Why is H2O(g)'s enthalpy of formation magnitude smaller than that of H2O(l)?

[Stella Huang 2C]

Hi! I was looking in Achieve's standard enthalpy of formation list and saw that H2O(g)'s standard enthalpy of formation is –241.8 kJ/mol, whereas H2O(l)'s standard enthalpy of formation is –285.8 kJ/mol.

I was wondering if anyone had any insights on what factor(s) determines the magnitude of the enthalpy of formation. I guess my assumption was that with phase changes taken into account (more energy/heat has to be put in to form a gas as opposed to a liquid), the magnitude of the standard enthalpy of formation of gas would be higher than that of a liquid. Any clarification would be appreciated :-)

[Bridget Vause 1A]

Hi!

Your thinking is exactly right! Because you have to add heat to H2O(l) to turn it into a gas (which is an endothermic process), the enthalpy of formation will be slightly less exothermic than that of H2O(l).

Hope this helps! :)

[Stella Huang 2C]

Hi!

Your thinking is exactly right! Because you have to add heat to H2O(l) to turn it into a gas (which is an endothermic process), the enthalpy of formation will be slightly less exothermic than that of H2O(l).

Hope this helps! :)

Hey Bridget! I appreciate your response. That makes a lot of sense in that because we must apply heat to convert liquid to gas, the enthalpy of formation will be less negative for gas compared to liquid.

[Colin Wong 1C]

H2O (g) enthalpy of formation is smaller than H20 (l) because head is needed in order to turn h20 into a gas, this would make the process "less" endothermic, and smaller than the liquid h20