# Question about calculating enthalpy change

$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$

$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$

Both $$\Delta H$$'s given by the above formulas should be equivalent. However, I found one special case that the above 2 formulas give 2 different results. $$\ce{2 H+ + 2 e- → H2}$$ $$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)= 0-0=0$$ $$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)<0$$ because there is only bond formation, no bond break. How to explain this difference? Can I say the heat is released during this process, because $$\Delta H$$ is smaller than 0?

• Where in the calculation is $\Delta H_f$ of the electron? – Karsten Theis Jul 1 at 2:38
• I think the $$\Delta Hf$$ of the electron should be zero – king Jul 1 at 2:40

The bond dissociation energy of the $$\ce{H-H}$$ bond is reported as 436 kJ/mol. The relevant reaction is:
$$\ce{H2(g) -> 2H.(g)}$$
Note that this reaction is not the same as $$\ce{H2 -> 2H+ + 2e-}$$. The homolytic cleavage reaction produces hydrogen atoms, not hydrogen ions. If we look up the standard enthalpy of formation for hydrogen atoms, it is +218 kJ/mol. Since the standard enthalpy of formation for $$\ce{H2}$$ is defined as zero, using the enthalpies of formation we also get $$\Delta H = +436$$ kJ/mol for this reaction.