0
$\begingroup$

In my data book, there's a list of common compounds and their molar enthalpies of formation -- at 298.15 K. What's the meaning of this given temperature value (is it the final temperature of the compound in question?) and how does it affect the molar enthalpy of formation of a substance?

What, for example, does it mean when the molar enthalpy of formation of H2O(g) at 298.15 K is given as -241.8 kJ/mol, even though water should be in a liquid state at 298.15 K? And what would the molar enthalpy of formation of H2O(g) be at, for instance, 500 K?

$\endgroup$
5
  • $\begingroup$ Seems you forgot about water vapor. $\endgroup$
    – Mithoron
    Sep 17, 2015 at 23:41
  • $\begingroup$ Sorry, I still don't understand. Isn't water vapour H2O(g)? $\endgroup$ Sep 17, 2015 at 23:45
  • $\begingroup$ Yes, so there's no problem with it at 25C $\endgroup$
    – Mithoron
    Sep 17, 2015 at 23:47
  • $\begingroup$ So is the standard temperature listed in these molar enthalpy tables the temperature of the system upon losing/gaining the given amount of enthalpy? $\endgroup$ Sep 17, 2015 at 23:50
  • $\begingroup$ It's simply enthalpy in this temp. you have to pick temp. so... Oh whatever, I need to go to sleep $\endgroup$
    – Mithoron
    Sep 17, 2015 at 23:56

1 Answer 1

3
$\begingroup$

Most enthalpies (or themodynamic observables whatever their type) are not constant across the temperature range — not even across the range of one phase. So the enthalpy of vapourisation of water is different a $273\,\mathrm{K}$ from what it is at $295\,\mathrm{K}$ or $350\,\mathrm{K}$.

Of course, standard conditions have been defined. However, the definition of standard conditions does not explicitly include temperature! Likely because some wanted $273\,\mathrm{K}$, others wanted $295\,\mathrm{K}$. So the temperature at which a certain entropy is valid must always be specified.

Some books put a ‘standard temperature’ at the top and call that their specification.

$\endgroup$
1

Your Answer

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

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