To find the entropy change \(\Delta S\) of the reaction \(CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\), we use the Gibbs free energy relation:
[tex]\[
\Delta G = \Delta H - T \Delta S
\][/tex]
Given values:
[tex]\[
\Delta G = 130.5 \, \text{kJ/mol}
\][/tex]
[tex]\[
\Delta H = 178.3 \, \text{kJ/mol}
\][/tex]
Temperature at \(25.0^\circ C\) which is:
[tex]\[
T = 25.0 + 273.15 = 298.15 \, \text{K}
\][/tex]
First, rearrange the Gibbs free energy equation to solve for \(\Delta S\):
[tex]\[
\Delta S = \frac{\Delta H - \Delta G}{T}
\][/tex]
We need to convert \(\Delta H\) and \(\Delta G\) from kJ to J to match the units of entropy (J/(mol·K)):
[tex]\[
\Delta H = 178.3 \, \text{kJ/mol} \times 1000 \, \frac{\text{J}}{\text{kJ}} = 178300 \, \text{J/mol}
\][/tex]
[tex]\[
\Delta G = 130.5 \, \text{kJ/mol} \times 1000 \, \frac{\text{J}}{\text{kJ}} = 130500 \, \text{J/mol}
\][/tex]
Now, plug these values into the rearranged equation:
[tex]\[
\Delta S = \frac{178300 \, \text{J/mol} - 130500 \, \text{J/mol}}{298.15 \, \text{K}}
\][/tex]
[tex]\[
\Delta S = \frac{47800 \, \text{J/mol}}{298.15 \, \text{K}}
\][/tex]
[tex]\[
\Delta S \approx 160.3 \, \text{J/(mol·K)}
\][/tex]
Therefore, the entropy change \(\Delta S\) for the reaction is \(160.3 \, \text{J/(mol·K)}\). The correct answer is:
[tex]\[
160.3 \, \text{J/(mol·K)}
\][/tex]
