Question

How much heat energy is required to convert 19.8 g of solid ethanol at –114.5 °C to gasesous ethanol at 175.0 °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 2.45J/g • °C, and that of gaseous ethanol is 1.43J/g. °C.

Answer 1

Ti = -114.5 oC
Tf = 175.0 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
= 19.8/46.068
= 0.4298 mol

Heat required to convert solid to liquid at -114.5 oC
Q1 = n*Hfus
= 0.4298 mol *4600 J/mol
= 1977.0774 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)
= 19.8 g * 2.45 J/g.oC *(78.4--114.5) oC
= 9357.579 J

Hvap = 38.56KJ/mol =
38560J/mol

Heat required to convert liquid to gas at 78.4 oC
Q3 = n*Hvap
= 0.4298 mol *38560 J/mol
= 16573.0659 J

Cg = 1.43 J/g.oC

Heat required to convert vapour from 78.4 oC to 175.0 oC
Q4 = m*Cg*(Tf-Ti)
= 19.8 g * 1.43 J/g.oC *(175-78.4) oC
= 2735.1324 J

Total heat required = Q1 + Q2 + Q3 + Q4
= 1977.0774 J + 9357.579 J + 16573.0659 J + 2735.1324 J
= 30643 J
= 30.64 KJ
Answer: 30.64 KJ