Sure, let's calculate the change in entropy ([tex]\(\Delta S_{rxn}\)[/tex]) for the given reaction step by step:
The reaction is:
[tex]\[ \text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{H}_2\text{O(l)} + \text{NaCl(aq)} \][/tex]
The given entropy ([tex]\(S\)[/tex]) values are:
- [tex]\( S_{\text{NaOH}} = 49.8 \, \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{HCl}} = 56.5 \, \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{H}_2\text{O}} = 69.91 \, \text{J/mol K} \)[/tex]
- [tex]\( S_{\text{NaCl}} = 115.5 \, \text{J/mol K} \)[/tex]
To find the change in entropy for the reaction ([tex]\(\Delta S_{rxn}\)[/tex]), we use the formula:
[tex]\[ \Delta S_{rxn} = \left( \sum S_{\text{products}} \right) - \left( \sum S_{\text{reactants}} \right) \][/tex]
1. Calculate the sum of the entropy of the reactants:
[tex]\[
S_{\text{reactants}} = S_{\text{NaOH}} + S_{\text{HCl}} = 49.8 \, \text{J/mol K} + 56.5 \, \text{J/mol K} = 106.3 \, \text{J/mol K}
\][/tex]
2. Calculate the sum of the entropy of the products:
[tex]\[
S_{\text{products}} = S_{\text{H}_2\text{O}} + S_{\text{NaCl}} = 69.91 \, \text{J/mol K} + 115.5 \, \text{J/mol K} = 185.41 \, \text{J/mol K}
\][/tex]
3. Find the change in entropy for the reaction:
[tex]\[
\Delta S_{rxn} = S_{\text{products}} - S_{\text{reactants}} = 185.41 \, \text{J/mol K} - 106.3 \, \text{J/mol K} = 79.11 \, \text{J/mol K}
\][/tex]
4. Round [tex]\(\Delta S_{rxn}\)[/tex] to the nearest whole number:
[tex]\[
\Delta S_{rxn} \approx 79 \, \text{J/mol K}
\][/tex]
Thus, [tex]\(\Delta S_{rxn} = 79 \, \text{J/mol K}\)[/tex].
