The enthalpy of vaporization of liquid water using the data:

(i). $H_{2} (g) + (\frac{1}{2}) O_{2} (g) \to H_{2} O (l)$

$∆ H = - 285.77 k J / m o l$

(ii). $H_{2} (g) + (\frac{1}{2}) O_{2} (g) \to H_{2} O (g)$ ,

$∆ H = - 241.84 k J / m o l$ in kJ/mol is

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Enthalpy of vapourisation of H 2 O is H 2 O l → H 2 O g i.e. (ii)-(i) Enthalpy of vaporization of liquid water = 285.77 - 241.84 = + 43.93 k J m o l - 1

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