0.9 Energetics of chemical reactions  (Page 7/7)

It is tempting to use the heat of to calculate the energy of an O-H bond. Since breaking the two O-H bonds in water requires $926.9\frac{\mathrm{kJ}}{\mathrm{mol}}$ , then we might infer that breaking a single O-H bond requires $\frac{926.9\frac{\mathrm{kJ}}{\mathrm{mol}}}{2}=463.5\frac{\mathrm{kJ}}{\mathrm{mol}}$ . However, the reaction

has $\Delta (H^{°})=492\frac{\mathrm{kJ}}{\mathrm{mol}}$ . Therefore, the energy required to break an O-H bond in $H_{2}O$ is not the same as the energy required to break the O-H bond in the $OH$ diatomic molecule. Stated differently, it requires more energy to break the first O-H bond in water than is required to break thesecond O-H bond.

In general, we find that the energy required to break a bond between any two particular atoms depends upon themolecule those two atoms are in. Considering yet again oxygen and hydrogen, we find that the energy required to break the O-H bond inmethanol ( $C{H}_{3}OH$ ) is $437\frac{\mathrm{kJ}}{\mathrm{mol}}$ , which differs substantially from the energy of . Similarly, the energy required to break a single C-H bond in methane( $C{H}_4$ ) is $435\frac{\mathrm{kJ}}{\mathrm{mol}}$ , but the energy required to break all four C-H bonds in methane is $1663\frac{\mathrm{kJ}}{\mathrm{mol}}$ , which is not equal to four times the energy of one bond. As anothersuch comparison, the energy required to break a C-H bond is $400\frac{\mathrm{kJ}}{\mathrm{mol}}$ in trichloromethane ( $HC{\mathrm{Cl}}_3$ ), $414\frac{\mathrm{kJ}}{\mathrm{mol}}$ in dichloromethane ( $H_{2}C{\mathrm{Cl}}_2$ ), and $422\frac{\mathrm{kJ}}{\mathrm{mol}}$ in chloromethane ( $H_{3}C\mathrm{Cl}$ ).

These observations are somewhat discouraging, since they reveal that, to use bond energies to calculate the heatof a reaction, we must first measure the bond energies for all bonds for all molecules involved in that reaction. This is almostcertainly more difficult than it is desirable. On the other hand, we can note that the bond energies for similar bonds in similarmolecules are close to one another. The C-H bond energies in the three chloromethanes above illustrate this quite well. We canestimate the C-H bond energy in any one of these chloromethanes by the average C-H bond energy in the three chloromethanes molecule,which is $412\frac{\mathrm{kJ}}{\mathrm{mol}}$ . Likewise, the average of the C-H bond energies in methane is $\frac{1663\frac{\mathrm{kJ}}{\mathrm{mol}}}{4}=416\frac{\mathrm{kJ}}{\mathrm{mol}}$ and is thus a reasonable approximation to the energy required to break a single C-H bond in methane.

By analyzing many bond energies in many molecules, we find that, in general, we can approximate the bondenergy in any particular molecule by the average of the energies of similar bonds. These average bond energies can then be used toestimate the heat of a reaction without measuring all of the required bond energies.

Consider for example the combustion of methane to form water and carbon dioxide:

We can estimate the heat of this reaction by using average bond energies. We must break four C-H bonds at anenergy cost of approximately $\times (4, 412\frac{\mathrm{kJ}}{\mathrm{mol}})$ and two $O_2$ bonds at an energy cost of approximately $\times (2, 496\frac{\mathrm{kJ}}{\mathrm{mol}})$ . Forming the bonds in the products releases approximately $\times (2, 743\frac{\mathrm{kJ}}{\mathrm{mol}})$ for the two C=O double bonds and $\times (4, 463\frac{\mathrm{kJ}}{\mathrm{mol}})$ for the O-H bonds. Net, the heat of reaction is thus approximately $\Delta (H^{°})=1648+992-1486-1852=-698\frac{\mathrm{kJ}}{\mathrm{mol}}$ . This is a rather rough approximation to the actual heat ofcombustion of methane, $-890\frac{\mathrm{kJ}}{\mathrm{mol}}$ . Therefore, we cannot use average bond energies to predictaccurately the heat of a reaction. We can get an estimate, which may be sufficiently useful. Moreover, we can use these calculationsto gain insight into the energetics of the reaction. For example, is strongly exothermic, which is why methane gas (the primary component in natural gas) isan excellent fuel. From our calculation, we can see that the reaction involved breaking six bonds and forming six new bonds. Thebonds formed are substantially stronger than those broken, thus accounting for the net release of energy during thereaction.

Review and discussion questions