Enthalpy of Vaporization

Practice Problems


Problem 1

The enthalpy of vaporization of water is ΔHvap = 44.01 kJ mol–1 at standard temperature. How much heat (in kJ) is required to vaporize exactly two moles of water at that temperature?

Solution

\[\begin{align} 2~\mathrm{mol} \left ( \dfrac{44.01~\mathrm{kJ}}{\mathrm{mol}} \right ) = 88.02~\mathrm{kJ} \end{align}\]


Timed Assessments


Heat of Vaporization - q

VIDEO

Answers
  1. 1.13 × 103 kJ
  2. 6.09 × 102 kJ
  3. 2.16 × 102 kJ

Heat of Vaporization - V

VIDEO

Answers
  1. 1.80 L
  2. 1.60 L
  3. 1.56 L


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