Thermochemistry. How many kilojoules are needed?

Thermochemistry ....

q = moles x molar heat capacity x ΔT + moles x molar heat of vaporization

q = 1.056 mol x 0.1124 kJ/molK x 25K + 1.056 mol x 38.56 kJ/mol

q = 43.7 kJ

The 2015 kJ comes from using 48.7 MOLES of C2H5OH, not 48.7 grams of the alcohol. So which is it? Moles or grams?

q = moles x molar heat capacity x ΔT + moles x molar heat of vaporization

q = 48.7 mol x 0.1124 kJ/molK x 25K + 48.7 mol x 38.56 kJ/mol

q = 2015 kJ

The first step is to calculate the number of moles of ethanol.

n = 48.7 ÷ 46.1

This is approximately 1.06 moles. Use the following equation to calculate the amount of heat that is required to increase the temperature of the ethanol.

Q = n * Heat capacity * ∆ T

Q = (48.7 ÷ 46.1) * 112.4 * 25 = 136,847 ÷ 46.1

To convert to kilojoules, divide by 1,000.

Q = 136,847 ÷ 46,100

Rounded to three significant digits, this is 2.97 kilojoules. Use the following equation to calculate the amount of heat that is required to vaporize the ethanol.

Q = n * Heat of vaporization

Q = (48.7 ÷ 46.1) * 38.56 = 1,877.872 ÷ 46.1

Rounded to three significant digits, this is 40.7 kilojoules. The total amount of heat energy is approximately 43.9 kilojoules. I hope this is helpful for you.

No. of moles of ethanol = (48.7/46.1) mol

Total heat required

= (Heat required to raise the temperature from 0°C to 25°C) + (Latent heat required for vaporization)

= [(48.7/46.1) mol] × [(112.4/1000) kJ/(mol K)] × [(25 - 0)°C] + [(48.7/46.1) mol] × (38.56 kJ/mol)

= 43.7 kJ