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How much heat energy is required to convert 51.4 g of solid ethanol at -114.5 °C to gaseous ethanol at 135.7 °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.
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1 Answer

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Final answer:

The heat energy required to convert 51.4 g of solid ethanol at -114.5 °C to gaseous ethanol at 135.7 °C is 148.75 kJ.

Step-by-step explanation:

To convert the solid ethanol at -114.5 °C to gaseous ethanol at 135.7 °C, we need to consider the different phase changes it undergoes. First, we need to calculate the heat energy required to raise the temperature of the solid ethanol from -114.5 °C to its melting point of -114.5 °C. We can use the specific heat capacity of the solid ethanol to calculate this heat energy:



Q1 = mass * specific heat capacity * temperature change



Q1 = 51.4 g * 2.45 J/g °C * (0 °C - (-114.5 °C))



Q1 = 51.4 g * 2.45 J/g °C * 114.5 °C



Q1 = 14,159.15 J



Next, we need to calculate the heat energy required to melt the solid ethanol at its melting point. We can use the molar heat of fusion to calculate this:



Q2 = moles * molar heat of fusion



Since we know the molar mass of ethanol is 46 g/mol, we can calculate the number of moles:



moles = mass / molar mass



moles = 51.4 g / 46 g/mol



moles = 1.12 mol



Q2 = 1.12 mol * 4.60 kJ/mol



Q2 = 5.15 kJ



Finally, we need to calculate the heat energy required to vaporize the liquid ethanol at its boiling point. We can use the molar heat of vaporization to calculate this:



Q3 = moles * molar heat of vaporization



Q3 = 1.12 mol * 38.56 kJ/mol



Q3 = 43.15 kJ



The total heat energy required is the sum of Q1, Q2, and Q3:



Total heat energy = Q1 + Q2 + Q3



Total heat energy = 14,159.15 J + 5.15 kJ + 43.15 kJ



Total heat energy = 48,454.15 J + 48.30 kJ



Total heat energy = 148.75 kJ

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