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 vapourising two moles of water under the same conditions ? What is the standard enthalpy of vapourisation for water ?

18.0 g of water completely vaporises at 100°C and 1 bar pressure and

1.	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 vapourising two moles of water under the same conditions ? What is the standard enthalpy of vapourisation for water ?
Answer» SOLUTION :Quantity of water = 18.0 g, pressure 1 bar As we know that, `18.0 g H_(2) O = 1 "MOLE" H_(2) O` ENTHALPY change for vapourising l mole of `H_(2) O = 40.79 "kJ mol"^(-1)` `therefore` Enthalpy change for vapourising 2 moles of `H_(2) O = 2 xx 40.79 "kJ" = 81.358 "kJ"` Standard enthalpy of vaporisation at 100°C and 1 bar pressure, `Delta_("vap") H^(@) = + 40.79 "kJ mol"^(-1)`