18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol^–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?

o The enthalpy or ∆H of a reaction is defined as the change in energy (more specifically heat energy) for one mole of a substance involved in a process.

It is given that, the enthalpy change for 18g of water (at 100^⁰ and 1 bar pressure) is =40.79 kJ mol^–1 in the process of vapourisation.

And for water, 18g is =1 mole (For H₂O, the calculation of molar mass: 2×1 + 16 = 18g)

Hence, enthalpy change of vapourisation for 1 mole = 40.79 kJ mol^–1 enthalpy change of vapourisation for 2 moles of water = (40.79 × 2) = 81.58kJ mol^–1

o Standard enthalpy of a process or ∆H⁰ is the enthalpy change of the process when involved substances are in their standard states.

Hence, for water ∆H_{vapourisation,} ⁰ will be equal to =40.79 kJ mol^–1