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 f...
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