Thus for benzene the number of moles in the vapor

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Unformatted text preview: d Benzene (l) Benzene (v) Toluene (l) Toluene (v) nin (moles) 30 0 70 0 nout (moles) 14.5 15.5 48.7 21.3 (kJ/mole) (kJ/mole) We need to pick a reference state. Let’s pick the liquid compounds at 25°C as our reference. Thus all of the inlet enthalpies are 0. For the liquids in the out side of the table you would get 0.1265 0.1265 101 0.1488 0.1488 101 25 25 9.61 11.31 For the vapors take the liquid at 25°C, increase the temperature to their normal boiling point, add the heat of vaporization then bring the vapor to 101°C. Thus you get 0.1265 80.1 . 25 0.1265 30.765 30.765 0.07406 101 . 80.1 0.07406 39.28 . 0.1488 33.47 0.1488 110.62 0.09418 . 25 33.47 0.09418 101 110.62 45.30 Thus the enthalpy table becomes Compound Benzene (l) Benzene (v) Toluene (l) Toluene (v) nin (moles) 30 0 70 0 (kJ/mole) 0 0 nout (moles) 14.5 15.5 48.7 21.3 (kJ/mole) 9.61 39.28 11.31 45.30 The energy balance then gives Δ 14.5 9.61 15.5 39.28 48.7 11.31 2260 21.3 45.30 0 The positive sign indicates that heat must be added to the system to reach this state. Problem #4 115 113 11...
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