Chapter05rw-final

Chapter05rw-final - 5 5.1 5.2(a(b(c 5.3(a(b Thermochemistry...

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Unformatted text preview: 5 5.1 5.2 (a) (b) (c) 5.3 (a) (b) Thermochemistry Visualizing Concepts The book’s potential energy is due to the opposition of gravity by an object of mass m at a distance d above the surface of the earth. Kinetic energy is due to the motion of the book. As the book falls, d decreases and potential energy changes into kinetic energy. The first law states that the total energy of a system is conserved. At the instant before impact, all potential energy has been converted to kinetic energy, so the book’s total kinetic energy is 85 J, assuming no transfer of energy as heat. The internal energy, E, of the products is greater than that of the reactants, so the diagram represents an increase in the internal energy of the system. ΔE for this process is positive, +. If no work is associated with the process, it is endothermic. For an endothermic process, the sign of q is positive; the system gains heat. This is true only for system (iii). In order for ΔE to be less than 0, there is a net transfer of heat or work from the system to the surroundings. The magnitude of the quantity leaving the system is greater than the magnitude of the quantity entering the system. In system (i), the magnitude of the heat leaving the system is less than the magnitude of the work done on the system. In system (iii), the magnitude of the work done by the system is less than the magnitude of the heat entering the system. None of the systems has ΔE < 0. In order for ΔE to be greater than 0, there is a net transfer of work or heat to the system from the surroundings. In system (i), the magnitude of the work done on the system is greater than the magnitude of the heat leaving the system. In system (ii), work is done on the system with no change in heat. In system (iii) the magnitude of the heat gained by the system is greater than the magnitude of the work done on the surroundings. ΔE > 0 for all three systems. No. This distance traveled to the top of a mountain depends on the path taken by the hiker. Distance is a path function, not a state function. Yes. Change in elevation depends only on the location of the base camp and the height of the mountain, not on the path to the top. Change in elevation is a statfunction, not a path function. (c) 5.4 (a). (b) 106 5 Thermochemistry 5.5 (a) (b) Solutions to Exercises w = –PΔV. Since ΔV for the process is (–), the sign of w is (+). ΔE = q + w. At constant pressure, ΔH = q. If the reaction is endothermic, the signs of ΔH and q are (+). From (a), the sign of w is (+), so the sign of ΔE is (+). The internal energy of the system increases during the change. (This situation is described by the diagram (ii) in Exercise 5.3.) The temperature of the system and surroundings will equalize, so the temperature of the hotter system will decrease and the temperature of the colder surroundings will increase. The system loses heat by decreasing its temperature, so the sign of q is (–). The process is exothermic. If neither volume nor pressure of the system changes, w = 0 and ΔE = q = ΔH. The change in internal energy is equal to the change in enthalpy. 5.6 (a) (b) 5.7 Cooling 1 kg H2O releases more heat than cooling 1 kg Al . Equal masses of H2O and Al are cooled by the same number of °C. Under these circumstances, the amount of heat released on cooling is proportional to the specific heat capacities of the substances. Specific heat capacity is expressed as J/g-K or energy per mass and temperature change. Since mass and temperature change are equal for the two substances, the one with the larger specific heat capacity loses the most heat. From Table 5.2, the specific heat capacity of H2O(l) is 4.18 J/g-K and of Al(s) is 0.90 J/g-K, so 1 kg H2O releases more heat than 1 kg Al. (a) (b) N 2 (g) + O 2 (g) → 2NO(g). Since ΔV = 0, w = 0. ΔH = 90.37 kJ for production of 1 mol of NO(g). The definition of a formation reaction is one where elements combine to form one mole of a single product. The enthalpy change for such a reaction is the enthalpy of formation. ΔH A = ΔH B + ΔH C . The net enthalpy change associated with going from the initial state to the final state does not depend on path. The diagram shows that the change can be accomplished via reaction A, or via two successive reactions, B then C, with the same net enthalpy change. ΔHA = ΔHB + ΔHC because ΔH is a state function, independent of path. ΔH Z = ΔH X + ΔH Y . The diagram indicates that Reaction Z can be written as the sum of reactions X and Y. Hess’s law states that the enthalpy change for the net reaction Z is the sum of the enthalpy changes of the steps X and Y, regardless of whether the reaction actually occurs via this path. The diagrams are a visual statement of Hess’s law. 5.8 5.9 (a) (b) (c) 5.10 No. The standard enthalpy of formation of a compound is the change in enthalpy for the reaction that forms one mole of the compound from elements in their standard states. The equation C(graphite) + 4H(g) + O(g) → CH3OH(l) shows elemental carbon (graphite) in its standard state, but elemental hydrogen and oxygen exist as diatomic gases H2(g) and O2(g) at standard conditions. ΔH for this equation does not equal ΔH o for CH3OH(l). f 107 5 Thermochemistry The Nature of Energy 5.11 Solutions to Exercises An object can possess energy by virtue of its motion or position. Kinetic energy, the energy of motion, depends on the mass of the object and its velocity. Potential energy, stored energy, depends on the position of the object relative to the body with which it interacts. (a) (b) (c) The kinetic energy of the ball decreases as it moves higher. As the ball moves higher and opposes gravity, kinetic energy is changed into potential energy. The potential energy of the ball increases as it moves higher. The heavier ball would go half as high as the tennis ball. At the apex of the trajectory, all initial kinetic energy has been changed into potential energy. The magnitude of the change in potential energy is m g Δh, which is equal to the energy initially imparted to the ball. If the same amount of energy is imparted to a ball with twice the mass, m doubles so Δh is half as large. Analyze. Given: mass and speed of ball. Find: kinetic energy. 5.12 5.13 (a) Plan. Since 1 J = 1 kJ - m 2 /s 2 , convert g → kg to obtain E k in joules. Solve. Ε k = 1/ 2 mv 2 = 1 / 2 × 45 g × 2 1 kg 84 kg - m 2 ⎛ 61 m ⎞ ×⎜ = 84 J ⎟= 1000 g ⎝ 1 s ⎠ 1 s2 Check. 1/2(45 × 3600/1000) ≈ 1/2(40 × 4) ≈ 80 J (b) (c) 83.72 J × 1 cal = 20 cal 4.184 J As the ball hits the sand, its speed (and hence its kinetic energy) drops to zero. Most of the kinetic energy is transferred to the sand, which deforms when the ball lands. Some energy is released as heat through friction between the ball and the sand. Plan. Convert lb → kg, mi/hr → m/s. Solve. 850 lb × 1 kg = 385.49 = 385 kg 2.205 lb 5.14 (a) 66 mi 1.6093 km 1000 m 1 hr 1 min × × × × = 29.504 = 30 m/s 1 hr 1 mi 1 km 60 min 60 sec E k = 1/2 mv 2 = 1/2 × 385.49 kg × (29.504) 2 m 2 /s 2 = 1.7 × 10 5 J (b) (c) E k is proportional to v 2 , so if speed decreases by a factor of 2, kinetic energy decreases by a factor of 4. Brakes stop a moving vehicle, so the kinetic energy of the motorcycle is primarily transferred to friction between brakes and wheels, and somewhat to deformation of the tire and friction between the tire and road. 5.15 Analyze. Given: heat capacity of water = 1 Btu/lb - °F Find: J/Btu J J J → → Plan. heat capacity of water = 4.184 J/g- o C; Btu g - oC lb- o F 108 5 Thermochemistry 453.6 g 5 o C 1 lb - o F × × = 1054 J/Btu 1 Btu lb 9 oF Solutions to Exercises This strategy requires changing °F to °C. Since this involves the magnitude of a degree on each scale, rather than a specific temperature, the 32 in the temperature relationship is not needed. 100 °C = 180 °F; 5 °C = 9 °F Solve. 5.16 (a) 4.184 J g -o C × Analyze. Given: 1 kwh; 1 watt = 1 J/s; 1 watt • s = 1 J. Find: conversion factor for joules and kwh. Plan. kwh → wh → ws → J Solve. 1 kwh × 1000 w 60 min 60 s 1J × × × = 3.6 × 10 6 J 1 kw h min 1 w - s 1 kwh = 3.6 × 10 6 J (b) Analyze. Given: 100 watt bulb. Find: heat in kcal radiated by bulb or person in 24 hr. Plan. 1 watt = 1 J/s; 1 kcal = 4.184 × 10 3 J; watt → J/s → J → kcal. 100 watt = Solve. 100 J 60 sec 60 min 1 kcal × × × 24 hr × = 2065 = 2.1 × 10 3 kcal 3 1s min hr 4.184 × 10 J 24 hr has 2 sig figs, but 100 watt is ambiguous. The answer to 1 sig fig would be 2 × 10 3 kcal. Check. (1 × 10 2 × 6 × 10 1 × 6/10 3 ) ≈ 6 3 × 10 ≈ 2000 kcal 5.17 (a) (b) 5.18 (a) In thermodynamics, the system is the well-defined part of the universe whose energy changes are being studied. A closed system can exchange heat but not mass with its surroundings. The system is open, because it exchanges both mass and energy with the surroundings. Mass exchange occurs when solution flows into and out of the apparatus. The apparatus is not insulated, so energy exchange also occurs. Closed systems exchange energy but not mass, while isolated systems exchange neither. If the system is defined as shown, it can be closed by blocking the flow in and out, but leaving the flask full of solution. Work is a force applied over a distance. The amount of work done is the magnitude of the force times the distance over which it is applied. w = F × d. Heat is the energy transferred from a hotter object to a colder object. Heat is transferred from one object (system) to another until the two objects (systems) are at the same temperature. (b) 5.19 (a) (b) 5.20 (a) (b) 109 5 Thermochemistry 5.21 (a) (b) 5.22 (a) (b) Solutions to Exercises Gravity; work is done because the force of gravity is opposed and the pencil is lifted. Mechanical force; work is done because the force of the coiled spring is opposed as the spring is compressed over a distance. Electrostatic attraction; no work is done because the particles are held stationary. Magnetic attraction; work is done because the nail is moved a distance. The First Law of Thermodynamics 5.23 (a) (b) (c) In any chemical or physical change, energy can be neither created nor destroyed, but it can be changed in form. The total internal energy (E) of a system is the sum of all the kinetic and potential energies of the system components. The internal energy of a closed system (where no mass exchange with surroundings occurs) increases when work is done on the system by the surroundings and/or when heat is transferred to the system from the surroundings (the system is heated). 5.24 (a) (b) ΔE = q + w The quantities q and w are negative when the system loses heat to the surroundings (it cools), or does work on the surroundings. 5.25 Analyze. Given: heat and work. Find: magnitude and sign of ΔE. Plan. In each case, evaluate q and w in the expression ΔE = q + w. For an exothermic process, q is negative; for an endothermic process, q is positive. Solve. (a) (b) (c) q is positive because the system absorbs heat and w is negative because the system does work. ΔE = 105 kJ – 29 kJ = 76 kJ. The process is endothermic. ΔE = 1.50 kJ – 657 J = 1.50 kJ – 0.657 kJ = 0.843 = 0.84 kJ. The process is endothermic. q is negative because the system releases heat, and w is negative because the system does work. ΔE = –57.5 kJ – 22.5 kJ = –80.0 kJ. The process is exothermic. 5.26 In each case, evaluate q and w in the expression ΔE = q + w. For an exothermic process, q is negative; for an endothermic process, q is positive. (a) (b) (c) q is positive and w is negative. ΔE = 850 J – 382 J = 468 J. The process is endothermic. q is negative and w is essentially zero. ΔE = –3140 J. The process is exothermic. q is negative and w is zero. ΔE = –6.47 kJ. The process is exothermic. 5.27 Analyze. How do the different physical situations (cases) affect the changes to heat and work of the system upon addition of 100 J of energy? Plan. Use the definitions of heat and work and the First Law to answer the questions. 110 5 Thermochemistry (a) (b) Solutions to Exercises Solve. If the piston is allowed to move, case (1), the heated gas will expand and push the piston up, doing work on the surroundings. If the piston is fixed, case (2), most of the electrical energy will be manifested as an increase in heat of the system. Since little or no work is done by the system in case (2), the gas will absorb most of the energy as heat; the case (2) gas will have the higher temperature. In case (2), w ≈ 0 and q ≈ 100 J. In case (1), a significant amount of energy will be used to do work on the surroundings (–w), but some will be absorbed as heat (+q). (The transfer of electrical energy into work is never completely efficient!) (c) ΔE is greater for case (2), because the entire 100 J increases the internal energy of the system, rather than a part of the energy doing work on the surroundings. κ Q 1Q 2 5.28 For two oppositely charged particles, the sign of E e l is negative; the r2 closer the particles, the greater the magnitude of E e l. (a) (b) (c) The potential energy becomes less negative as the particles are separated (r increases). Ε el = ΔE for the process is positive; the internal energy of the system increases as the oppositely charged particles are separated. Work is done on the system to separate the particles so w is positive. We have no direct knowledge of the change in q, except that it cannot be large and negative, because overall ΔE = q + w is positive. A state function is a property of a system that depends only on the physical state (pressure, temperature, etc.) of the system, not on the route used by the system to get to the current state. Internal energy and enthalpy are state functions; heat is not a state function. Work is not a state function. The amount of work required to move from state A to state B depends on the path or series of processes used to accomplish the change. Independent. Potential energy is a state function. Dependent. Some of the energy released could be employed in performing work, as is done in the body when sugar is metabolized; heat is not a state function. Dependent. The work accomplished depends on whether the gasoline is used in an engine, burned in an open flame, or in some other manner. Work is not a state function. 5.29 (a) (b) (c) 5.30 (a) (b) (c) Enthalpy 5.31 (a) Change in enthalpy (ΔH) is usually easier to measure than change in internal energy (ΔE) because, at constant pressure, ΔH = q. The heat flow associated with a process at constant pressure can easily be measured as a change in temperature. Measuring ΔE requires a means to measure both q and w. 111 5 Thermochemistry (b) 5.32 (a) (b) Solutions to Exercises If ΔH is negative, the enthalpy of the system decreases and the process is exothermic. When a process occurs under constant external pressure, the enthalpy change (ΔH) equals the amount of heat transferred. ΔH = q p . ΔH = q p . If the system absorbs heat, q and ΔH are positive and the enthalpy of the system increases. 5.33 At constant pressure, ΔE = ΔH – PΔV. In order to calculate ΔE, more information about the conditions of the reaction must be known. For an ideal gas at constant pressure and temperature, PΔV = RTΔn. The values of either P and ΔV or T and Δn must be known to calculate ΔE from ΔH. At constant volume (ΔV = 0), ΔE = q v . According to the definition of enthalpy, H = E + PV, so ΔH = ΔE + Δ(PV). For an ideal gas at constant temperature and volume, ΔPV = VΔP = RTΔn. For this reaction, there are 2 mol of gaseous product and 3 mol of gaseous reactants, so Δn = –1. Thus VΔP or Δ(PV) is negative. Since ΔH = ΔE + Δ(PV), the negative Δ(PV) term means that ΔH will be smaller or more negative than ΔE. Analyze/Plan. q = –79 kJ (heat is given off by the system), w = –18 kJ (work is done by the system). Solve. 5.34 5.35 ΔE = q + w = –79 kJ – 18 kJ = –97 kJ. ΔH = q = –79 kJ (at constant pressure). Check. The reaction is exothermic. 5.36 The gas is the system. If 378 J of heat is added, q = +378 J. Work done by the system decreases the overall energy of the system, so w = –56 J. ΔE = q + w = 378 J – 56 J = 322 J. ΔH = q = 378 J (at constant pressure). 5.37 (a) (b) CH 3 CO O H (l) + 2O 2 (g) → 2H 2 O(l) + 2CO 2 (g) Analyze. How are reactants and products arranged on an enthalpy diagram? Plan. The substances (reactants or products, collectively) with higher enthalpy are shown on the upper level, and those with lower enthalpy are shown on the lower level. ΔH = – 871.7 kJ CH3 COOH(l) + 2 O2 (g) ΔH = −871.7 kJ 2 H 2O(l) + 2 CO2 (g) Solve. For this reaction, ΔH is negative, so the products have lower enthalpy and are shown on the lower level; reactants are on the upper level. The arrow points in the direction of reactants to products and is labeled with the value of ΔH. 5.38 (a) ZnCO 3 (s) → ZnO(s) + CO 2 (g) ΔH = 71.5 kJ 112 5 Thermochemistry 5.39 Plan. Consider the sign of ΔH. 5.40 Solutions to Exercises Solve. Since ΔH is negative, the reactants, 2Cl(g) have the higher enthalpy. Plan. Consider the sign of an enthalpy change that would convert one of the substances into the other. Solve. (a) (b) (c) CO 2 (s) → CO 2 (g). This change is sublimation, which is endothermic, +ΔH. CO 2 (g) has the higher enthalpy. H 2 → 2H. Breaking the H–H bond requires energy, so the process is endothermic, +ΔH. Two moles of H atoms have higher enthalpy. H 2 O(g) → H 2 (g) + 1/2 O 2 (g). Decomposing H 2 O into its elements requires energy and is endothermic, +ΔH. One mole of H 2 (g) and 0.5 mol O 2 (g) at 25°C have the higher enthalpy. N 2 (g) at 100° → N 2 (g) at 300°. An increase in the temperature of the sample requires that heat is added to the system, +q and +ΔH. N 2 (g) at 300° has the higher enthalpy. Solve. (d) 5.41 Analyze/Plan. Follow the strategy in Sample Exercise 5.4. (a) (b) Exothermic (ΔH is negative) 2.4 g Mg × 1 mol Mg −1204 kJ × = −59 kJ heat transferred 24.305 g Mg 2 mol Mg Check. The units of kJ are correct for heat. The negative sign indicates heat is evolved. (c) − 96.0 kJ × 2 mol MgO 40.30 g MgO × = 6.43 g MgO produced − 1204 kJ 1 mol Mg Check. Units are correct for mass. (100 × 2 × 40/1200) ≈ (8000/1200) ≈ 6.5 g (d) 2MgO(s) → 2Mg(s) + O 2 (g) ΔH = +1204 kJ This is the reverse of the reaction given above, so the sign of ΔH is reversed. 7.50 g MgO × 1 mol MgO 1204 kJ × = +112 kJ heat absorbed 40.30 g MgO 2 mol MgO Check. The units are correct for energy. (~9000/80) ≈ 110 kJ) 5.42 (a) The reaction is endothermic, so heat is absorbed by the system during the course of reaction. 45.0 g CH 3 OH × 25.8 kJ × 1 mol CH 3 OH 90.7 kJ × = 127 kJ heat transferred (absorbed) 32.04 g CH 3 OH 1 mol CH 3 OH (b) (c) 2 mol H 2 2.016 g H 2 × = 1.15 g H 2 produced 90.7 kJ 1 mol H 2 The sign of ΔH is reversed for the reverse reaction: ΔH = −90.7 kJ 113 5 Thermochemistry (d) 5.43 50.9 g CO × Solutions to Exercises 1 mol CO −90.7 kJ × = −165 kJ heat transferred (released) 28.01 g CO 1 mol CO Analyze. Given: balanced thermochemical equation, various quantities of substances and/or enthalpy. Plan. Enthalpy is an extensive property; it is “stoichiometric.” Use the mole ratios implicit in the balanced thermochemical equation to solve for the desired quantity. Use molar masses to change mass to moles and vice versa where appropriate. Solve. (a) 0.200 mol AgCl × −65.5 kJ = −13.1 kJ 1 mol AgCl Check. Units are correct; sign indicates heat evolved. (b) 2.50 g AgCl × 1 mol AgCl −65.5 kJ × = −1.14 kJ 143.3 g AgCl 1 mol AgCl Check. Units correct; sign indicates heat evolved. (c) 0.150 mmol AgCl × 1 × 10 −3 mol + 65.5 kJ × = 0.009825 kJ = 9.83 J 1 mmol 1 mol AgCl Check. Units correct; sign of ΔH reversed; sign indicates heat is absorbed during the reverse reaction. 5.44 (a) (b) (c) 0.632 mol O 2 × 8.57 g KCl × −89.4 kJ = −18.83 = −18.8 kJ 3 mol O 2 1 mol KCl −89.4 kJ × = −5.1386 = −5.14 kJ 74.55 g KCl 2 mol KCl Since the sign of ΔH is reversed for the reverse reaction, it seems reasonable that other characteristics would be reversed, as well. If the forward reaction proceeds spontaneously, the reverse reaction is probably not spontaneous. Also, we know from experience that KCl(s) does not spontaneously react with atmospheric O 2 (g), even at elevated temperature. 5.45 Analyze. Given: balanced thermochemical equation. Plan. Follow the guidelines given in Section 5.4 for evaluating thermochemical equations. Solve. (a) When a chemical equation is reversed, the sign of ΔH is reversed. CO 2 (g) + 2H 2 O(l) → CH 3 OH(l) + 3/2 O 2 (g) (b) ΔH = 726.5 kJ Enthalpy is extensive. If the coefficients in the chemical equation are multiplied by 2 to obtain all integer coefficients, the enthalpy change is also multiplied by 2. 2CH 3 OH(l) + 3O 2 (g) → 2CO 2 (g) + 4H 2 O(l) ΔH = 2(–726.5) kJ = –1453 kJ (c) (d) The exothermic forward reaction is more likely to be thermodynamically favored. Vaporization (liquid → gas) is endothermic. If the product were H 2 O(g), the reaction would be more endothermic and would have a smaller negative ΔH. (Depending on temperature, the enthalpy of vaporization for 2 mol H 2 O is about +88 kJ, not large enough to cause the overall reaction to be endothermic.) 114 5 Thermochemistry 5.46 (a) (b) 3C 2 H 2 (g) → C 6 H 6 (l) Solutions to Exercises ΔH = –630 kJ C 6 H 6 (l) → 3C 2 H 2 (g) ΔH = 630 kJ ΔH for the formation of 3 mol of acetylene is 630 kJ. ΔH for the formation of 1 mol of C2H2 is then 630 kJ/3 = 210 kJ. The exothermic reverse reaction is more likely to be thermodynamically favored. (c) If the reactant is in the higher enthalpy gas phase, the overall ΔH for the reaction has a smaller positive value. Calorimetry The specific heat of water to four significant figures, 4.184 J/g - K, will be used in many of the following exercises; temperature units of K and °C will be used interchangeably. 5.47 (a) J/mol-K or J/mol-°C. Heat capacity is the amount of heat in J required to raise the temperature of an object or a certain amount of substance 1 °C or 1 K. Molar heat capacity is the heat capacity of one mole of substance. J g- C o (b) or J g-K Specific heat is a particular kind of heat capacity where the amount of substance is 1 g. (c) 5.48 To calculate heat capacity from specific heat, the mass of the particular piece of copper pipe must be known. Analyze. Both objects are heated to 100°C. The two hot objects are placed in the same amount of cold water at the same temperature. Object A raises the water temperature more than object B. Plan. Apply the definition of heat capacity to heating the water and heating the objects to determine which object has the greater heat capacity. Solve. (a) Both beakers of water contain the same mass of water, so they both have the same heat capacity. Object A raises the temperature of its water more than object B, so more heat was transferred from object A than from object B. Since both objects were heated to the same temperature initially, object A must have absorbed more heat to reach the 100° temperature. The greater the heat capacity of an object, the greater the heat required to produce a given rise in temperature. Thus, object A has the greater heat capacity. Since no information about the masses of the objects is given, we cannot compare or determine the specific heats of the objects. (b) 115 5 Thermochemistry 5.49 (a) (b) (c) (d) 4.184 J 4.184 J or 1g - K 1 g - oC 75.40 J 4.184 J 18.02 g H 2 O × = mol - o C 1 g - o C 1 mol H 2 O 185 g H 2 O × 4.184 J = 774 J/ o C 1 g - oC 10.00 kg H 2 O × Solutions to Exercises Plan. Manipulate the definition of specific heat to solve for the desired quantity, paying close attention to units. Cs = q/(m × Δt). Solve. 1000 g 4.184 J 1 kJ × × × ( 46.2 o C − 24.6 o C) = 904 kJ o 1 kg 1 g - C 1000 J Check. (10 × 4 × 20) ≈ 800 kJ; the units are correct. Note that the conversion factors for kg → g and J → kJ cancel. An equally correct form of specific heat would be kJ/kg - °C 5.50 (a) (b) 5.51 In Table 5.2, Hg(l) has the smallest specific heat, so it will require the smallest amount of energy to heat 50.0 g of the substance 10 k. 50.0 g Hg(l) × 10 K × 0.14 J = 70 J g- K Analyze/Plan. Follow the logic in Sample Exercise 5.5. 1.05 kg Fe × Solve. 1000 g 0.450 J × × (88.5 o C − 25.0 o C) = 3.00 × 10 4 J (or 30.0 kJ) 1 kg g-K 2.42 J × ( 40.5 o C − 13.1o C) = 4.11 × 10 3 J g-K 5.52 5.53 62.0 g ethylene glycol × Analyze. Since the temperature of the water increases, the dissolving process is exothermic and the sign of ΔH is negative. The heat lost by the NaOH(s) dissolving equals the heat gained by the solution. Plan/Solve. Calculate the heat gained by the solution. The temperature change is 47.4 – 23.6 = 23.8°C. The total mass of solution is (100.0 g H 2 O + 9.55 g NaOH) = 109.55 = 109.6 g. 109.55 solution × 4.184 J 1 kJ × 23.8 o C × = 10.909 = 10.9 kJ o 1000 J 1g - C This is the amount of heat lost when 9.55 g of NaOH dissolves. The heat loss per mole NaOH is 40.00 g NaOH −10.909 kJ × = −45.7 kJ/mol 9.55 g NaOH 1 mol NaOH ΔH = q p = −45.7 kJ/mol NaOH Check. (–11/9 × 40) ≈ –45 kJ; the units and sign are correct. 5.54 (a) Following the logic in Solution 5.53, the dissolving process is endothermic, ΔH is positive. The total mass of the solution is (60.0 g H 2 O + 3.88 g NH 4 NO 3 ) = 63.88 = 63.9 g. The temperature change of the solution is 23.0 – 18.4 = 4.6°C. The heat lost by the surroundings is 116 5 Thermochemistry 63.88 g solution × Solutions to Exercises 4.184 J 1 kJ × 4.6 o C × = 1.229 = 1.2 kJ 1000 J 1 g - oC Thus, 1.2 kJ is absorbed when 3.88 g NH 4 NO 3 (s) dissolves. 80.05 g NH 4 NO 3 + 1.229 kJ × = +25.36 = +25 kJ/mol NH 4 NO 3 3.88 NH 4 NO 3 1 mol NH 4 NO 3 (b) This process is endothermic, because the temperature of the surroundings decreases, indicating that heat is absorbed by the system. Solve. 5.55 Analyze/Plan. Follow the logic in Sample Exercise 5.7. q b omb = –q r xn; ΔT = 30.57°C – 23.44°C = 7.13°C q bomb = 7.854 kJ × 7.13 o C = 56.00 = 56.0 kJ 1o C At constant volume, q v = ΔE. ΔE and ΔH are very similar. ΔH rxn ≈ ΔE rxn = q rxn = −q bomb = ΔH rxn = −56.0 kJ = −25.454 = −25.5 kJ/g C 6 H 4 O 2 2.200 g C 6 H 4 O 2 −25.454 kJ 108.1 g C 6 H 4 O 2 × = −2.75 × 10 3 kJ/mol C 6 H 4 O 2 1 g C 6 H 4 O 2 1 mol C 6 H 4 O 2 5.56 (a) (b) C 6 H 5 OH(s) + 7O 2 (g) → 6CO 2 (g) + 3H 2 O(l) q b omb = –q r xn; ΔT = 26.37°C – 21.36°C = 5.01°C q bomb = 11.66 kJ × 5.01o C = 58.417 = 58.4 kJ 1o C At constant volume, q v = ΔE. ΔE and ΔH are very similar. ΔH rxn ≈ ΔE rxn = q rxn = −q bomb = −58.417 kJ = −32.454 = −32.5 kJ/g C 6 H 5 OH 1.800 g C 6 H 5 OH ΔH rxn = − 32.454 kJ 94.11 g C 6 H 5 OH − 3.054 × 10 3 kJ × = 1 g C 6 H 5 OH 1 mol C 6 H 5 OH mol C 6 H 5 OH = –3.05 × 10 3 kJ/mol C 6 H 5 OH 5.57 Analyze. Given: specific heat and mass of glucose, ΔT for calorimeter. Find: heat capacity, C, of calorimeter. Plan. All heat from the combustion raises the temperature of the calorimeter. Calculate heat from combustion of glucose, divide by ΔT for calorimeter to get kJ/°C. Solve. (a) (b) C total = 2.500 g glucose × 15.57 kJ 1 × = 14.42 = 14.4 kJ/ o C 1 g glucose 2.70 o C Qualitatively, assuming the same exact initial conditions in the calorimeter, twice as much glucose produces twice as much heat, which raises the calorimeter temperature by twice as many °C. Quantitatively, 117 5 Thermochemistry 5.000 g glucose × Solutions to Exercises 15.57 kJ 1o C × = 5.40 o C 1 g glucose 14.42 kJ Check. Units are correct. ΔT is twice as large as in part (a). The result has 3 sig figs, because the heat capacity of the calorimeter is known to 3 sig figs. 5.58 (a) (b) (c) C = 1.640 g C 6 H 5 COOH × 26.38 kJ 1 × = 8.740 = 8.74 kJ/ o C 1 g C 6 H 5 COOH 4.95 o C 8.740 kJ 1 × 4.68 o C × = 30.99 = 31.0 kJ/g sample o 1.320 g sample C If water is lost from the calorimeter, there is less water to heat, so the same amount of heat (kJ) from a reaction would cause a larger increase in the calorimeter temperature. The calorimeter constant, kJ/°C, would decrease, because °C is in the denominator of the expression. Hess’s Law 5.59 Hess’s Law is a consequence of the fact that enthalpy is a state function. Since ΔH is independent of path, we can describe a process by any series of steps that add up to the overall process and ΔH for the process is the sum of the ΔH values for the steps. (a) Analyze/Plan. Arrange the reactions so that in the overall sum, B, appears in both reactants and products and can be canceled. This is a general technique for using Hess’s Law. Solve. A→B B→C A→C ΔH = +30 kJ ΔH = +60 kJ ΔH = +90 kJ 5.60 (b) Check. The process of A forming C can be described as A forming B and B forming C. 5.61 Analyze/Plan. Follow the logic in Sample Exercise 5.8. Manipulate the equations so that “unwanted” substances can be canceled from reactants and products. Adjust the Solve. corresponding sign and magnitude of ΔH. P4 O 6 (s) → P4 (s) + 3O 2 (g ) P4 (s) + 5O 2 (g ) → P4 O10 (s) P4 O 6 (s) + 2O 2 (g ) → P4 O10 (s) ΔH = 1640.1 kJ ΔH = −2940.1 kJ ΔH = −1300.0 kJ Check. We have obtained the desired reaction. 118 5 Thermochemistry 5.62 3H 2 (g) + 3/2 O 2 (g) 3H 2 (g) + O 3 (g ) → 3H 2 O(g ) → 3H 2 O(g ) O 3 (g) → 3 / 2 O 2 (g ) ΔH = −867.7 kJ Solutions to Exercises ΔH = 3/2( −483.6 kJ) ΔH = 1/ 2( −284.6 kJ) 5.63 Analyze/Plan. Follow the logic in Sample Exercise 5.9. Manipulate the equations so that “unwanted” substances can be canceled from reactants and products. Adjust the Solve. corresponding sign and magnitude of ΔH. C 2 H 4 (g ) → 2 H 2 (g ) + 2C(s) 2C(s) + 4F2 (g ) → 2CF4 (g ) 2H 2 (g ) + 2F2 (g ) → 4HF(g) C 2 H 4 (g ) + 6F2 (g ) → 2CF4 (g ) + 4HF(g) ΔH = −52.3 kJ ΔH = 2( −680 kJ) ΔH = 2( −537 kJ) ΔH = −2.49 × 10 3 kJ Check. We have obtained the desired reaction. 5.64 N 2 O(g ) N 2 ( g ) + O 2 (g ) → N 2 (g ) + 1/2 O 2 (g) → 2NO(g) ΔH = 1/ 2 ( −163.2 kJ) ΔH = 1/ 2(113.1 kJ) ΔH = 180.7 kJ ΔH = 155.7 kJ NO 2 (g) → NO(g) + 1/2 O 2 (g) N 2 O(g ) + NO 2 (g ) → 3NO(g ) Enthalpies of Formation 5.65 (a) Standard conditions for enthalpy changes are usually P = 1 atm and T = 298 K. For the purpose of comparison, standard enthalpy changes, ΔHº, are tabulated for reactions at these conditions. Enthalpy of formation, ΔH f , is the enthalpy change that occurs when a compound is formed from its component elements. Standard enthalpy of formation, ΔH o , is the enthalpy change that accompanies f formation of one mole of a substance from elements in their standard states. º Tables of ΔH f are useful because, according to Hess’s law, the standard enthalpy of any reaction can be calculated from the standard enthalpies of formation for the reactants and products. ΔH o = ΣΔH o (products) − ΣΔH o (reactants) rxn f f (b) (c) 5.66 (a) (b) The standard enthalpy of formation for any element in its standard state is zero. Elements in their standard states are the reference point for the enthalpy of formation scale. 6C(s) + 6H 2 (g) + 3O 2 (g) → C 6 H 1 2O 6 (s) 1/2 N 2 (g) + 3/2 H 2 (g) → NH 3 (g) S (s) + O 2 (g) → SO 2 (g) Rb(s) + 1/2 Cl 2 (g) + 3/2 O 2 (g) → RbClO 3 (s) N 2 (g ) + 2H 2 (g ) + 3 / 2 O 2 (g ) → NH 4 NO 3 (s) ΔH o = −46.19 kJ f ΔH o = −296.9 kJ f ΔH o = −392.4 kJ f ΔH o = −365.6 kJ f (c) 5.67 (a) (b) (c) (d) 119 5 Thermochemistry 5.68 (a) (b) (c) (d) 5.69 1/2 H 2 (g) + 1/2 Br 2 (l) → HBr(g) Ag(s) + 1/2 N 2 (g) + 3/2 O 2 (g) → AgNO 3 (s) 2Fe(s) + 3/2 O 2 (g) → Fe2O3(s) 2C(s) + 2H 2 (g) + O2(g) → CH3COOH(l) Solutions to Exercises ΔH o = −36.23 kJ f ΔH o = −124.4 kJ f º ΔH f = −822.16 kJ º ΔH f = −487.0 kJ o Plan. ΔH rxn = ΣnΔH o (products) − ΣnΔH o (reactants). Be careful with coefficients, f f states, and signs. Solve. o ΔH rxn = ΔH o Al 2 O 3 (s) + 2ΔH o Fe(s) − ΔH o Fe 2 O 3 − 2ΔH o Al(s) f f f f o ΔH rxn = ( −1669.8 kJ) + 2(0) − ( −822.16 kJ) − 2(0) = −847.6 kJ 5.70 Use heats of formation to calculate ΔHº for the combustion of butane. C 4 H10 (l) + 13 / 2 O 2 (g ) → 4CO 2 (g ) + 5H 2 O(l) o ΔH rxn = 4ΔH o CO 2 (g ) + 5ΔH o H 2 O(l) − ΔH o C 4 H10 (l) − 13 / 2 ΔH o O 2 (g ) f f f f o ΔH rxn = 4( −393.5 kJ) + 5( −285.83 kJ) − ( −147.6 kJ) − 13 / 2(0) = −2855.6 = −2856 kJ/mol C 4 H10 5.00 g C 4 H10 × 5.71 1 mol C 4 H10 − 2855.6 kJ × = −246 kJ 58.123 g C 4 H10 1 mol C 4 H10 o Plan. ΔH rxn = ΣnΔH o (products) − ΣnΔH o (reactants). Be careful with coefficients, states f f and signs. Solve. (a) o ΔH rxn = 2 ΔH o SO 3 (g ) − 2 ΔH o SO 2 (g ) − ΔH o O 2 (g ) f f f = 2(–395.2 kJ) – 2(–296.9 kJ) – 0 = –196.6 kJ (b) o ΔH rxn = ΔH o MgO(s) + ΔH o H 2 O(l) − ΔH o Mg(OH) 2 (s) f f f = –601.8 kJ + (–285.83 kJ) – (–924.7 kJ) = 37.1 kJ (c) o ΔH rxn = 4ΔH o H 2 O(g ) + ΔH o N 2 (g) − ΔH o N 2 O 4 (g ) − 4 ΔH o H 2 (g ) f f f f = 4(–241.82 kJ) + 0 – (9.66 kJ) – 4(0) = –976.94 kJ (d) o ΔH rxn = ΔH o SiO 2 (s) + 4 ΔH o HCl(g) − ΔH o SiCl 4 (g ) − 2 ΔH o H 2 O(l) f f f f = –910.9 kJ + 4(–92.30 kJ) – (–640.1 kJ) – 2(–285.83 kJ) = –68.3 kJ 5.72 (a) º º º º º ΔH rxn = 2ΔH f Br2 (l) + 2ΔH f H 2 O(l) − 4ΔH f HBr(g ) − ΔH f O 2 (g) = 2(0) + 2(–285.83 kJ) – 4(–36.23 kJ) – 0 = –426.74 kJ º º º º º ΔH rxn = ΔH f Na 2SO 4 (s) + ΔH f H 2 O(g) − 2ΔH f NaOH(s) − ΔH f SO 3 (g) (b) = −1387.1 kJ + (−241.82 kJ) − 2(−425.6 kJ) − (−395.2 kJ) = −382.5 kJ (c) o ΔH rxn = ΔH o CCl 4 (l) + 4ΔHo HCl(g) − ΔH o CH 4 (g ) − 4 ΔH o Cl 2 (g ) f f f f = –139.3 kJ + 4(–92.30 kJ) − (–74.8 kJ) – 4(0) = –433.7 kJ (d) º º º º º ΔH rxn = 2ΔH f FeCl3 (s) + 3ΔH f H 2 O(g) − ΔH f Fe 2 O 3 (g ) − 6ΔH f HCl(g) = 2(–400 kJ) + 3(–241.82 kJ) – (–822.16 kJ) – 6(–92.30 kJ) = –149.5 = –150 kJ 120 5 Thermochemistry 5.73 Solutions to Exercises Analyze. Given: combustion reaction, enthalpy of combustion, enthalpies of formation for most reactants and products. Find: enthalpy of formation for acetone. Plan. Rearrange the expression for enthalpy of reaction to calculate the desired enthalpy of formation. Solve. o ΔH rxn = 3ΔHo CO 2 (g ) + 3ΔHo H 2 O(l) − ΔHo C 3 H 6 O(l) + 4 ΔHo O 2 (g ) f f f f o C H O(l) − 4(0) − 1790 kJ = 3( −393.5 kJ) + 3( −285.83 kJ) − ΔH f 3 6 ΔHo C 3 H 6 O(l) = −248 kJ f 5.74 o ΔH rxn = ΔH o Ca(OH) 2 (s) + ΔH o C 2 H 2 (g ) − 2 ΔH o H 2 O(l) − ΔH o CaC 2 (s) f f f f − 127.2 kJ = −986 .2 kJ + 226.7 kJ − 2( −285.83 kJ) − ΔH o CaC 2 ( s) f o for CaC ( s) = −60.6 kJ ΔH f 2 5.75 (a) (b) (c) C 8 H 1 8(l) + 25/2 O 2 (g) → 8CO 2 (g) + 9H 2 O(g) 8C(s) + 9H 2 (g) → C 8 H 1 8(l) ΔH ° = –5064.9 kJ º ΔH f = ? Plan. Follow the logic in Solution 5.73 and 5.74. Solve. o ΔH rxn = 8 ΔH o CO 2 (g ) + 9ΔH o H 2 O(g) − ΔH o C 8 H 18 (l) − 25 / 2 ΔH o O 2 (g ) f f f f o C H (l) − 25 / 2(0) − 5064.9 kJ = 8( −393.5 kJ) + 9( −241.82 kJ) − ΔH f 8 18 ΔH o C 8 H 18 (l) = 8( −393.5 kJ) + 9( −241.82 kJ) + 5064.9 kJ = −259.5 kJ f 5.76 (a) 10C(s) + 4H 2 (g) → C 1 0H 8 (s) C 1 0H 8 (s) + 12O 2 (g) → 10CO 2 (g) + 4H 2 O(l) formation combustion, ΔH = –5154 kJ ° (b) o ΔH rxn = 10 ΔHo CO 2 (g ) + 4ΔHo H 2 O(l) − ΔHo C10 H 8 (s) − 12 ΔHo O 2 (g ) f f f f − 5154 = 10( −393.5 kJ) + 4( −285.83 kJ) − ΔHo C10 H 8 (s) − 12(0) f o C H (s) = 10( −393.5 kJ) + 4( −285.83 kJ) + 5154 kJ = 76 kJ ΔH f 10 8 Check. The result has 0 decimal places because the heat of combustion has 0 decimal places. 5.77 (a) (b) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) o ΔH rxn = 2ΔHo CO 2 (g ) + 3ΔH o H 2 O(g) − ΔH o C 2 H 5 OH(l) − 3ΔH o O 2 (g ) f f f f = 2(−393.5 kJ) + 3(−241.82 kJ) − (−277.7 kJ) − 3(0) = −1234.76 = −1234.8 kJ (c) Plan. The enthalpy of combustion of ethanol [from part (b)] is −1234.8 kJ/mol. Change mol to mass using molar mass, then mass to volume using density. Solve. −1234.76 kJ 1 mol C 2 H 5 OH 0.789 g 1000 mL × × × = − 21,147 = − 2.11 × 10 4 kJ/L mol C 2 H 5 OH 46.06844 g mL L Check. (1200/50) ≈ 25; 25 × 800 ≈ 20,000 (d) Plan. The enthalpy of combustion corresponds to any of the molar amounts in the equation as written. Production of −1234.76 kJ also produces 2 mol CO2. Use this relationship to calculate mass CO2/kJ. 121 5 Thermochemistry Solutions to Exercises 2 mol CO 2 44.0095 g CO 2 × = 0.071284 g CO 2 /kJ emitted − 1234.76 kJ mol Check. The negative sign associated with enthalpy indicates that energy is emitted. 5.78 (a) (b) CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(g) o ΔH rxn = ΔH o CO 2 (g ) + 2ΔH o H 2 O(g) − ΔH o CH 3 OH(l) − 3 / 2 ΔHo O 2 (g ) f f f f = −393.5 kJ + 2(−241.82 kJ) −(−238.6 kJ) − 3/2(0) = −638.54 = −638.5 kJ (c) 1 mol CH 3 OH 0.791 g 1000 mL −638.54 kJ × × × mol CH 3 OH 32.04 g mL L = 1.58 × 10 4 kJ/L produced (d) 1 mol CO 2 44.0095 g CO 2 × = 0.06892 g CO 2 /kJ emitted − 638.54 kJ mol Foods and Fuels 5.79 (a) (b) Fuel value is the amount of heat produced when 1 gram of a substance (fuel) is combusted. The fuel value of fats is 9 kcal/g and of carbohydrates is 4 kcal/g. Therefore, 5 g of fat produce 45 kcal, while 9 g of carbohydrates produce 36 kcal; 5 g of fat are a greater energy source. Fats are appropriate for fuel storage because they are insoluble in water (and body fluids) and have a high fuel value. For convenience, assume 100 g of chips. 12 g protein × 14 g fat × 17 kJ 1 Cal × = 48.76 = 49 Cal 1 g protein 4.184 kJ 5.80 (a) (b) 38 kJ 1 Cal × = 127.15 = 130 Cal 1 g fat 4.184 kJ 17 kJ 1 Cal × = 300.67 = 301 Cal 1 g carbohydrates 4.184 kJ 74 g carbohydrates × total Cal = (48.76 + 127.15 + 300.67) = 476.58 = 480 Cal % Cal from fat = 127.15 Cal fat × 100 = 26.68 = 27% 476.58 total Cal (Since the conversion from kJ to Cal was common to all three components, we would have determined the same percentage by using kJ.) (c) 5.81 25 g fat × 17 kJ 38 kJ ; = x g protein × g protein g fat x = 56 g protein Plan. Calculate the Cal (kcal) due to each nutritional component of the Campbell’s® soup, then sum. Solve. 122 5 Thermochemistry 9 g carbohydrates × 1 g protein × 7 g fat × 17 kJ = 153 or 2 × 10 2 kJ 1 g carbohydrate 17 kJ = 17 or 0.2 × 10 2 kJ 1 g protein Solutions to Exercises 38 kJ = 266 or 3 × 10 2 kJ 1 g fat total energy = 153 kJ + 17 kJ + 266 kJ = 436 or 4 × 10 2 kJ 436 kJ × 1 kcal 1 Cal × = 104 or 1 × 10 2 Cal/serving 4.184 kJ 1kcal Check. 100 Cal/serving is a reasonable result; units are correct. The data and the result have 1 sig fig. 5.82 Calculate the fuel value in a pound of M&M® candies. 96 fat × 38 kJ = 3648 kJ = 3.6 × 10 3 kJ 1 g fat 17 kJ = 5440 kJ = 5.4 × 10 3 kJ 1 g carbohydrate 320 g carbohydrate × 21 g protein × 17 kJ = 357 kJ = 3.6 × 10 2 kJ 1 g protein total fuel value = 3648 kJ + 5440 kJ + 357 kJ = 9445 kJ = 9.4 × 10 3 kJ/lb 42 g 9445 kJ 1 lb × × = 874.5 kJ = 8.7 × 10 2 kJ/serving lb 453.6 g serving 874.5 kJ 1 kcal 1 Cal × × = 209.0 Cal = 2.1 × 10 2 Cal/serving serving 4.184 kJ 1 kcal Check. 210 Cal is the approximate food value of a candy bar, so the result is reasonable. 5.83 Plan. g → mol → kJ → Cal 16.0 g C 6 H12 O 6 × Solve. 1 mol C 6 H12 O 6 2812 kJ 1 Cal × × = 59.7 Cal 180.2 g C 6 Η 12 O 6 mol C 6 Η 12 O 6 4.184 kJ Check. 60 Cal is a reasonable result for most of the food value in an apple. 5.84 177 mL × 1.0 g wine 0.106 g ethanol 1 mol ethanol 1367 kJ 1 Cal × × × × 1 mL 1 g wine 46.1 g ethanol 1 mol ethanol 4.184 kJ = 133 = 1.3 × 10 2 Cal Check. A “typical” 6 oz. glass of wine has 150–250 Cal, so this is a reasonable result. Note that alcohol is responsible for most of the food value of wine. 5.85 Plan. Use enthalpies of formation to calculate molar heat (enthalpy) of combustion using Hess’s Law. Use molar mass to calculate heat of combustion per kg of hydrocarbon. Solve. 123 5 Thermochemistry Propyne: C 3 H 4 (g) + 4O 2 (g) → 3CO 2 (g) + 2H 2 O(g) (a) Solutions to Exercises o ΔH rxn = 3(–393.5 kJ) + 2(–241.82 kJ) – (185.4 kJ) – 4(0) = –1849.5 = –1850 kJ/mol C 3 H 4 1000 g C 3 H 4 1 mol C 3 H 4 −1849.5 kJ × × = −4.616 × 10 4 kJ/kg C 3 H 4 1 mol C 3 H 4 40.065 g C 3 H 4 1 kg C 3 H 4 (b) Propylene: C 3 H 6 (g) + 9/2 O 2 (g) → 3CO 2 (g) + 3H 2 O(g) (a) o ΔH rxn = 3(–393.5 kJ) + 3(–241.82 kJ) – (20.4 kJ) – 9/2(0) = –1926.4 = –1926 kJ/mol C 3 H 6 (b) 1000 g C 3 H 6 1 mol C 3 H 6 −1926.4 kJ × × = −4.578 × 10 4 kJ/kg C 3 H 6 1 mol C 3 H 6 42.080 g C 3 H 6 1 kg C 3 H 6 Propane: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) (a) o ΔH rxn = 3(–393.5 kJ) + 4(–241.82 kJ) – (–103.8 kJ) – 5(0) = –2044.0 = –2044 kJ/mol C 3 H 8 1000 g C 3 H 8 1 mol C 3 H 8 −2044.0 kJ × × = −4.635 × 10 4 kJ/kg C 3 H 8 1 mol C 3 H 8 44.096 g C 3 H 8 1 kg C 3 H 8 (b) (c) 5.86 These three substances yield nearly identical quantities of heat per unit mass, but propane is marginally higher than the other two. o ΔH rxn = ΔHo CO 2 (g ) + 2ΔHo H 2 O(g) − ΔHo CH 4 (g ) − 2ΔHo O 2 (g ) f f f f = –393.5 kJ + 2(–241.82 kJ) – (–74.8 kJ) – 2(0) kJ = –802.3 kJ o ΔH rxn = ΔH o CF4 (g ) + 4 ΔH o HF(g) − ΔH o CH 4 (g ) − 4 ΔH o F2 (g ) f f f f = –679.9 kJ + 4(–268.61 kJ) – (–74.8 kJ) – 4(0) kJ = –1679.5 kJ The second reaction is twice as exothermic as the first. The “fuel values” of hydrocarbons in a fluorine atmosphere are approximately twice those in an oxygen º atmosphere. Note that the difference in ΔH ° values for the two reactions is in the ΔH f º for the products, since the ΔH f for the reactants is identical. Additional Exercises 5.87 (a) mi/hr → m/s 1050 mi 1.6093 km 1000 m 1 hr × × × = 469.38 = 469.4 m/s hr 1 mi 1 km 3600 s (b) Find the mass of one N 2 molecule in kg. 28.0134 g N 2 1 kg 1 mol × × = 4.6518 × 10 − 26 23 1 mol 1000 g 6.022 × 10 molecules = 4.652 × 10 − 26 kg E k = 1/2 mv 2 = 1/2 × 4.6518 × 10 – 26 kg × (469.38 m/s) 2 = 5.1244 × 10 − 21 kg - m 2 = 5.124 × 10 − 21 J s2 124 5 Thermochemistry (c) 5.88 (a) (b) (c) 5.89 Solutions to Exercises 5.1244 × 10 −21 J 6.022 × 10 23 molecules × = 3086 J/mol = 3.086 kJ/mol molecule 1 mol E p = mgh = 52.0 kg × 9.81 m/s 2 × 10.8 m = 5509.3 J = 5.51 kJ E k = 1 / 2 mv ; v = (2E k / m) 2 1/2 ⎛ 2 × 5509 kg - m 2 / s 2 =⎜ ⎜ 52.0 kg ⎝ ⎞ ⎟ ⎟ ⎠ 1/ 2 = 14.6 m/s Yes, the diver does work on entering (pushing back) the water in the pool. In the process described, one mole of solid CO 2 is converted to one mole of gaseous CO 2 . The volume of the gas is much greater than the volume of the solid. Thus the system (that is, the mole of CO 2 ) must work against atmospheric pressure when it expands. To accomplish this work while maintaining a constant temperature requires the absorption of additional heat beyond that required to increase the internal energy of the CO 2 . The remaining energy is turned into work. Like the combustion of H 2 (g) and O 2 (g) described in Section 5.4, the reaction that inflates airbags is spontaneous after initiation. Spontaneous reactions are usually exothermic, –ΔH. The airbag reaction occurs at constant atmospheric pressure, ΔH = q p ; both are likely to be large and negative. When the bag inflates, work is done by the system on the surroundings, so the sign of w is negative. Freezing is an exothermic process (the opposite of melting, which is clearly endothermic). When the system, the soft drink, freezes, it releases energy to the surroundings, the can. Some of this energy does the work of splitting the can. (a) (b) q = 0, w > 0 (work done to system), ΔE > 0 Since the system (the gas) is losing heat, the sign of q is negative. The changes in state described in cases (a) and (b) are identical and ΔE is the same in both cases. The distribution of energy transferred as either work or heat is different in the two scenarios. In case (b), more work is required to compress the gas because some heat is lost to the surroundings. (The moral of this story is that the more energy lost by the system as heat, the greater the work on the system required to accomplish the desired change.) 5.90 5.91 5.92 5.93 ΔE = q + w = +38.95 kJ – 2.47 kJ = +36.48 kJ ΔH = q p = +38.95 kJ 5.94 If a function sometimes depends on path, then it is simply not a state function. Enthalpy is a state function, so ΔH for the two pathways leading to the same change of state pictured in Figure 5.9 must be the same. However, q is not the same for the both. Our conclusion must be that ΔH ≠ q for these pathways. The condition for ΔH = q p (other than constant pressure) is that the only possible work on or by the system is pressurevolume work. Clearly, the work being done in this scenario is not pressure-volume work, so ΔH ≠ q, even though the two changes occur at constant pressure. Plan. Change mass of the iceberg to moles, then apply the enthalpy change for melting. 1 million = 1 × 106. Solve. 5.95 125 5 Thermochemistry 1.25 × 10 6 metric tons ice × Solutions to Exercises 1000 kg 1000 g 1 mol H 2 O 6.01 kJ × × × = 4.17 × 1011 kJ metric ton kg 18.02 g H 2 O mol 5.96 (a) Plan. Calculate the total volume of water associated with 0.50 inches rain over an area of one square mile. Use density to change volume to mass, then apply the enthalpy change. Volume = area × height = 1 mi2 × 0.50 in; for the purpose of sig figs, assume 1 mi2 is an exact number. Note that there are many equivalent ways to obtain the rain volume in mL. Solve. 1 mi × 5280 ft 12 in × = 6.336 × 10 4 in 1 mi 1 ft (6.336 × 104)2 in2 × 0.50 in = 2.007 × 109 = 2.0 × 109 in3 H2O 2.007 × 10 9 in 3 × 16.4 cm 3 1 mL × = 3.292 × 10 10 = 3.3 × 10 10 mL 1 in 3 1 cm 3 At 25°C, the density of H2O is 0.99707 g/mL 3.292 × 1010 mL × 0.99707 g = 3.282 × 1010 = 3.3 × 1010 g H2O mL 1 mol H 2 O 44.0 kJ × = 8.014 × 1010 = 8.0 × 1010 kJ 18.02 g H 2 O mol H 2 O 3.282 × 1010 g H2O × 8.0 × 1010 kJ (this is 80 billion kJ!) are released when enough water condenses to produce 0.50 rainfall over an area of one square mile. (b) 8.014 × 1010 kJ × 1 ton dynamite 4.2 × 10 6 kJ = 1.908 × 104 = 1.9 × 104 ton dynamite The energy released by this rainstorm is equivalent to explosion of 19,000 tons of dynamite. 5.97 Find the heat capacity of 1.7 × 10 3 gal H 2 O. C H 2 O = 1.7 × 10 3 gal H 2 O × 1 × 10 3 cm 3 4 qt 1g 4.184 J 1L × × × × 3 1 gal 1.057 qt 1L 1 cm 1 g - oC = 2.692 × 10 7 J/ ° C = 2.7 × 10 4 kJ/ ° C; then, 1 kg 2.692 × 10 7 J 1 g - o C 1 brick × × × = 1.8 × 10 4 or 18,000 bricks o 3 0.85 J 1 × 10 g 1.8 kg 1C Check. (1.7 × ~16 × 10 6 )/(~1.6 × 10 3 ) ≈ 17 × 10 3 bricks; the units are correct. 5.98 (a) q Cu = 0.385 J × 121.0 g Cu × ( 30.1o C − 100.4 o C) = −3274.9 = −3.27 × 10 3 J g-K 4.184 J × 150.0 g H 2 O × ( 30.1o C − 25.1o C) = 3138 = 3.1 × 10 3 J g-K The negative sign indicates the 3.27 × 10 3 J are lost by the Cu block. (b) q H 2O = The positive sign indicates that 3.14 × 10 3 J are gained by the H 2 O. 126 5 Thermochemistry (c) 0.137 × 10 3 J × 1 = 27.4 = 3 × 10 J/K. 5.0 o C Solutions to Exercises The difference in the heat lost by the Cu and the heat gained by the water is 3.275 × 10 3 J – 3.138 × 10 3 J = 0.137 × 10 3 J = 1 × 10 2 J. The temperature change of the calorimeter is 5.0°C. The heat capacity of the calorimeter in J/K is Since q H 2O is known to one decimal place, the difference has one decimal place and the result has 1 sig fig. If the rounded results from (a) and (b) are used, 0.2 × 10 3 J C calorimeter = = 4 × 10 J/K. 5.0 o C (d) q H 2 O = 3.275 × 10 3 J = 4.184 J × 150.0 g × ( ΔT) g-K ΔT = 5.22°C; T f = 25.1°C + 5.22°C = 30.3°C 5.99 (a) From the mass of benzoic acid that produces a certain temperature change, we can calculate the heat capacity of the calorimeter. 0.235 g benzoic acid 1.642 C change observed o × 26.38 kJ = 3.7755 = 3.78 kJ/ o C 1 g benzoic acid Now we can use this experimentally determined heat capacity with the data for caffeine. 1.525 o C rise 3.7755 kJ 194.2 g caffeine × × = 4.22 × 10 3 kJ/mol caffeine o 0.265 g caffeine 1 mol caffeine 1C (b) The overall uncertainty is approximately equal to the sum of the uncertainties due to each effect. The uncertainty in the mass measurement is 0.001/0.235 or 0.001/0.265, about 1 part in 235 or 1 part in 265. The uncertainty in the temperature measurements is 0.002/1.642 or 0.002/1.525, about 1 part in 820 or 1 part in 760. Thus the uncertainty in heat of combustion from each measurement is 4220 4220 4220 4220 = 18 kJ; = 16 kJ; = 5 kJ; = 6 kJ 760 235 265 820 The sum of these uncertainties is 45 kJ. In fact, the overall uncertainty is less than this because independent errors in measurement do tend to partially cancel. 5.100 Plan. Use the heat capacity of H 2 O to calculate the energy required to heat the water. Use the enthalpy of combustion of CH 4 to calculate the amount of CH 4 needed to provide this amount of energy, assuming 100% transfer. Assume H 2 O(l) is the product of combustion at standard conditions. Solve. 1.00 kg H 2 O × 1000 g 4.184 J × × (90.0 o C − 25.0 o C) = 2.720 × 10 5 J = 272 kJ required o 1 kg g- C CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) o ΔH rxn = ΔH o CO 2 (g ) + 2ΔH o H 2 O(l) − ΔHo CH 4 (g ) − ΔH o O 2 (g ) f f f f 127 5 Thermochemistry 2.720 × 10 2 kJ × Solutions to Exercises = –393.5 kJ + 2(–285.83 kJ) – (–74.8 kJ) – (0) = –890.4 kJ 1 mol CH 4 16.04 g CH 4 × = 4.899 = 4.90 g CH 4 − 890.4 kJ mol CH 4 5.101 (a) Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g) o ΔH rxn = ΔH o Mg(OH) 2 (s) + ΔH o H 2 (g) − 2ΔHo H 2 O(l) − ΔHo Mg(s) f f f f = −924.7 kJ + 0 – 2(−285.83 kJ) − 0 = −353.04 = −353.0 kJ (b) Use the specific heat of water, 4.184 J/g-°C, to calculate the energy required to heat the water. Use the density of water at 25°C to calculate the mass of H2O to be heated. (The change in density of H2O going from 15°C to 85°C does not substantially affect the strategy of the exercise.) Then use the ‘heat stoichiometry’ in (a) to calculate mass of Mg(s) needed. 25 mL × 0.997 g H 2 O 4.184 J 1 kJ × × 70 o C × = 7.300 kJ = 7.3 kJ required o mL 1000 J g- C 1 mol Mg 24.305 g Mg × = 0.5026 g = 0.50 g Mg needed 353.04 kJ 1 mol Mg 7.300 kJ × 5.102 (a) For comparison, balance the equations so that 1 mole of CH 4 is burned in each. CH 4 (g) + O 2 (g) → C(s) + 2H 2 O(l) CH 4 (g) + 3/2 O 2 (g) → CO(g) + 2H 2 O(l) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ΔH ° = –496.9 kJ ΔH ° = –607.4 kJ ΔH ° = –890.4 kJ (b) o ΔH rxn = ΔH o C(s) + 2 ΔH o H 2 O(l) − ΔH o CH 4 (g ) − ΔHo O 2 (g ) f f f f = 0 + 2(–285.83 kJ) – (–74.8) – 0 = –496.9 kJ o ΔH rxn = ΔHo CO(g) + 2 ΔHo H 2 O(l) − ΔHo CH 4 (g ) − 3 / 2 ΔHo O 2 (g ) f f f f = (–110.5 kJ) + 2(–285.83 kJ) – (–74.8 kJ) – 3/2(0) = –607.4 kJ o ΔH rxn = ΔHo CO 2 (g ) + 2 ΔHo H 2 O(l) − ΔHo CH 4 (g ) − 2 ΔHo O 2 (g ) f f f f = –393.5 kJ + 2(–285.83 kJ) – (–74.8 kJ) – 2(0) = –890.4 kJ (c) Assuming that O 2 (g) is present in excess, the reaction that produces CO 2 (g) represents the most negative ΔH per mole of CH 4 burned. More of the potential energy of the reactants is released as heat during the reaction to give products of lower potential energy. The reaction that produces CO 2 (g) is the most “downhill” in enthalpy. 2 B(s) + 3/2 O 2 (g ) → B 2 O 3 ( s) ΔH° = 1/ 2( −2509.1 kJ) ΔH° = 3/2( −571.7 kJ) ΔH° = −( −2147.5 kJ) ΔHo = +35.4 kJ f 5.103 (a) 3H 2 (g ) + 3 / 2 O 2 (g ) → 3H 2 O(l) B 2 O 3 (s) + 3H 2 O(l) → B 2 H 6 (g ) + 3O 2 (g ) 2 B(s) + 3H 2 (g ) → B 2 H 6 (g ) 128 5 Thermochemistry (b) Solutions to Exercises If, like B 2 H 6 , the combustion of B 5 H 9 produces B 2 O 3 as the boron-containing product, the heat of combustion of B 5 H 9 in addition to data given in part (a) would enable calculation of the heat of formation of B 5 H 9 . The combustion reaction is: B 5 H 9 (l) + 6O 2 (g) → 5/2 B 2 O 3 (s) + 9/2 H 2 O(l) 5 / 4 [4B(s) + 3O 2 (g ) → 2B 2 O 3 (s)] 9 / 4 [2H 2 (g ) + O 2 (g ) → 2H 2 O(l)] 5 / 2 B 2 O 3 (s) + 9 / 2 H 2 O(l) → B 5 H 9 (l) + 6O 2 (g ) ΔH° = 5 / 4( −2509.1 kJ) ΔH° = 9 / 4 ( −571.7 kJ) ΔH° = − ( heat of combustion) ΔHo of B H (l) 5B(s) + 9/2 H 2 (g ) → B 5 H 9 (l) 59 f o B H (l) = − [heat of combustion of B H (l)] − 3136.4 kJ − 1286 kJ ΔH f 5 9 59 We need to measure the heat of combustion of B 5 H 9 (l). 5.104 For nitroethane: 1.052 g C 2 H 5 NO 2 1 mol C 2 H 5 NO 2 1368 kJ × × = 19.17 kJ/cm 3 1 mol C 2 H 5 NO 2 75.072 g C 2 H 5 NO 2 1 cm 3 For ethanol: 0.789 g C 2 H 5 OH 1 mol C 2 H 5 OH 1367 kJ × × = 23.4 kJ/cm 3 1 mol C 2 H 5 OH 46.069 g C 2 H 5 OH 1 cm 3 For methylhydrazine: 0.874 g CH 6 N 2 1 mol CH 6 N 2 1305 kJ × × = 24.8 kJ/cm 3 1 mol CH 6 N 2 46.072 g CH 6 N 2 1 cm 3 Thus, methylhydrazine would provide the most energy per unit volume, with ethanol a close second. 5.105 (a) 3C 2 H 2 (g) → C 6 H 6 (l) ΔHo = ΔHo C 6 H 6 (l) − 3ΔHo C 2 H 2 (g ) = 49.0 kJ − 3( 226.77 kJ) = −631.31 = − 631.3 kJ rxn f f (b) Since the reaction is exothermic (ΔH is negative), the product, 1 mole of C 6 H 6 (l), has less enthalpy than the reactants, 3 moles of C 2 H 2 (g). The fuel value of a substance is the amount of heat (kJ) produced when 1 gram of the substance is burned. Calculate the molar heat of combustion (kJ/mol) and use this to find kJ/g of fuel. C 2 H 2 (g) + 5/2 O 2 (g) → 2CO 2 (g) + H 2 O(l) o ΔH rxn = 2 ΔHo CO 2 (g ) + ΔH o H 2 O(l) − ΔHo C 2 H 2 (g ) − 5 / 2 ΔH o O 2 (g ) f f f f (c) = 2(–393.5 kJ) + (–285.83 kJ) – 226.77 kJ – 5/2 (0) = –1299.6 kJ/mol C 2 H 2 1 mol C 2 H 2 −1299.6 kJ × = 49.916 = 50 kJ/g 1 mol C 2 H 2 26.036 g C 2 H 2 C 6 H 6 (l) + 15/2 O 2 (g) → 6CO 2 (g) + 3H 2 O(l) o ΔH rxn = 6 ΔH o CO 2 (g ) + 3ΔH o H 2 O(l) − ΔH o C 6 H 6 (l) − 15 / 2 ΔH o O 2 (g ) f f f f 129 5 Thermochemistry 1 mol C 6 H 6 −3267.5 kJ × = 41.830 = 42 kJ/g 1 mol C 6 H 6 78.114 g C 6 H 6 Solutions to Exercises = 6(–393.5 kJ) + 3(–285.83 kJ) – 49.0 kJ – 15/2 (0) = –3267.5 kJ/mol C 6 H 6 5.106 The reaction for which we want ΔH is: 4NH 3 (l) + 3O 2 (g) → 2N 2 (g) + 6H 2 O(g) Before we can calculate ΔH for this reaction, we must calculate ΔH f for NH 3 (l). We know that ΔH f for NH 3 (g) is –46.2 kJ, and that for NH 3 (l) → NH 3 (g), ΔH = 23.2 kJ Thus, ΔH v ap = ΔH f NH 3 (g) – ΔH f NH 3 (l). 23.2 kJ = –46.2 kJ – ΔH f NH 3 (l); ΔH f NH 3 (l) = –69.4 kJ/mol Then for the overall reaction, the enthalpy change is: ΔH r xn = 6ΔH f H 2 O(g) + 2ΔH f N 2 (g) – 4ΔH f NH 3 (l) – 3ΔH f O 2 = 6(–241.82 kJ) + 2(0) – 4(–69.4 kJ) – 3(0) = –1173.3 kJ 4 − 1173.3 kJ 1 mol NH 3 0.81 g NH 3 1000 cm 3 1.4 × 10 kJ × = × = 4 mol NH 3 17.0 g NH 3 1L L NH 3 1 cm 3 (This result has two significant figures because the density is expressed to two figures.) 2CH 3 OH(l) + 3O 2 (g) → 2CO 2 (g) + 4H 2 O(g) ΔH = 2(–393.5 kJ) + 4(–241.82 kJ) – 2(–239 kJ) – 3(0) = –1276 kJ 1 mol CH 3 OH 0.792 g CH 3 OH 1000 cm 3 1.58 × 10 4 kJ − 1276 kJ × × × = 2 mol CH 3 OH 32.04 g CH 3 OH 1L L CH 3 OH 1 cm 3 In terms of heat obtained per unit volume of fuel, methanol is a slightly better fuel than liquid ammonia. 5.107 1,3-butadiene, C 4 H 6 , MM = 54.092 g/mol (a) C 4 H 6 (g) + 11/2 O 2 (g) → 4CO 2 (g) + 3H 2 O(l) o ΔH rxn = 4 ΔHo CO 2 (g ) + 3ΔHo H 2 O(l) − ΔHo C 4 H 6 (g ) − 11/ 2 ΔHo O 2 (g ) f f f f = 4(–393.5 kJ) + 3(–285.83 kJ) – 111.9 kJ + 11/2 (0) = –2543.4 kJ/mol C 4 H 6 (b) (c) −2543.4 kJ 1 mol C 4 H 6 × = 47.020 → 47 kJ/g 1 mol C 4 H 6 54.092 g %H = 6(1.008) × 100 = 11.18% H 54.092 1-butene, C 4 H 8 , MM = 56.108 g/mol (a) C 4 H 8 (g) + 6O 2 (g) → 4CO 2 (g) + 4H 2 O(l) o ΔH rxn = 4 ΔHo CO 2 (g ) + 4 ΔHo H 2 O(l) − ΔHo C 4 H 8 (g ) − 6 ΔHo O 2 (g ) f f f f = 4(–393.5 kJ) + 4(–285.83 kJ) – 1.2 kJ – 6(0) = –2718.5 kJ/mol C 4 H 8 130 5 Thermochemistry (b) (c) 1 mol C 4 H 8 −2718.5 kJ × = 48.451 → 48 kJ/g 1 mol C 4 H 8 56.108 g C 4 H 8 %H = 8(1.008) × 100 = 14.37% H 56.108 Solutions to Exercises n-butane, C 4 H 1 0(g), MM = 58.124 g/mol (a) C 4 H 1 0(g) + 13/2 O 2 (g) → 4CO 2 (g) + 5H 2 O(l) o ΔH rxn = 4ΔHo CO 2 (g) + 5ΔH o H 2 O(l) − ΔH o C 4 H 10 (g ) − 13 / 2 ΔHo O 2 (g ) f f f f = 4(–393.5 kJ) + 5(–285.83 kJ) – (–124.7 kJ) – 13/2(0) = –2878.5 kJ/mol C 4 H 1 0 (b) (c) (d) 1 mol C 4 H 10 −2878.5 kJ × = 49.523 → 50 kJ/g 1 mol C 4 H 10 58.124 g C 4 H 10 %H = 10(1.008) × 100 = 17.34% H 58.124 It is certainly true that as the mass % H increases, the fuel value (kJ/g) of the hydrocarbon increases, given the same number of C atoms. A graph of the data in parts (b) and (c) (see below) suggests that mass % H and fuel value are directly proportional when the number of C atoms is constant. 5.108 (a) C 6 H 1 2O 6 (s) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(l) o ΔH rxn = 6ΔH o CO 2 (g ) + 6ΔHo H 2 O(l) − ΔH o C 6 H 12 O 6 (s) − 6ΔH o O 2 (g ) f f f f = 6(–393.5 kJ) + 6(–285.83 kJ) – (–1273 kJ) – 6(0) = –2803 kJ/mol C 6 H 1 2O 6 C 1 2H 2 2O 1 1(s) + 12O 2 (g) → 12CO 2 (g) + 11H 2 O(l) o ΔH rxn = 12 ΔH o CO 2 (g ) + 11ΔH o H 2 O(l) − ΔH o C 12 H 22 O 11 (s) − 12 ΔH o O 2 (g ) f f f f = 12(–393.5 kJ) + 11(–285.83 kJ) – (–2221 kJ) – 12(0) = –5645 kJ/mol C 1 2H 2 2O 1 1 (b) 1 mol C 6 H 12 O 6 15.55 kJ −2803 kJ × =− → 16 kJ/g C 6 H 12 O 6 (fuel value) 1 mol C 6 H 12 O 6 180.2 g C 6 H 12 O 6 1 g C 6 H 12 O 6 131 5 Thermochemistry (c) Solutions to Exercises 1 mol C 12 H 22 O11 16.49 kJ −5645 kJ × =− → 16 kJ/g C12 H 22 O 11 1 mol C12 H 22 O 11 342.3 g C 12 H 22 O11 1 g C12 H 22 O 11 (fuel value) The average fuel value of carbohydrates (Section 5.8) is 17 kJ/g. These two carbohydrates have fuel values (16 kJ/g), slightly lower but in line with this average. (More complex carbohydrates supply more energy and raise the average value.) 5.109 ΔE p = m g Δh. Be careful with units. 1 J = 1 kg - m 2 / s 2 200 lb × 1 kg 9.81 m 45 ft 1 yd 1m × × × × × 20 times 2 2.205 lb time 3 ft 1.0936 yd s = 2.441 × 10 5 kg - m 2 / s 2 = 2.441 × 10 5 J = 2.4 × 10 2 kJ 1 Cal = 1 kcal = 4.184 kJ 2.441 × 10 2 kJ × 1 Cal = 58.34 = 58 Cal 4.184 kJ If all work is used to increase the man’s potential energy, 20 rounds of stair-climbing will not compensate for one extra order of 245 Cal fries. In fact, more than 58 Cal of work will be required to climb the stairs, because some energy is required to move limbs and some energy will be lost as heat (see Solution 5.92). 5.110 Plan. Use dimensional analysis to calculate the amount of solar energy supplied per m 2 in 1 hr. Use stoichiometry to calculate the amount of plant energy used to produce sucrose per m 2 in 1 hr. Calculate the ratio of energy for sucrose to total solar energy, per m 2 per hr. Solve. 1 W = 1 J/s, 1 kW = 1 kJ/s 3 1.0 kW 1.0 kJ/s 1.0 kJ 60 s 60 min 3.6 × 10 kJ = =2× × = 1 hr m2 m2 m - s 1 min m 2 - hr 5645 kJ 1 mol sucrose 0.20 g sucrose × × = 3.298 = 3.3 kJ/m 2 - hr mol sucrose 342.3 g sucrose m 2 - hr for sucrose production 3.298 kJ for sucrose × 100 = 0.092% sunlight used to produce sucrose 3.6 × 10 3 kJ total solar 5.111 (a) 6CO 2 (g) + 6H 2 O(l) → C 6 H 1 2O 6 (s) + 6O 2 (g), ΔH° = 2803 kJ This is the reverse of the combustion of glucose (Section 5.8 and Solution 5.108), so ΔH° = –(–2803) kJ = +2803 kJ. 5.5 × 10 16 g CO 2 1 mol CO 2 2803 kJ × × = 5.838 × 10 17 = 5.8 × 10 17 kJ yr 44.01 g CO 2 6 mol CO 2 132 5 Thermochemistry (b) 1 W = 1 J/s; 1 W - s = 1 J Solutions to Exercises 5.838 × 10 17 kJ 1000 J 1 yr 1d 1 hr 1min 1 W - s × × × × × × yr kJ 365 d 24 hr 60 min 60 s J 1 MW × = 1.851 × 10 7 MW = 1.9 × 10 7 MW 1 × 10 6 W 1.9 × 10 7 MW × 1 plant 10 3 MW = 1.9 × 10 4 = 19 ,000 nuclear power plants Integrative Exercises 5.112 (a) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ΔH o = ΔH o CO 2 (g ) + 2 ΔHo H 2 O(l) − ΔH o CH 4 (g ) − 2ΔHo O 2 (g ) f f f f = –393.5 kJ + 2(–285.83 kJ) – (–74.8 kJ) – 2(0) = –890.36 = –890.4 kJ/mol CH 4 −890.36 kJ 1000 J 1 mol × × = 1.4785 × 10 −18 23 mol CH 4 1 kJ 6.022 × 10 molecules CH 4 = 1.479 × 10 18 J/molecule (b) 1eV = 96.485 kJ/mol 8 keV × 1000 eV 96.485 kJ 1 mol 1000 J × × × = 1.282 × 10 −15 = 1 × 10 −15 J/X - ray 1 keV eV - mol 6.022 × 10 23 kJ The X-ray has approximately 1000 times more energy than is produced by the combustion of 1 molecule of CH 4 (g). 5.113 The situation in Figure 4.3 is a more complex version of that pictured in Exercise 5.28 and discussed in Solution 5.28. An ionic solid is an orderly arrangement of closely spaced ions, oppositely charged particles. The potential energy of any pair of these ions is described as E e l = κ Q 1 Q 2 /r 2 . Separating these oppositely charged particles leads to an increase in the energy of the system, + ΔE. Work is done to the NaCl by the water molecules. Since both NaCl and water are part of the system, the net amount of work is zero. Since ΔE = q + w, ΔE = q and both are positive. The dissolving process typically takes place at constant atmospheric pressure, so ΔH = q and ΔH is also positive. To verify this conclusion, carry out the dissolution of NaCl in a constant pressure calorimeter. Begin with 1 L of H 2 O, record the temperature, add 0.1 mol NaCl, and dissolve completely; record the final temperature. If ΔH is positive, the temperature will decrease. 5.114 (a),(b) Ag + (aq) + Li(s) → Ag(s) + Li + (aq) ΔH o = ΔH o Li + (aq ) − ΔH o Ag + (aq ) f f = –278.5 kJ – 105.90 kJ = –384.4 kJ Fe(s) + 2Na + (aq) → Fe 2 +(aq) + 2Na(s) ΔHo = ΔHo Fe 2 + (aq ) − 2ΔHo Na + (aq ) f f = –87.86 kJ – 2(–240.1 kJ) = +392.3 kJ 133 5 Thermochemistry 2K(s) + 2H 2 O(l) → 2KOH(aq) + H 2 (g) ΔH o = 2 ΔH o KOH(aq) − 2 ΔH o H 2 O(l) f f Solutions to Exercises = 2(–482.4 kJ) – 2(–285.83 kJ) = –393.1 kJ (c) (d) Exothermic reactions are more likely to be favored, so we expect the first and third reactions be spontaneous and the second reaction to be nonspontaneous. In the activity series of metals, Table 4.5, any metal can be oxidized by the cation of a metal below it on the table. Ag + is below Li, so the first reaction will occur. Na + is above Fe, so the second reaction will not occur. H + (formally in H 2 O) is below K, so the third reaction will occur. These predictions agree with those in part (c). 5.115 (a) ΔH o = ΔH o NaNO 3 (aq ) + ΔH o H 2 O(l) − ΔH o HNO 3 (aq ) − ΔH o NaOH(aq) f f f f ΔH° = –446.2 kJ – 285.83 kJ – (–206.6 kJ) – (–469.6 kJ) = –55.8 kJ ΔH o = ΔH o NaCl(aq) + ΔH o H 2 O(l) − ΔH o HCl(aq) − ΔH o NaOH(aq) f f f f ΔH° = –407.1 kJ – 285.83 kJ – (–167.2 kJ) – (–469.6 kJ) = –56.1 kJ ΔH o = ΔH o NH 3 (aq ) + ΔH o Na + (aq ) + ΔH o H 2 O(l) − ΔH o NH 4 + (aq ) − ΔH o NaOH(aq) f f f f f = –80.29 kJ – 240.1 kJ – 285.83 kJ – (–132.5 kJ) – (–469.6 kJ) = –4.1 kJ (b) H + (aq) + OH – (aq) → H 2 O(l) is the net ionic equation for the first two reactions. NH 4 + (aq) + OH – (aq) → NH 3 (aq) + H 2 O(l) (c) The ΔH° values for the first two reactions are nearly identical, –55.8 kJ and –56.1 kJ. The spectator ions by definition do not change during the course of a reaction, so ΔH° is the enthalpy change for the net ionic equation. Since the first two reactions have the same net ionic equation, it is not surprising that they have the same ΔH°. Strong acids are more likely than weak acids to donate H + . The neutralization of the two strong acids is energetically favorable, while the third reaction is not. NH 4 + (aq) is probably a weak acid. mol Cu = M × L = 1.00 M × 0.0500 L = 0.0500 mol g = mol × / = 0.0500 × 63.546 = 3.1773 = 3.18 g Cu (b) (c) The precipitate is copper(II) hydroxide, Cu(OH) 2 . CuSO 4 (aq) + 2KOH(aq) → Cu(OH) 2 (s) + K 2 SO 4 (aq), complete Cu 2 +(aq) + 2OH – (aq) → Cu(OH) 2 (s), net ionic (d) The temperature of the calorimeter rises, so the reaction is exothermic and the sign of q is negative. (d) 5.116 (a) 134 5 Thermochemistry q = −6.2 o C × 100 g × Solutions to Exercises 4.184 J = −2.6 × 10 3 J = −2.6 kJ 1 g -o C The reaction as carried out involves only 0.050 mol of CuSO 4 and the stoichiometrically equivalent amount of KOH. On a molar basis, ΔH = −2.6 kJ = −52 kJ for the reaction as written in part (c) 0.050 mol 5.117 (a) AgNO 3 (aq) + NaCl(aq) → NaNO 3 (aq) + AgCl(s) net ionic equation: Ag + (aq) + Cl – (aq) → AgCl(s) ΔH o = ΔH o AgCl(s) − ΔH o Ag + (aq) − ΔH o Cl − (aq ) f f f ΔH° = –127.0 kJ – (105.90 kJ) – (–167.2 kJ) = –65.7 kJ (b) ΔH° for the complete molecular equation will be the same as ΔH° for the net ionic equation. Na + (aq) and NO 3 – (aq) are spectator ions; they appear on both sides of the chemical equation. Since the overall enthalpy change is the enthalpy of the products minus the enthalpy of the reactants, the contributions of the spectator ions cancel. ΔH o = ΔHo NaNO 3 (aq ) + ΔHo AgCl(s) − ΔHo AgNO 3 (aq ) − ΔHo NaCl(aq) f f f f o AgNO (aq ) = ΔH o NaNO (aq ) + ΔH o AgCl(s) − ΔH o NaCl(aq) − ΔH o ΔH f 3 3 f f f o AgNO (aq ) = −446.2 kJ + ( −127.0 kJ) − ( −407.1 kJ) − ( −65.7 kJ) ΔH f 3 (c) ΔH o AgNO 3 (aq ) = −100.4 kJ/mol f 5.118 (a) 21.83 g CO 2 × 4.47 g H 2 O × 12.01 g C 1 mol CO 2 1 mol C × × = 5.9572 = 5.957 g C 44.01 g CO 2 1 mol CO 2 1 mol C 1.008 g H 1 mol H 2 O 2 mol H × × = 0.5001 = 0.500 g H 18.02 g H 2 O 1 mol H 2 O mol H The sample mass is (5.9572 + 0.5001) = 6.457 g (b) 5.957 g C × 0.500 g H × 1 mol C = 0.4960 mol C; 0.4960/0.496 = 1 12.01 g C 1 mol H = 0.496 mol H; 0.496/0.496 = 1 1.008 g H The empirical formula of the hydrocarbon is CH. (c) Calculate ΔH o for 6.457 g of the sample. o 6.457 g sample + O 2 (g) → 21.83 g CO 2 (g) + 4.47 g H 2 O(g), ΔH comb = –311 kJ o ΔH comb = ΔHo CO 2 (g ) + ΔHo H 2 O(g) − ΔHo sample − ΔHo O 2 (g ) ΔHo sample = ΔHo CO (g ) + ΔHo H O(g) − ΔH o 2 2 comb 1 mol CO 2 − 393.5 kJ ΔHo CO 2 (g) = 21.83 g CO 2 × × = −195.185 = −195.2 kJ 44.01 g CO 2 mol CO 2 ΔHo H 2 O(g) = 4.47 g H 2 O × 1 mol H 2 O −241.82 kJ × = 59.985 = −60.0 kJ 18.02 g H 2 O mol H 2 O 135 5 Thermochemistry Solutions to Exercises ΔHo sample = −195.185 kJ − 59.985 kJ − ( −311 kJ) = 55.83 = 56 kJ 13.02 g 55.83 kJ × = 112.6 = 1.1 × 10 2 kJ/CH unit 6.457 g sample CH unit (d) The hydrocarbons in Appendix C with empirical formula CH are C 2 H 2 and C6H6. substance C 2 H 2 (g) C 6 H 6 (g) C 6 H 6 (l) sample ΔH º /mol f 226.7 kJ 82.9 kJ 49.0 kJ ΔH º /CH unit f 113.4 kJ 13.8 kJ 8.17 kJ 1.1 × 10 2 kJ º The calculated value of ΔH f / CH unit for the sample is a good match with acetylene, C 2 H 2 (g). 5.119 (a) CH 4 (g) → C(g) + 4H(g) CH 4 (g) → C(s) + 2H 2 (g) (i) reaction given (ii) reverse of formation The differences are: the state of C in the products; the chemical form, atoms, or diatomic molecules, of H in the products. (b) i. ΔH o = ΔH o C(g) + 4ΔH o H(g) − ΔH o CH 4 (g ) f f f = 718.4 kJ + 4(217.94) kJ – (–74.8) kJ = 1665.0 kJ ii. ΔH o = ΔH o CH 4 = −( −74.8) kJ = 74.8 kJ f The rather large difference in ΔH° values is due to the enthalpy difference between isolated gaseous C atoms and the orderly, bonded array of C atoms in graphite, C(s), as well as the enthalpy difference between isolated H atoms and H 2 molecules. In other words, it is due to the difference in the enthalpy stored in chemical bonds in C(s) and H 2 (g) versus the corresponding isolated atoms. (c) CH 4 (g) + 4F 2 (g) → CF 4 (g) + 4HF(g) ΔH° = –1679.5 kJ The ΔH° value for this reaction was calculated in Solution 5.86. 3.45 g CH 4 × 1.22 g F2 × 1 mol CH 4 × 0.21509 = 0.215 mol CH 4 16.04 g CH 4 1 mol F2 = 0.03211 = 0.0321 mol F2 38.00 g F2 There are fewer mol F 2 than CH 4 , but 4 mol F 2 are required for every 1 mol of CH 4 reacted, so clearly F 2 is the limiting reactant. 0.03211 mol F2 × −1679.5 kJ = −13.48 = −13.5 kJ heat evolved 4 mol F2 136 ...
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This note was uploaded on 04/04/2009 for the course CHEM 121 taught by Professor Wyzlouzil during the Spring '07 term at Ohio State.

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