How much heat energy is required to convert 16.2 g of solid ethanol at -114.5 °C to gasesous ethanol at 191.9 °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.
Ti = -114.5 oC
Tf = 191.9 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
= 16.2/46.068
= 0.3517 mol
Heat required to convert solid to liquid at -114.5 oC
Q1 = n*Hfus
= 0.3517 mol *4600 J/mol
= 1617.6088 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)
= 16.2 g * 2.45 J/g.oC *(78.4--114.5) oC
= 7656.201 J
Hvap = 38.56KJ/mol =
38560J/mol
Heat required to convert liquid to gas at 78.4 oC
Q3 = n*Hvap
= 0.3517 mol *38560 J/mol
= 13559.7812 J
Cg = 1.43 J/g.oC
Heat required to convert vapour from 78.4 oC to 191.9 oC
Q4 = m*Cg*(Tf-Ti)
= 16.2 g * 1.43 J/g.oC *(191.9-78.4) oC
= 2629.341 J
Total heat required = Q1 + Q2 + Q3 + Q4
= 1617.6088 J + 7656.201 J + 13559.7812 J + 2629.341 J
= 25463 J
= 25.5 KJ
Answer: 25.5 KJ
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