Is the magnitude of ionisation enthalpy and ionisation energy is same?

In my textbook it is written that ionisation energy and ionisation enthalpy are two different quantities. ionisation energy is the amount of energy provided to extract an electron from the outermost shell of a neutral or ionic gaseous atom and ionisation enthalpy is the change in energy of the 1 mole of gaseous atoms or ions when the outermost electron is removed from the atoms or ions. Since electron is in bounded state in an atom so the net potential energy of the electron nucleus system is negative and to free up the electron from the bounded system we need to change the negative potential energy of the system to zero or positive value by providing an extra amount of energy from our side which is called ionisation energy and since we are PROVIDING the energy mathematically it is written with the positive sign (by definition of ionisation energy), ionisation energy is always positive. But ionisation enthalpy is something different it is the difference in the energy of the state of the system before and after ionisation it can be positive or negative depending on the stability of product with respect to reactant. The question arises here is that are ionisation energy and ionisation enthalpy same in the magnitude. 




I also had the same doubt, and [IUPAC's Gold Book](https://goldbook.iupac.org/terms/view/I03199) gives the definition of **ionization energy**, but not for **ionization enthalpy**. Searching in journals also didn't give me any hit so this is what I did, I grabbed Physical Chemistry by Atkins, that is a book that talks about both of them.


In table $2.4$ of Chapter $2$ he introduces all the enthalpies of transition. The process of ionization is defined and well known to us


$$ \ce{X(g) -> X^+(g) + e^-(g)} \quad \Delta\_\mathrm{ion} H(\ce{X}) \tag{1} $$


In the following page he does that old classic example of calculating lattice enthalpy $\Delta H\_\mathrm{L}$ using a **Born-Haber cycle**. He uses potassium chloride to illustrate. In the path, you need the ionization enthalpy of potassium $\Delta\_\mathrm{ion} H(\ce{K)}$. I have the book so I give you the screenshot:


[![enter image description here](https://i.stack.imgur.com/3GVza.jpg)](https://i.stack.imgur.com/3GVza.jpg)


and we clearly see that $\Delta\_\mathrm{ion} H(\ce{K)} = 418 \; \pu{kJ mol^-1}$. When we go back to the appendix to search for the ionization enthalpies, he doesn't like that term, and names the table "Ionization Energies". The value listed in Table $9.3$ is $I(\ce{K}) = \pu{418.8 kJ mol^-1}$. Thus, at least for Atkins, they just seem to be the same or $\Delta\_\mathrm{ion} X = I(\ce{X}).$


As an additional check, I saw if he did the same for electron affinity. Again, [IUPAC's Gold Book](https://goldbook.iupac.org/terms/view/E01977) defines it, but in the same chapter we are introduced to the **electron gain enthalpy** as


$$ \ce{X(g) + e^-(g) -> X^-(g) } \quad \Delta\_\mathrm{eg} H(\ce{X}) \tag{2}$$


Eq. (2) is the "inverse" of electron affinity.


In the same calculation of the image, Atkins claims that $\Delta\_\mathrm{eg}H(\ce{Cl}
) = \pu{-349 kJ mol^-1}$. What happens when we go to table $9.4$ of the appendix? He disregards the term and lists the **electron affinities**. The value there is $E\_\mathrm{ea}(\ce{Cl}) = \pu{+348.7 kJ mol^-1}$ (see the possitive sign). Once more, for Atkins they are just the same $-\Delta\_\mathrm{eg}H(\ce{X}) = E\_\mathrm{ea}(\ce{X})$ (see the negative sign).


The question as "why multiply entities with no necessity remains". I like more the enthalpy notation of Atkins, though.


**References**


* Atkins, P. and De Paula, J., "Physical Chemistry", 9th ed, W.H. Freeman and Company New York (2016).


 0

I'm trying to understand this myself, this is what I think.


Enthalpy can be defined as **Final Energy of the system - Initial Energy of the system** so here Ionisation Enthalpy must mean **Energy of system after Ionisation - Energy of system before Ionisation**. This is the same as saying how much energy you have to put into or take out of the system to remove an electron which is the definition of Ionisation energy, so both are equivalent terms.


Note:- Ionisation by itself is always endothermic as you have to give energy to the electron to excite it, allowing it to escape the hold of the nucleus so Ionisation Enthalpy is always positive.


 -1

Ionization enthalpy is defined as the energy required to remove the electron at $0\ \mathrm{K}$, whereas the ionization energy is defined at any temperature.
so, generally, they are not equal quantities.


 -2