Question

From data below, calculate the total heat (in J) associated with the conversion of 0.655 mol ethanol gas (C2H6O) at 351°C and 1 atm to liquid ethanol at 25.0°C and 1 atm. (Pay attention to the sign of the heat.)

Boiling point at 1 atm 78.5°C

cgas 1.43 J/g°C

cliquid 2.45 J/g°C

H°vap 40.5 kJ/mol

Answer 1

Using step by step process

Molar mass of ethanol = 2 * 12 + 6 * 1 + 1 * 16 = 46 gm/mol

Mass of ethanol(0.655 mol) = 0.655 mol * 46 gm/mol = 30.13 gms

Step 1: Heat required to cool the ethanol gas from temperature of 351C to 78.5C

Heat = mass * specific heat (g) * (Change in temperature)

=> 30.13 * 1.43 * (78.5-351)

=> -11740.90J

Step 2: adding heat of vaporization for 0.655 moles of ethanol

heat of vaporization required = 0.655 mol * (-40 KJ/mol) = -26200 J

Step3: cooling the liquid from 78.5 to 25C

Heat = mass * specific heat (l) * (Change in temperature)

=> 30.13 * 2.45 * (25 - 78.5)

=> -3949.28J

Total heat in J = -11740.90 - 26200 - 3949.28

=> -33991.61 J