Question

How much heat energy is required to convert 21.1 g of solid ethanol at -114.5 °C to gaseous ethanol at 191.5 °C? The molar heat of fusion of ethanol is 4.60 kJ/mol and its molar heat of vaporization is 38.56 kJ/mol. Ethanol has a normal melting point of -114.5 °C and a normal boiling point of 78.4 °C. The specific heat capacity of liquid ethanol is 2.45 J/g·°C and that of gaseous ethanol is 1.43 J/g·°C.

________kJ

Answer 1

Ti = -114.5 oC
Tf = 191.5 oC
here
Hfus= 4.6KJ/mol =
4600J/mol


Lets convert mass to mol
Molar mass of C2H5OH = 46.068 g/mol
number of mol
n= mass/molar mass
= 21.1/46.068
= 0.458 mol

Heat required to convert solid to liquid at -114.5 oC
Q1 = n*Hfus
= 0.458 mol *4600 J/mol
= 2106.8855 J

Cl = 2.45 J/g.oC

Heat required to convert liquid from -114.5 oC to 78.4 oC
Q2 = m*Cl*(Tf-Ti)
= 21.1 g * 2.45 J/g.oC *(78.4--114.5) oC
= 9971.9655 J

Hvap = 38.56KJ/mol =
38560J/mol

Heat required to convert liquid to gas at 78.4 oC
Q3 = n*Hvap
= 0.458 mol *38560 J/mol
= 17661.1965 J

Cg = 1.43 J/g.oC

Heat required to convert vapour from 78.4 oC to 191.5 oC
Q4 = m*Cg*(Tf-Ti)
= 21.1 g * 1.43 J/g.oC *(191.5-78.4) oC
= 3412.5663 J

Total heat required = Q1 + Q2 + Q3 + Q4
= 2106.8855 J + 9971.9655 J + 17661.1965 J + 3412.5663 J
= 33153 J
= 33.2 KJ
Answer: 33.2 KJ