[Pinoylegacy]

What is the molar heat of vaporization of water given the following thermochemical equations?

H2(g)+ 1/2 O2(g) -> H2O (g) + 241.8 kJ/mol
H2(g)+ 1/2 O2(g) -> H2O (L) + 285.8  kJ/mol

So i tried the equation and what i was thinking is all you have to do is subtract the energy from each equation but i'm not 100% sure. The answer i got was 44.0kJ/mol. Can somebody tell me if i'm doing it right?

[Schrödinger]

44 kJ/mol is correct. As you can see from the equations, a certain amount of energy is released in both cases. But in the first reaction (where steam is produced) only 241.8 kJ/mol of energy is released whereas 285.8 kJ/mol of energy is released in the second.

Since the reactants are the same in both the reactions, you can safely say that a lesser amount of energy is released in the first reaction, because some amount of heat is absorbed by water to vaporize.

Had water not absorbed 44 kJ/mol, then the first reaction would have been the same as the second and the energy released would have been 285.8kJ/mol.
So, molar heat of vaporization of water = 44kJ/mol



Alternatively, just reverse the second equation and add the first equation to this reversed one... and voila!!