Heat of formation of aqueous H⁺

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Where am I going wrong here?

Your chemical reaction is not balanced. You wrote

$$\ce{1/2 H2(g) -> H(g) -> H+(g) -> H+(aq)}$$

but it should be

$$\ce{1/2 H2(g) -> H(g) -> H+(g) + e- (?)-> H+(aq) + e- (?)}$$

I put a question mark for the physical state of the electron because it's not clear where that electron should go or what enthalpy of formation we would assign it. You could instead involve a second element in your reaction, e.g.:

$$\ce{H2 + Cl2 -> 2H+ + 2Cl-}$$

You could assign an enthalpy of formation for the mixture of hydrogen and chloride ions, but you would not know how to assign it to the individual components of the mixture.

In my book it is given that by convention heat of formation of aqueous H+ is taken to be zero.

Yes, that's what we do. It would equally be possible to assign it a value of 100 kJ/mol. All the enthalpies of formation of ions would change (for example, that of chloride ions would by 100 kJ/mol lower to compensate). However, all the calculations of enthalpies of reactions would have the same result because it all cancels out in a balanced equation. Setting the enthalpies of formation of elements to zero is also a bit arbitrary (not once we specify that to measure them, we are making the compounds from the elements).

Updating the definition

Instead of saying the enthalpy of formation of a compound is the enthalpy of making that compound from the elements, we could widen the definition to say:

"The enthalpy of formation of a species is the enthalpy of the reaction making the species where all other reactants and products are either elements or the hydrogen ion""

This definition would include ions. For chloride ions, the following reaction would work:

$$\ce{1/2H2 + 1/2Cl2 -H+ -> Cl-}$$

For sodium ions, the following reaction would work:

$$\ce{Na + H+ - 1/2H2 -> Na+}$$

Why the hydrogen ion?

Again, it is arbitrary and any other ion would have been fine as well. The hydrogen ion is special for two reasons, though. It is the smallest ion, and it is fairly common in reactions, so it saves some arithmetic effort when doing calculations.

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himanshu
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Updated on August 01, 2022

Comments

  • himanshu
    himanshu over 1 year

    Heat of formation of aqueous $\ce{H+}$ is taken to be zero, by convention. Heat of formation is energy released or absorbed when 1 mole of a compound is formed from its elements in their standard states.

    $$\ce{1/2 H2(g) -> H(g) -> H+(g) -> H+(aq)}$$

    Since heat of formation of $\ce{H+(aq)}$ is assigned to be zero, this would imply that this reaction has $ΔH = 0$. Wouldn't this imply that:

    $$\text{Enthalpy of atomization} + \text{Ionization energy} + \text{Heat of hydration} = 0~?$$

    However, this doesn't seem to be the case:

    $$435 + 1312 - 1091 = \pu{656 kJ/mol}$$

    Where am I going wrong here?