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Karsten ♦
Karsten ♦
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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).

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).

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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Source Link
Karsten ♦
Karsten ♦
  • 42.3k
  • 8
  • 75
  • 194

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).

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

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

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).

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).

Source Link
Karsten ♦
Karsten ♦
  • 42.3k
  • 8
  • 75
  • 194

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

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

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).