Confusion regarding 1st and 2nd electron gain enthalpy

$$\ce{O}$$ has the 1st electron gain enthalpy $$\pu{-141 kJ mol-1}$$. $$-ve$$ value implies that energy is released when electron is added to an isolated atom. This also means that if $$\pu{141 kJ}$$ energy is supplied to $$\pu{1 mol}\ \ce{O-}$$ it forms $$\pu{1 mol}\ \ce{O}$$.

The second electron gain enthalpy has value $$\pu{+780 kJ mol-1}$$. This implies that energy is to be supplied to form $$\ce{O^2-}$$. But if we supply $$\pu{780 kJ}$$ energy to $$\ce{O-}$$, would not the 1st added electron be removed; since it takes $$\pu{141 kJ}$$ of energy to be removed?

What am I missing?

Edit $$1$$- You guys are very rude. Seriously, $$3$$ downvotes but $$0$$ comments for a relatively well written question? Get over your attitude people.

• @Poutnik Are Ionization enthalpy and Electron Gain Enthalpy not different? Jun 27, 2022 at 14:20
• @Poutnik Yes, I know they are different! My question was about electron affinities, but your comment referenced ionization energies. Jun 27, 2022 at 14:26
• @Poutnik Very sorry for the misunderstanding. Please see the edit. Jun 27, 2022 at 14:31
• Useful links for text and formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial. Jun 27, 2022 at 14:39
• E.g. write $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ to get $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ (all eventually with double dollars in the display mode like $$\ce{H2SO4}$$. Jun 27, 2022 at 14:40

• Thanks for the answer. I understand that $O^{2-}$ is unstable, but how can it be formed, since addition of the second electron requires an energy more than that required to remove the 1st added electron? Jun 28, 2022 at 4:51
• But is it not true that if we add the first electron affinity to the $O^-$ it becomes $O$? Jun 28, 2022 at 5:02