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HW 10 key - rm l0 6 18 a Total heat must bring both the...

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Unformatted text preview: rm l0 6,, 18) a. Total heat must bring both the container and the water to 100.0C. The specific heat capacity of stainless steel is 0.51 Jlg*C and that of water is 4.184 Jlg*C. q = mCSAT q = (0.51 Jlg*C)x(500.0 g)x(1000-25.0C) + (4,184 J/g*C)x(450.0 g)x(100C-25.0C) =19,125 J + 141,210 J = 160,335 J = 160.3 kJ of heat b. voqwale. = qHZO/qmtxtoo =141,210 £160,335 J *100 = 88.1% c. the percentage is 94.3% in 6.17. This indicates that the copper kettle is more effiecient in heating the water. 20) The heat gain from the water wilt be equal to the heat loss of the metal. Therfore, (mCsAT)H20 = "(mCsAT)metal (4.184 Jlg*C)x(50.7 g)x(25.TC-220) = (Cs)x(20.0 g)x(25.7C—100C) Solve for C5: C5 = 0.538 JIg*C 34) Sulfur dioxide is a nonlinear molecule and if the vibrational motions are not accessible, it will have molar heat capacity given by CV,m = 3/2R +3/2R = 3R (see page 233 in book). 9(Temperature) = hvvfb/k, k = 1.38066 x 10'23 JIK, h = 6.626 x 10'34 J*s a1 = (6.626 x 10-34 J*s)*(3.5 x 1013 34)/(1.38066 x 10-23 J/K) = 1679 K 02 = (6.626 x 10-34 J*s)*(4.1 x1013 5'1)/(1.38066 x10“23 J/K) = 1976 K 93 = (6.626 X 10'34 J*S)*(1.6 x1013 5'1)/(1.38066 X 10'23 J/K) = 768 K a. The high temperature limit of molar heat capacity includes all degrees of freedom. This is 3 translations. 3 rotations and 3 vibrations for a total of 3R (0mm) + SR (1R for each vlbation) = E- b.for an isochoric process at 1000K, only one vibration is accessible, so CV = E. b. for an lsochoric process at RT, there are no vibrations accessible. so CV = E. 52) for an isobaric process AH = q. We know that AU = w + q, so q = AU — w = AH AH = AU -w= -65 kJ - {-48 kJ) = -17 kJ 112) S + 02 = 302 and S + 312 02 = 803. Aern = (-5.270 kJ/C)x(1.140 C) = -6.008 kJ AH°f(SOZ) = -296.83 lemol AH°f(803) =. -395.72 kJ/mol We have (0.6192 g S)x(1 moi 81' 32.06 g S) = 0.01931 mol 8 let x = mol 302 and y = mol 803_ Therefore, x+y = 0.01931. 24) 98) AH°f(SOZ)x(x) + AH°f(SO3)X(y) = 0.01931 Solve for x as a function of y and then plug it back into the above equation. This gives x = 0.01653 mol 802 and y = 0.002782 mol 803. xly = 5.92 P1 = 2.57 atm, T1 = 298 K, V. = 3.42 L. Assume Ideal Gas (IG). n = PV/RT = 0.359 moi Path A: Isothermal, reversible. P2 = ? atm, T2 = 298 K, V2 = 7.39 L w = -Pede = —(nRTN)dV = —nRT V In (V2N1) = (0.359 mol)x(8.3245 v Jlmol*K)x(298 K) in (T.39Ll3.42L) = -686 .1 q = -w = 686 J Path B: 1. cooled isochorically to 1.19 atm. w = —Pede = 0 2. isobaric expansion to 7.39 L. w = ~Pede = -(1.19 atm)x(7.39 L — 3.42 L) = 4.72 L*atm x(8.3145 JImoi*K)/(0.0B206 L*atmimol*K) = 478.7 J a. isothermal expansion from V1 = 7.00 L to V2 = 15.50 L for 1 mol of [6. at 298 K. w = -nRT In (V2/V1) = 4970 J b. irreversible adiabatic expansion for sample gas against 760 torr (1 atm). w = —PBXdV = -861.3 J c. for an adiabatic process q = 0. Therefore AU = w = -861.3 J. We also know that AU = nCVAT (see page 242). We need to know CV because we are solving for AT. The problem tells us that we have an ideal gas. So, we need to know whether or not it is monatomic, linear. or nonlinear. We typically assume monatomic if we have no reason to choose the others. (We could check this assumption if we were given the actual temperature change). Since we assume that it is monatomic, CV = 312 R or 12.5 J/K*mo| 108) 33) Therefore, AT = AUl(n*CV) = -BB1.3 JI'(1 mol*12.5 J/K*rno|) = -68.9 K Tf = 298.15 K — 68.9 K = 229.2 K step 1: q = +50 J, isochoric. step 2: q = —5 J at 1atm. AU 2 0 = qtot + wtot total work = -q = —(50—5) = -45 J = -Pede. Solve for ON. IN = wrl-Pex = (—45J/‘l atm)x(0.08206 L*atmlmol*K)/(8.3145 J/mol*K) = .444 L Therefore it is an exgansion. m = 325 9, T1 = 30.0, q to get liquid to vapor at 100.C Molar hear capacity of H20: 75 JIC*mol, Heat of vaporization: 40.7 kJ/mol. 1. raise temperature to 100.C q = nCde =(325g)x(1moi/18.02g)X(T5J/C*moi)x(70.C) = 94.69 kJ. 2. vaporize the water: (3259)x(1moi/18.029)x(40.7 kJimol) = 734.0 Rd. 3. total heat is 94.69kJ + 734.0 N = 828.74 kJ. 76) Enthalpies for reactions are given by the following: Aern : zm‘Hfoprod) " Z(flu-'foreact) a. AHnm = [AHf°(NO(g)) + ZAHf°(HN03(aq))] - [AHf°(H20(l)) + 3(AHf°(N02(g))] = -1 38.18 kJ b. AHm.l = [2AHf°(BF3(g)) + AHf°(CaO(5))] - [AHf°(3203(S)) + 3(AHf°(CaF2(S))] = 752.3 kJ c. AHDm = [2AHf°(H20m) + AHf°(KZS(aq))] - [ZAHf°(KOH(aq)) + (AHf°(HZS(aq))] =-38.72 kJ 90) AHorn : E(AHfoprod) ' z(‘."\‘l-lforeact) a. CH3OHm ----> CHSOHE) AHvapo = 200.66 kJ/mol - (238.86 kamol) = 38.2 kJImol b. Use Kirchhoff's law. AH°(T2) = AH°(T1) + (T2 - T1 )1116p AH°(64.7 0C) = (38.20kJ/mol) + (337.85 K -298.15 K)x(43.89 J1K*mol — 81.6 J1K*mol)x(1 kJIi 000 J) = 38.20 kJ/mol — 1.495 lemol = 36.7 lemol c. Table 6.3 gives AHVEPO of 35.3 kJ/mol. A cause of this error is that heat capacities are not entirely independent of temperature. 100) Desired reaction: CH4(9) + 312 02(9) ---> 00(9) + 2H2O{g) We can get this by rearranging the given equations in the following way. multiply equation a) by 314. (muliply alt compounds by 314) multiply equation b) by 314. (muliply all compounds by 314} multiply equation c) by -112. (muliply all compounds by 112, and reverse the reaction) The reaction enthalpies for each reaction will be affected the same way as the reactions themselves. This will give us the foltowing reactions: a) 314 0144(9) + 312 02(9) __> 314 (302(9) + 312 Hzom) AHO = 4302 kJ*(31'4) b) 314 CH4(Q) + 314 com) -——> 312 cow-1- 312 Hag) AHD = +247 kJ*(3/4) c) 312 Hug} + 112 com) ——-> 112 H20[g)+ 112 CHM) AHD = -206 kJ*(1/2) a) 314 01-14(9) + 312 02(9) ---> 314.99% + 312 1420(9) AHD = 601.5 kJ b) 31‘4- 11'4 CH4(g) + W%—-——> 31'2- CO(9) + 3,1214% AHO = +185.25 kJ OH + 312 02(9) ---> COG) + 2H20{g) H2 = -519 RJ 4(9) 104) we burned 1.00 g of each sample. CD of the calorimeter is 600 JIDC a) brand X: AH = 600 JIOC X (309.0 K - 300.2 K) = 5230 Jig brand ABC: AH = 600 JI°C x (307.5 K - 299.0 K) = 5100 Jig b) 30 g of brand X gives off 30 g x 5280 Jig x(1C/4,184J) = 38 Cal while 30 g of brand ABC gives 5100 Jig x(1Cl4,184J) x 30 g = 37 Cal. - 2:“; (MG; JHM .M/ P3 All 3mm Egg/mt rs Am jig-()aianm-g {>3 A 25 I577 mod/M ‘~ \ t H7 ltd/“.4 27 ‘l 4 k1 b 35% ml 1%. [.34 ‘ \ Agtfi‘qqu/mt) W I'll/[Ls +7OH51+319+ 307,1) + 577+ le 29sz 3649): [5454 ki/WM N...— N’ H m, 5,;th a“ tag/L45 9/. N15 1% @4. . ‘ ‘ I“ I UAW/wk .M \\ ’ M V "m“ 5’” 3 / . 1. N BCIL‘BH/WO , H”&”3'd gawkmw) 6 N === N {D , 3N3,“ -gmqumfl Mm ‘‘‘‘‘‘‘ WWW—V - H”; H'Cl/MOL N [‘3 ”ugly 47,» \Lz— \J-‘L’fj 'WM;MM§ :5 (7t W’KJ 466‘?be 666.555 ‘* 5 Mal-aha. ___W_F____.______ *Wf/ .4 4 (WW. I ”(W {are n bylt’b {Mr J g9) Wu; N==N Wig/M MN Iggy: EEC? >1NP3(3) F F x3 tsgwmlzs tNE‘N ;; I: P, ARI: :fum : N“ Px6 “(Haw/mane) .m_-_ [m DCH CH CH ,LchHlflcjr-fifiHgCH(DH)CH5C9) a“ H (H ‘,__ u u k: rt 1““ ‘3” “ “/1“ ”54-0ka M ché Law (6‘?) ' _ ((—1 "'" X ‘0‘” ( 53) H q ‘51-” 6—H”. 4 Sr: 6:44 Cit? C'CKZ Chi/{x7 C) 014qu Clggw—mp CHBUW Mug) (4, *2 E C-H (4m [KC/"H Cl—U umrba + H—u CI—u (6%) ...
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