To calculate [tex]\(\Delta S^\circ_{rxn}\)[/tex] for the given reaction:
[tex]\[4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)\][/tex]
we will follow these steps:
1. Identify the standard molar entropy ([tex]\(S^\circ\)[/tex]) values for each species:
- [tex]\(S^\circ_{NH_3(g)} = 192.8 \, \text{J/(mol·K)}\)[/tex]
- [tex]\(S^\circ_{O_2(g)} = 205.0 \, \text{J/(mol·K)}\)[/tex]
- [tex]\(S^\circ_{NO(g)} = 210.8 \, \text{J/(mol·K)}\)[/tex]
- [tex]\(S^\circ_{H_2O(g)} = 188.8 \, \text{J/(mol·K)}\)[/tex]
2. Identify the stoichiometric coefficients for each species:
- For [tex]\(NH_3\)[/tex]: 4
- For [tex]\(O_2\)[/tex]: 5
- For [tex]\(NO\)[/tex]: 4
- For [tex]\(H_2O\)[/tex]: 6
3. Calculate the total standard entropy of the reactants:
[tex]\[
\text{sum\_S\_reactants} = (4 \cdot 192.8) + (5 \cdot 205.0)
\][/tex]
- Contribution of [tex]\(NH_3\)[/tex]: [tex]\(4 \cdot 192.8 = 771.2 \, \text{J/K}\)[/tex]
- Contribution of [tex]\(O_2\)[/tex]: [tex]\(5 \cdot 205.0 = 1025.0 \, \text{J/K}\)[/tex]
- Total entropy of reactants:
[tex]\[
\text{sum\_S\_reactants} = 771.2 + 1025.0 = 1796.2 \, \text{J/K}
\][/tex]
4. Calculate the total standard entropy of the products:
[tex]\[
\text{sum\_S\_products} = (4 \cdot 210.8) + (6 \cdot 188.8)
\][/tex]
- Contribution of [tex]\(NO\)[/tex]: [tex]\(4 \cdot 210.8 = 843.2 \, \text{J/K}\)[/tex]
- Contribution of [tex]\(H_2O\)[/tex]: [tex]\(6 \cdot 188.8 = 1132.8 \, \text{J/K}\)[/tex]
- Total entropy of products:
[tex]\[
\text{sum\_S\_products} = 843.2 + 1132.8 = 1976.0 \, \text{J/K}
\][/tex]
5. Calculate the standard entropy change for the reaction ([tex]\(\Delta S^\circ_{rxn}\)[/tex]):
[tex]\[
\Delta S^\circ_{rxn} = \text{sum\_S\_products} - \text{sum\_S\_reactants}
\][/tex]
[tex]\[
\Delta S^\circ_{rxn} = 1976.0 - 1796.2 = 179.8 \, \text{J/K}
\][/tex]
The entropy change [tex]\(\Delta S^\circ_{rxn}\)[/tex] for the reaction is [tex]\(+179.8 \, \text{J/K}\)[/tex]. Thus, the correct answer is:
[tex]\[
\boxed{+178.8 \, \text{J/K}}
\][/tex]
