0.9 Energetics of chemical reactions  (Page 6/7)

We can use diatomic bond energies to calculate the heat of reaction $\Delta (H)$ for any reaction involving only diatomic molecules. We consider two simple examples. First, the reaction

is observed to be endothermic with heat of reaction $70\frac{\mathrm{kJ}}{\mathrm{mol}}$ . Note that this reaction can be viewed as consisting entirely of thebreaking of the $H_2$ bond followed by the formation of the $H\mathrm{Br}$ bond. Consequently, we must input energy equal to the bond energy of $H_2$ ( $436\frac{\mathrm{kJ}}{\mathrm{mol}}$ ), but in forming the $H\mathrm{Br}$ bond we recover output energy equal to the bond energy of $H\mathrm{Br}$ ( $366\frac{\mathrm{kJ}}{\mathrm{mol}}$ ). Therefore the heat of at constant pressure must be equal to difference in these bondenergies, $70\frac{\mathrm{kJ}}{\mathrm{mol}}$ .

Now we can answer the question, at least for this reaction, of where the energy "goes" during the reaction. Thereason this reaction absorbs energy is that the bond which must be broken, $H_2$ , is stronger than the bond which is formed, $H\mathrm{Br}$ . Note that energy is released when the $H\mathrm{Br}$ 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 second example is similar:

This reaction is exothermic with $\Delta (H^{°})=-103\frac{\mathrm{kJ}}{\mathrm{mol}}$ . In this case, we must break an $H_2$ bond, with energy $436\frac{\mathrm{kJ}}{\mathrm{mol}}$ , and a ${\mathrm{Br}}_2$ bond, with energy $193\frac{\mathrm{kJ}}{\mathrm{mol}}$ . Since two $H\mathrm{Br}$ molecules are formed, we must form two $H\mathrm{Br}$ bonds, each with bond energy $366\frac{\mathrm{kJ}}{\mathrm{mol}}$ . In total, then, breaking the bonds in the reactants requires $629\frac{\mathrm{kJ}}{\mathrm{mol}}$ , and forming the new bonds releases $732\frac{\mathrm{kJ}}{\mathrm{mol}}$ , for a net release of $103\frac{\mathrm{kJ}}{\mathrm{mol}}$ . This calculation reveals that the reaction is exothermic because,although we must break one very strong bond and one weaker bond, we form two strong bonds.

There are two items worth reflection in these examples. First, energy is released in a chemical reaction due tothe formation of strong bonds. Breaking a bond, on the other hand, always requires the input of energy. Second, does not actually proceed by the two-step process of breaking both reactant bonds, thus forming fourfree atoms, followed by making two new bonds. The actual process of the reaction is significantly more complicated. The details of thisprocess are irrelevant to the energetics of the reaction, however, since, as we have shown, the heat of reaction $\Delta (H)$ does not depend on the path of the reaction. This is another example of the utility of Hess' law.

We now proceed to apply this bond energy analysis to the energetics of reactions involving polyatomicmolecules. A simple example is the combustion of hydrogen gas discussed previously here . This is an explosive reaction, producing 483.6kJ per mole of oxygen.Calculating the heat of reaction from bond energies requires us to know the bond energies in $H_{2}O$ . In this case, we must break not one but two bonds:

The energy required to perform this reaction is measured to be $926.9\frac{\mathrm{kJ}}{\mathrm{mol}}$ . can proceed by a path in which we first break two $H_2$ bonds and one $O_2$ bond, then we follow the reverse of twice:

$$4H\left(g\right)+2O\left(g\right)\to 2H_{2}O\left(g\right)$$ $$2H_2\left(g\right)+O_2\left(g\right)\to 2H_{2}O\left(g\right)$$ Therefore, the energy of must be the energy required to break two $H_2$ bonds and one $O_2$ bond minus twice the energy of . We calculate that $\Delta (H^{°})=2436\frac{\mathrm{kJ}}{\mathrm{mol}}+498.3\frac{\mathrm{kJ}}{\mathrm{mol}}-2926.9\frac{\mathrm{kJ}}{\mathrm{mol}}=-483.5\frac{\mathrm{kJ}}{\mathrm{mol}}$ . It is clear from this calculation that is strongly exothermic because of the very large amount of energy released when two hydrogen atomsand one oxygen atom form a water molecule.