Thermochemistry

CHAPTER 6 THERMOCHEMISTRY Problem Categories Biological: 6.77, 6.97, 6.131. Conceptual: 6.31, 6.47, 6.48, 6.49, 6.50, 6.71, 6.89, 6.91, 6.96, 6.109, 6.111, 6.123, 6.129, 6.132. Descriptive: 6.74, 6.121, 6.122. Environmental: 6.86, 6.104, 6.105. Industrial: 6.25, 6.57, 6.127. Organic: 6.55, 6.81, 6.102, 6.117, 6.122. Difficulty Level Easy: 6.15, 6.16, 6.17, 6.18, 6.25, 6.26, 6.32, 6.33, 6.34, 6.45, 6.46, 6.48, 6.51, 6.53, 6.54, 6.55, 6.57, 6.59, 6.61, 6.64, 6.72, 6.78, 6.81, 6.91, 6.92, 6.93, 6.95, 6.97, 6.112, 6.121. Medium: 6.19, 6.20, 6.27, 6.28, 6.31, 6.35, 6.36, 6.37, 6.47, 6.49, 6.50, 6.52, 6.56, 6.58, 6.60, 6.62, 6.63, 6.71, 6.73, 6.74, 6.75, 6.76, 6.77, 6.79, 6.80, 6.82, 6.84, 6.89, 6.90, 6.94, 6.96, 6.98, 6.99, 6.101, 6.102, 6.103, 6.104, 6.106, 6.107, 6.109, 6.110, 6.111, 6.113, 6.114, 6.115, 6.118, 6.120, 6.122, 6.123, 6.126, 6.131, 6.132. Difficult: 6.38, 6.83, 6.85, 6.86, 6.87, 6.88, 6.100, 6.105, 6.108, 6.116, 6.117, 6.119, 6.124, 6.125, 6.127, 6.128, 6.129, 6.130, 6.133, 6.134, 6.135. 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 = −PΔV w = −(0)(5.4 − 1.6)L = 0 (b) w = −PΔV w = −(0.80 atm)(5.4 − 1.6)L = −3.0 L⋅atm 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 = −PΔV w = −(3.7 atm)(5.4 − 1.6)L = −14 L⋅atm 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 = −PΔV = −(0)(89.3 − 26.7)mL w = 0 CHAPTER 6: THERMOCHEMISTRY (b) 167 The external, opposing pressure is 1.5 atm, so w = −PΔV = −(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 × (c) 101.3 J = − 9.5 J 1 L ⋅ atm The external, opposing pressure is 2.8 atm, so w = −PΔV = −(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: ΔE = 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 (ΔE), 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: ΔE = 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 1.00 atm P The pressure-volume work done is then: w = − PΔV = − (1.00 atm)(10.3 L) = − 10.3 L ⋅ atm × 101.3 J = − 1.04 × 103 J 1 L ⋅ atm 168 CHAPTER 6: THERMOCHEMISTRY 6.20 Strategy: The work done in gas expansion is equal to the product of the external, opposing pressure and the change in volume. w = −PΔV 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 L⋅atm and J? 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, ΔV = Vf − Vi = 31 L − 0 L = 31 L Now, we substitute P and ΔV into Equation (6.3) of the text to solve for w. w = −PΔV = −(1.0 atm)(31 L) = −31 L⋅atm The problems asks for the work done in units of joules. The following conversion factor can be obtained from Appendix 2 of the text. 1 L⋅atm = 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. 6.25 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. The heat evolved per gram of ZnS roasted is: heat evolved = 6.26 879 kJ 1 mol ZnS × = 4.51 kJ / g ZnS 2 mol ZnS 97.46 g ZnS 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 of ΔH). 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: THERMOCHEMISTRY 169 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 given off is: (1.26 × 104 g NO2 ) × 6.27 1 mol NO2 114.6 kJ × = 1.57 × 104 kJ 46.01 g NO 2 2 mol NO2 We can calculate ΔE using Equation (6.10) of the text. ΔE = ΔH − RTΔn 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, Δn = +1. Looking at the equation given in the problem, it requires 483.6 kJ to decompose 2.0 moles of water (ΔH = 483.6 kJ). Substituting into the above equation: 3 ΔE = 483.6 × 10 J − (8.314 J/mol⋅K)(398 K)(+1 mol) 5 2 ΔE = 4.80 × 10 J = 4.80 × 10 kJ 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; ΔV = 0. w = −PΔV = −(1 atm)(0 L) = 0 −PΔV = 0, so ΔE° = ΔH° − PΔV ΔE = ΔH D ΔH = 3ΔH rxn = 3(−184.6 kJ/mol) = −553.8 kJ/mol D We need to multiply ΔH 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. ΔE° = ΔH° = −553.8 kJ/mol 6.31 (d). The definition of heat is the transfer of thermal energy between two bodies that are at different temperatures. 6.32 Specific heat = 6.33 q = mCusCuΔt = (6.22 × 10 g)(0.385 J/g⋅°C)(324.3 − 20.5)°C = 7.28 × 10 J = 728 kJ 6.34 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 = msΔt = (366 g)(0.139 J/g·°C)(12.0 − 77.0)°C = −3.31 × 10 J = −3.31 kJ The amount of heat liberated is 3.31 kJ. 170 CHAPTER 6: THERMOCHEMISTRY 6.35 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 = mAusAuΔt = (10.0 g)(0.129 J/g⋅°C)(tf − 18.0)°C where m and s are the mass and specific heat, and Δt = tfinal − tinitial. The heat lost by the iron sheet is given by: qFe = mFesFeΔt = (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? 6.36 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 6.2 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: qH 2O + qAl = 0 or qH 2O = − qAl The heat gained by water is given by: qH 2O = mH2O sH2O Δt = mH 2O (4.184 J/g ⋅°C)(24.9 − 23.4)°C where m and s are the mass and specific heat, and Δt = tfinal − tinitial. The heat lost by the aluminum is given by: qAl = mAl sAl Δt = (12.1 g)(0.900 J/g ⋅°C)(24.9 − 81.7)°C CHAPTER 6: THERMOCHEMISTRY 171 Substituting into the equation derived above, we can solve for mH 2O . qH 2O = − qAl mH 2O (4.184 J/g ⋅°C)(24.9 − 23.4)°C = − (12.1 g)(0.900 J/g ⋅°C)(24.9 − 81.7)°C (6.28)(mH 2O ) = 619 mH 2O = 98.6 g 6.37 The heat gained by the calorimeter is: q = CpΔt q = (3024 J/°C)(1.126°C) = 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 = msolnssolnΔt where m and s are the mass and specific heat of the solution and Δt = 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/g⋅°C)(tf − 20.48°C) 172 CHAPTER 6: THERMOCHEMISTRY 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: ⎯→ 2H2O(l) + BaCl2(aq) 2HCl(aq) + Ba(OH)2(aq) ⎯ There is exactly enough Ba(OH)2 to neutralize all the HCl. Note that 2 mole HCl Q 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 + 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/g⋅°C)(tf − 20.48°C) = −(−9.67 × 10 J) 3 4 3 (1.67 × 10 )tf − (3.43 × 10 ) = 9.67 × 10 J tf = 26.3°C 6.45 CH4(g) and H(g). All the other choices are elements in their most stable form ( ΔH fD = 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 ΔH fD (O 2 ) = 0, O2 is the more stable form of the element oxygen at this temperature. 6.47 H2O(l) → H2O(g) Endothermic D ΔH rxn = ΔH fD [H 2 O( g )] − ΔH fD [H 2 O(l )] > 0 D ΔH fD [H 2 O(l )] is more negative since ΔH rxn > 0. You could also solve the problem by realizing that H2O(l) is the stable form of water at 25°C, and therefore will have the more negative ΔH fD value. 6.48 (a) Br2(l) is the most stable form of bromine at 25°C; therefore, ΔH fD [Br2 (l )] = 0. Since Br2(g) is less stable than Br2(l), ΔH fD [Br2 ( g )] > 0. (b) I2(s) is the most stable form of iodine at 25°C; therefore, ΔH fD [I 2 ( s )] = 0. Since I2(g) is less stable than I2(s), ΔH fD [I2 ( g )] > 0. CHAPTER 6: THERMOCHEMISTRY 6.49 173 2H2O2(l) → 2H2O(l) + O2(g) Because H2O(l) has a more negative ΔH fD than H2O2(l). 6.50 Strategy: What is the reaction for the formation of Ag2O from its elements? What is the ΔHfD 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 ⎯→ Ag2O(s) O2(g) ⎯ 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: D ΔH rxn = ∑ nΔH fD (products) − ∑ mΔH fD (reactants) D ΔH rxn = [ΔH fD (Ag 2 O)] − [2ΔH fD (Ag) + 1 ΔH D (O )] f 2 2 D ΔH rxn = [ΔH fD (Ag 2 O)] − [0 + 0] D ΔH fD (Ag 2O) = ΔH rxn D In a similar manner, you should be able to show that ΔH fD (CaCl 2 ) = ΔH rxn for the reaction ⎯→ CaCl2(s) Ca(s) + Cl2(g) ⎯ 6.51 ΔH ° = [ΔH fD (CaO) + ΔH fD (CO 2 )] − ΔH fD (CaCO3 ) ΔH° = [(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, ΔH fD , of the species. Solution: We use the ΔHfD values in Appendix 3 and Equation (6.18) of the text. D ΔH rxn = ∑ nΔH fD (products) − ∑ mΔH fD (reactants) (a) + − HCl(g) → H (aq) + Cl (aq) D ΔH rxn = ΔH fD (H + ) + ΔH fD (Cl− ) − ΔH fD (HCl) − −74.9 kJ/mol = 0 + ΔH fD (Cl ) − (1)(−92.3 kJ/mol) ΔH fD (Cl − ) = − 167.2 kJ/mol (b) The neutralization reaction is: + and, − H (aq) + OH (aq) → H2O(l) D ΔH rxn = ΔH fD [H 2 O(l )] − [ΔH fD (H + ) + ΔH fD (OH − )] 174 CHAPTER 6: THERMOCHEMISTRY ΔH fD [H 2 O(l )] = − 285.8 kJ/mol (See Appendix 3 of the text.) D ΔH rxn = (1)(−285.8 kJ/mol) − [(1)(0 kJ/mol) + (1)(−229.6 kJ/mol)] = −56.2 kJ/mol 6.53 (a) ΔH ° = 2ΔH fD (H 2 O) − 2ΔH fD (H 2 ) − ΔH fD (O2 ) ΔH° = (2)(−285.8 kJ/mol) − (2)(0) − (1)(0) = −571.6 kJ/mol (b) ΔH ° = 4ΔH fD (CO 2 ) + 2ΔH fD (H 2 O) − 2ΔH fD (C2 H 2 ) − 5ΔH fD (O 2 ) ΔH° = (4)(−393.5 kJ/mol) + (2)(−285.8 kJ/mol) − (2)(226.6 kJ/mol) − (5)(0) = −2599 kJ/mol 6.54 (a) ΔH ° = [2ΔH fD (CO 2 ) + 2ΔH fD (H 2 O)] − [ΔH fD (C2 H 4 ) + 3ΔH fD (O 2 )] ] ΔH° = [(2)(−393.5 kJ/mol) + (2)(−285.8 kJ/mol)] − [(1)(52.3 kJ/mol) + (3)(0)] ΔH° = −1411 kJ/mol (b) ΔH ° = [2ΔH fD (H 2 O) + 2ΔH fD (SO2 )] − [2ΔH fD (H 2S) + 3ΔH fD (O 2 )] ΔH° = [(2)(−285.8 kJ/mol) + (2)(−296.1 kJ/mol)] − [(2)(−20.15 kJ/mol) + (3)(0)] ΔH° = −1124 kJ/mol 6.55 6.56 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 (c) −33.4 kJ 60.09 g × = − 2.01 × 103 kJ/mol 1g 1 mol D ΔH rxn = ∑ nΔH fD (products) − ∑ mΔH fD (reactants) The reaction is: ⎯→ H(g) + H(g) H2(g) ⎯ and, D ΔH rxn = [ΔH fD (H) + ΔH fD (H)] − ΔH fD (H 2 ) ΔH fD (H 2 ) = 0 D ΔH rxn = 436.4 kJ/mol = 2ΔH fD (H) − (1)(0) ΔH fD (H) = 6.57 436.4 kJ/mol = 218.2 kJ/mol 2 ΔH ° = 6ΔH fD (CO2 ) + 6ΔH fD (H 2 O) − [ΔH fD (C6 H12 ) + 9ΔH fD (O2 )] ΔH° = (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? CHAPTER 6: THERMOCHEMISTRY 6.58 175 Using the ΔH fD values in Appendix 3 and Equation (6.18) of the text, we write D ΔH rxn = [5ΔH fD (B2 O3 ) + 9ΔH fD (H 2 O)] − [2ΔH fD (B5 H9 ) + 12ΔH fD (O 2 )] ΔH° = [(5)(−1263.6 kJ/mol) + (9)(−285.8 kJ/mol)] − [(2)(73.2 kJ/mol) + (l2)(0 kJ/mol)] ΔH° = −9036.6 kJ/mol Looking at the balanced equation, this is the amount of heat released for every 2 moles of B5H9 reacted. We can use the following ratio 9036.6 kJ 2 mol B5 H9 to convert to kJ/g B5H9. The molar mass of B5H9 is 63.12 g, so heat released per gram B5 H9 = 6.59 The amount of heat given off is: (1.26 × 104 g NH3 ) × 6.60 1 mol B5 H9 9036.6 kJ × = 71.58 kJ / g B5 H 9 2 mol B5 H9 63.12 g B5 H9 1 mol NH3 92.6 kJ × = 3.43 × 104 kJ 17.03 g NH3 2 mol NH3 D ΔH rxn = ∑ nΔH fD (products) − ∑ mΔH fD (reactants) The balanced equation for the reaction is: ⎯→ CaO(s) + CO2(g) CaCO3(s) ⎯ D ΔH rxn = [ΔH fD (CaO) + ΔH fD (CO 2 )] − ΔH fD (CaCO3 ) D ΔH rxn = [(1)(−635.6 kJ/mol) + (1)( −393.5 kJ/mol)] − (1)( −1206.9 kJ/mol) = 177.8 kJ/mol 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. ΔH ° = 66.8 g CO2 × 6.61 1 mol CO2 177.8 kJ × = 2.70 × 102 kJ 44.01 g CO 2 1 mol CO2 ΔH° (kJ/mol) Reaction S(rhombic) + O2(g) → SO2(g) SO2(g) → S(monoclinic) + O2(g) S(rhombic) → S(monoclinic) −296.06 296.36 D ΔH 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). 176 CHAPTER 6: THERMOCHEMISTRY Solution: Here is the equation for the formation of C2H6 from its elements. D ΔH 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) D ΔH rxn = 2(−393.5 kJ/mol) = − 787.0 kJ/mol ⎯→ 2CO2(g) 2C(graphite) + 2O2(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 D ΔH rxn = 3(−285.8 kJ/mol) = − 857.4 kJ/mol ⎯→ 3H2O(l) O2(g) ⎯ 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) ⎯→ C2H6(g) + 2CO2(g) + 3H2O(l) ⎯ 7 2 O2(g) D ΔH rxn = 1 (3119.6 2 kJ/mol) = 1559.8 kJ/mol Adding Equations (d), (e), and (f) together, we have: ΔH° (kJ/mol) Reaction (d) ⎯→ 2CO2(g) 2C(graphite) + 2O2(g) ⎯ (e) 3H2(g) + (f) ⎯→ C2H6(g) + 2CO2(g) + 3H2O(l) ⎯ 3 2 −787.0 ⎯→ 3H2O(l) O2(g) ⎯ −857.4 7 2 ⎯→ C2H6(g) 2C(graphite) + 3H2(g) ⎯ 6.63 O2(g) 1559.8 ΔH° = −84.6 kJ/mol ΔH° (kJ/mol) Reaction CO2(g) + 2H2O(l) → CH3OH(l) + 3 2 O2(g) C(graphite) + O2(g) → CO2(g) 2H2(g) + O2(g) → 2H2O(l) C(graphite) + 2H2(g) + 1 2 O2(g) → CH3OH(l) 726.4 −393.5 2(−285.8) D ΔH rxn = − 238.7 kJ/mol D We have just calculated an enthalpy at standard conditions, which we abbreviate ΔH rxn . In this case, the reaction in question was for the formation of one mole of CH3OH from its elements in their standard state. D Therefore, the ΔH rxn that we calculated is also, by definition, the standard heat of formation ΔH fD of CH3OH (−238.7 kJ/mol). 6.64 The second and third equations can be combined to give the first equation. 2Al(s) + 3 2 ⎯→ Al2O3(s) O2(g ) ⎯ ⎯→ 2Fe(s) + Fe2O3(s) ⎯ 3 2 O2(g) ⎯→ 2Fe(s) + Al2O3(s) 2Al(s) + Fe2O3(s) ⎯ ΔH° = −1669.8 kJ/mol ΔH° = 822.2 kJ/mol ΔH° = −847.6 kJ/mol CHAPTER 6: THERMOCHEMISTRY 177 6.71 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. 6.72 Rearrange the equations as necessary so they can be added to yield the desired equation. 6.73 ⎯→ A 2B ⎯ −ΔH1 ⎯→ C A ⎯ ΔH2 ⎯→ C 2B ⎯ ΔH° = ΔH2 − ΔH1 The reaction corresponding to standard enthalpy of formation, ΔH fD , 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 D ΔH rxn provided. AgNO3(s) → AgNO2(s) + D ΔH rxn = ΔH fD (AgNO2 ) + 1 2 O2(g) 1 ΔH D (O ) f 2 2 − ΔH fD (AgNO3 ) 78.67 kJ/mol = ΔH fD (AgNO2) + 0 − (−123.02 kJ/mol) ΔH fD (AgNO 2 ) = − 44.35 kJ/mol 6.74 (a) D ΔH rxn = ∑ nΔH fD (products) − ∑ mΔH fD (reactants) D ΔH rxn = [4ΔH fD (NH3 ) + ΔH fD (N 2 )] − 3ΔH fD (N 2 H 4 ) D ΔH rxn = [(4)(−46.3 kJ/mol) + (0)] − (3)(50.42 kJ/mol) = − 336.5 kJ/mol (b) The balanced equations are: (1) ⎯→ N2(g) + 2H2O(l) N2H4(l) + O2(g) ⎯ (2) ⎯→ 2N2(g) + 6H2O(l) 4NH3(g) + 3O2(g) ⎯ The standard enthalpy change for equation (1) is: D ΔH rxn = ΔH fD (N 2 ) + 2ΔH fD [H 2 O(l )] − {ΔH fD [N 2 H 4 (l )] + ΔH fD (O2 )} D 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: D ΔH rxn = [2ΔH fD (N 2 ) + 6ΔH fD (H 2 O)] − [4ΔH fD (NH3 ) + 3ΔH fD (O2 )] D ΔH rxn = [(2)(0) + (6)(−285.8 kJ/mol)] − [(4)(−46.3 kJ/mol) + (3)(0)] = −1529.6 kJ/mol 178 CHAPTER 6: THERMOCHEMISTRY D We can now calculate the enthalpy change per kilogram of each substance. ΔH rxn above is in units of kJ/mol. We need to convert to kJ/kg. D N 2 H 4 (l ): ΔH 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 D NH3 ( g ): ΔH rxn = 1 mol NH3 −1529.6 kJ 1000 g × × = − 2.245 × 104 kJ/kg NH 3 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. 6.75 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 = − PΔV = − (1 atm)(−98 L) = 98 L ⋅ atm × ΔH = ΔE + PΔV or 101.3 J = 9.9 × 103 J = 9.9 kJ 1 L ⋅ atm ΔE = ΔH − PΔV Using ΔH 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 −PΔV as 9.9 kJ (for which we just solved): ΔE = −185.2 kJ + 9.9 kJ = −175.3 kJ 6.76 The reaction is, 2Na(s) + Cl2(g) → 2NaCl(s). First, let's calculate ΔH° for this reaction using ΔH fD values in Appendix 3. D ΔH rxn = 2ΔH fD (NaCl) − [2ΔH fD (Na) + ΔH fD (Cl2 )] D ΔH 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 D Finally, we can use the ΔH rxn calculated above to find the amount of heat released when 0.0530 mole of Cl2 reacts. −822.0 kJ 0.0530 mol Cl2 × = − 43.6 kJ 1 mol Cl2 The amount of heat released is 43.6 kJ. 6.77 (a) D Although we cannot measure ΔH rxn for this reaction, the reverse process, is the combustion of glucose. D We could easily measure ΔH rxn for this combustion in a bomb calorimeter. → 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ⎯⎯ CHAPTER 6: THERMOCHEMISTRY (b) 179 D We can calculate ΔH rxn using standard enthalpies of formation. D ΔH rxn = ΔH fD [C6 H12 O6 ( s)] + 6ΔH fD [O 2 ( g )] − {6ΔH fD [CO 2 ( g )] + 6ΔH fD [H 2 O(l )]} D ΔH rxn = [(1)(−1274.5 kJ/mol) + 0] − [(6)(−393.5 kJ/mol) + (6)(−285.8 kJ/mol)] = 2801.3 kJ/mol D ΔH rxn has units of kJ/1 mol glucose. We want the ΔH° change for 7.0 × 10 14 calculate how many moles of glucose are in 7.0 × 10 following strategy to solve the problem. 14 kg glucose. We need to kg glucose. You should come up with the kg glucose → g glucose → mol glucose → kJ (ΔH°) ΔH ° = (7.0 × 1014 kg) × 1 mol C6 H12 O6 1000 g 2801.3 kJ × × = 1.1 × 1019 kJ 1 kg 180.2 g C6 H12 O6 1 mol C6 H12 O6 6.78 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. 6.79 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 × 10−3 mol H 2 22.99 g Na 2 mol Na Using the ideal gas equation, we write: V = nRT (7.4 × 10−3 mol)(0.0821 L ⋅ atm / K ⋅ mol)(273 K) = = 0.17 L H 2 P (1 atm) ΔV = 0.17 L w = − PΔV = − (1.0 atm)(0.17 L) = − 0.17 L ⋅ atm × 6.80 ⎯→ HBr(g) H(g) + Br(g) ⎯ 101.3 J = − 17 J 1 L ⋅ atm D ΔH 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) ⎯→ HBr(g) Br2(g) ⎯ ⎯→ HBr(g) H(g) + Br(g) ⎯ 6.81 D ΔH rxn = 1 (−436.4 2 kJ/mol) = − 218.2 kJ/mol D ΔH rxn = 1 (−192.5 2 kJ/mol) = − 96.25 kJ/mol D ΔH rxn = 1 (−72.4 2 kJ/mol) = − 36.2 kJ/mol ΔH° = −350.7 kJ/mol Using the balanced equation, we can write: D ΔH rxn = [2ΔH fD (CO 2 ) + 4ΔH fD (H 2 O)] − [2ΔH fD (CH3OH) + 3ΔH fD (O 2 )] −1452.8 kJ/mol = (2)(−393.5 kJ/mol) + (4)(−285.8 kJ/mol) − (2) ΔH fD (CH3OH) − (3)(0 kJ/mol) 477.4 kJ/mol = −(2) ΔH fD (CH3OH) ΔH fD (CH3OH) = −238.7 kJ/mol 180 CHAPTER 6: THERMOCHEMISTRY 6.82 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/g⋅°C)(28.4 − 24.0)°C + (12.4 J/°C)(28.4 − 24.0)°C = 0 3 3 (−3.11 × 10 )smetal (g⋅°C) = −1.53 × 10 J smetal = 0.492 J/g⋅°C 6.83 The reaction is: 2CO + 2NO → 2CO2 + N2 The limiting reagent is CO (NO is in excess). ΔH ° = [2ΔH fD (CO2 ) + ΔH fD (N 2 )] − [2ΔH fD (CO) + 2ΔH fD (NO)] ΔH° = [(2)(−393.5 kJ/mol) + (1)(0)] − [(2)(−110.5 kJ/mol) + (2)(90.4 kJ/mol)] = −746.8 kJ/mol 6.84 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 ⎯→ CO(g) O2(g) ⎯ This reaction corresponds to the standard enthalpy of formation of CO, so we use the value of −110.5 kJ/mol (see Appendix 3 of the text). Calculate the standard enthalpy for the reaction: ⎯→ CO(g) + H2(g) C(s) + H2O(g) ⎯ D ΔH rxn = [ΔH fD (CO) + ΔH fD (H 2 )] − [ΔH fD (C) + ΔH fD (H 2 O)] D ΔH 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. 6.85 Let's start with the combustion of methane: CH4(g) + O2(g) → CO2(g) + 2H2O(l) D ΔH rxn = ΔH fD (CO 2 ) + 2ΔH fD (H 2 O) − ΔH fD (CH 4 ) D ΔH rxn = (1)(−393.5 kJ/mol) + (2)(−285.8 kJ/mol) − (1)(−74.85 kJ/mol) = − 890.3 kJ/mol Now, let's calculate the heat produced by the combustion of water gas. We will consider the combustion of H2 and CO separately. We can look up the ΔH fD of H2O(l) in Appendix 3. H2(g) + 1 2 O2(g) → H2O(l) D ΔH rxn = − 285.8 kJ/mol CHAPTER 6: THERMOCHEMISTRY 181 D For the combustion of CO(g), we use ΔH fD values from Appendix 3 to calculate the ΔH rxn . CO(g) + 1 2 D ΔH rxn = ? O2(g) → CO2(g) D ΔH rxn = ΔH fD (CO 2 ) − ΔH fD (CO) − 1 ΔH D (O ) f 2 2 D ΔH rxn = (1)(−393.5 kJ/mol) − (1)(−110.5 kJ/mol) = − 283.0 kJ/mol D The ΔH rxn values calculated above are for the combustion of 1 mole of H2 and 1 mole of CO, which equals 2 moles of water gas. The total heat produced during the combustion of 1 mole of water gas is: (285.8 + 283.0)kJ/mol = 284.4 kJ / mol of water gas 2 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. 6.86 First, calculate the energy produced by 1 mole of octane, C8H18. C8H18(l) + 25 2 ⎯→ 8CO2(g) + 9H2O(l) O2(g) ⎯ D ΔH rxn = 8ΔH fD (CO2 ) + 9ΔH fD [H 2 O(l )] − [ΔH fD (C8 H18 ) + 25 ΔH D (O )] f 2 2 D ΔH rxn = [(8)(−393.5 kJ/mol) + (9)(−285.8 kJ/mol)] − [(1)(−249.9 kJ/mol) + ( 25 )(0)] 2 = −5470 kJ/mol D The problem asks for the energy produced by the combustion of 1 gallon of octane. ΔH 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: ΔH ° = −5470 kJ 1 mol octane 2660 g × × = − 1.274 × 105 kJ / gal 1 mol octane 114.2 g octane 1 gal The combustion of hydrogen corresponds to the standard heat of formation of water: H2(g) + 1 2 ⎯→ H2O(l) O2(g) ⎯ D Thus, ΔH rxn is the same as ΔH fD 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 Finally, use the ideal gas law to calculate the volume of gas corresponding to 445.8 moles of H2 at 25°C and 1 atm. L ⋅ atm ⎞ ⎛ (445.8 mol) ⎜ 0.0821 (298 K) nH2 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 25°C! 182 CHAPTER 6: THERMOCHEMISTRY 6.87 The reaction for the combustion of octane is: C8H18(l) + 25 2 O2(g) → 8CO2(g) + 9H2O(l) D ΔH rxn for this reaction was calculated in problem 6.86 from standard enthalpies of formation. D ΔH rxn = − 5470 kJ/mol Heat/gal of octane = 1 mol C8 H18 0.7025 g 3785 mL 5470 kJ × × × 1 mol C8 H18 114.2 g 1 mL 1 gal 5 Heat/gal of octane = 1.27 × 10 kJ/gal gasoline The reaction for the combustion of ethanol is: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) D ΔH rxn = 2ΔH fD (CO 2 ) + 3ΔH fD (H 2 O) − ΔH fD (C2 H5 OH) − 3ΔH fD (O 2 ) D ΔH rxn = (2)(−393.5 kJ/mol) + (3)(−285.8 kJ/mol) − (1)(−277.0 kJ/mol) = −1367 kJ/mol Heat/gal of ethanol = 1 mol C2 H5OH 0.7894 g 3785 mL 1367 kJ × × × 1 mol C2 H5 OH 46.07 g 1 mL 1 gal 4 Heat/gal of ethanol = 8.87 × 10 kJ/gal ethanol For ethanol, what would the cost have to be to supply the same amount of heat per dollar of gasoline? For 5 gasoline, it cost $4.50 to provide 1.27 × 10 kJ of heat. $4.50 1.27 × 105 kJ 6.88 × 8.87 × 104 kJ = $3.14 / gal ethanol 1 gal ethanol The combustion reaction is: C2H6(l) + 7 2 ⎯→ 2CO2(g) + 3H2O(l) O2(g) ⎯ The heat released during the combustion of 1 mole of ethane is: D ΔH rxn = [2ΔH fD (CO2 ) + 3ΔH fD (H 2 O)] − [ΔH fD (C2 H6 ) + 7 ΔH D (O )] f 2 2 D ΔH rxn = [(2)(−393.5 kJ/mol) + (3)(−285.8 kJ/mol)] − [(1)(−84.7 kJ/mol + ( 72 )(0)] = −1560 kJ/mol The heat required to raise the temperature of the water to 98°C is: 5 q = mH 2O sH 2O Δt = (855 g)(4.184 J/g⋅°C)(98.0 − 25.0)°C = 2.61 × 10 J = 261 kJ The combustion of 1 mole of ethane produces 1560 kJ; the number of moles required to produce 261 kJ is: 261 kJ × 1 mol ethane = 0.167 mol ethane 1560 kJ The volume of ethane is: Vethane nRT = = P L ⋅ atm ⎞ ⎛ (0.167 mol) ⎜ 0.0821 (296 K) ⋅ K ⎟⎠ mol ⎝ = 4.10 L ⎛ 1 atm ⎞ ⎜ 752 mmHg × ⎟ 760 mmHg ⎠ ⎝ CHAPTER 6: THERMOCHEMISTRY 6.89 As energy consumers, we are interested in the availability of usable energy. 6.90 The heat gained by the liquid nitrogen must be equal to the heat lost by the water. 183 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 = qH 2O = mH 2O sH 2O Δt 2 4 q H 2O = (2.00 × 10 g)(4.184 J/g⋅°C)(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. ΔH 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 6.91 The evaporation of ethanol is an endothermic process with a fairly high ΔH° value. When the liquid evaporates, it absorbs heat from your body, hence the cooling effect. 6.92 Recall that the standard enthalpy of formation ( ΔH fD ) is defined as the heat change that results when 1 mole D of a compound is formed from its elements at a pressure of 1 atm. Only in choice (a) does ΔH rxn = ΔH fD . In choice (b), C(diamond) is not the most stable form of elemental carbon under standard conditions; C(graphite) is the most stable form. 6.93 −3 w = −PΔV = −(1.0 atm)(0.0196 − 0.0180)L = −1.6 × 10 L⋅atm Using the conversion factor 1 L⋅atm = 101.3 J: w = ( −1.6 × 10−3 L ⋅ atm) × 101.3 J = − 0.16 J L ⋅ atm 0.16 J of work are done by water as it expands on freezing. 6.94 (a) No work is done by a gas expanding in a vacuum, because the pressure exerted on the gas is zero. (b) w = −PΔV w = −(0.20 atm)(0.50 − 0.050)L = −0.090 L⋅atm Converting to units of joules: w = − 0.090 L ⋅ atm × 101.3 J = − 9.1 J L ⋅ atm 184 CHAPTER 6: THERMOCHEMISTRY (c) 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. L ⋅ atm ⎞ ⎛ (0.020 mol) ⎜ 0.0821 (273 + 20)K nRT mol ⋅ K ⎟⎠ ⎝ V = = = 2.4 L P 0.20 atm The amount of work done is: w = −PΔV = (0.20 atm)(2.4 − 0.050)L = −0.47 L⋅atm Converting to units of joules: w = − 0.47 L ⋅ atm × 6.95 101.3 J = − 48 J L ⋅ atm The equation corresponding to the standard enthalpy of formation of diamond is: → C(diamond) C(graphite) ⎯⎯ Adding the equations: → CO2(g) C(graphite) + O2(g) ⎯⎯ ΔH° = −393.5 kJ/mol → C(diamond) + O2(g) CO2(g) ⎯⎯ ΔH° = 395.4 kJ/mol → C(diamond) C(graphite) ⎯⎯ ΔH° = 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? 6.96 (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) 6.97 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: → 2C2H5OH(l) + 2CO2(g) C6H12O6(s) ⎯⎯ D ΔH rxn = [2ΔH fD (C2 H5 OH) + 2ΔH fD (CO 2 )] − ΔH fD (C6 H12 O6 ) D ΔH rxn = [(2)(−276.98 kJ/mol) + (2)(−393.5 kJ/mol)] − (1)(−1274.5 kJ/mol) = −66.5 kJ/mol 6.98 4Fe(s) + 3O2(g) → 2Fe2O3(s). This equation represents twice the standard enthalpy of formation of Fe2O3. From Appendix 3, the standard enthalpy of formation of Fe2O3 = −822.2 kJ/mol. So, ΔH° for the given reaction is: D ΔH rxn = (2)(−822.2 kJ/mol) = − 1644 kJ/mol CHAPTER 6: THERMOCHEMISTRY 185 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 ΔH° 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. 6.99 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 × 6.100 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 1°C is: 4.184 J 1g 1000 mL × × × 1°C = 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 20 The amount of heat needed to raise the temperature of 7.2 × 10 (7.2 × 1020 L) × L of water is: 4184 J = 3.0 × 1024 J 1L Finally, we can calculate the number of atomic bombs needed to produce this much heat. (3.0 × 1024 J) × 6.101 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 P 0.995 atm w = −PΔV = −(0.995 atm)(10.6 L) = −10.5 L⋅atm 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. 186 CHAPTER 6: THERMOCHEMISTRY 6.102 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 = mwaterswaterΔt 4 qwater = (2000 g)(4.184 J/g⋅°C)(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. ΔHrxn 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 × × = − 52.49 kJ 122.1 g benzoic acid 1 mol benzoic acid And, qcal = −qrxn − qwater qcal = 52.49 kJ − 32.0 kJ = 20.5 kJ To calculate the heat capacity of the bomb calorimeter, we can use the following equation: qcal = CcalΔt Ccal = 6.103 qcal 20.5 kJ = = 5.35 kJ/ °C Δt (25.67 − 21.84)°C From thermodynamic data in Appendix 3 of the text, we can calculate the amount of heat released per mole of H2 and CH4. H2(g) + ½ O2(g) → H2O(l) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH° = −285.8 kJ/mol ΔH° = −890.3 kJ/mol We know that 2354 kJ of heat is released from the combustion of 25.0 g of the gaseous mixture of H2 and CH4. mH 2 + mCH 4 = 25.0 g and mCH 4 = 25.0 g − mH 2 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. ⎛ 1 mol H 2 1 mol CH 4 285.8 kJ ⎞ ⎛ 890.3 kJ ⎞ × × ⎜ mH 2 × ⎟ + ⎜ (25.0 − mH 2 ) × ⎟ = 2354 kJ 2.016 g H 1 mol H 16.04 g CH 1 mol CH 4 ⎠ ⎝ 2 2⎠ ⎝ 4 141.8 mH 2 + (1.39 × 103 ) − 55.50 mH 2 = 2354 86.30 mH2 = 964 mH 2 = 11.2 g mCH4 = 25.0 g − 11.2 g = 13.8 g CHAPTER 6: THERMOCHEMISTRY 6.104 First, let’s calculate the standard enthalpy of reaction. D ΔH rxn = 2ΔH fD (CaSO4 ) − [2ΔH fD (CaO) + 2ΔH fD (SO 2 ) + ΔH fD (O2 )] = (2)(−1432.7 kJ/mol) − [(2)(−635.6 kJ/mol) + (2)(−296.1 kJ/mol) + 0] = −1002 kJ/mol 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. 1 mol SO 2 −1002 kJ × = − 5.2 × 106 kJ 64.07 g SO2 2 mol SO 2 (6.6 × 105 g SO 2 ) × 6.105 Volume of room = (2.80 m × 10.6 m × 17.2 m) × 1000 L 1 m3 = 5.10 × 105 L PV = nRT PV (1.0 atm)(5.10 × 105 L) = = 2.04 × 104 mol air RT (0.0821 L atm / K ⋅ mol)(32 + 273)K nair = mass air = (2.04 × 104 mol air) × 29.0 g air = 5.9 × 105 g air 1 mol air Heat to be removed from air: q = mairsairΔt 5 q = (5.9 × 10 g)(1.2 J/g⋅°C)(−8.2°C) 6 3 q = −5.8 × 10 J = −5.8 × 10 kJ Conservation of energy: qair + qsalt = 0 qair + nΔHfus = 0 qair + msalt ΔH fus = 0 Msalt msalt = −qair Msalt ΔH fus msalt = −(−5.8 × 103 kJ)(322.3 g/mol) = 2.5 × 104 g 74.4 kJ/mol msalt = 25 kg 6.106 First, we need to calculate the volume of the balloon. V = 4 3 4 1000 L πr = π(8 m)3 = (2.1 × 103 m3 ) × = 2.1 × 106 L 3 3 3 1m 187 188 CHAPTER 6: THERMOCHEMISTRY (a) We can calculate the mass of He in the balloon using the ideal gas equation. nHe ⎛ ⎞ 1 atm ⎜ 98.7 kPa × ⎟ (2.1 × 106 L) 2 ⎜ ⎟ × 1.01325 10 kPa PV ⎠ = = ⎝ = 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 = −PΔV ⎛ ⎞ 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 6.107 The heat produced by the reaction heats the solution and the calorimeter: qrxn = −(qsoln + qcal) qsoln = msΔt = (50.0 g)(4.184 J/g⋅°C)(22.17 − 19.25)°C = 611 J qcal = CΔt = (98.6 J/°C)(22.17 − 19.25)°C = 288 J −qrxn = (qsoln + qcal) = (611 + 288)J = 899 J The 899 J produced was for 50.0 mL of a 0.100 M AgNO3 solution. 50.0 mL × 0.100 mol Ag + = 5.00 × 10−3 mol Ag + 1000 mL soln On a molar basis the heat produced was: 899 J 5.00 × 10−3 mol Ag + = 1.80 × 105 J/mol Ag + = 180 kJ/mol Ag + + The balanced equation involves 2 moles of Ag , so the heat produced is 2 mol × 180 kJ/mol = 360 kJ Since the reaction produces heat (or by noting the sign convention above), then: + qrxn = −360 kJ/mol Zn (or −360 kJ/2 mol Ag ) 6.108 (a) The heat needed to raise the temperature of the water from 3°C to 37°C can be calculated using the equation: q = msΔt 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 = msΔt = (1.0 × 10 g)(4.184 J/g⋅°C)(37 − 3)°C = 1.4 × 10 J = 1.4 × 10 kJ CHAPTER 6: THERMOCHEMISTRY (b) 189 We need to calculate both the heat needed to melt the snow and also the heat needed to heat liquid water form 0°C to 37°C (normal body temperature). The heat needed to melt the snow is: 1 mol 6.01 kJ × = 2.7 × 102 kJ 18.02 g 1 mol (8.0 × 102 g) × The heat needed to raise the temperature of the water from 0°C to 37°C is: 2 5 2 q = msΔt = (8.0 × 10 g)(4.184 J/g⋅°C)(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 6.109 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, 1 mu 2 2 q = Thus the amount of heat generated must be proportional to the braking distance, d: d∝q 2 d∝u 2 2 Therefore, as u increases to 2u, d increases to (2u) = 4u which is proportional to 4d. 6.110 (a) ΔH ° = ΔH fD (F− ) + ΔH fD (H 2 O) − [ΔH fD (HF) + ΔH fD (OH − )] ΔH° = [(1)(−329.1 kJ/mol) + (1)(−285.8 kJ/mol)] − [(1)(−320.1 kJ/mol) + (1)(−229.6 kJ/mol) ΔH° = −65.2 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. − − ΔH° = −65.2 kJ/mol − ΔH° = +56.2 kJ/mol ⎯→ F (aq) + H2O(l) HF(aq) + OH (aq) ⎯ + ⎯→ H (aq) + OH (aq) H2O(l) ⎯ + − ⎯→ H (aq) + F (aq) HF(aq) ⎯ ΔH° = −9.0 kJ/mol 6.111 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. 6.112 The equation we are interested in is the formation of CO from its elements. C(graphite) + 1 2 ⎯→ CO(g) O2(g) ⎯ ΔH° = ? Try to add the given equations together to end up with the equation above. ⎯→ CO2(g) C(graphite) + O2(g) ⎯ ⎯→ CO(g) + CO2(g) ⎯ C(graphite) + 1 2 1 2 O2(g) ⎯→ CO(g) O2(g) ⎯ ΔH° = −393.5 kJ/mol ΔH° = +283.0 kJ/mol ΔH° = −110.5 kJ/mol 190 CHAPTER 6: THERMOCHEMISTRY We cannot obtain ΔH fD for CO directly, because burning graphite in oxygen will form both CO and CO2. 6.113 Energy intake for mechanical work: 0.17 × 500 g × 3000 J = 2.6 × 105 J 1g 5 2.6 × 10 J = mgh 2 −2 1 J = 1 kg⋅m s 2.6 × 105 kg ⋅ m 2 s 2 = (46 kg)(9.8 m/s 2 )h 2 h = 5.8 × 10 m 6.114 (a) mass = 0.0010 kg Potential energy = mgh 2 = (0.0010 kg)(9.8 m/s )(51 m) Potential energy = 0.50 J (b) Kinetic energy = 1 mu 2 = 0.50 J 2 1 (0.0010 kg)u 2 = 0.50 J 2 2 3 2 2 u = 1.0 × 10 m /s u = 32 m/s (c) q = msΔt 0.50 J = (1.0 g)(4.184 J/g°C)Δt Δt = 0.12°C 6.115 For Al: (0.900 J/g⋅°C)(26.98 g) = 24.3 J/°C This law does not hold for Hg because it is a liquid. 6.116 The reaction we are interested in is the formation of ethanol from its elements. 2C(graphite) + 1 2 ⎯→ C2H5OH(l) O2(g) + 3H2(g) ⎯ 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: ⎯→ C2H5OH(l) + 3O2 (g) 2CO2(g) + 3H2O(l) ⎯ ΔH° = +1367.4 kJ/mol We need to add equations to add C (graphite) and remove H2O from the reactants side of the equation. We write: CHAPTER 6: THERMOCHEMISTRY ⎯→ C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ⎯ ΔH° = +1367.4 kJ/mol ⎯→ 2CO2(g) 2C(graphite) + 2O2(g) ⎯ ΔH° = 2(−393.5 kJ/mol) 3H2(g) + 3 2 ⎯→ 3H2O(l) O2(g) ⎯ 2C(graphite) + 6.117 1 2 C6H6(l) + 15 2 (b) C2H2(g) + 5 2 (c) C(graphite) + O2 → CO2(g) (d) H2(g) + 1 2 ΔH° = 3(−285.8 kJ/mol) ΔH fD = −277.0 kJ/mol ⎯→ C2H5OH(l) O2(g) + 3H2(g) ⎯ (a) O2(g) → 6CO2(g) + 3H2O(l) ΔH° = −3267.4 kJ/mol O2(g) → 2CO2(g) + H2O(l) ΔH° = −1299.4 kJ/mol O2(g) → H2O(l) 191 ΔH° = −393.5 kJ/mol ΔH° = −285.8 kJ/mol Using Hess’s Law, we can add the equations in the following manner to calculate the standard enthalpies of formation of C2H2 and C6H6. C2H2: − (b) + 2(c) + (d) 2C(graphite) + H2(g) → C2H2(g) ΔH° = +226.6 kJ/mol Therefore, ΔH fD (C2 H 2 ) = 226.6 kJ/mol C6H6: − (a) + 6(c) + 3(d) 6C(graphite) + 3H2(g) → C6H6(l) ΔH° = 49.0 kJ/mol Therefore, ΔH fD (C6 H 6 ) = 49.0 kJ/mol Finally: 3C2H2(g) → C6H6(l) ΔHrxn = (1)(49.0 kJ/mol) − (3)(226.6 kJ/mol) = −630.8 kJ/mol 6.118 Heat gained by ice = Heat lost by the soft drink mice × 334 J/g = −msdssdΔt mice × 334 J/g = −(361 g)(4.184 J/g⋅°C)(0 − 23)°C mice = 104 g 6.119 The heat required to heat 200 g of water (assume d = 1 g/mL) from 20°C to 100°C is: q = msΔt 4 q = (200 g)(4.184 J/g⋅°C)(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. 192 CHAPTER 6: THERMOCHEMISTRY Use standard enthalpies of formation (see Appendix 3) to calculate the heat of combustion of methane. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH° = −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 The volume of 0.15 mole CH4 at 1 atm and 20°C 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 × 10−3 ×⎜ ×⎜ × = ⎟ ⎟ 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 × 6.120 $3.1 × 10−3 = $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 Pa⋅m 3 1 Pa⋅m = 1 N m 2 3 m = 1 N⋅m = 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 = 9.07 × 10 g) q = mCusCuΔt 8 6 9.9 × 10 J = (9.07 × 10 g)(0.385 J/g°C)(tf − 21°C) 6 (3.49 × 10 )tf = 1.06 × 10 9 tf = 304°C 6.121 Energy must be supplied to break a chemical bond. By the same token, energy is released when a bond is formed. CHAPTER 6: THERMOCHEMISTRY 6.122 (a) ⎯→ Ca(OH)2(s) + C2H2(g) CaC2(s) + 2H2O(l) ⎯ (b) The reaction for the combustion of acetylene is: 193 ⎯→ 4CO2(g) + 2H2O(l) 2C2H2(g) + 5O2(g) ⎯ We can calculate the enthalpy change for this reaction from standard enthalpy of formation values given in Appendix 3 of the text. D ΔH rxn = [4 ΔH fD (CO 2 ) + 2ΔH fD (H 2 O)] − [2ΔH fD (C2 H 2 ) + 5ΔH fD (O2 )] D ΔH rxn = [(4)(−393.5 kJ/mol) + (2)(−285.8 kJ/mol)] − [(2)(226.6 kJ/mol) + (5)(0)] D ΔH 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 D calculated above as a conversion factor to determine the amount of heat Now, we can use the ΔH rxn obtained when 1.16 moles of C2H2 are burned. 1.16 mol C2 H 2 × 2599 kJ = 1.51 × 103 kJ 2 mol C2 H 2 6.123 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. 40°F 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. 6.124 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 (ΔV = 0); therefore, no work will result from the change. w = −PΔV = 0 From Equation (6.4) of the text, it follows that the change in energy is equal to the heat change. ΔE = q + w = qv = −5153 kJ/mol To calculate ΔH, we rearrange Equation (6.10) of the text. ΔE = ΔH − RTΔn ΔH = ΔE + RTΔn 194 CHAPTER 6: THERMOCHEMISTRY To calculate ΔH, Δn 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) Δn = 10 − 12 = −2 ΔH = ΔE + RTΔn ΔH = − 5153 kJ/mol + (8.314 J/mol ⋅ K)(298 K)(−2) × 1 kJ 1000 J ΔH = −5158 kJ/mol Is ΔH equal to qp in this case? 6.125 Let's write balanced equations for the reactions between Mg and CO2, and Mg and H2O. Then, we can D calculate ΔH rxn for each reaction from ΔH fD values. (1) 2Mg(s) + CO2(g) → 2MgO(s) + C(s) (2) Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g) D For reaction (1), ΔH rxn is: D ΔH rxn = 2ΔH fD [MgO( s )] + ΔH fD [C( s )] − {2ΔH fD [Mg( s )] + ΔH fD [CO2 ( g )]} D ΔH rxn = (2)(−601.8 kJ/mol) + (1)(0) − [(2)(0) + (1)(−393.5 kJ/mol)] = − 810.1 kJ/mol D For reaction (2), ΔH rxn is: D ΔH rxn = ΔH fD [Mg(OH) 2 ( s )] + ΔH fD [H 2 ( g )] − {ΔH fD [Mg( s )] + 2ΔH fD [H 2 O(l )]} D ΔH 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. 6.126 We know that ΔE = q + w. ΔH = q, and w = −PΔV = −RTΔn. Using thermodynamic data in Appendix 3 of the text, we can calculate ΔH. 2H2(g) + O2(g) → 2H2O(l), ΔH = 2(−285.8 kJ/mol) = −571.6 kJ/mol Next, we calculate w. The change in moles of gas (Δn) equals −3. w = −PΔV = −RTΔn 3 w = −(8.314 J/mol⋅K)(298 K)(−3) = +7.43 × 10 J/mol = 7.43 kJ/mol ΔE = q + w ΔE = −571.6 kJ/mol + 7.43 kJ/mol = −564.2 kJ/mol Can you explain why ΔE is smaller (in magnitude) than ΔH? CHAPTER 6: THERMOCHEMISTRY 6.127 (a) 195 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 From the equation for the production of Ca(OH)2, we have 1 mol H2O Q 1 mol CaO Q 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 27.7 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 = msΔt Rearranging, we get Δt = Δt = q ms 1.809 × 106 J (2.056 × 103 g)(1.20 J/g ⋅°C) = 733°C and the final temperature is Δt = tfinal − tinitial tfinal = 733°C + 25°C = 758°C A temperature of 758°C is high enough to ignite wood. (b) The reaction is: CaO(s) + H2O(l) → Ca(OH)2(s) D ΔH rxn = ΔH fD [Ca(OH) 2 ] − [ΔH fD (CaO) + ΔH fD (H 2 O)] D ΔH rxn is given in the problem (−65.2 kJ/mol). Also, the ΔH fD values of CaO and H2O are given. Thus, we can solve for ΔH fD of Ca(OH)2. −65.2 kJ/mol = ΔH fD [Ca(OH)2 ] − [(1)(−635.6 kJ/mol + (1)(−285.8 kJ/mol)] ΔH fD [Ca(OH)2 ] = − 986.6 kJ/mol 196 CHAPTER 6: THERMOCHEMISTRY 6.128 First, we calculate ΔH for the combustion of 1 mole of glucose using data in Appendix 3 of the text. We can then calculate the heat produced in the calorimeter. Using the heat produced along with ΔH 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) D ΔH rxn = (6)(−393.5 kJ/mol) + (6)(−285.8 kJ/mol) − (1)(−1274.5 kJ/mol) = − 2801.3 kJ/mol The heat produced in the calorimeter is: (3.134°C)(19.65 kJ/°C) = 61.58 kJ Let x equal the mass of glucose in the sample: x g glucose × 1 mol glucose 2801.3 kJ × = 61.58 kJ 180.2 g glucose 1 mol glucose x = 3.961 g % glucose = 6.129 (a) (b) (c) (d) (e) (f) q − − − + + + 3.961 g × 100% = 96.21% 4.117 g w − + − − 0 + ΔE − 0 − + + + ΔH − 0 − + + + In (b), the internal energy of an ideal gas depends only on temperature. Since temperature is held constant, ΔE = 0. Also, ΔH = 0 because ΔH = ΔE + Δ(PV) = ΔE + Δ(nRT) = 0. 6.130 (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. (b) 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) CHAPTER 6: THERMOCHEMISTRY 197 17.55 kJ of heat is released when 0.4196 g of the hydrocarbon undergoes combustion. We can now ° calculate the enthalpy of combustion ( ΔH rxn ) for the above reaction in units of kJ/mol. Then, from the enthalpy of combustion, we can calculate the enthalpy of formation of C6H6. 78.11 g C6 H 6 −17.55 kJ × × 2 mol C6 H 6 = − 6534 kJ/mol 0.4196 g C6 H 6 1 mol C6 H 6 ° ΔH rxn = (12)ΔH f° (CO2 ) + (6)ΔH f° (H 2 O) − (2)ΔH f° (C6 H 6 ) −6534 kJ/mol = (12)(−393.5 kJ/mol) + (6)(−285.8 kJ/mol) − (2)ΔH f° (C6 H 6 ) ΔH f° (C6 H 6 ) = 49 kJ/mol 6.131 If the body absorbs all the heat released and is an isolated system, the temperature rise, Δt, is: q = msΔt Δt = q 1.0 × 107 J = = 48°C ms (50, 000 g)(4.184 J/g ⋅°C) 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. 6.132 6.133 (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. Begin by using Equation (6.20) of the text, ΔHsoln = U + ΔHhydr, where U is the lattice energy. + − + − (1) Na (g) + Cl (g) → Na (aq) + Cl (aq) (2) Na (g) + I (g) → Na (aq) + I (aq) (3) K (g) + Cl (g) → K (aq) + Cl (aq) + − + + − − + − ΔHhydr = (4.0 − 788) kJ/mol = −784.0 kJ/mol ΔHhydr = (−5.1 − 686) kJ/mol = −691.1 kJ/mol ΔHhydr = (17.2 − 699) kJ/mol = −681.8 kJ/mol Adding together equation (2) and (3) and then subtracting equation (1) gives the equation for the hydration of KI. (2) (3) (1) + − + − Na (g) + I (g) → Na (aq) + I (aq) + − + − K (g) + Cl (g) → K (aq) + Cl (aq) + − + − Na (aq) + Cl (aq) → Na (g) + Cl (g) + − + − K (g) + I (g) → K (aq) + I (aq) ΔH = −691.1 kJ/mol ΔH = −681.8 kJ/mol ΔH = +784.0 kJ/mol ΔH = −588.9 kJ/mol We combine this last result with the given value of the lattice energy to arrive at the heat of solution of KI. ΔHsoln = U + ΔHhydr = (632 kJ/mol − 588.9 kJ/mol) = 43 kJ/mol 198 CHAPTER 6: THERMOCHEMISTRY 6.134 A→B B→C C→D D→A w = 0, because ΔV = 0 w = −PΔV = −(2 atm)(2 − 1)L = −2 L⋅atm w = 0, because ΔV = 0 w = −PΔV = −(1 atm)(1 − 2)L = +1 L⋅atm The total work done = (−2 L⋅atm) + (1 L⋅atm) = −1 L⋅atm 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. 6.135 C (graphite) → C (diamond) H = E + PV ΔH = ΔE + PΔV ΔH − ΔE = 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 − ΔE = PΔV = (50,000 atm)(0.0034119 L/mol − 0.0053378 L/mol) ΔH − ΔE = − 96.295 L ⋅ atm 101.3 J × = − 9.75 × 103 J / mol mol 1 L ⋅ atm Answers to Review of Concepts Section 6.2 (p. 233) Section 6.3 (p. 238) Section 6.4 (p. 245) Section 6.5 (p. 252) (a) Isolated system. (b) Open system. (c) Closed system. Gas in the fixed volume container: q > 0, w = 0. Gas in the cylinder with a movable piston: q > 0, w < 0. (b) Al because it has a larger specific heat. Section 6.6 (p. 258) Look at Equation (6.18) of the text. The negative sign before ΔH fD for reactants means that D (exothermic reactants with positive ΔH fD values will likely result in a negative ΔH rxn reaction).