Combining Bond Enthalpies and Enthalpies of Formation

Can I mix formation enthalpies and bond enthalpies in the same calculation?


I *think* the answer is no, since they are relative measurements, and possibly relative to different things, but I haven't seen it stated explicitly anywhere.




You can combine this data. You just need to find a sensible reaction cycle, and you have to make sure the temperatures (usually 298 K) match. 


To give a simple example, consider $\ce{CO2(g)}$. 


What is the $\ce{CO}$ bond enthalpy $\mathrm{\epsilon\_{C=O}}$in $\ce{CO2(g)}$? 


It is one half of the enthalpy for the reaction 


$$\ce{ CO2(g)->C(g) + 2O(g)}~~~~~\mathrm{\Delta H^\circ = 2 \epsilon\_{C=O}}$$


The enthalpy of formation of $\ce{CO2(g)}$ is equal to the enthalpy of combustion of graphite:


$$\ce{C(s) + O2(g) -> CO2(g)} ~~~~~~\mathrm{\Delta \_f H^\circ = -393.51}\pu{kJ mol^{-1}}$$


The heat of formation of $\ce{C(g)}$ is: 


$$\ce{C(s) -> C(g) }~~~~~~\mathrm{\Delta \_f H^\circ = 716.68}\pu{kJ mol^{-1}}$$


We also need a value of the $\ce{OO}$ bond enthalpy in $\ce{O2}$: 


$$\ce{O\_2(g) -> 2O(g) }~~~~\mathrm{\Delta H^\circ = \epsilon\_{O=O} = 498.3}\pu{kJ mol^{-1}}$$


Adding things up we obtain $\mathrm{\epsilon\_{C=O}}$: 


$$\mathrm{\epsilon\_{C=O}} = \frac12(393.51+716.68+498.3)\pu{kJ mol^{-1}= 804.2 kJ mol^{-1}}$$


 0