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18.0 g of water completely vaporises at ...

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 ?

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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)`
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