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Ammonia [tex]$\left( NH _3(g), \Delta H_f = -45.9 \text{ kJ/mol} \right)$[/tex] reacts with oxygen to produce nitrogen and water [tex]$\left( H _2 O(g), \Delta H_f = -241.8 \text{ kJ/mol} \right)$[/tex] according to the equation below:

[tex]\[ 4 NH _3(g) + 3 O _2(g) \rightarrow 2 N_2(g) + 6 H _2 O(g) \][/tex]

What is the enthalpy of the reaction? Use [tex]$\Delta H_{\text{rxn}} = \sum \left( \Delta H_{\text{f,products}} \right) - \sum \left( \Delta H_{\text{f,reactants}} \right)$[/tex].

A. [tex]\(-3,572.1 \text{ kJ}\)[/tex]
B. [tex]\(-1,267.2 \text{ kJ}\)[/tex]
C. [tex]\(1,267.2 \text{ kJ}\)[/tex]
D. [tex]\(3,572.1 \text{ kJ}\)[/tex]



Answer :

GinnyAnswer
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].

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