4
$\begingroup$

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:

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

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:

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

Is the discrepancies between the two thermochemical equations simply due to measurement inaccuracies?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ The enthalpy change is a function of temperature. It could be that the data has been reported at different temperatures (e.g at 298K and 373K). In that case $\Delta _r H (\text{at 298 K}) \neq \Delta _r H (\text{at 373 K}) $ $\endgroup$
    – getafix
    Oct 11, 2016 at 13:34
  • $\begingroup$ @getafix Could you resolve the discrepancies? $\endgroup$
    – DHMO
    Oct 11, 2016 at 13:39
  • 1
    $\begingroup$ You forgot to include the (hypothetical) enthalpy change for cooling the vapor back down to 25 C. $\endgroup$
    – Chet Miller
    Oct 12, 2016 at 3:28

1 Answer 1

Reset to default
6
$\begingroup$

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:

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

Improve this answer
$\endgroup$
6
  • $\begingroup$ If I raise the water from $298.15\ \mathrm{K}$ to the boiling point, then convert to steam using the $2258\ \mathrm{kJ\ kg^{-1}}$, and then use the specific heat capacity of steam to cool it to $298.15\ \mathrm{K}$, why do I obtain a different result? $\endgroup$
    – DHMO
    Oct 11, 2016 at 15:56
  • $\begingroup$ @DHMO I am not sure what you tried. Anyway, note that the specific heat capacity of liquid water and steam also depend on temperature. Thus, for a precise calculation, you cannot use constant values for the heat capacity; you would have to integrate the values over the considered temperature changes. Therefore, for precise engineering calculations, the enthalpy balance based on values taken from so-called steam tables is usually preferred. $\endgroup$
    – user7951
    Oct 11, 2016 at 16:12
  • $\begingroup$ If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha. The corresponding results for a temperature of $T=298.15\ \mathrm K$ at equilibrium can be obtained using the input “water boiling at 298.15 K”. $\endgroup$
    – user7951
    Oct 11, 2016 at 16:13
  • $\begingroup$ Suggestion: Calculate the enthalpy change for the process $\ce{H2O} (\mathrm{l; 298~K}) \ce{-> H2O} (\mathrm{l; 373~K}) \ce{-> H2O} (\mathrm{g; 373~K}) \ce{-> H2O} (\mathrm{g; 298~K})$ for comparison. (I expect $\Delta \approx 0$ when compared to the above value.) $\endgroup$
    – Jan
    Oct 11, 2016 at 21:04
  • 2
    $\begingroup$ As I expected: the question’s ‘missing’ enthalpy derives from forgetting to cool the steam back down to room temperature. $\endgroup$
    – Jan
    Oct 11, 2016 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.