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 $\mathrm{\epsilon_{C=O}}$:

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

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