# Standard Enthalpy of Formation of Refrence state(A consequence or assumption) and H+(aq)

Ok, this has been bugging me for a while now. Our teacher taught us that the standard enthalpy of formation of the reference states of elements is "assumed" to be zero. However, I've read in many places that it's a mere consequence of the definition of standard enthalpy of formation because when an element is formed from its reference state, the reaction is a null reaction, and consequently, the change in enthalpy is zero. While the second argument seemed very logical to me, in many textbooks and websites I found that the enthalpy change of formation of H+(aq) is "assumed" to be zero by convention. Now I'm left confused about whether the prior argument about reference state formation enthalpy is correct or not. Furthermore, how can one assume a "change" in something to be zero? I've never seen any instances where a change in something is assumed to be a value because they can be calculated irrespective of whether you know the absolute values of parameters (e.g., heat, internal energy, potential, etc.), but the absolute value is assumed to be something because it poses no problems when calculating changes in the parameters using those values as the reference values cancels out. Any help is appreciated.

• Consider formatting guides for texts and formulas/equations/expressions including mhchem \ce{} specifics. Commented May 19 at 20:09
• You need to set the enthalpy of formation for one ion if you want to define the enthalpy of formation for ions. This is because you can't form an ion from elements, see chemistry.stackexchange.com/q/112302
– Karsten
Commented May 19 at 22:42
• For the elements, both absolute Gibbs free energies for phases as well as relative to a standard (usually but not always the stable phase at STP) have been compiled for use in phase diagram modeling. In no case that I know of is the entropy at 0K given as zero. Commented May 20 at 0:17

It would be bad if you set the enthalpy of formation for both diamond and graphite to zero because one can react to form the other (and the enthalpy of reaction is non-zero). However, it is fine to set the enthalpy of formation of one ion to zero as well. The convention is to do that for $$\ce{H+ (aq)}$$.
$$\ce{H2(g) + Cl2(g) -> 2H+ (aq) + 2 Cl- (aq)}$$
This also works for cations: $$\ce{Na(s) + H+ (aq) -> Na+ (aq) + 1/2 H2 (g)}$$