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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?

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3 Answers 3

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

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    $\begingroup$ Do you mind explaining what the dots infront of NH2 and H represent? $\endgroup$
    – Chris28
    Dec 30, 2019 at 14:54
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    $\begingroup$ The dots represent the fact that the species are free radicals (they have unpaired electrons). Strictly speaking, they could have put a dot in front of H in all 4 options and N in option 1. $\endgroup$
    – atbm
    Dec 31, 2019 at 7:39
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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).

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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

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    $\begingroup$ B is not correct because of the formation of new bonds in the products. For the correct (but not listed) alternative, see my answer. $\endgroup$
    – Karsten
    Aug 23, 2023 at 12:46
  • $\begingroup$ oh yeah ur right the N2 forms a new triple bond. But this question was actually an IB Chemistry question and they marked B as the correct answer. So technically all of the given answers are wrong right? $\endgroup$
    – user136808
    Aug 29, 2023 at 2:28

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