How do you find the bond enthalpy from an equation?

Question

Which equation represents the $\ce{N-H}$ bond enthalpy in $\ce{NH3}?$

$$ \begin{align} &\textbf{A.} &\ce{NH3(g) &-> N(g) + 3 H(g)}\\ &\textbf{B.} &\ce{1/3 NH3(g) &-> 1/3 N2(g) + H(g)}\\ &\textbf{C.} &\ce{NH3(g) &-> 1/2 N2(g) + 3/2 H2(g)}\\ &\textbf{D.} &\ce{NH3(g) &-> .NH2(g) + .H(g)} \end{align} $$

From what I've learnt, I understand that the bond enthalpy is defined as the energy required to break one mole of a specific bond.

In the question above, I opted for answer C as it was the only one with the products in the form of $\ce{N2}$ and $\ce{H2}.$ But since there are three $\ce{N-H}$ bonds in $\ce{NH3},$ I am unsure about answer B.

Can you clarify this?

Answer 1

Option D is the only one corresponding to the definition "energy required to break one mole of a specific bond" and nothing more.

Answer 2

If you want to have a "none of the above" answer, you could also consider $$\ce{1/3 NH3(g) -> 1/3 N(g) + H(g)}$$

or written more conventionally, one third of the reaction enthalpy of

$$\ce{NH3(g) -> N(g) + 3 H(g)}$$

The atoms $\ce{N}$ and $\ce{H}$ are radicals but are shown without dots here.

This would be the average of the three N-H bond dissociation energies. The key is not to form new bonds (i.e. no $\ce{N#N}$ or $\ce{H-H}$ as products), and to cleave bonds in a homolytic manner (one electron of the bond remains on each atom).

Answer 3

Nope, the answer is B from my understanding. This is because the definition of bond enthalpy is the energy required to break "one mole" of a bond. One mole of $\ce{NH3}$ has three N-H bonds, so for you to have one N-H bond you would need 1/3 moles of NH3