Alright, let's solve the problem step by step using the enthalpies of formation provided.
### 1. Write Down Given Data
- Enthalpy of formation of [tex]\( NH_3(g) \)[/tex] ([tex]\(\Delta H_{f,\ NH_3}\)[/tex]): -45.9 \text{ kJ/mol}\)
- Enthalpy of formation of [tex]\( H_2O(g) \)[/tex] ([tex]\(\Delta H_{f,\ H_2O}\)[/tex]): -241.8 \text{ kJ/mol}\)
### 2. Balanced Chemical Equation
[tex]\[
4 \text{NH}_3(g) + 3 \text{O}_2(g) \rightarrow 2 \text{N}_2(g) + 6 \text{H}_2O(g)
\][/tex]
### 3. Calculate the Enthalpy of the Reactants
The enthalpy change for the reactants includes only [tex]\( NH_3 \)[/tex], as [tex]\( O_2 \)[/tex] in its standard state has zero enthalpy of formation.
[tex]\[
\Delta H_{\text{reactants}} = (\text{moles of } NH_3) \times (\Delta H_{f,\ NH_3})
\][/tex]
[tex]\[
\Delta H_{\text{reactants}} = 4 \text{ moles} \times (-45.9 \text{ kJ/mol}) = -183.6 \text{ kJ}
\][/tex]
### 4. Calculate the Enthalpy of the Products
The enthalpy change for the products includes only [tex]\( H_2O \)[/tex].
[tex]\[
\Delta H_{\text{products}} = (\text{moles of } H_2O) \times (\Delta H_{f, \ H_2O})
\][/tex]
[tex]\[
\Delta H_{\text{products}} = 6 \text{ moles} \times (-241.8 \text{ kJ/mol}) = -1450.8 \text{ kJ}
\][/tex]
### 5. Using the Formula for the Enthalpy of Reaction
The enthalpy change of the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) is given by:
[tex]\[
\Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}}
\][/tex]
Plug in the values:
[tex]\[
\Delta H_{\text{reaction}} = -1450.8 \text{ kJ} - (-183.6 \text{ kJ})
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -1450.8 \text{ kJ} + 183.6 \text{ kJ} = -1267.2 \text{ kJ}
\][/tex]
### 6. Conclusion
The enthalpy change for the reaction is [tex]\(-1267.2 \text{ kJ}\)[/tex]. Therefore, the correct answer is:
[tex]\[
\boxed{-1267.2 \text{ kJ}}
\][/tex]
This matches the option: [tex]\(-1,267.2 \text{ kJ}\)[/tex].
