Ethanol (C2H5OH) melts at -114 C and boils at 78 C. The enthalpy of fusion of et

Question

Ethanol (C2H5OH) melts at -114 C and boils at 78 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/gK are 2.3 J/gK respectively.

a. How much heat is required to convert 39.0 g of ethanol at 35 C to the vapor phase at 78 C? (I think the answer is 36 kJ)

b. How much heat is required to convert 39.0 g of ethanol at -169 C to the vapor phase at 78 C?

Please be as detailed as possible, thank you!

Explanation / Answer

a)

39 g at 35°C to vpaor at 78°C...

Q1 = m*c*(Tf-Ti) = 39*2.3*(78-35) = 3857.1J

Q2 = m´*LH = 39 g /46 g/mol *38.56 kJ/mol *1000 J/kJ= 32692.173 J

TotalQ = 3857.1+32692.173 = 36549.273J = 36.5 kJ

b.

we need only, heat form -169°C to liquid at -114°C; then liquid from -114°C to 35°C, and add all previous heat

so

Q1 = m*C*(Tf-Ti) = 39*0.97 *(-114 - -169) = 2080.65

Q2 = m*LHfusion = 39 / 46 * 5.02 *1000 = 4256.08

Q3 = m*Cliq*(Tf-Ti) = 39 * 2.3 * (35--169) = 18298.8

Q4 from previous --> 36549.273J J

total Q= 2080.65+4256.08 +18298.8+36549.273 = 61184.803 J = 61.18 kJ

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