (238.7 kJ/mol). 6.83 The original volume of ammonia is: (a) (b) t  (286  298)C  12C q  mst  (17.03 g)(0.0258 J/gC)(12C)  5.27 J E  q  w  5.27 J  (2.20  103 J)  2.21  103 J  2.21 kJ 6.87 The reaction for the combustion of octane is: C8H18(l)  O2(g)  8CO2(g)  9H2O(l) for this reaction was calculated in problem 6.86 from standard enthalpies of formation. Heat/gal of octane  1.27  105 kJ/gal gasoline The reaction for the combustion of ethanol is: C2H5OH(l)  3O2(g)  2CO2(g)  3H2O(l)  (2)(393.5 kJ/mol)  (3)(285.8 kJ/mol)  (1)(277.0 kJ/mol)  1367 kJ/mol Heat/gal of ethanol  8.87  104 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 $1.20 to provide 1.27  105 kJ of heat. 5 Chem 101 Recitation 9 Chapter 6 Chang 9 th 6.119 The heat required to heat 200 g of water (assume d  1 g/mL) from 20C to 100C is: q  mst q  (200 g)(4.184 J/gC)(100  20)C  6.7  104 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: 2(6.7  104 J)  1.3  105 J  1.3  102 kJ. Use standard enthalpies of formation (see Appendix 3) to calculate the heat of combustion of methane. CH4(g)  2O2(g)  CO2(g)  2H2O(l) H  890.3 kJ/mol The number of moles of methane needed to produce 1.3  102 kJ of heat is: The volume of 0.15 mole CH4 at 1 atm and 20C is: Since we have the volume of methane needed in units of liters, let's convert the cost of natural gas per 15 ft3 to the cost per liter. The cost for 3.6 L of methane is: 6