Ethanol (C_2H_5OH) melts at -114 degree C and boils at 78 degree C. The enthalpy

Question

Ethanol (C_2H_5OH) melts at -114 degree C and boils at 78 degree C. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heat of solid and liquid ethanol are 0.97 J/g middot K are 2.3 J/g middot K respectively. How much heat is required to convert 48.5 g of ethanol at 20 degree C to the vapor phase at 78 degree C? Express your answer using two significant figures. How much heat is required to convert 48.5 g of ethanol at -145 degree C to the vapor phase at 78 degree C? Express your answer using two significant figures.

Explanation / Answer

A)

Q required for m = 48.5 g

from 20°C to 78°C

we need:

increase liquid ethanol from 20 to 78

evaporate liquid ethanol to vapor at 78

Q1 = m*Cp*(Tf-Ti)

Q1 = (48.5)(2.3)(78-20) = 6469.9 J

Q2 = n*LH

n = mass/MW = 48.5/46 = 1.054

Q2 = 1.054*38.56 = 40.64kJ = 40642.24 J

Qt = Q1+Q2 = 6469.9 + 40642.24= 47112.14 J

Q = 47.11 kJ

2)

qtotal = mcT + nHfus + mcT + nHvap
qtotal = 48.5g*0.97J/gK * (-114+ 145) + (48.5/46 moles * 5.020 J/mole) + 48.5g*2.3J/g*K *(78 + 114K) + 48.5/46 moles *38,560J/mole
qtotal = 63,536.94

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