G Ethanol boils at 78.4 °C with \DeltaΔHvap = 38.6 kJ/mol. A 0.200-mol sample of ethanol is heated from some colder temperature up to 78.4 °C, which requires 1.05 kJ of heat, and then vaporized. What will be the total amount of heat required (for both the heating and the vaporizing)?

8.77 kilo Joules  will be the total amount of heat required for both the heating and the vaporizing.

Explanation:

Moles of ethanol of ethanol = 0.200 mol

Heat required to heat 0.200 moles of ethanol = Q = 1.05 kJ

Enthalpy of vaporization of ethanol =

Heat required to vaporize 0.200 moles of ethanol = Q'

Total heat required to fore heating and the vaporizing :

= Q + Q' = 1.05 kJ + 7.72 kJ = 8.77 kJ

8.77 kilo Joules  will be the total amount of heat required for both the heating and the vaporizing.

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