ch8 - 8 8.1 Convert lb to kg. 8.2 8.3 Thermochemistry:...

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8 Thermochemistry: Chemical Energy 8.1 Convert lb to kg. kg 1043 = g 1000 kg 1 x lb 1 g 453.59 x lb 2300 Convert mi/h to m/s. m/s 24.6 = s 3600 h 1 x km 1 m 1000 x mi 0.62137 km 1 x h mi 55 1 kg m 2 /s 2 = 1 J; E = ½mv 2 = ½(1043 kg)(24.6 m/s) 2 = 3.2 x 10 5 J E = 3.2 x 10 5 J x kJ 10 x 3.2 = J 1000 kJ 1 2 8.2 (a) and (b) are state functions; (c) is not. 8.3 ΔV = (4.3 L 8.6 L) = 4.3 L w = PΔV = (44 atm)( 4.3 L) = +189.2 L atm w = (189.2 L atm)(101 atm L J ) = +1.9 x 10 4 J The positive sign for the work indicates that the surroundings does work on the system. Energy flows into the system. 8.4 w = PΔV = (2.5 atm)(3 L 2 L) = 2.5 L atm w = (2.5 L atm) atm L J 101 = 252.5 J = 250 J = 0.25 kJ The negative sign indicates that the expanding system loses work energy and does work on the surroundings. 8.5 (a) w = PΔV is positive and PΔV is negative for this reaction because the system volume is decreased at constant pressure. (b) PΔV is small compared to ΔE. ΔH = ΔE + PΔV; ΔH is negative. Its value is slightly more negative than ΔE. 8.6 ΔH o = 484 H mol 2 kJ 2 PΔV = (1.00 atm)(5.6 L) = 5.6 L atm PΔV = (5.6 L atm)(101 atm L J ) = 565.6 J = 570 J = 0.57 kJ w = PΔV = 570 J = 0.57 kJ ΔH = H mol 0.50 kJ 121 _ 2 ΔE = ΔH PΔV = 121 kJ (0.57 kJ) = 120.43 kJ = 120 kJ 8.7 ΔV = 448 L and assume P = 1.00 atm 173

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w = PΔV = (1.00 atm)(448 L) = 448 L atm w = (448 L atm)(101 atm L J ) = 4.52 x 10 4 J w = 4.52 x 10 4 J x J 1000 kJ 1 = 45.2 kJ 8.8 (a) C 3 H 8 , 44.10 amu; ΔH o = 2219 kJ/mol C 3 H 8 15.5 g x H C mol 1 kJ 2219 - x H C g 44.10 H C mol 1 8 3 8 3 8 3 = 780. kJ 780. kJ of heat is evolved. (b) Ba(OH) 2 8 H 2 O, 315.5 amu; ΔH o = +80.3 kJ/mol Ba(OH) 2 8 H 2 O 4.88 g x O H 8 ) Ba(OH mol 1 kJ 80.3 x O H 8 ) Ba(OH g 315.5 O H 8 ) Ba(OH mol 1 2 2 2 2 2 2 = +1.24 kJ 1.24 kJ of heat is absorbed. 8.9 CH 3 NO 2 , 61.04 amu q = = NO CH mol 4 kJ 2441.6 x NO CH g 61.04 NO CH mol 1 x NO CH g 100.0 2 3 2 3 2 3 2 3 1.000 x 10 3 kJ 8.10 q = (specific heat) x m x ΔT = (4.18 C g J o )(350 g)(3 o C 25 o C) = 3.2 x 10 4 J q = 3.2 x 10 4 J x J 1000 kJ 1 = 32 kJ 8.11 q = (specific heat) x m x ΔT; specific heat = C) g)(10 (75 J 96 = T x m q o = 0.13 J/(g o C) 8.12 25.0 mL = 0.0250 L and 50.0 mL = 0.0500 L mol H 2 SO 4 = (1.00 mol/L)(0.0250 L) = 0.0250 mol H 2 SO 4 mol NaOH = (1.00 mol/L)(0.0500 L) = 0.0500 mol NaOH NaOH and H 2 SO 4 are present in a 2:1 mol ratio. This matches the stoichiometric ratio in the balanced equation. q = (specific heat) x m x ΔT m = (25.0 mL + 50.0 mL)(1.00 g/mL) = 75.0 g J 2790 = C) 25.0 _ C g)(33.9 )(75.0 C g J (4.18 = q o o o mol H 2 SO 4 = 0.0250 L x 1.00 L mol H 2 SO 4 = 0.0250 mol H 2 SO 4 Heat evolved per mole of H 2 SO 4 SO H mol / J 10 x 1.1 = SO H mol 0.0250 J 10 x 2.79 = 4 2 5 4 2 3 Because the reaction evolves heat, the sign for ΔH is negative. ΔH = 1.1 x 10
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This note was uploaded on 08/27/2008 for the course CHEM 1280 taught by Professor Schmidt during the Fall '07 term at Toledo.

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ch8 - 8 8.1 Convert lb to kg. 8.2 8.3 Thermochemistry:...

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