I was studying bond enthalpy and came across the example of CH4. All 4 bonds have different bond enthalpies because we go about it like we first remove the 1st hydrogen, then 2nd and so on. But my question is why don't we assume that we break all the 4 bonds at a particular time t simultaneously. Is it possible and if not, why not? Pls do put something that proves your point in a solid way and pls do give references if any. Pls don't make guesses. Thank you very much in advance ☺️

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    $\begingroup$ You misunderstand something. In $\ce{CH4}$, the bond enthalpy on all four $\ce{C-H}$ bonds is equal. But this value is not the same for the three $\ce{C-H}$ when removing one H from $\ce{CH^+_3}$. And so on. But already the first step is so costly that you need high energy density as in a EI mass spectrometer to observe this rather than in the standard round bottom flask. Look up the ionization energies, IE; e.g., for $\ce{CH^+_3}$ this is $\ge \pu{13.2 eV}$ (reference) (in excess of $\pu{1200 kJ/mol}$). $\endgroup$
    – Buttonwood
    Nov 24 at 12:32
  • $\begingroup$ @Buttonwood it is somehow trivial but if the question stand up and is reputed suitable I would transform the comment in the answer. $\endgroup$
    – Alchimista
    Nov 24 at 13:43
  • $\begingroup$ @Buttonwood I would change $\ce{CH3+}$ to $\ce{CH3^.}$ $\endgroup$ Nov 24 at 13:49
  • $\begingroup$ @IvanNeretin Given the small difference in electronegativity (2.55 for carbon, 2.50 for hydrogen in Pauling's scale, reference), the homolysis leading to $\ce{CH^*_3}$ indeed appears more probable than the heterolysis. $\endgroup$
    – Buttonwood
    Nov 24 at 14:01