Discrepancies between specific heat of vaporization of water and enthalpy of formation of water and steam According to Wikipedia, the enthalpy of formation of water is $-285.8~\mathrm{kJ/mol}$ while the enthalpy of formation of steam is $-241.818~\mathrm{kJ/mol}$, implying the following:

$$\ce{H2O(l) -&gt; H2O(g)}\qquad (\Delta H=44.0~\mathrm{kJ/mol)}$$



Let us derive the enthalpy of the above reaction using the specific heat capacity of water and the specific heat of vaporization of water instead.

$1~\mathrm{mol}$ of water weighs $0.018~\mathrm{kg}$.

To raise $1~\mathrm{mol}$ of water from $25~\mathrm{^\circ C}$ to its boiling temperature requires $(0.018~\mathrm{kg}) \times (4200~\mathrm{J~kg^{-1}~K^{-1}}) \times (75~^\circ\mathrm{C}) = 5.67~\mathrm{kJ}$.

To turn that amount of water to steam requires $(0.018~\mathrm{kg}) \times (2258~\mathrm{kJ~kg^{-1}}) = 40.644~\mathrm{kJ}$.

Adding these two terms give $46.314~\mathrm{kJ}$, implying the following:

$$\ce{H2O(l) -&gt; H2O(g)}\qquad (\Delta H=46.314~\mathrm{kJ/mol)}$$



Is the discrepancies between the two thermochemical equations simply due to measurement inaccuracies?
According to “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL, the standard molar enthalpy of formation at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\mathrm fH_\mathrm m^\circ=-285.8\ \mathrm{kJ\ mol^{-1}}$ for liquid water,

and

$\Delta_\mathrm fH_\mathrm m^\circ=-241.8\ \mathrm{kJ\ mol^{-1}}$ for gaseous water.

(These values are in agreement with the values that you have found on Wikipedia.)

Both values refer to a temperature of $T=298.15\ \mathrm K$; therefore, you do not have to raise the temperature of the liquid water to its boiling point. Nevertheless, you may do this; however, then you would have to cool down the steam back to $T=298.15\ \mathrm K$ later.

Instead, you can directly use the enthalpy of vaporization at the given temperature. Note that the enthalpy of vaporization depends on temperature, and that your value for the specifc enthalpy of vaporization of $\Delta_\text{vap}h=2258\ \mathrm{kJ\ kg^{-1}}$ refers to the boiling point at normal pressure, i.e. to a temperature of approximately $T=100\ \mathrm{^\circ C}$.

According to REFPROP – NIST Standard Reference Database 23, Version 9.0, the molar enthalpy of vaporization at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\text{vap}H_\mathrm m=43.987\ \mathrm{kJ\ mol^{-1}}$,

which is in good agreement with the difference of the values for the molar enthalpy of formation:

$-241.8\ \mathrm{kJ\ mol^{-1}}-\left(-285.8\ \mathrm{kJ\ mol^{-1}}\right)=44.0\ \mathrm{kJ\ mol^{-1}}$