chap7-odd

chap7-odd - Chapter 7 Thermodynamic Processes and

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Unformatted text preview: Chapter 7 Thermodynamic Processes and Thermochemistry _______________________._____—-———-———-—-—-—-—-—— 7.1 The work done on a gas in a change of volume at constant pressure is given by w = — extAl/C1 The problem states values for the external pressure and for the final and initial volumes. Substituting gives: in : upextav : —(50.0 atm)(974 L A 542 L): —2.16 x 103 L atm where. as ever, the change in a quantity (in this case the volume) is the final value minus the initial. To convert to joules, multiply by the proper unit factor:2 1. r. w z ~2.16 ><104 Latin x (M I —-2.19 ><106 J 1Latm the nitrogen does positive work on its surroundings. The Tip. Negative work is performed on the nitrogen; ourselves with the gas looking at the surroundings or in difference is a matter of point of view. Do we put the surroundings looking at the gas? 7.3 A ball of mass M falls 3. distance Ah under the influence of gravity. It experiences a change in potential Ah, where Ah is the change in height and g is the acceleration of gravity. The (non- ground. According to the problem. the total energy of the ball 1 energy instead is converted in internal energy that goes to energy equal to Mg bouncing) ball stops dead when it hits the does not change at impact. All of the potentia heat up the ball. This is expressed mathematically: McaAT-l- i’l/ngh. : 0 where c5 is the specific heat capacity of the ball. Cancel out the M’s and solve for Ah: csA'l' 9 Ah: which equals 1.00 K) and c5 equals 0.850 x 103 J K"1kg‘1. Note that In this problem, AT equals 100°C ( cancellation of units. Also, 9 is 9.81 m S”. Substituting gives: c5 is put on a per—kilogram basis to aid the r a —-1 -1 (0800 x10 J K kg )(1.00 K) : _86.6 J kg_1m_182 h: —__———-———~—-—-——-— A 9.81 ms—2 __________,.__.____.——-——-—— 1This is text equation 7.1. 2From text page 202. 67 68 Solutions Manual T0 ACCOMPANY Principles of Modern Chemistry, 5th Edition By its definition a joule equals a kg rugs—2. Therefore, in the above cluster of units all but the meter cancel out: Ah. is —86.6 m. The negative sin simly means that the final height of the ball is less than the initial height. The ball falls a distance of Mdown. 7.5 The molar heat capacity of a substance equals its specific heat capacity multiplied by its molar mass. Here is a sample calculation for lithium: cp : csM = 3.57 J K-lg-‘(sg4 g moi-1) = 24.8 J K"1mol‘1 The full set of values in the group: Li[s) Nafs} K(s) Rbfs) Cs(s) 24.8 28.3 29.6 31.0 32.? JK" mol— Beyond sodium there is a steady increase of about 1.3 J K‘imol‘1 for every element. Extrapolation of the trend assigns francium a molar heat capacity of about 33.5 J K‘lmol‘1 . Although the trend is small, it is distinct. Indeed, the molar heat capacities of the metallic elements are remarkably constant. This constancy is the law of Dulong and Petit (see 7.7). 7 .7 Again, the molar heat capacity of a substance equals its specific heat capacity multiplied by its molar mass. The calculations proceed as in 7.5 with the results: 7.9 a) During the heating process, heat flows from the surroundings to the system. Therefore, . Since the container is rigid, it neither expands nor contracts. Hence AV is zero. Conseuently, no pressure- volume work is performed on the system. No other type of work is possible, so . Then, by the first law, AB is positive . h) During the cooling process, the heat absorbed by the system is us. No work can be done on the system ‘w = 0 . The energy of the system is therefore lowered AE < 0 . e) Since no work was done in either step 1 or step 2, ml + 1152 : 0 . Nothing in the problem suggests that the system is in the same thermodynamic state after it is coo e ack to its original temperature. All that is stated is that the temperature is the same. Other variables, such as the internal energy, might be greatly affected by the heating—cooling cycle. Hence. AE1 + All}; is not necessarily zero. Tip. A trap in this problem is to assume. without justification, that the system in the container is ideal gas (for which the internal energy depends only on the temperature). A related trap is to assume that any changes brought on by the heating are exactly reversed by the coolin. This is not the case when an egg is boiled so why should it be true here? Ali that can be said is AEl + AEg = q; + ()9 . The two sides of this equation could be positive, negative, or zero. 7.11 Let the system under consideration consist of have sub—systems: the metal and the water. If the mixing of hot metal and cool water takes place in a well—insulated container (which prevents leaks of heat}, then the heat absorbed by the system equals zero. The system is the sum of the two sub‘systems. Therefore: gsySIUIQm+9w For both sub—systems, the amount of heat gained equals the specific heat capacity times the mass times the temperature change: gm 'i' qw : MWC5,WATW ‘l' Mmcs,mATm : 0 Chapter 7 Thermodynamic Processes and Thermochemistry 69 Solving for the specific heat capacity of the metal: —ch, WAT“r (100.0 g) 4.18 J K‘lg‘1(6.39°C} = _—._._' : —'—““—-—"—-———-——_ = .4 65"“ MmATm (61.0 g)(—93.61°C) 0 68 J K 5 Tip. We do not bother to convert “C to K. A change of one degree Celsius is identical to a change of one kelvin. The Kelvin and Celsius scales have the same size increments and differ only in the location of their ZEI’OSA 7.13 Body 1 and body 2 are originally at different. temperatures. They are brought into thermal contact with each other and held in thermal isolation from other objects. Then: 91 + 92 = MicuATi + M2Cs2AT2 = 0 If the masses of the two bodies are equal, then M1 = M2, and: c, _ _AT2 s2 — ATI The last equation shows that the specific heat capacities of the two bodies are inVersely proportional to the temperature changes they undergo in this experiment. Tip. The minus sign in the answer reflects the fact that the AT’s of body 1 and body 2 opposite signs; one warms up while the other cools down. h- l CslATl I —C52AT2 from which ('5 are always of A 7.15 The difference in temperature AT between water at its boiling point and melting point is 100°C. The heat needed to bring 1.00 g of water at 0°C to 100°C equals: q = Mani”: (1.00 g)(4.18 J(°C)'1g’1)(100°C) = 418 .1 The amount of heat needed to melt 1.00 g of ice is, according to the statement of Lavoisier and Laplace, 3/4 of this amount or 314 J . More recent experiments set the amount of heat to melt 1.00 g of ice at 333 J. 7.17 The 0.500 mol of neon expands against a constant pressure of 0.100 atm. Define the system as the neon. Before the expansion, the volume of the system is 11.20 L (calculated using the ideal—gas equation with 11 equal 0.500 mol at 1.00 atm and 273 K). The expanded volume is 43.08 L (calculated from the idealwgas equation with P 2 0.200 atm, n = 0.500 rnol, and T : 210 The. gas expands against a constant pressure (of 0.100 atm). The work done on the system is w z — mnv : —0.100 atm(43.08 — 11.20) L = The gas cools from 273 to 210 K. Since it is an ideal inonatomic gas: the change in its internal energy is directly proportional to the change in its temperature; the constant. of proportionality is n(%)R, the heat capacity at constant volume: AE : nchT I n AT Su bstitutiug gives: AE = 0.500 mol 0.08206 L atm mol'lK'l) (—63 K): —3.88 L atm By the first law: q = AE — w = —3.88 Latm~ (—3.19 Latin) 2 The three answers can also be given in joules (1 L atm : 101.325 .1) w: —323J AEz—393J qz—TOJ —-——-———-—-—-—————————_______________________________ 70 Solutions Manual To ACCOMPANY Principles of Modern Chemistry, 5th Edition 7.19 a) The statement of the problem gives the initial quantity (2.00 mol), pressure (3.00 atm), and temper— ature (350 K) of the ideal monatomic gas. The initial volume of the gas is V 2 nRT/P : 19.15 L. The final volume is twice this original volume or 38.3 L . The change in volume AV equals 38.30 — 19.15 = 19.15 L. b) The adiabatic expansion occurs against a constant pressure of 1.00 atm. Under that circumstance. the work done on the gas is 1. 2 . w = -—PAV = —1.00(19.15)Latm >< = —1.94 >-<103 .1 1 Latm The expansion is adiabatic so by definition, and: AE:q+w=O—1.94XIO3J= —1.94x103J c) Any change in the internal energy of an ideal gas causes a change in temperature in direct proportion: AE : chAT Solving for AT and substituting the various values: AE_ -1.94x103J ncv — 2.00 mol(3/2)8.3145 J K"1mol"1 Thus, _T2. the final temperature, is T; + AT 2 350 + (-47.8) = 272 K . AT: 2 —77.8 K 7.21 The system consists of the 6.00 mol of argon. The change in internal energy of this monatomic gas (assuming ideality) is AE = nan? = (6.00 mol)8.3145JK'1mol'1) (150 K) =11.2 ><103 J The change is adiabatic which means that . From the first law: w:AE—q=11.2x1033—0= +11.2x 103.1 The work done on the argon is 11.2 x 103 J , all of which goes to increase its internal energy. 7.23 The balanced equation tells the enthalpy change taking place during the production or consumption of a specific number of moles of product or reactant. All that is necessary is to put these enthalpy changes on abasisofmass. I» ——~—-—. as; x (————fi?§.:°t§:ao) -—- 7.25 Only 119.0 J of the measured 121.3 J of heat comes from the reaction of the 0.00288 mol of Br2(l). The rest of the heat (2.34 J) is added mechanicallya by breaking the capsule and stirring the liquid. The amount of heat evolved from 1.00 mol of Br2(l) is “9'03 ~— molx — 10 J 3See text Figure 73, text page 204. Chapter 7 Thermodynamic Processes and Thermochemistry ' 71 7.27 The vaporization is (30(1) —> CO(g). For this change, AHvap is 6.04 in] moi—1. The following series of conversions provides the answer: 1 mol CO 6.04 kJ 7.29 The 36.0 g ice cube contains 2.00 mol of H20 because 1.00 mol of water equals 18.0 g of water. The cube is put in contact with 360 g of 20°C water. At —10°C, the ice is well below its melting point. It must heat up before it can start to melt. Warming the ice from —10°C to 0°C absorbs: q = ncpAT = (2.00 mol)(38 J K“1mol_1)(10 K): 760 J Melting the ice at 0°C gives water at 0°C and absorbs: q = mam = (2.00 mol)(6007 J moi—1) = 12 014 J On the other hand, cooling 360 g (20.0 mol) of water from 20°C to 0°C would absorb: q 2 mpg} — T.) = (20.0 mol)(75 J K'lmol'IM—QO K) : —3.0 x 10“ J This result can be rephrased: cooling 20.0 moi of water from 20°C to 0°C retfiiires removal of +3.0 x 104 .1. Since the ice cube absorbs only 12774 J by warming up and then melting to liquid water at 0°C, 1‘}, the final temperature of the mixture, must be above 0°C. No heat is lost to the surroundings. The heat absorbed in warming and melting the ice, and then warming the melt-water to the actual I} can therefore be added with the heat absorbed in cooling the 200° water to 7} to equal zero: 12 774 J + (2.00 mol)(75 J K-1m01-1)m — 0) + (20.0 mol)(75 J K‘lmol—‘M’Tf — 20.0): 0 q for warm water q for ice Solving gives :1} z . Tip. A source of difficulty in this problem is the wrong concept that ice is always at 0°C. Like any other material, ice comes to the temperature of its surroundings. q for meltwater 7.31 Multiply the equation and AH for the combustion of methane by 2. Reverse the equation for the combustion of ketene and multiply its AH by —1: - AH : —1604.6 kJ AH = 981.1kJ 2 CH4(yl + 4 02(9) -+ 2 002(9) + 4 H20(g) 2 002(9) + Hgom) —> esteem + 202(g) Adding these two equations gives the desired equation: 2 cute) + 202(9) -+ cagcom + 3H20(g) AH = By Hess’s law, the enthalpy of the total reaction equals the sum of the enthalpies of the two reactions that are added. Tip. How does one know which equations to reverse or double in problems like this? Manipulate to put the correct number of moles of each substance on the correct side of the final equation. Thus, the ketene equation had to be reversed because ketene is among the products in the target equation. W 72 Solutions Manual T0 ACCOMPANY Principles of Modern Chemistry, 5th Edition 7.33 The conversion C(gr) —> C(dia) is endothermic (positive AH). Therefore, one pound of diamonds contains more enthalpy than one pound of graphite. Both diamond and graphite give the same product (carbon dioxide) when burned. When burned, the pound of will give off more heat. 7.35 A reaction enthalpy is calculated by summing the enthalpies of formation of the products and subtracting the enthalpies of formation of the reactants: ‘ N2H4(ll + 3 02(9) 4 2 Norah?) + 2 H200) AH°:2 33.18 +2 —285.83 —1 50.63 —3 0 = —555.93 M i i < i i J i), Noah?) "200) “21140) 02(3) In the preceding equation, all of the AHf’s are in kJ mol”. All are multiplied by the number of moles of the substance represented in the balanced equation. 7 .37 a) As in the preceding: 2ZnS(s) + 302(9) m} 221108) + 2502(g) AH°:2 —348.28 +2 —296.83 —2 —205.98 —3 0 = —878.26 M i i i i i J i), 2110(3) 502(9) ZnS(s) 03(9) t b) Compute the chemical amount of ZnS (in moles) and multiply it by the molar AH" to get the amount of heat absorbed in the roasting of the 300 metric tons of ZnS. It is known that 2 mol of ZnS(s) has a AH° of —878.26 kJ. Hence: _ o _ _____ ...._*_ _.___.. = _ ‘ 7 qp _ AH _ 3.0 metric ton 2113 x ( metric to“) x (97.456 8 ) x (2 “101an ) 1.35 x 10 kJ I . 7.39 a) The balanced equation is CaClfis) —> Ca2+(aq) + 2Cl‘(aq). Combine the enthalpies of formation as follows: AH" : 2 -—167.16 _1 542.83 —1 —795.8 2 -——81.4 kJ Cl—(aq) Ca2+(aql CaCl;[a) b) Compute AH" for the dissolution of 20.0 g of CaCl2(s) 1 mol CaClg —81.35 kJ o — ~---—--—--— 2 — '. AH _ 20.0 g ca012 X (110.98 g CflClg) X (1 mol (33512) 1466 k] The process of dissolution absorbs —14.66 M. The immediate surroundings of the dissolution (the water) therefore must absorb +1466 id. The temperature change of the water equals the heat it absorbs divided by its heat capacity: _ q _ 14.66x 103.1 AT“ cpM _ 418 J K-1 The final temperature is T; = 200°C + 351°C = 55.1°C. 7.41 The balanced equation is CsH12(l)+ 902(9) -> 6 002(9) + 6 H200). Set up a calculation of a standard enthalpy of this combustion reaction in terms of standard enthalpies of formation of the products and reactants. The standard enthalpy of combustion is known, but one of the AHf’s is not known: = 35.1 K = 351°C AH“ = _3923.? Id : 6 (—393.51)+6 (—285.83)-—1./_\1-1f'3 (cerium — 9 (0) w “w V 002(9) H=0(9) 02(9) fl Irena 5925' for car d)' Next state ——u—--w+o—‘—~A—-—-———.-—.._.._.' ~WW... ___..._..._.,..... A “a; m Chapter 7 Thermodynamic Processes and Thermochemistry 73 The standard enthalpies of formation are all in M mol—1. All are therefore multiplied by the number of moles of each substance appearing in the balanced equation. Soiving gives the AH;) of liquid cyclohexane as —152.3 kJ mol'1 . 7.43 a) The equation is C10H3(s) +1202(g) —+ 10 002(9) + 4H20(l) . b) The amount of heat evolved (—q) in the combustion of 0.6410 g of naphthalene was observed to equal 25.79 kJ. Since the combustion was performed at constant volume, no Work was done on the system {in "—- 0). Therefore, AE = q + w = "25.79 kJ + 0 = —25.79 M. Put AE on a molar basis _25.79 kJ 128.17 g owns) _1 A :- __._____ __________ z _ E (0.6410 g CmHs) X ( 1 mol (3,038 5157 kJ mol The temperature is essentially 25°C both before and after the reaction. Hence, the AE" in the combustion of 1.000 mol of naphthalene is —5157 kl . c) To calculate AH" use the definition as” = AE° + A(PV) As diseussed on text page 215, this leads to = as“ + (AnglRT for cases like this one. The Aug is the change in the number of moles of gases during the reaction. The combustion of 1.000 mol of naphthalene produces 10.00 mol of gases, while consuming 12.00 mol. Hence: (AnglRT = (—2.00 m01)(s.3145 J films-11098.15 K): 41.96 k3 AH“ = AE° + (Ang)RT : —5157 — 4.96 : —5162 kJ (1) Use AH" : BAH? (products) -— Z AHf" (reactants) to write: AH" : ~5162 kJ : 10 (fi393.5l)+4 (—285.83) —12 {0)w1AH€(naphthalene(s)) ‘—v--’ ‘---/—’ ‘-v-’ (302(9) H200) 02(9) where each term on the right consists of a AHFo in kJ mol"1 multiplied by the number of moles in the balanced equation. Solving gives the AH,” of solid naphthalene as +84 kJ mol‘1 at 25°C. 7.45 Write an equation for the formation of CClgF(g) from the “naked atoms”: C(51) + 301(9) + F(g} —> CCl3F(g) From the average bond enthalpies4 estimate the AH° for this reaction as: AH“ = 1 (~441)+3 (—328) : 4425 1d RH \_V_../ C—F C—Cl Next, write equations that show the preparation of the naked atoms from the elements in their standard states. Each of these atomization processes has an associated enthalpy derived from the data in the text:5 0(3) —> C(g) AH° = 715.7 kJ 3/2 (My) —> 301(9) AH° = 365.1 kJ 1/2 F2(g) —> F(g) as” = 79.0 M 4Text Table 7.3, text page 219. 5The atomization enthalpies in Table 7.3 are per mole of atom formed. Accordingly, each is multiplied by the number of moles of the atom involved. 74 Solutions Manual T0 ACCOMPANY Principles of Modern Chemistry, 5th Edition Combine the four reactions to arrive at the formation reaction of CC]3F(y) from the elements in their standard states: 0(9) + 301(9) + F(g) a; ccsro) AH” = 4425 1d 0(3) —-> C(g) Ah” = 716.7 kJ 3/2 012(9) —> 301(9) AH" = 365.1 kJ 1/2 F2(g) —> Po) AH° = 79.0 H 0(5) + 3/2012“) +1/2F2(g) —+ CCl3F[g) AH“ z ~264 Id The AH; is —264 kJ mol"1 because the reaction involves 1 mol of CCl3F(g). 7.47 The reaction is the combustion of propane in oxygen: CaHgfg) + 5 02(5)) —> 3 (302(9) + 4 HgOLq) As this reaction proceeds, bonds are both broken and formed. Broken are 2 mol of CWC bonds, 8 mol of CfiH bonds, and 5 mol of 0:0 double bonds Formed are 6 mol of C20 double bonds and 8 mo] of OfiH bonds, The net enthalpy of the reaction approximately equals: an AH a: 6 (728.) +8 (463) —5 (498) —8 {413) M2 (348) .: —1.58 ><103 kJ V V \_.V-z V v C=O 0—H 0:0 C—H C—C Tip. The anSWer “1.582 x 103 kJ is correct according to the rules for significant digits.6 In View of the fact that bond enthalpies are only approximately constant7 —1.58 x 103 k3 is a more sensible answer. 7.49 The Lewis structures are: 3% >644 Br / :Cl I. Cl \ >3—P~Br : )3—(31 :Cl Br Tip. Boron tribromide and boron trichloride are octet-deficient molecules. 7.51 The system is the 2.00 mol of ideal gas. In an isothermal change, no change of temperature occurs. (AT : O). The internal energy of an ideal gas depends only on its temperature which means that . As for the enthalpy: AH = AE+A(PV) : 0+ Amer) : 0+nRAT= E ~_._______H___ 6Terri; page A.6. 7Texc page 220. Chapter 7' Thermodynamic Processes and Thermochemistry 75 The expansion is reversible. Hence: g V2 _ _ 3.3145 J 36-90 The first law requires that AB = q + w. Hence q equals +6.87 kl. 7.53 During any adiabatic process q = 0. During this reversible adiabatic expansion of an ideal gas: 'IIVI'Tml = Tz‘éfif—l where 7 is cP/cv and the subscripts refer the initial and final states of the gas. In this problem, V1 is 20.0 L, V2 is 60.0 L, ‘y is 5/3, and T1 is 300 K. Solving for T2 and substituting gives; V1 "‘1 200 L 2/3 T2 .—. T; = (300 K) (gob—L) =144.22 K = 144 K Meanwhile, the AE of the ideal gas depends solely on its change in temperature: AE : chAT 2 (2.00 mol) 8.3145 JI{“1mol'1)(—l55.78‘l{) : —3.89 1d This number also equals to, the work done on the gas, because AE = q + w and q is zero in this process. Finally, AH of an ideal gas also depends entirely on AT: AH = ncpAT .: (2.00 mol) 3.3145 J K’lmol‘l) (—155.78 K}: —6.48 kJ Tip. Notice that AH : 7AE for this reversible adiabatic process. 7.55 The law of Dulong and Petit states that all metals have a molar heat capacity of approximately 25 J K“mol”’. The molar heat capacity equals the specific heat capacity of a substance multiplied by its molar mass. Hence: c : 65M 3 25 J K“1mol_1 The experimental specific heat capacity of indium is 0.233 .1 K"g“1. A molar mass of 75 g rnol‘1 combines with this number to give a molar heat capacity for indium of only 17.7 J K’lmol‘l. This violates the law of Dulong and Petit badly. The modern value ofM for indium (114.8 g moi—1) gives a c that is consistent with the law of Dulong and Petit. Tip. For solids and liquids the distinction between up and eV is unimportant. especially in an approximate relationship. For this reason there is no p or v subscript on c in this problem. 7.57 a) The system is the 2.00 mol of argon gas. The work done on the system is —PAV. Since the gas is ideal and P is constant, PAV : nRAT for the system. In this case AT is given as —100 K. Then: to = —nRAT = —(2.00 mol)(8.3145 J K-lmor‘)(—100 K): b) The process goes on at constant pressure so the heat absorbed is qp: q 2 nc AT: (2.00 mol £8.3145J K‘lmoi—1 —100 K = —4.16 x103 J P P 2 c) Use the first law of thermodynamics: A5 = q + w = —4157+ 1663 = ——2494 J. This rounds off to —2.49 x 103 kJ . Note the use of un-rounded answers from parts a and b in the addition. d) The AH of a system always equals qp. Hence, AH is $4.16 M . Solutions Manual TO ACCOMPANY Principles of Modern Chemistry, 5th Edition of water. Hence the heat absorbed in 1.00 x 103 g. Therefore: csM : (4.18.1 K-lg-1)(1.00 x103 g) = 0‘194K 7.61 Use the molar mass of glucose ( Cnggoe) as a unit-factor to obtain the chemical amount of glucose in the candy bar. Then use the molar enthalpy of combustion of glucose as a unit-factor to obtain the heat absorbed: 1 mol Ca H1205 = 1 . ———____ -_.__.___ z _ I ‘1 4 3 g CGHIQOS X (180.16 g Csngoa) X (1 mol cannot.) 223 8 k‘] dings of the reaction (which are the pgpeon’s body) therefore equals at is absorbed by 50 kg of water: The heat absorbed by the surroun +223.8 kJ. When this amount of he , q 223.8x103J , : —~— 2 —'_'——~——-—__ 2 1.11\ M cs 4 (4.13 J K—lg-1)(5o x103 g) 7.63 Determine which liquid is a better coolant nea ' ‘ ' r the bellmg pomt by comparing their specific heat capacities. He(l) absorbs 4.25 J of heat per gram as it heats up by I K. Ngfl) absorbs only 1.95 J of heat per gram as it warms by the same amount. Therefore, HeU) is a better coolant near the boiling point. At their boiling point, the two liquids cool by vaporization; N2(l) is better because it absorbs much more heat per gram in vaporization than HeU). 7.65 a) The combustion of isooctane is represented: and takes place at constant systems: the combustion reaction, the calorimeter body, , the system neither gains nor loses heat because the bom add up to zero: qsys I 01120 + Gealorimeter + qcombusuon = 0 constant V The heat absorbed by a system (or sub-system) in a change at constant volume equals: qv : cVAT or q" = McsAT pacity or specific heat capacity is available. The problem gives the heat capacity cv of the calorimeter (48 J K‘I) and the specific heat capacity of water (4.184 J K‘lg'“1]. The AT of the calorimeter equals 28.670 -— 20.450 2 8220" C, which also equals 8.220 K. The AT of the 750 g of water also equals 8.220 K. because the water and calorimeter are in thermal contact. Insert these numbers for the q’s of the water and calorimeter: (48.1 K‘1)(8.22 K) + (750 g)(4.184 J K-‘gr1)(s.22 K) +qcombustiun : o h—W ‘-—-—-_,—____/ calorimeter Chapter 7 Thermodynamic Processes and Thermochemi'stry 7? Solving for the last q gives w2.62 x 10‘1 J. This equals the heat absorbed by this combustion reaction at constant volume. At constant volume. zero work is done by or upon the combustion reaction. Hence: AEcombustion 2 q ‘l' w = (Iv + 0 = —2-62 X 104 J c) The molar mass of CSHIS is 114.23 g mol—1. The combustion of an entire mole of isooctane absorbs more heat than the combustion of 0.542 g: _ 4 AE : igéi‘glgfi x 114.23 g mol—1 = —5.52 x 106 J mol—1 = ~5520 kJ mol“ d) By definition, AH = AE + A(PV). If the gases in the combustion reaction are ideal and the liquids 1 have negligible volume, then A(PV) : (AnflRT, where Ang is the change in the number of moles of gas. The balanced equation for the combustion of 1 mol of isooctane shows that Ans : 8 — 12.5 : —4.5 mol. Therefore: (Ang)RT = (—4.5 mol)(0.008315 kJ mol-1K*1)(298 K) 2 _11.15 kJ Although the temperature rises from 20450" to 28.670°C, taking it as a constant 25°C (298 K) causes little error. Complete the calculation as follows: l_. l l AH : AE + A(PV) : —5520+(~11.15): e) The standard enthalpy of the combustion reaction written above is given by the equation 1:; Ah” 2 8 AHflCOflgfl + 9 AHf°(HgO(l)) — AHflisooctane) The AH obtained from the bomb calorimetry experiment does not equal the AH° of this reaction at 298.15 K, but should approximate it closely. Insert the AH on the left in the preceding, substitute AHf‘s from text Appendix D on the right, and solve for AHflisooctane) 25 -5530 Id : 8 (—393.51)+9 (—285.83)—AHf(isoocta.ne) — m (0.00) W W 2 W C02(54) H200) I 02(9) AHfi’fisooctane) : ---190 kJ mol—1 7.67 a) To get A5” in the combustion of 1 mol of acetylene, combine AHfi’s as follows: AH“ = 2 (_393.51)+1 (—241.82)-1 (mam—g (0.00): 71255.57 H W \_‘,_/ ‘u—V—z W (302(9) Hzolgl CeHelgl 02(9) b) The total heat capacity of the mixture of the two gases equals the molar heat capacity of the first multiplied by the number of moles of the first plus the molar heat capacity of the second multiplied by the number of moles of the second: mp = (2.00 mol) (37 J K“1mol_1)+(1.00 mol) (36 J K‘lmol‘l) = 110 J K’1 \——v—/ haw—J CO; H20 c) Assume for convenience that 1.00 mol of C2H2(g) is burned. Then the product gases, which are 2.00 mol of 002(9) and 1.00 mol of H20(g), absorb 1255.57 k} of heat. For these gases, which comprise the flame: _i_1.25557x106J_ 4 _ o AT_nCP_ “MPH _1.14x10K_114OOC If the temperature before combustion is 25°C, the maximum flame temperature is 11400°C . F 78 Solutions Manual To ACCOMPANY Principles of Modern Chemistry, 5th Edition 7.69 Define the system as the contents of the engine cylinder. Before the explosive combustion of the octane, the temperature is 600 K, the volume is 0.150 L, and the pressure is 12.0 atm. Apply the ideal-gas equation to the mixed contents of the cylinder before the combustion: _ PV _ (12.0 atm)(0.l5U L) enre: Deane mr—*~—_‘_‘“—————"—: i: . "h f' n t + "' RT (0.08206 L atm moi—IK-l)(600 K) 0 0 6 m 36 56 “"“°' Also, the cylinder holds octane and air in a l-to—SO molar ratio: 80“octane : flair Solving these simultaneous equations gives: nocume = 0.4514 mmol and nail. = 36.11 mmol According to the problem, the system does not change its volume during the actual combustion of the fuel. so in is zero. Furthermore, q is zero (the combustion happens so fast that there is no time for heat to be lost or gained). Since in and I] both equal zero, AE of the system equals zero. Imagine the combustion to occur in two stages: a: the reaction goes at a constant temperature of 600 K; b: the product gases heat up at constant volume. The sum of these two changes is the overall change within the cylinder. Therefore: ABS” "—- 0 = 13Ea + AEb which means AB“ 2 —AEb The problem offers data pertaining to enthalpy changes, not energy changes, in the two steps. Deal with this by substituting for the AEa and AE;J in terms of AH’s: AH. _ A(PV)a = —- (as. v- Ami/t) Step :1 involves ideal gases, takes place at a constant temperature, and involves change in the chemical amount of gas. Therefore A(PV),I equals AngRT, where Arts is the change in the chemical amount of gases during the reaction. Step (2 is the heating of the ideal gases inside the cylinder. The term A(PV).», therefore equals nanerRAT where “after” refers to the chemical amount of gases present after the reaction. Also, for the change in temperature that comprises step 5, AH}, is equal to nanercpAT, as long as the molar heat capacity CD is independent of temperature. Substitution of these relations gives: AHG — AngRT : —(nafteGCAT — narterRAT) In this equation T is 600 K, and AT is the temperature change during the heating. The plan is to compute all the other quantities and then substitute in this equation to get AT and, from it, the final temperature. The cylinder contains 36.11 mmol of air and 0.4514 mmol of octane before the reaction. Air is 80% N2 and 20% 03 on a molar basis. Therefore nN, : 0.80(36.11) mmol no, 2 0.20(36.11) mmol noctane : 0.4514 mmol before the reaction. The octane burns according to CBHlflig) + 12% 02(9) —> s 002(9) + 9 mow) so that after the reaction the amounts of the different gases are: nN, : O.80(36.11) mmol no, = 0.20(36.11)- 12.5(0.4514) mmol 1".ch2 = 8(0.4514) mmol nmo = 9(0.4514) mmol -. (WWW—W. “,._.._...,...=.._........W,_...... tvmm .‘ . . Chapter 7 Thermodynamic Processes and Thermochemistry 79 Note that N; does not react, and that the octane, the limiting reactant, is all used up. Addition and subtraction confirm that nbefore = 36.56 mmol “after 2 38.14 mmol and Ans = +0.158 mmol The molar enthalpy of combustion of gaseous octane at 600 K can be approximately using the AHf’s of the products and reactants at 298.15 K as follows: AH = 9 (—2418) +8 (—3935) —1 (—57.4)—12.5 (0.00): "5266.8 kJ H—d W W v moo) c020) names) 020;) This is not AH“, the enthalpy of the combustion reaction in the cylinder. Only 0.4514 mmol of octane burns, SO AHa = —5266.8 kJ mol'1 x (0.4514 x 10"3 moi) = —2.377 kJ = —2377 J The composite heat capacity of the contents of the cylinder after the reaction is the sum of the non values for the four product gases, as in 7.671): nanercp = (0.00158 mol) (35.2 J K-lmol—1)+(o.0289 mol) (29.8 J K-‘moi-l) ‘_v——-’ ‘—-——n—-’ 02 N2 + (0.00406 moi) (38.9 J K-lmol-1)+(0.00361moi)(45.5 J K"1mol‘1)= 1.24 J K-1 hum—H &——-fi.,-—/ H30 CO: Now, solve the equation derived previously for AT and make the various substitutions: AT: AHa-An RT : “afterR "- “aftercp 4377 J * (0.00158 mol)(8.3145 J K-lmoi-1)(600 K) (0.03814 mol)(8.3145 J K—lniol—1)—1.24 J K’1 2 2580 K The maximum temperature inside the cylinder is 600 + 2580 = 3180 K. This equals 2910°C . 7.71 a) The gases trapped inside the cylinder of the “one-lung” engine have volume V1 when the piston is fully withdrawn but a smaller volume V2 when the piston is thrust home. The compression ratio is 8 : I so V1 = 81/2. The area of the base of the engine’s cylinder is arr-2, where r is the radius of the base. The volume of a cylinder is the area of its base times its height h: V1 = Ah and V2 : A(h #1200 cm) which employs the (given) fact that full compression shortens h by 12.00 cm. Because 1“ is 5.00 cm, the area A is 78.54 c1112. Substituting for V1 and V2 in terms of A and It gives: Ah = 8A(h _ 12.00 cm) The A’s cancel, allowing solution for h. The result is 13.714 cm. With 11 known it is easy to compute V1 and V2, which equal 1.077 L and 0.1347 L respectively. The temperature and pressure of the fuel mixture are 353 K (80°C) and 1.00 atm when the mixture enters the cylinder with fully withdrawn piston (V1). Assuming the fuel mixture is an ideal gas: (1.00 atm)(1.077 L) mix ure 3 .....__..__.__—________ = ' 2 n t L atm moi-1K-1)(353 0 mo] The molar ratio of air to fuel (031113) is 62.5 to 1. Then: flfueg + mm = 0.0372 mol and nair = 62.5mm; 80 Solutions Manual To ACCOMPANY Principles of Modern Chemistry, 5th Edition Solving these simultaneous equations establishes that at the start the cylinder contains 0.0366 mol of air and 5.86 x 10"4 mol of octane fuel. During the compression stroke, the system undergoes an irreversible adiabatic compression to one-eighth of its initial volume. None of the relationships that goVern reversible adiabatic processes apply. Assume however, as advised in the problem, that the compression is near to reversible. If it is, then: 35 J K‘l‘inol‘1 26.7 J K‘lmol‘1 21-31 111/3“ 2: TZVJ'I where 7 : The temperature after the compression stroke is V1 *‘1 1077 L “'31 31 b) The. compressed gases occupy a volume of 0135 L just before they are ignited, as calculated above. c) The pressure of the compressed fuel mixture just before ignition is P2. Compute it by applying the ideal-gas equation to the system with T2 = 673 K. V2 : 0.1347 L, and n = 0.0372 mol. It equals 15.3 atm. Alternatively, estimate P2 using the formula for a reversible adiabatic change: P—P fl 7—100atm 1'0771‘31— 2" I v; ‘ ' 0.1347 ‘ (1) AH for the combustion of gaseous octane at 600 K is ~5266.8 kJ mol—1, as estimated in 7.69. This number is preferable to AHD : —5530 kJ rnol’1 for the combustion of liquid isooctane that was obtained in 7.65. The latter is for combustion of a diiferent compound (isooctane) in a different form (liquid, not gaseous) at a different. temperature (298 K not 600 K) to give a differth product (liquid water, not water vapor). The combustion mixture inside the cylinder contains 5.86 x 10‘4 mol of octane. Consequently, the AH of combustion in this system equals: —5266.8 kJ AH: ( lmol ) x (5.86 x 10“1 mol) = 73.09 M After the combustion, the cylinder contains (30;, 1-190 and unreactecl 02 and N3, all gases. The balanced chemical equation shows that the combustion consumes 5.86 ><10‘4 mol ofoctane and 12.5 x (5.86 x10“4) mol of 0-3 to produce 8 x (5.86 x 10—4) mol of C02 and 9 X (5.86 x 10‘4) mol of H20. The effect of the reaction is to increase the chemical amount of gases in the cylinder by 3.5 X (5.86 x 10‘”) mol. This is Aug for the reaction. The original quantity of gases is 0.0372 mol. After the combustion there is 0.0393 mol of gases. The energy (not enthalpy) releaSed from the reaction all goes to heat up the gaseous contents of the cylinder as long as noheat escapes to the cylinder walls and no work is done until the power stroke starts. Therefore: _ Alum“ — AngRT AT nH. — ncp In this equation, which is derived in 7.69, every quantity but AT is known: ~3090 J — (0.002051mol)(8.3145 J K‘Imol*1)(673 K] _ 2960 K [0.0393 mol)(8.3145 J K‘1niol‘1} — (0.0393 mol)(35 J K‘lmol“) — The temperature inside the cylinder rises by 2960 K to a maximum of 3630 K . “Lawn—w mm .p—P-u—nmw W Chapter 7’ Thermodynamic Processes and Thermochemistry 81 applies. In this case, T1 is 3630 K. The ratio V1 / V2 is 1 to 8 because now the initial state is the small volume state just before the expansion stroke of the piston. The exponent 7 - 1 is still 0.31, as previously established. Substituting gives: 031 T2 2 (3630 K) = 1900 K This is the temperature of the exhaust gases. 7.73 This oxidation of the 00(9) can be represented 00(9) + 1/2 02(9) —> (302(9) The standard enthalpy of this reaction at 25° is AH° = AH?(002(91)- 1/2 AHHOzUD - AHHCOQJ) = 493.5 kJ — 1/2 (0 Id) — (—1105 U) = —283.0 M Suppose that 1 g of air passes over the catalyst in the tube and that no heat is lost from the system when the CO in this air burns. Assume also that the actual AH for the reaction equals AH°. Then qsys : Qreaction + (lair 3 0 m 7100 + maircairAT t "'00 —283.0 x 103 J : M00 ( 1 mol (30 W mco (—2830 X 10"3 J 28.0 g mol’1 1 mol CO ) + maircsAT ) +1 g(1.01 J g_1K“1)3.2 K Solving the last equation for the only unknown, which is mco, the mass of the carbon monoxide, gives. 3.2 x 10"1 g. The mass percentage of CO in the air is .- —4 M x 100% = 0.003296 lgair 7.75 Solid (321(0ng dissolves in aqueous HCl according to Ca(OH)2(s) + 2 H30+(aq) —> Ca2+(aq) + 4 H200!) This event can be viewed as the dissolution of the solid followed by the neutralization of the OH’ ion. Use AH? values from text Appendix D to compute its AH" AH° —_~ 1 AH;(Ca2+(aq)) +4 AHf(H20(l)) — 1 AH;(Ca(0H)2(s)) — 2 AH:(H30+(aq)) = —542.83 + 4(—285.83) — (—986.98) - 2(-285.83) = —127.51kJ in which the standard units of mol and kJ mol"1 are omitted for compactness. The greatest change in temperature occurs when the system is thermally insulated from its surroundings. Under that condition, the system as a whole absorbs zero heat as the subsystems “reaction” and “solution” exchange heat 0 = sts : qreaction ‘l' geolution = AH 'l’ mcsAT The q of the reaction should be quite close to Ali” because the reactants and products are in standard states. The rise in temperature does afi'ect AH°, but the efl'ect is surely slight. The mass of the solution can “L‘ng 82 Solutions Manual T0 ACCOMPANY Principles of Modem Chemistry, 5th Edition be computed from its volume and its density. Approximate the density as 1.0 g mL‘l, which is the density of pure water at 25°C. Then —127.51x103 J 1.0 x103 g = . H ————.— 1 __.___ . -—1 —1 O 0 5 mol Ca(0 )2(1molca(OHJ2)+ L ( L ) [4184 J g K )AT 0 = —63.7 x 103 J + (4134 J K-lmT Solving gives AT = 15 K. The maximum temperature is 25" + 15° : . 7.77 The chemical amount of the silane at the T and P stated in the problem equals: n 7 _ at; g (0.658 atm](0.250 L) 5‘?“ ‘ RT ‘ (0.08206 L atm mol—lK-1)(298 K) The combustion of this much silane at constant volume (in a bomb calorimeter) absorbs -—9.757 kJ of heat at 25°C (which is the same as evolving +9.575 kJ). That is, AE : qV = ~9.757 in]. This is a standard AE (a AE°) if the reactants and products are in standard states. Assume that they are. The standard molar energy of combustion of silane is then —9.757 kJ —1 E0 : m = —-1450 kJ mol Next, compute the AH° of the combustion of silane The balanced equation given in the problem shows that 3 mol of gaseous reactants gives 0 mol of gaseous products: “ AH° = AE° + RTAng = —1450.4 kJ + (0.0083145 kJ K-imoi-lmssis K)(—3 mol) = —1458 Id 2 6.727 x 10—3 mol AH° for the combustion of silane equals the sum of the standard enthalpies of formation of the products minus the sum‘of the standard enthalpies of formation of the reactants. Taking values from Appendix D: AH" : —1458 Id :1(—910.94) + 2 (—285.83) — 1 (AH?) W V-—-—.,-—/ \—v—/ SiO; quartz H:O{t) SiHa (9) (AH?) = —25 kJ mol-1 ‘—-v—’ SiH1[g) Compute AB? of silane from AH;, and the known Ang in the formation of 1 mol of silane from its elements: AB? = AH;3 — RTDnig = —25 kJ — [0.008315 kJ K‘lmol'1)(298.15 K)(—1 mol) : —23 kJ 7.79 Substances with the strongest intermolecular forces have the highest enthalpies of vaporization. Liquid KBr has strong ion-ion forces holding its ions together. It has the highest AHvap. NH3 has dipole-dipole attractions, which are stronger than the Weak dispersion (van der Waals) forces that maintain the liquid in Ar and He. The dispersion forces should be stronger in Ar than in He because Ar has a larger molar mass. Therefore: He < Ar < NH3 < KCl . 7.81 a) Lewis structures for carbonic acid show two 0%H single bonds, two CVO single bonds and one C=O double bond: Chapter 7 Thermodynamic Processes and Thermochemistry 83 b) Imagine the reaction to proceed by the breaking of the five bonds in H2003 foliowed by the making of the four bonds in H—O—H plus O=C=O. The enthalpy of bond breaking is positive; the enthalpy of bond making is negative. Take bond enthaipies from text Table 7.3,8 and combine them accordingly: AH:—2 463 —-2 728 +2 463 +2 351 +1 728: —26kJ r) ( ) ( ) < ) ( J 0—-H C=O 0—H O—C C20 ‘4,-_»_.-y...i.w “mph WW,“ u . _ .wwnm... A- 8Text page 219. ...
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