# Bond energy and enthalpy of formation

Is the following equation true? $$\ce{∆_fH} = \ce{∆_{BE_{reactants}}H} - \ce{∆_{BE_{products}}H}$$ where $$BE$$ stands for Bond energy and $$f$$ stands for formation.

I read that, $$\ce{∆_fH} = \ce{∆_{products}H} - \ce{∆_{reactants}H} = \ce{-∆_{BE}H}$$ But how is the first equation true?

Is it because of this $$\ce{∆_f_{reactants}H} = \ce{-∆_{BE}_{reactants}H}$$ ? When we substitute this in the second equation we get the first. Is it?

## 1 Answer

This equation is valid. Recall that the enthalpy of formation can be written as: $$\Delta H_{break} - \Delta H_{form}$$. When you translate this into bond energies, bonds breaking corresponds to the reactants and bond forming corresponds to the products. Thus,

$$\Delta_f H = \Delta_{BE, react} H - \Delta_{BE, prod} H$$

as you have. I don't think your second equation is correct however. If you're looking at enthalpies of products and reactants, the correct equation should be:

$$\Delta_f H = \Delta_{f, prod} H - \Delta_{f, react} H$$

• second equation i actually meant to mention what you mentioned. Thanks! – Inquisitive Oct 1 '19 at 14:08