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This alternative pathway involving breaking and making bonds is very revealing. Now we can answer the question for this reaction of where the energy "goes" during the endothermic reaction. The reason this reaction absorbs energy is that the bond which must be broken, H 2 , is stronger than the bond which is formed, HBr. Note that energy is released when the HBr bond is formed, but the amount of energy released is less than the amount of energy required to break the H 2 bond in the first place. The molecule with the weaker bond, HBr, is higher in energy than the molecule with the stronger bond, H 2 .

A second example is similar:

H 2 (g) + Br 2 (g) → 2 HBr(g)

This reaction is exothermic with ΔHº = -103 kJ/mol. In this case, we must break an H 2 bond, with energy 436 kJ/mol, and a Br 2 bond, with energy 193 kJ/mol. Since two HBr molecules are formed, we must form two HBr bonds, each with a bond energy of 366 kJ/mol. In total, then, breaking the bonds in the reactants requires 629 kJ/mol, and forming the new bonds releases 732 kJ/mol, for a net release of -103 kJ/mol. This calculation reveals that we have to compare not just the strengths of the bonds broken and formed but also the number of bonds broken and formed. In this case, even though the HBr bond is weaker than the H 2 bond, there are two HBr bonds formed and this releases more energy, making the reaction exothermic.

Observation 3: bond energies in polyatomic molecules

So far in this study, we have discussed the energies of substances in two different ways. One is to compare the energy of the substance to the energies of the elements which make it up, what we have called the enthalpy of formation. The other is to compare the energy of the substance to the separated atoms which make it up, what we have called the bond energy. These two would seem to be closely related. Since it would be helpful to understand that relationship, we need to look at data which connect the two.

We have actually already considered one reaction where the connection should be relatively easy. [link] above involves the formation of HBr from elemental H 2 and Br 2 . Since two HBr molecules are formed, the energy of [link] is just double the enthalpy of formation of HBr. We have also calculated the energy of [link] from the bond energies of HBr, H 2 and Br 2 . In this case, it is clear that the formation energy of HBr is easy to understand in terms of the relative strengths of the bonds of the reactants and products.

It seems that this should be generally true. Let’s look at some additional data, this time involving a reaction with molecules which contain more than two atoms. A simple example is the combustion of hydrogen gas discussed previously:

2 H 2 (g) + O 2 (g) → 2 H 2 O(g)

This is an explosive reaction, producing 483.6 kJ per mole of oxygen. Note that this is also the formation reaction for H 2 O(g). Calculating the enthalpy of formation of H 2 O(g) from bond energies requires us to know the bond energies in H2O. In this case, we must break not one but two bonds:

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Source:  OpenStax, Concept development studies in chemistry 2013. OpenStax CNX. Oct 07, 2013 Download for free at http://legacy.cnx.org/content/col11579/1.1
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