Question

at its boiling point, the vaporization of 0.579 mol ch4 (l) requires 4.76 kj of heat. a. what is the enthalpy of vaporization of methane? b. an electrical heater was immersed in a flask of boiling ethanol (c2h5oh) and 22.45 g of ethanol was vaporized when 21.2 kj of energy was supplied. what is the enthalpy of vaporization of ethanol?

Answer 1

The enthalpy of vaporization of methane is 8.22KJ and enthalpy of vaporization of ethane is 44.25KJ.

We know, the enthalpy of vaporization is defined as the amount of energy required by one mole of compound to get completely evaporated.

a. So, it is given that 0.579 moles of methane takes 4.76 KJ of energy.

So, 1 mole of methane will take 4.76KJ/0.579 energy.

Hence, the enthalpy of vaporization of methane is 8.22 KJ.

b. The amount of energy required to evaporate 22.45 grams of ethane is 21.2 KJ.

We know already,

Moles = mass of substance given/molar mass of substance.

Molar mass of ethane = 46g/mol.

So,

Moles of ethane = 22.45/46

Hence, the energy required to evaporate one mole of ethane = 21.2 ×46/22.45

Energy required = 44.25 KJ.

Hence, the enthalpy of evaporation of ethane is 44.25KJ.

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