Question

the molar enthalpy of vaporization of methanol is 35.2 kj/mol at c. how much energy is required to evaporate 1 kg of methanol at c?

Answer 1

To calculate the energy required to evaporate 1 kg of methanol at a given temperature, we need to use the molar enthalpy of vaporization and the molar mass of methanol.

Given:

Molar enthalpy of vaporization (ΔHvap) = 35.2 kJ/mol

Temperature (T) = 100 °C (or 373 K)

Molar mass of methanol (M) = 32.04 g/mol

Step 1: Convert the mass of methanol from grams to moles.

1 kg of methanol = 1000 g

Number of moles (n) = mass (m) / molar mass (M)

n = 1000 g / 32.04 g/mol = 31.22 mol

Step 2: Calculate the total energy required using the molar enthalpy of vaporization.

Energy required = ΔHvap * n

Energy required = 35.2 kJ/mol * 31.22 mol

Step 3: Calculate the energy required in kilojoules (kJ).

Energy required = 35.2 kJ/mol * 31.22 mol = 1098.304 kJ

Therefore, approximately 1098.304 kJ of energy is required to evaporate 1 kg of methanol at 100 °C.