Comparing formula for enthalpy change with bond dissociation energy and formation enthalpy

Question

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 1

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.

Answer 2

Given the reaction $\ce{A -> B}$ you know that $\ce{A}$ and $\ce{B}$ must have the same atomic composition, i.e. they consist of the same number of atoms for each element, because if this wasn't the case there would be other reactants or products in the reaction equation (the reaction is some sort of rearrangement). So, $\ce{A}$ and $\ce{B}$ only differ in the way the atoms are linked together, i.e. some bonds that are present in $\ce{A}$ might not be present in $\ce{B}$ and vice versa.

Now, Hess' law, which states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps, leads you to the equation

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

where $\Delta H_F$ is the enthalpy of formation. This $\Delta H_F$ is simply the energy difference between the target compound and the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature.

\begin{equation} \Delta H_F = H_{\text{molecule}} - H_{\text{elements}} \end{equation}

So, $\Delta H_F$ (mostly) describes how much energy is gained by forming bonds in the molecule. Now, since $\ce{A}$ and $\ce{B}$ have the same atomic composition, i.e. they have the same $H_{\text{elements}}$, you see that the energy difference between them must come from the energy difference of the bonds in $\ce{A}$ and $\ce{B}$. The part of bonding energy that comes from bonds that both $\ce{A}$ and $\ce{B}$ have in common cancels out and what remains is

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

So, yes, both methods are equivalent.