To determine the enthalpy change for the reaction, we follow these steps:
### Step 1: Identify the enthalpies of formation for the reactants and products
Given data:
- For [tex]\( \text{NH}_3 \)[/tex], ΔH = -45.9 kJ
- For [tex]\( \text{H}_2\text{O} \)[/tex], ΔH = -241.8 kJ
- For [tex]\( \text{NO} \)[/tex], ΔH = 91.3 kJ
### Step 2: Write down the balanced chemical equation
[tex]\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 6 \text{H}_2\text{O}(g) + 4 \text{NO}(g) \][/tex]
### Step 3: Apply the stoichiometric coefficients to the enthalpies of formation
Calculate the sum of the enthalpies of formation for the reactants and products.
#### Reactants
Sum of the enthalpies of formation for the reactants:
[tex]\[ 4 \times \Delta H_{\text{NH}_3} = 4 \times (-45.9 \, \text{kJ}) = -183.6 \, \text{kJ} \][/tex]
#### Products
Sum of the enthalpies of formation for the products:
[tex]\[ 6 \times \Delta H_{\text{H}_2\text{O}} + 4 \times \Delta H_{\text{NO}} = 6 \times (-241.8 \, \text{kJ}) + 4 \times 91.3 \, \text{kJ} \][/tex]
Calculating each term:
[tex]\[ 6 \times (-241.8 \, \text{kJ}) = -1450.8 \, \text{kJ} \][/tex]
[tex]\[ 4 \times 91.3 \, \text{kJ} = 365.2 \, \text{kJ} \][/tex]
Sum of the products:
[tex]\[ -1450.8 \, \text{kJ} + 365.2 \, \text{kJ} = -1085.6 \, \text{kJ} \][/tex]
### Step 4: Calculate the enthalpy change for the reaction
Use the formula:
[tex]\[ \Delta H_{\text{reaction}} = \sum \Delta H_{\text{f,products}} - \sum \Delta H_{\text{f,reactants}} \][/tex]
Substitute the sums:
[tex]\[ \Delta H_{\text{reaction}} = -1085.6 \, \text{kJ} - (-183.6 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1085.6 \, \text{kJ} + 183.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -902.0 \, \text{kJ} \][/tex]
The enthalpy change for the reaction is [tex]\(-902 \, \text{kJ}\)[/tex].
### Final Answer
The enthalpy change for the reaction is [tex]\(-902 \, \text{kJ}\)[/tex]. Therefore, the correct answer is [tex]\(-902 \, \text{kJ}\)[/tex].
