Bond enthalpy

In the vapour state, hydrogen and iodine undergo the following reaction.
H2(g) + I2(g) (reversible arrows) 2HI(g)

reaction 2.1Write an equation, including state symbols, for the bond enthalpy of
I - I.

The answer is I-I(g) -> 2I(g)

Why is it not I2-> I +I because doesn't bond enthalpy show the bonds breaking?

Thanks.

Same thing? I think they may have written I2 as I-I to emphasise the bond, and 2I is the way to write I + I.

So is this saying that the 2I bond is broken to form I + I? And why do they have to be gaseous?? Thanks.

Iodine gas, which is two Iodine atoms joined with a covalent bond is written as $I_2$. 2I implies two separate iodine atoms (or radicals).

So this equation is suggesting that after bond breaking, two iodine atoms are formed?

Yep.

And elements in the bond enthalpy equation are always gaseous?

Yep.