I learnt that given a reaction: $$\ce{A -> B}$$ the enthalpy change is given by:
$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$
By convention, we keep the minus sign in the equation, and keep all bond enthalpies positive, regardless of whether bonds are made or broken. In this case, the bonds in the reactants are broken, and the bonds in the products are formed. I was also given the equation,
$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$
Are both $\Delta H$'s given by the above formulas equivalent?
Answer
Yes. The two methods are equivalent.
You can try this with the dimerization of $\ce{NO_2}$ into $\ce{N_2O_4}$. In this reaction we form one $\ce{N-N}$ bond. Every other bond stays the same.
$\ce{2NO_2 <=> N_2O_4}$
If we calculate the enthalpy of the reaction through bond energies, we get the negative of the $\ce{N-N}$ bond energy. Why? Because we only formed one bond in this entire reaction. Every other bond stays the same - we start with four resonance stabilized $\ce{N-O}$ bonds and end with four resonance stabilized $\ce{N-O}$ bonds. The net change in bonds is the formation of a $\ce{N-N}$ bond. Here is a picture of the process (note that this picture doesn't show any resonance stabilization but there is; the "pi" bonding electrons are delocalized despite what the picture implies).
Bond energy is a positive value; the negative of this value implies that the dimerization reaction is exothermic - which is obviously true! We are going from a radical species - nitrogen dioxide - into a species with no unpaired electrons. Radicals are reactive!
We can also calculate this using heats of formation for both nitrogen tetraoxide and nitrogen dioxide and we'd get the same enthalpy of reaction.
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