Question

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)?

Answer 1

Answer:

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 = $\Delta H_{vap}=38.6 kJ/mol$

Heat required to vaporize 0.200 moles of ethanol = Q'

$Q'=\Delta H_{vap}\times 0.200 mol=38.6 kJ/mol\times 0.200 mol=7.72 kJ$

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.