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My Chemistry textbook defines bond enthalpy as "amount of energy required to break one mole of bonds of a particular type between two atoms in gaseous state". Can someone confirm to me whether

a particular type

is related to bond order, to the atoms involved, both, or something else?

Also, should not the bond enthalpy of all diatomic molecules be large as one mole of bonds to be broken. But this does not happens. Why?

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closed as unclear what you're asking by Mithoron, M.A.R. ಠ_ಠ, bon, Jon Custer, andselisk Oct 4 '17 at 16:47

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  • $\begingroup$ Well, it is pretty large. $\endgroup$ – Ivan Neretin Oct 4 '17 at 15:37
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Bond energies are usually given for each bond-type. You can see this table of values, for example (that's a great site). In order to clasify them we could say that particular type means clasify the bond according to the atoms and to bond-order. So you have C=O, C-O, N=C, N-C, and so on.

If you check the site, look that those values are positive, meaning that the energy -as you said- is the necessary to break the bond. If you need to estimate the difference in enthalpy of formation of:

$$\ce{A<=>B}$$

you will need to change sign of bond-enthalpy values (because usually entalphy is of formation).

The energy is calculated per mol, as you say. And it is a very big value, bigger than enthalpy of vibration, rotation or translation.

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  • $\begingroup$ So by different type we mean different kind of atoms and bond orders and not single,double,triple bonds etc?Thanks $\endgroup$ – user52631 Oct 4 '17 at 15:24

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