Why is I₂ + H₂ → 2HI exothermic if breaking bonds requires more energy?

[PPonte]

I2 (g) + H2 --> 2HI dH = - 10 kJ/mol

In this reaction the energy involved in the breaking of bonds in H2 and I2 is great in quantity than the energy released in the formation of bonds in HI.

My book says that this sentence is true, but I do not agree.
The breaking of bonds is an endoenergetic process and the formation of bonds is an exoenergetic process. If the reaction is exoenergetic this means that the exoenergetic process involves more energy than the endoenergetic process.
We can see it by other way. If the energy involved in the breaking of bonds in H2 and I2 is great in quantity than the energy released in the formation of bonds in HI this means that the reactional systhem obtained energy and thus the enthalpy of the products is higher than the enthalpy of the reactants, which is an caractheristic of an endoenergetic reaction and not of an exoenergetic reaction.
Am I right?

 

[Hootenanny]

If the enthalpy change is negative, this indicated an exothermic reaction, energy is being release into the surroundinga. The energy associated with bonds being made is greater than that of the bonds being broken.

 

[PPonte]

Exactly! I think I am right.

 

[Hootenanny]

According to my data the bond enthalpies are as follows:
$$H-H = 436 kJ\cdot mol^{-1}$$
$$I-I = 151 kJ\cdot mol^{-1}$$
$$H-I = 298 kJ\cdot mol^{-1}$$
This gives the enthalpy of broken bonds to be 587 and bonds made of 596 which means:
$$\Delta H = -9 kJ\cdot mol^{-1}$$
Which would agree with your comments. It seems your book has a typo!

 

[PPonte]

It seems so. But has less typos than the other books I had.
Thank you very much for your help!