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 ***** of water under the same conditions? What is the standard enthalphy of vapourisation for water?
Enthalpy of a reaction is the energy change per mole for the process.
18 g of H2O = 1 mole (ΔHvap = 40.79 kJ moE1 )
Enthalpy change for vapourising 2 ***** of H2O = 2 x 40.79 = 81.58 kJ ΔH°vap = 40.79 kJ mol-1
