# Question 7aa26

Mar 9, 2016

#### Explanation:

The most important thing to remember about the standard enthalpy of formation is that it represents the enthalpy change of reaction when one mole of a compound is formed from its constituent elements in their stable form under standard conditions.

In order for the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}^{\circ}$, to be the same as the standard enthalpy of formation, $\Delta {H}_{f}^{\circ}$, the reaction must lead to the formation of one mole of a given compound.

Take a look at the options given to you. The first balanced chemical equation is

${\text{N"_text(2(g]) + 3"H"_text(2(g]) -> color(red)(2)"NH}}_{\textrm{3 \left(g\right]}}$

Notice that this reaction results in the formation of $\textcolor{red}{2}$ moles of ammonia. In order for the enthalpy changes to match, you need the reaction to result in the formation of $1$ mole of ammonia.

This reaction will have an enthalpy change of reaction, $\Delta {H}_{\text{rxn a}}^{\circ}$, which characterizes the formation of two moles of ammonia.

This means that you can use this as a conversion factor to find the enthalpy change of reaction when $1$ mole of ammonia is formed

1 color(red)(cancel(color(black)("mole NH"_3))) * (DeltaH_"rxn a"^@)/(color(red)(2)color(red)(cancel(color(black)("moles NH"_3)))) = (DeltaH_a^@)/color(red)(2)#

This means that you have

$\Delta {H}_{f}^{\circ} = \frac{\Delta {H}_{\text{rxn a}}^{\circ}}{\textcolor{red}{2}}$

This should confirm the fact that you must look for a reaction that results in the formation of $1$ mole of ammonia.

Reaction (b) results in the formation of one mole of ammonia, but the chemical equation given is not balanced, which means that it is not valid.

Moreover, you can't have atomic nitrogen, $\text{N}$, as a reactant, since nitrogen exists as a diatomic molecule, ${\text{N}}_{2}$, in its most stable form.

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{N"_text((g]) + 3"H"_text(2(g]) -> "NH}}_{\textrm{3 \left(g\right]}}}}} \to$ NOT valid

Option (d) is out of the question, since it describes the decomposition of ammonia, not its formation.

This leaves you with option (c). Once again, this reaction results in the formation of $1$ mole of ammonia, but this time the equation is also balanced, and thus valid

$\frac{1}{2} {\text{N"_text(2(g]) + 3/2"H"_text(2(g]) -> "NH}}_{\textrm{3 \left(g\right]}}$

This time, you have

$\Delta {H}_{\text{rxn c}}^{\circ} = \Delta {H}_{f}^{\circ}$

Here's how you can interpret this equation - when one atom of nitrogen, which is denoted by half of a nitrogen molecule, $\frac{1}{2} {\text{N}}_{2}$, reacts with three atoms of hydrogen, which are denoted by $\frac{3}{2}$ of a hydrogen molecule, $\frac{3}{2} {\text{H}}_{2}$, one mole of ammonia is formed.

This reaction has a standard enthalpy change of reaction equal to the standard enthalpy change of formation, $\Delta {H}_{f}^{\circ}$, for ammonia.