18.0 g of water completely vaporises 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 vaporising two moles of water under the same conditions ? What is the standard enthalpy of vaporisation for water ?

Explanation

Given that, quantity of water = 18.0 g, pressure =1 bar
As we know that, 18.0 g `H_(2)O` = 1mole. `H_(2)O`
Enthalpy change for vaporizing 1 mole of `H_(2)O = 40.79 kJ mol^(-1)`
Enthalpy change for vaporizing 2 moles of `H_(2)O = 2 xx 40.79 kJ = 81.350` kJ
Standard enthalpy of vaporization at `100^(@)` C and 1 bar pressure `Delta_("vap")H^(-) = +40.79 kJ mol^(-1)`