If the standard enthalpy of formation needs to be determined at $25ºC$ and 1 atm of pressure then why is it possible for the enthalpy of formation to be calculated for the liquid, solid, and/or gas states for the same substance?

For instance (at $25ºC$): $$H_2O(l): -285.8$$ $$H_2O(g): -241.818$$

  • $\begingroup$ The -241.818 is for the hypothetical ideal gas state of water vapor at 25 C and 1 atm. This is not really an equilibrium state of water, but it gives correct values of enthalpy for calculations when water is part of the gas phase at higher temperatures. $\endgroup$ Dec 8 '20 at 2:53
  • $\begingroup$ Ok, I think I understand. Thanks! $\endgroup$ Dec 8 '20 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.