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]
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]
