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Sagot :
To calculate the change in standard entropy, [tex]\(\Delta S_{r}\)[/tex], for the reaction
[tex]\[4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g)\][/tex]
given the standard molar entropies:
- [tex]\( S_{\text{NH}_3} = 192.8 \ \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{O}_2} = 205.0 \ \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{H}_2\text{O}} = 188.83 \ \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{NO}} = 210.8 \ \text{J/mol K} \)[/tex],
we follow these steps:
1. Identify the coefficients of each compound in the balanced equation:
- For [tex]\( \text{NH}_3 \)[/tex]: coefficient is 4
- For [tex]\( \text{O}_2 \)[/tex]: coefficient is 5
- For [tex]\( \text{NO} \)[/tex]: coefficient is 4
- For [tex]\( \text{H}_2\text{O} \)[/tex]: coefficient is 6
2. Calculate the total standard entropy of the reactants:
[tex]\[ S_{\text{reactants}} = ( \text{coefficient of } \text{NH}_3 \times S_{\text{NH}_3} ) + ( \text{coefficient of } \text{O}_2 \times S_{\text{O}_2} ) \][/tex]
Substituting the values, we get:
[tex]\[ S_{\text{reactants}} = (4 \times 192.8) + (5 \times 205.0) = 771.2 + 1025.0 = 1796.2 \ \text{J/K} \][/tex]
3. Calculate the total standard entropy of the products:
[tex]\[ S_{\text{products}} = ( \text{coefficient of } \text{NO} \times S_{\text{NO}} ) + ( \text{coefficient of } \text{H}_2\text{O} \times S_{\text{H}_2\text{O}} ) \][/tex]
Substituting the values, we get:
[tex]\[ S_{\text{products}} = (4 \times 210.8) + (6 \times 188.83) = 843.2 + 1132.98 = 1976.18 \ \text{J/K} \][/tex]
4. Calculate the change in standard entropy ([tex]\(\Delta S_{r}\)[/tex]):
[tex]\[ \Delta S_{r} = S_{\text{products}} - S_{\text{reactants}} \][/tex]
Substituting the values calculated:
[tex]\[ \Delta S_{r} = 1976.18 - 1796.2 = 179.98 \ \text{J/K} \][/tex]
5. Round the final result to the nearest whole number:
[tex]\[ \Delta S_{r} \approx 180 \ \text{J/K} \][/tex]
Therefore, the change in standard entropy [tex]\(\Delta S_{r}\)[/tex] for the given reaction is approximately [tex]\(\boxed{180 \ \text{J/K}}\)[/tex].
[tex]\[4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g)\][/tex]
given the standard molar entropies:
- [tex]\( S_{\text{NH}_3} = 192.8 \ \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{O}_2} = 205.0 \ \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{H}_2\text{O}} = 188.83 \ \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{NO}} = 210.8 \ \text{J/mol K} \)[/tex],
we follow these steps:
1. Identify the coefficients of each compound in the balanced equation:
- For [tex]\( \text{NH}_3 \)[/tex]: coefficient is 4
- For [tex]\( \text{O}_2 \)[/tex]: coefficient is 5
- For [tex]\( \text{NO} \)[/tex]: coefficient is 4
- For [tex]\( \text{H}_2\text{O} \)[/tex]: coefficient is 6
2. Calculate the total standard entropy of the reactants:
[tex]\[ S_{\text{reactants}} = ( \text{coefficient of } \text{NH}_3 \times S_{\text{NH}_3} ) + ( \text{coefficient of } \text{O}_2 \times S_{\text{O}_2} ) \][/tex]
Substituting the values, we get:
[tex]\[ S_{\text{reactants}} = (4 \times 192.8) + (5 \times 205.0) = 771.2 + 1025.0 = 1796.2 \ \text{J/K} \][/tex]
3. Calculate the total standard entropy of the products:
[tex]\[ S_{\text{products}} = ( \text{coefficient of } \text{NO} \times S_{\text{NO}} ) + ( \text{coefficient of } \text{H}_2\text{O} \times S_{\text{H}_2\text{O}} ) \][/tex]
Substituting the values, we get:
[tex]\[ S_{\text{products}} = (4 \times 210.8) + (6 \times 188.83) = 843.2 + 1132.98 = 1976.18 \ \text{J/K} \][/tex]
4. Calculate the change in standard entropy ([tex]\(\Delta S_{r}\)[/tex]):
[tex]\[ \Delta S_{r} = S_{\text{products}} - S_{\text{reactants}} \][/tex]
Substituting the values calculated:
[tex]\[ \Delta S_{r} = 1976.18 - 1796.2 = 179.98 \ \text{J/K} \][/tex]
5. Round the final result to the nearest whole number:
[tex]\[ \Delta S_{r} \approx 180 \ \text{J/K} \][/tex]
Therefore, the change in standard entropy [tex]\(\Delta S_{r}\)[/tex] for the given reaction is approximately [tex]\(\boxed{180 \ \text{J/K}}\)[/tex].
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