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5 Thermochemistry Solutions to Exercises 5.107 The reaction for which we want is: + -> Before we can calculate for this reaction, we must calculate for We know that for NH₃(g) is -46.2kJ/mol, and that for NH₃(g), AH = 23.2kJ/mol Thus, 23.2 kJ = = Then for the overall reaction, the enthalpy change is: = H₂O(g) + N₂ - - O₂ = 6(-241.82 kJ) + 2(0) 4(-69.4 - 3(0) = -1173.3 kJ -1173.3kJ 1 mol 1000 cm³ - 1.4x 10⁴ kJ 4 mol X 17.0 1L LNH₃ (This result has two significant figures because the density is expressed to two figures.) AH = 2(-393.5 kJ) + 4(-241.82 - 2(-239 - 3(0) = -1276 kJ -1276kJ 0.792 g 1000 = - 2 mol 32.04 1L L CH₃OH In terms of heat obtained per unit volume of fuel, methanol is a slightly better fuel than liquid ammonia. 5.108 1,3-butadiene, (a) C₄H₆(g) + = CO₂(g) + - C₄H₆(g) - O₂(g) = 4(-393.5 kJ) + kJ) 111.9 + 11/2 (0) = -2543.4 kJ/mol (b) 1 mol C₄H₆ 54.092 C₄H₆ = 47.020 -> 47kJ/g (c) (a) + = 4(-393.5 kJ) + - 6(0) = -2718.5 kJ/mol C₄H₈ (b) -2718.5 1 mol C₄H₈ = 48.451 -> C₄H₈ 56.108 133