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The decomposition of calcium carbonate, [tex]CaCO_3(s) \rightarrow CaO(s) + CO_2(g)[/tex], has the following values for free energy and enthalpy at [tex]25.0^{\circ} C[/tex]:

[tex]\[
\begin{array}{l}
\Delta G = 130.5 \, \text{kJ/mol} \\
\Delta H = 178.3 \, \text{kJ/mol}
\end{array}
\][/tex]

What is the entropy of the reaction? Use [tex]\Delta G = \Delta H - T \Delta S[/tex].

A. [tex]-160.3 \, \text{J/(mol \cdot K)}[/tex]

B. [tex]-47.8 \, \text{J/(mol \cdot K)}[/tex]

C. [tex]160.3 \, \text{J/(mol \cdot K)}[/tex]

D. [tex]1,912 \, \text{J/(mol \cdot K)}[/tex]



Answer :

GinnyAnswer
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]

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