Nitrogen reacts with hydrogen to form ammonia according to the equation below.

[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \][/tex]

Standard Enthalpies of Formation:

[tex]\[
\begin{tabular}{|c|c|}
\hline
Substance & [tex]$\Delta H_f \ (kJ/mol)$[/tex] \\
\hline
[tex]$C_2H_2(g)$[/tex] & -26.7 \\
\hline
[tex]$NH_3(g)$[/tex] & -46.19 \\
\hline
[tex]$HBr(g)$[/tex] & 236.23 \\
\hline
[tex]$HCl(g)$[/tex] & -92.30 \\
\hline
[tex]$HF(g)$[/tex] & -268.6 \\
\hline
[tex]$HI(g)$[/tex] & 25.9 \\
\hline
[tex]$NaCl(s)$[/tex] & -411.0 \\
\hline
\end{tabular}
\][/tex]

Based on the equation and the information in the table, what is the enthalpy of the reaction?

[tex]\[ \Delta H_{\text{reaction}} = -46.19 \ kJ \][/tex]



Answer :

To determine the enthalpy change of the reaction (\(\Delta H_{\text{rxn}}\)) for the formation of ammonia from nitrogen and hydrogen, we can use the concept of standard enthalpies of formation (\(\Delta H_f^\circ\)).

The balanced chemical equation for the reaction is:
[tex]\[ \text{N}_2 (g) + 3\text{H}_2 (g) \rightarrow 2\text{NH}_3 (g) \][/tex]

The standard enthalpy change of the reaction (\(\Delta H_{\text{rxn}}\)) can be determined using the standard enthalpies of formation of the reactants and products. This is given by the formula:
[tex]\[ \Delta H_{\text{rxn}}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \][/tex]

From the table of standard enthalpies of formation:
- \(\Delta H_f^\circ (\text{NH}_3(g)) = -46.19 \text{ kJ/mol}\)
- \(\Delta H_f^\circ (\text{N}_2(g)) = 0 \text{ kJ/mol}\) (since nitrogen is in its elemental form)
- \(\Delta H_f^\circ (\text{H}_2(g)) = 0 \text{ kJ/mol}\) (since hydrogen is in its elemental form)

Now, applying these values to the reaction:

Products:
[tex]\[ 2\text{NH}_3(g) \][/tex]
Since there are 2 moles of \(\text{NH}_3\), the total enthalpy contribution from \(\text{NH}_3\) is:
[tex]\[ 2 \times \Delta H_f^\circ (\text{NH}_3) = 2 \times (-46.19 \text{ kJ/mol}) \][/tex]

Reactants:
[tex]\[ \text{N}_2(g) + 3\text{H}_2(g) \][/tex]
The enthalpy contribution from \(\text{N}_2\) and \(\text{H}_2\) is zero since their standard enthalpies of formation are zero:
[tex]\[ \Delta H_f^\circ (\text{N}_2) + 3 \times \Delta H_f^\circ (\text{H}_2) = 0 + 3 \times 0 = 0 \][/tex]

Putting it all together:
[tex]\[ \Delta H_{\text{rxn}}^\circ = [2 \times (-46.19 \text{ kJ/mol})] - [0 + 0] \][/tex]
[tex]\[ \Delta H_{\text{rxn}}^\circ = -92.38 \text{ kJ} \][/tex]

Therefore, the enthalpy change for the reaction is:
[tex]\[ \Delta H_{\text{rxn}} = -92.38 \text{ kJ} \][/tex]