Ch7 Solutions - Ill Chapter 7 Thennochemistry IE" |...

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Unformatted text preview: \ Ill: Chapter 7: Thennochemistry IE" |:'-' 50.0 g x 4.18%(216—232) °c = 9.2 x101 = 45.0 g x sp.ht.x(27.6- 30.0)°c B 9.2x 102 J _ .ht. = —— = 0.23 J ' "C”1 for A Sp 75.0 g x 524°C g g gm =375 gx4.13%(87-26)°c=9.5gx10‘ J=—qim g gm =-9.5§x10‘ J:465gx0449-j,'—C(s:r—T,)=1.t;1§><10‘.I—2.03§x102 r; g —9.5§x 10‘ —l.81§x10‘ __ -11.3§x10‘ 4.03;; x101 $2.031; x 102 The number of significant figures in the final answer is limited by the two significant figures for the given temperatures. T, = =S.4fix103'C or 545 °C heat lost by steel = heat gained by water —mx0.50—l—(51.5—133)°C=66m=125 mLxl'oo gx4.18—-;'—— (51.5~23.2)°c g'C I g C 4 66m=1.4sx10‘ J m= W = 2.2 x 10’ gstainlesssteel. The precision of this method of determining mass is limited by the fact that some heat leaks out of the system. When we deal with temperatures far above (or far below) room temperature, this assumption becomes less and less valid. Furthermore, the precision of the method is limited to two significant figures by the Specific heat of the steel. If the two specific heats were known more precisely, then the temperature difference would determine the final precision of the method. It is unlikely that we could readily measure temperatures more precisely than i0.01°C , without expensive equipment. The mass of steel in this case would be measurable to four significant figures, to :l:0.1 g. This is hardly comparable to modern analytical balances which typically measure such masses to :t0.l mg. heat lost by Mg = heat gained by water J . «1.13ng (i; -20.0 c) X lL lcrn i 1000 — 1.00 kgng 3]].024; (sq, —40.0'C) = 1.00 Lx 1 kg g'C —1.024><103 T), +4.10x10‘ =4.18><103 T), —8.36x10“ 4.10x10‘+8.36x10‘ =(4.18x103 +1.024x103)Tf —> 12.46x10‘ =5.20x103 T, 1000 cm' 1.00 g] 12.46x10‘ T =—-——~= 24.0“C " 5.2Dx103 heat gained by the water = heat lost by the brass 150.0 gx4.1sg_icx(i} —22.4-c)=—[15.2 cm3x 3.40 g l .. 0385—”— T ~163 C I cm3 ] ° ( f ) g C =1.40><10‘ +8.01x 103 = 326°C 6.2752102 + 49.2 azrxioli} ~1.40x10‘ =—-49.2 T). +3.01x10’; r, 179 Chapter 7: Themochemistry _9_. heat lost by copper = heat gained by glycerol . .26 —74.8 gx0385 J x(3l.l‘C-—l43.2°C)=165 min gxsp.ht.x(3l.l°C—24.8°C) 3°C 1 mL 3 3.23x103 =1.3x103 x (515.111.) sp.ht.= 1331-12—= 2.5 J g" 'C" 1.3x 10 . - - 92.1 g .. _ 1111 =2.51"c' _+————fl=2.3 IOZJml'C‘ moar ea capacity g x 1 molCaI-l 03 x 10. The additional water simply acts as a heat transfer medium The essential relationship 13 heat lost by iron = heat gained by water (of unknown mass) 00 — 1.23 kgx' 0" 5.0449—(25.—6 -.'=685)C xgH0x4.18—(2s.6—18.5)C 1kg 3'0 EC 11 2.37x10‘J=29.1x x=3:”—"-1°—=29§ gH20xfl=8flx102 111131120 29.: 1.00 gI-le Heats of reaction 1000 g l 1111:11tCa(0H)2 552 kJ x#——~—x 1 kg 74.09 gCa(0H) 1 molCa(0H)2 g. heat=283 kgx 12. heatenergy=l.00 galx-'—_-x x—-——-—x x><——--— x———-—x———- 1L lmL 114.2 gC H111 lmoleHn heatenergy = 1.23 X 105 k] Mx“2377 1‘] = 65.59 1d, 58.123 gC4H1° 11ml Ciflm Mxfl=365x103 k1, 22.414 Lm 01H". 1 "‘01 CnHw (1:) Use the ideal gas equation to determine the amount of propane in moles and multiply this amount by 28'}? 16] heat produced per mole. 13;: (a) heat evolved = 1.325 g C‘Hm x (11) heat evolved = 28.4 L5“, C4HIo x [738mmngx—M7GJE1UIEH )x 12.6L heat evolved— — 0 08206Latm g >< fl =1.45><103k.l _'—fi_ x (273.2 + 23.6) K 1 “1°”an molK —29.4 1-1] 44.10 g c 11 14. a =———r~— ”24:4 22x 103 kJ ( ) q 0.584 gC3Hs x 111101 03118 1mm C H“ —5.27 11] 152.24 g CmHmO =-———#—t M=_5_90 103 11; 1C 1-] (b) q 0.136 g 0,011.60 x 16161 (3,011.60 x "m '° "0 —sa.3 1a 1 1111. 53.03 cu C0 - (c) g: ><—~---3LF)-’—=—l.82><103 kJ!moI(CH3)2CO —-———x 2.35mL(CH,]1CO 0.791g lmollCHJZCO 180 = 2.49 1:105 kJ of heat evolved. Chapter 7: Thermochemistry 20. 22. (a) (b) We first compute the heat produced by this reaction, then determine the value of A H in knmot KOH. gm.” =(0.205 +559) gx4.18 ”5—C(24.4 °C —23.5 °c)= 2x102 Jheat=—qm 3 2x102] “—10%? H_ I AH—- lmolKOI-I- 5x10 kJImol 0.205 _—-— gx 56.1 g KOH The A T here is known to just one significant figure (0.9 °C). Doubling the amount of KOH should give a temperature change known to two significant figures (1.6 °C) and using twenty times the mass of KOH should give a temperature change known to three significant figures (16.0 ”C). This would require 4.10 g KOH rather than the 0.205 g KOH actually used, and would increase the precision from one part in five to one part in 500, or ~02 %. Note that as the mass of KOH is increased and the mass of H20 stays constant, the assumption of a constant specific heat becomes less valid. First we must determine the heat absorbed by the solute during the chemical reaction, :1”... This is the negative of the heat lost by the solution, 9%. Since the solution (water plus solute) actually gives up heat, the temperature of the solution drops. _qnm l L x 2.50 molKI X 20.3 it] heat ofreaction= 150.0 mL X —— —-—---—— =7.61 kJ= 1000 mL 1 Lsoln 1 molKI q” 3 eqm =[150.0mLx1'30 gJX 2'? JxAT AT=%=—l4°c ““1“ 3 C 150.0mLx ' gx ' lmL g°C final T = initial T + A T = 23.5'C —14'C =10°C x 2?5x=8.8x10‘+63x; Let x be the mass, (in grams), of NH 4Cl added to the water. heat = mass x sp.ht. x A T 1 4. . x—meINH‘CI x—-—1 7" k3 x1000 J =— [1400 mLxloo g]+x]4.13i(10.—25)'C 53.49 gNH‘Cl 1 motNHpi 1 k1 1 mL g'C & 8.3x10“ x- 275—63 = 4.2x10‘ g NH‘CI Our final value is approximate because of the assumed density (1.00 g/mL). The solution’s density probably is a bit larger than 1.00 g/mL. Many aqueous solutions are somewhat more dense than water. heat=500mLx # x W x fl: _1_6x102kj 1000 mL 1 Lsoin 1 molNaOH = heat of reaction = - heat absorbed by solution OR gm =—qm 5 AT: 1.63;) J 400] =74°c final T=21°C+74°C=95 °C 500.mLx ' gx ‘ ImL 3°C 182 Chapter 7: 'i‘hennochemisu'y 2_3. We assume that the solution volumes are additive; that is, that 200.0 mL of solution is formed. Then we compute the heat needed to warm the solution and the cup, and finally A H for the reaction. heat=[200.0 mLx 1'02 g)4.02—J—{27.8"c—21.1'c)+10.J—(2';.s °c—21.1°c)=5.6x10’ J 1 mL g'C '0 AH _=flx_‘.5J—=—ss lemol (—55.6 kJimoltothreesignificant figures) 24. Neutralization reaction: NaOH (aq) + HCl (sq) -> NaCl (sq) + H200) Since NaOH and HCl react in a one-to-one molar ratio, and since there is twice the volume of NaOH solution as HCi solution, but the [HCl] is not twice the [NaOI-I], the HCl solution is the limiting reagent. heatrei 1:25.00 mLx 1L x136 molHClxl moleOx —55.84 kJ =-2.6D Id 1000 mL 1 L i molHCl 1 mol H10 = heat of reaction = — heat absorbed by solution or gm = - th 3- AT: 1601"; I 398] =s.54°c AT=TM-T: TM=AT+E 75.00mLx ' gx ' lmL g °C TM =8.54°C+24.72 °C=3326 °C Enthalpy Changes and States of Matter 2; 911200): 911206) _m(SP-hto)nzomATH200)=molnzomAHm H200) I 18.015gI-I0 J m __ 3.50 1H0 —-—~—1— 4.ls4-— .° =—————— . ( mo 2 X lmolHZO )( g“(3)0500 C) (18.015gH10x601XI03m01) lmolI-Izo 13.2 x103 J = m(333.§ J g") Hence, in = 39.6 g 26' fimwm=qmbym lmolHO J J - 5.00 H0x__._2_—— —40.6 103____ + 5.00 4. 4___ ‘- .° K S 2 18.015gHzox X 11101) ( 3X 18 gnCXTf 1000 (3)] = (100.0 gx4.1s4—g—Jo—é)(Tf—25.o °C) 1126;; J — 20.0273E5 (13+ 209; J = “M313 (To —10,4g0 J 113% 1+ 1mm] +20921=41340ic (Tf)+20.9;oic (n) or 23.3x1031=4391(rf) Tfr- 54.2 °c 183 Chapter 7: Thermochemisu'y 2. Assume 1120(1) density = 1.00 g mL'1 (at 285 DC) —qlostby w: = gamma by water + gm Water —[(125 g)(0.50 —~{E)(100 °C —525 “C)] = [(75.0 g)(4.184g+c)(100.0 °C # 28.5 °C)] + nazoAH°wp g massflxo 26562.5 J = 224m .1 + nnzoAH°m (Note: nugo = ) molar 111.215er0 1 mol H20 18.015 g H10 ”11120: 1.3; g H20 E 2 g H20( 1 Sig. fig.) 1 . = 4 . 103 —— 4125 8 J (mHZoX )( 0 6 x mol) 23- -qlost by ban = 4m: 'we J 1 mol H20 3 J — 0.5 — ° —25°C = AH" = ——-——— 6.01x10— [(125 g)( 0 g “C ){0 C 5 )1 111120 fus (mnzoxlsms g Hzox ml) 32% J = mnzo(333£ J g"); maze = 93.1 g H20 5 93 g H20. Calorimetry heat absorbed 5228 cal 4.184 J 1 k] 29. h a1 ' = —— = x = 4.98 kJ/“C "“ e capmty AT 4.39“C x 1 cal 1000 J 30. Heat absorbed by calorimeter = gm x moles = heat capacity x AT or AT = W heat capaclty 1 l [muzgx 4.134 fiHOJZGSg x W] (a) AT: "1° c ‘ g ‘ 1° 2 “ =1.390'c 5.136kJ!°C T! = 1'; + AT = 22.43'C +1.390'C = 23.82°C 24443 1.351111“ 0.805g X lmolC4H30 mo] lmL 72.11gC4HsO (1)) AT: =7.l7°C 5.136 kJI°C TI = 22.43'C + 117°C = 29.60°C . 4.728ka‘C 27.19—2 .29 ' 9-1; (a) heat=heatcapxm= ><( 3 )c=15.6kagxylose mass mass 1.183 g AH: heat given off! g x M(gi 11101) = —15.6 H x W 1 g CsHmO, 1 me] = —2.34x 103 kJImol csnmos (b) C,Hmo, (g) + 502 (g) —> 5002 (g) + 511200) AH = -234 ><10a 1d 184 Chapter 7: Themochemistry E The temperature should increase as the result of an exothermic combustion reaction. 38. A r =1.227 3 01211120,, x em... = —“————“1"m01cwn,4o 1 mol CHHROH 5.65x10’ kJ 1°C X—-—-—-—)( 342.3 g CanOn I molCuHuO” 3.87 k] =5.23"C '1 123°C X 4'63 “PC = —5.65 x103 k] x molCIDHHO . 7 c H 0 139 g "’ " ><150.2gc,.,H..0 Pressure Volume Work 42. 43. 44. (a) (b) (C) w = -PAV= -l.23 atm x (3.37 L— 5.62 L) x[ latm -PAV=3.SL 748 H ——-—— . x( mmg)[760 g = -3.4t_i L atm or -3.4 L atm 1L kPa= l 3, hence, 101.325 kPa]x[ 1J -3.4i_iLatrnx[ ]=-3.42x 1021 or -3.5 x1021 I aim lLkPa lea] 4.184J .142 x 102 J x [ ]= 453.5 cal or —33 cal 101.325 kPa]x[ lJ =280J 1 atm lLkPa That is, 280. I of work is' done on the gas by the surroundings. When the Ne(g) sample expands into an evacuated vessel it does not push aside any matter, hence no work is done. Yes, the gas from the aerosol does work. The gas pushes aside the atmosphere. (3) (b) (c) (a) (b) (c) No pressure-volume work done (no gases are formed or consumed). 2 1401(3) —> N204(g) An“ = -1 mole. Work is done on the system by the surroundings (compression). CaC03(s) -—> (330(3) + C02(g). Formation of a gas, An!” = +1 mole, results in an expansion. The system does work on the surroundings. 2 N0(g) + 02(g) —) 2 N02(g) Aug” = —1 mole. Work is done on the system by the surroundings (compression). MgClz(aq) + 2 NaOI-I(aq) —> Mg(OH)2(s) + 2 NaCl(aq) Ans” = 0, no pressure-volume work is done. CuSO4(s) + 5 H20(g) —> CuSO.;- 5 H20(s) M335 = -5 moles. Work is done on the system by the surroundings (compression). First Law of Thermodynamics 55: (a) (b) (e) AU=q+w=+SSJ+(-58J)=O AU=q+w=+125J+(—687.l)=—562.l 230mx(4.134il)=1171._21=1.1710 210:4 +w=—l.17kJ+ l.25kJ=0.08 k1 ca l86 Chapter 7: Thermochemistry 46. 48. 50. (a) AU=q+w=+23SJ+1281=363J (b) AU=q+w=—14SJ+98 1:471 (c) AU=q+w=0kJ+—l.0?kJ=—1.07k.l (a) Yes, the gas does work (w = negative value). (b) Yes, the gas exchanges energy with the surroundings, it absorbs energy. (c) The temperature of the gas stays the same if the process is isothermal. (:1) AU for the gas must equal zero by definition (temperature is not changing). (a) Yes, the gas does work (w = negative value). (b) The internal energy of the gas decreases (energy is expended to do work). (c) The temperature of the gas should decrease, as it cannot attain thermal equilibrium with its surroundings. This situation is impossible. An ideal gas expanding isothermally means that AU=0=q+w,orw=—q, notw=—2q. If a gas is compressed adiabatically, the gas will get hotter. Raise the temperature of the surroundings to an even higher temperature and heat will be transferred to the gas. Relating AH and AU g 52. 54. According the First Law of Thermodynamics, the answer is (c). Both (a) q, and (b) qp are heats of chemical reaction carried out under conditions of constant volume and constant pressure respectively. Both AU and AH incorporate terms related to work as well as heat. (in) 0.111000) + 6 02(g) -> 4C02(g) + 5 H200) mm = —2 mol, AH< AU (b) Cal-11205(s) + 6 02(g) —> 6 (302(3) + 6 H200) Ans... = 0 mol, AH: AU (c) NH4N03(s) —> 2 H200) + N20(g) M3,, = +1 mol, MI> AU cansoa) + 9,2 02(g) —> 3 cons) + 4 H200) Ans” = —1.5 mol (8) AU= _33_41E x W: _2[)og_k:]_ g 1 mol 031-130 mol (b) AH= AU—w, = AU— (—PAV) = AU— @me7) = AU+ AnguRT -3 AH: aces—“— + (—1.5 mol)(WXZQ'SJS K) = 4012i mol K mol mol cmH.40(1)+ 13 02(3) —> 10 001g) + 7 H200) An,as = —3 mol Qbombr'qtr = AU=— 5.65 x103 1:] = AU AH= AU —w; where w = 9.3an?) = —(—3 mol)( AH=—5.65 x 10310 —7.4kJ=—5.66 x 103 k} 3.3145 x 10'3 kJ Kmol )(298.15 K) = +7.4 id 187 Chapter 7: Thermochemisuy Hess’s Law g The formation reaction for NH3(g) is %N2(g) + g- H2 (g) —) NH3(g) . The given reaction is two-thirds the reverse of the formation reaction. The sign of the enthalpy is changed and it is multiplied by two-thirds. Thus, the enthalpy of the given reaction is r(—46.11kl)x%= +3034 1:]. & —(1) co(g) a C(graphite)+§01{g) AB" = +110.54 k1 +(2) C(graphite)+o,(g) —> co,(g) AH“ = —393.51 R] C0(g) +%02(g) —> C02(g) AB" = —282.97 kJ fl —(3) 3 001(g)+4H20(l)—)C3Ha(g)+5 02(g) AH” = +2219.1kJ +(2) c3114 (g)+4 02(g)—)3 C02(g)+2H20(1)AH° = 493? kJ 2(1) 2 H1(g)+02(g) —> 2 H300) AH“ = —571.6 kl chug) +2H2(g) —+ C3H3(g) AH“ = —290. id 58. The 2'“1 reaction is the only one in which N0(g) appears; it must be run twice to produce 2N0(g). 2 NH3(s)+% 02 (g)-> 2 N10(s)+3 H100) ZXAH; The 151 reaction is the only one that eliminates NI-I;(g); it must be run twice to eliminate 2N1-l3. N2 (g)+3 I-i2 (g) —>2 NH3 (g) 2x AH: We triple and reverse the third reaction to eliminate 3H2(g). 3 H20(l)-—>3 H2(g)+«g— 02 (g) —3><AH§ ResultN2 (g)+02(g)—)2 N0(g) ABM =2:<Alil’,+2><AI—l’z-»3><1‘.‘tl51’3 a 2HCt(g) + CZHAg) + to; (g) —> CZH4012(1) + 1120(1) AH' = 418.? kJ Clz(g)+ H200) --> ZHCl(g)+-}01(g) AH“ = 0.s(+202.4) = +1012 kJ CzH‘(g)+C11(g) —> CIH4C12(1) AH' = —217.5 1:; 60. N1H4(l)+02(g) -a> N2(g)+2I-I20(l) AH“ = —622.2 kJ 2H202 (1) —> 2H2 (g)+202 (3) AH“ m -2(—187.8 kl) = +3756 kJ 2H2(g) +01(g) —» 21-1200) AH“ = 2(—2ss.8 kl) = —5?1.6 k] 14,114(1)+2H,0,(1)—>N,(g)+4H20(1) AH' =—818.2 kJ §_1_. C0(g)+§02(g) —> C02 (g) AH“ = -283.0 k] . 3C(graphitc)+6H, (g) —> 3cm (g) AH’ = 3(—74.81)= —224.43 Id 1 2H2 (g)+02 (g) A 2H20(l) AH’ = 2(—2ss.s) = -—571.6 1o l 3C0(g)—>%02(g)+3C(graphite) AH“ =3(+110.5)=+331.5k.T l 4CO(g)+8H2 (g) —> C02(g)+3CH4{g)+2H20(1) AH“ = 447.5 k] 188 Chapter 7: Themochemistry 68. 72. 74. ziio(s)+so1 (g) 4» ZnS(S)+%01 (g); AH“ = —(—373.2 kJ)!2 = +4391 kJ 439.1 10 = AH;[ZnS(s)]+%AH;[OZ(g)]— AH;[2n0(s)]— AI-I;[SOl(g)] 439.1 k] = AH; [ZnS(s)] +%(0.00 kJ)—(—348.3 kJ) —(—296.8 1a) AH; [2103(3)] = (439.1 — 343.3 4 296.8)kJ = -—206.0 kJimol Most clearly established in Figure 7-18 is the point that the enthalpies of formation for alkane hydrocarbons are negative and that they become more negative as the length of the hydrocarbon chain increases (~20 11:] per added CH2 unit). For three hydrocarbons of comparable chain length, Csz, CgHA, and Csz, we can also infer that the one having only single bonds (CgHa) has the most negative enthalpy of formation. The presence of a carbon-to-carbon double bond (Col-I4) makes the enthalpy of formation more positive, while the presence of a triple bond (CgHz), makes it more positive still. AH“ = 4M? [140(3)] + M1302 (g)] — 29-11301: (3)] - 29H?[H20(l)] = 4 (—92.31) + (0.00) — 2 (0.00) —2 (—285.8)= +2024 kJ AH“ = 2 A H; [Fe(s)] + 3 A H; [002(g)] — A H; [Fe,0,(s)] — 3 A H; [co(g)] = 2 (0.00) + 3 (—393.5) — (4524.2) —3 (—110.5)= —24.8 H Balanced equation: c2H50H(1) + 301 (g) -) 2002 (g) + 3H,0(1) AH‘ = 2AHF[002(s)]+SAHPEHzoflfl-AHHCIHSOHOHA- SAHEIOI 0.1)] = 2 (—393.5) + 3 (435.3) — (—277.7) — 3 (0.00) = 4366.7 kJ First we must determine the value of AH; for C5H12(l) . Balanced combustion equation: csnn (1) + 1302 (g) —> 500, (g)+ 6H20(l) AH“ = —3509 k1 = 5 A H; [1:0, (g)]+ 6AH; [H,o(1)]—AH; [c,Hu (1)] ~8A H; [0, (g)] = 5(—393.5)+6(—285.8)eA H; [coin (1)]—s(0.00) = —3682 kl *1; H; [0,Hu (1)] AH; [CSH12 (1)] = -3682 + 3509 = —173 kamoi Then use this value to determine the value of AH” for the reaction in question. AH = A H;[csHo(1)] + s A H:[H200)] — 5 A H;[00(g)] — u A H:[H3(g)] = — 173 + 5 (-285.8) - 5 (—110.5) — 11 (0.00) = — 1050 1:1 AH = —397.3 k] = A H;[CC1;(g)] 44A H;[Hc1(g)]— AH;[CH;(g)]—4 A H;[Cl,(g)] = A H; [001; (g)] + (4 (—92.31)—(—74.s1)— 4 (0.00))kJ = A H; [ccn (g)]- 294.4 U A H; [con (3)] 2 (4397.3 + 294.4) 1d = — 102.9 kamol AH' = —8326 k] = 12A H; [00, (g)]+14 AH; [H20(g)]—2A H; [0611” (1)]—19 A Hr" [02 (g)] AH‘ = (12 (—393.5) + 14(—2ss.s)) k] — 2AH; [csnu (1)] —19(0 H) = —s723 k] — 2A H; [cfinH (1)] AHF[03H14(1)]= w = _199 kJ/mol 2 190 Chapter 7: Themochemistry fl. Potential energy = nigh = 7.26 kg ><9.81 ms'2 x 168 m = 1,20 X 10“ J. This potential energy is 82. converted entirely into kinetic energy just before the object hits, and this kinetic energy is converted entirely into heat when the object strikes. heat 1.20x10‘I D m_massxs .ht. _ lOOOg 0.471 _3'5 C p 7.26 kgx x 1kg 3°C This large of a temperature rise is unlikely as some of the kinetic energy will be converted into forms other than heat, such as sound and the fracturing of the object along with the surface it strikes. In addition, some heat energy would be transferred to the surface. heat heat =11: [heat cap. +(mass H20x4.184 4%)} heat cap. =—At-——(mass HIOx 4.184 I 8 ) 3 °C The heat of combustion of anthracene is —- 7067 kJ/mol, meaning that burning one mole of anthracene releases +7067 U of heat to the calorimeter. lmolCun x 70mm 1.354 chHm x heat cap. = 173‘233C"H'" 1mm C“H'° — 933.5 g x4 134 x10"’—“—- (3 5.63 - 24.8?) °C ' 3 °C 443904.115) kJ/°C=0.87S we heat=(27.19—25.01)°C[0.875 kJ/°C+(968.6gH20x4.184x10‘3ng" °C")]=10.7k] q = —10.3 kJ x192.1 g CGHsO, mm 1.053 g canso, 1 mol cargo, ==-l.95><103 kJ/mol CaHaO, heat absorbed by calorimeter and water = -heat of reaction =—1.143 g (3.14.302 x ‘26'42 k1 Magnum 1 g 0‘,sz x 1k] 4.184] g°C heat absorbed by calorimeter = 3.033x104 J -2.61x10“ J = 4.2>~:103 J 4.2x10’J 30.25°C—~24.96°C heat absorbed by water =1131 g H20 x x (30.25 °C — 24.96°C) = 2.61x10“ J heat capacity of the calorimeter = =‘i'.9><102 11°C 4.184J g°C heat absorbed by calorimeter = 7.9 x102 me x (29.31°c — 24.98°C) = 3.8 x103 J 2.35x10“ J+3.8x103 J lkJ x 0.395;; 10001 heat absorbed by water = I 162 3 H20 x x (29.31%: — 24.98 “0) = 2.3Sx10‘ J heat of combustion = = 30.5ng mass coal=2.15x 109 Lab. lgcoalx—IEE—xl—mflfll 30.5kJ IOOOg lOOOkg = 70.5 metric tons 192 Chapter 7: Thennochemistry rammngxfl—xmou PV ?60 mmHg n z _ = —____..._.—_._— RT 298.2Kx 0.08206Latm moi—1 16‘ heat available from cbmbustion = 0.00400 mol X ~890.3 klfmol = -3.56 k] lr'noleH:0(s)x 6.01 k] _3 18k] 18.02 g ice 1 mol H20(s) Since 3.56 kJ of heat was available, and only 3.18 it] went into melting the ice, the combustion must be incomplete. 03) One possible reaction is: 4 CH 4 (g) + 7 0; (g)~—> 2 CO2 (g) + 2C0(g) + 8H 20(1) , but this would generate but 3.00 U of energy under the stated conditions. The molar production of C02 (g) must be somewhat greater than that of C0(g). = 0004000101 CH4 heat used to melt ice = 9.53 3 ice x 89. (a) The heat of reaction would be smaller (less negative) if the H20 were obtained as a gas rather than as a liquid. (b) The reason why the heat of reaction would be less negative is because some of the 1410.9 k] of heat produced by the reaction will be needed to convert the H20 from liquid to gas. (c) AH° = ZAH‘} [002 001+ 2M2”f 0120an - Mr”f [€2H4(g)}— 3111?} [02(3)] = 20-3935) + 2(-241 .3) - (52.26) « 3(000) = —1 322.910 E. First we compute the amount of butane in the cylinder before and after some is withdrawn; the difference is the amount of butane withdrawn. T = (26.0 + 273.2)K = 299.2 K PV 2.35 atmx 200.0L RT 0.08206 Latin 11101 K x 299.2 K _ PV _ l.1031.m><200.01.. n2 = 8.96 mol butane ' E ” 0.08206Latm 11101" K" x 299.2K amount withdrawn = 19.1 mol - 8 .96 me] = 10.1 mol butane Then we compute the enthalpy change for one mole of butane and from that the heat produced by burning the withdrawn butane. C.Hm(g) + %Oz(g)—>4C02(g) + 5H20(l) :3 Hf... = 4111*? [002(g)]+ SAH°riH20(l)l — AH‘: [CtHlD(g)]— 9M1°rl02(g)] = 4...
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