To determine the standard entropy change, [tex]\(\Delta S^{\circ}\)[/tex], for the reaction:
[tex]\[ 4 \, \text{Cr (s)} + 3 \, \text{O}_2 \, \text{(g)} \rightarrow 2 \, \text{Cr}_2 \text{O}_3 \, \text{(s)} \][/tex]
we need to follow these steps:
1. List the standard molar entropies [tex]\(S^{\circ}\)[/tex] of the reactants and products:
[tex]\[
\begin{align*}
S^{\circ} (\text{Cr (s)}) &= 23.77 \, \text{J/K·mol} \\
S^{\circ} (\text{O}_2 \, \text{(g)}) &= 205.138 \, \text{J/K·mol} \\
S^{\circ} (\text{Cr}_2 \text{O}_3 \, \text{(s)}) &= 80.65 \, \text{J/K·mol}
\end{align*}
\][/tex]
2. Identify the coefficients in the balanced chemical equation:
[tex]\[
\begin{align*}
n_{\text{Cr}} &= 4 \\
n_{\text{O}_2} &= 3 \\
n_{\text{Cr}_2 \text{O}_3} &= 2
\end{align*}
\][/tex]
3. Calculate the total entropy of the products:
The only product is [tex]\( \text{Cr}_2 \text{O}_3 \)[/tex], and there are 2 moles of it.
[tex]\[
\sum S^{\circ}(\text{products}) = 2 \times S^{\circ}(\text{Cr}_2 \text{O}_3) = 2 \times 80.65 = 161.3 \, \text{J/K·mol}
\][/tex]
4. Calculate the total entropy of the reactants:
There are 4 moles of [tex]\( \text{Cr (s)} \)[/tex] and 3 moles of [tex]\( \text{O}_2 \, \text{(g)} \)[/tex].
[tex]\[
\sum S^{\circ}(\text{reactants}) = 4 \times S^{\circ}(\text{Cr (s)}) + 3 \times S^{\circ}(\text{O}_2 \, \text{(g)}) = 4 \times 23.77 + 3 \times 205.138 = 95.08 + 615.414 = 710.494 \, \text{J/K·mol}
\][/tex]
5. Calculate the standard entropy change, [tex]\(\Delta S^{\circ}\)[/tex]:
[tex]\[
\Delta S^{\circ} = \sum S^{\circ} (\text{products}) - \sum S^{\circ} (\text{reactants})
\][/tex]
[tex]\[
\Delta S^{\circ} = 161.3 \, \text{J/K·mol} - 710.494 \, \text{J/K·mol} = -549.194 \, \text{J/K·mol}
\][/tex]
Therefore, the standard entropy change for the reaction is [tex]\(-549.19 \, \text{J/K·mol}\)[/tex], which corresponds to option E:
E. [tex]\(-549.19 \, \text{J/K·mol}\)[/tex]
