Question

18.0 g of water completely vaporises at `100^(@)C` and 1 bar pressure and the enthalpy change in the process is40.79 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 ?

Answer 1

`18.0` g of water `=` 1 mole
`:. ` Standard enthalpy of vaporisation at`100^(@)C` and 1 bar pressure `=40.79 kJ mol^(-1)`
Enthalpy change in vaporisationof 2 moles of water `= 2 xx 40.79 kJ = 81. 58 kJ`