question_answer 55)

  18.0 g of water completely vaporises at \[100{}^\circ C\]and 1 bar pressure and the enthalpy change in the process is\[40.79\text{ }kJ\text{ }mo{{l}^{-1}}\]. What will be the enthalpy change for vaporizing two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

Answer:

  \[\Delta H_{vap}^{{}^\circ }=+40.79kJ\,mo{{l}^{-1}}\] Enthalpy of vaporisation for two moles of water \[=2\times 40.79kJ=81.58kJ\]