Solutions_Ch9

Solutions_Ch9 - ‘ CHAPTER 9 ENERGY ENTHALPY AND AH = AE...

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Unformatted text preview: ‘ CHAPTER 9 , ENERGY, ENTHALPY, AND @ AH = AE + PAV; from this equation, AH >AE when AV > 0, AH <AE when AV < 0, and AH = 513 when AV = 0. Concentrate on the moles of gaseous products versus the moles of gaseous reactants to predict AV for a reaction. a. There are 2 moles of gaseous reactants converting to 2 moles of gaseous products, so AV 2 0. For this reaction, AH =AE. b- There are 4 moles of gaseous reactants converting to 2 moles of gaseous products, so 4V < 0 and AH <AE. c. There are 9 moles of gaseous reactants converting to 10 moles of gaseous products, so AV > 0 andAH >£_\E. ’69 88.0 g N20 x 1 mol N20 44.02 g N20 : 2.00 moi N20 At constant pressure, qp =AH. AH = ncpnr = (2.00 mol)(38-70 J 00 [mo] ‘)(55°c - 165°C) AH=—-8510 J=—8.51kJ= qP w x—Pdv =—nRAT =—(2.00 mol)(8.3145 J K 1 mol 1)(~110. K) = 1830 J = 1.83 k] E=q+w=—-8.51kJ+1.83 kJ=-6.68 kJ . . J Heat grim-1 ch x 150.0g X(18.3°C-15.0°C)=2100.i A comrnon error in calorimetry problems is Sign errors. Keeping all quantities positive helps to eliminate sign errors. Heat 1055 S x g x18'3oC)’ S = J 56.7 °C x 150.0g =0.25J°c‘g‘ G8) a. Heat gain by calorimeter = heat loss by CH; = 6.79 g CH4X MX 802 kJ 16.04 g moi = 340. kJ Heat capacity of calorimeter = 340' k] = 31.5 kJ/°C 10.8 °C b. Heatios 13 CH =h ' ' = ° 31'5k'} — s y 2 2 eat gain by calorimeter 16.9 C x —532 kJ ABM!) 2 —532 M X 26.04g 12.6 ngI-I2 moiCzH2 =~1.10x103kJ/mor @ NO+03—‘>N02+Oz AH=~199kJ 3/2 02—5 03 AH =—1/2 (-427 k1) 0.7 1/2 02 AH = —112(495 k1) N0(g) + 0(g)-7N02(g) &H = —233 kl C5; P4010 -—7 R; + 5 02 AH =*(-2967.V3 1d) 10 PC13 + 5 02 —=: 10 (31390 AH = 10(——285.7 Id) 5 PC3156 6 m3 + 6 c12 AH = —6(«84.2 kJ) P4+6c12-—?4Pc13 AH=—1225.6k.1 moms) + 6 PCls(g) a 10 C13P0(g) AH =—-610.1 kJ 3 A115) + 511111004199 2112039) + moms) + 3 N019 + 6 11-20(11) A He 2E m0] {- 242 k1}+3 mo] 90. k3 Jr lmol(r704 kl +1molC—1676 1:12] 1110] I mo] mol ‘ mol , FED-101 (—295 1(1)] =r2677 k1 . mo] 92 Id 9 " 0! mo 9k 16 2 [ms 2k] H?! =( 70. C2H4(g) + 3 02(g) #22 002(g) + 2 H200) 411° = —141 1 1d AH° = -1411.1 k] = 2(~393.5)k1 + 2(~285.8) kJ—A Hg, cin —— 1411.1 id = ~1358.6 kJ—a H2! Cam , A H; c2114 = 52.5 kJ/mol ...
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Solutions_Ch9 - ‘ CHAPTER 9 ENERGY ENTHALPY AND AH = AE...

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