# Why can the standard enthalpy of formation be calculated for different states of matter when the temperature should be constant?

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

• 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. Dec 8 '20 at 2:53
• Ok, I think I understand. Thanks! Dec 8 '20 at 3:06