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Problem set 2 Solutions

# Problem set 2 Solutions - The standard enthalpy of...

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Problem set 2 Solutions 7.26 The reaction absorbs negative heat at constant P, so its ΔH is -670 J. (-670 J/1.00 g CuCl 2 )(134.452 g CuCl 2 /1 mol CuCl 2 ) = -90.0x10 3 J mol -1 7.28 In this process a liquid freezes, so ΔH of the liquid is negative: (-28.8 kJ/1 mol NaCl)(1 mol NaCl/58.44 g NaCl)(56.2x10 3 g NaCl) = -2.77x10 4 kJ

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7.48 The reaction forming one mole of ethane from ethylene and hydrogen in their standard states can be imagined as breaking all the bonds in the reactants, which are one mole of gaseous ethylene and one mole of gaseous hydrogen, and then constructing one mole of ethane from the resulting gaseous atoms. ΔHº = 4(413)kJ + 1(615)kJ + 1(436)kJ – 6(413)kJ – 1(348)kJ = -123 kJ C-H C=C H-H C-H C-C

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7.74 (a). C 4 H 8(g) + 6O 2(g) 4CO 2(g) + 4H 2 0 (g) The standard enthalpy of formation of isobutene ΔHº
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Unformatted text preview: The standard enthalpy of formation of isobutene ΔHº f = is given by;-2528 kJ = 4(-393.5 kJ) + 4(-241.8 kJ) - ΔHº f ΔHº f = -13 kJ mol-1 (b). When 0.50 mol of isobutene is burned at constant pressure q p = 0.50(-2528) kJ = -1.26x10 3 kJ If this is done adiabatically, then the heat will be absorbed by the substances present at the conclusion of the reaction and by the reaction vessel; their temperature will rise. There are present 2 mol CO 2 , 2 mol H 2 O and 5 mol O 2 [(2x37) + (2x34) + (5x29) + 700] J K-1 = 987 J K-1 The temperature rise will be about 1.26x10 6 J/987 J K-1 = 1.28x10 3 K Since the initial temperature was nearly 300 K, the final temperature will be near 1600 K, or about 1300ºC....
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