To find the enthalpy of reaction for the decomposition of calcium carbonate ([tex]\( \text{CaCO}_3(s) \)[/tex]), we use the enthalpies of formation for the substances involved in the reaction. The given reaction is:
[tex]\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \][/tex]
We are provided with the following enthalpies of formation ([tex]\(\Delta H_f\)[/tex]):
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Compound} & \Delta H_f (\text{kJ/mol}) \\
\hline
\text{CaO}(s) & -157.3 \\
\hline
\text{CaCO}_3(s) & -1207.1 \\
\hline
\text{CO}_2(g) & -393.5 \, (\text{corrected value}) \\
\hline
\end{array}
\][/tex]
The enthalpy change for the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) can be calculated using the enthalpies of formation of the reactants and products:
[tex]\[
\Delta H_{\text{reaction}} = \left[ \Delta H_f(\text{products}) \right] - \left[ \Delta H_f(\text{reactants}) \right]
\][/tex]
Substitute the given values:
[tex]\[
\Delta H_{\text{reaction}} = \left[ \Delta H_f(\text{CaO}(s)) + \Delta H_f(\text{CO}_2(g)) \right] - \Delta H_f(\text{CaCO}_3(s))
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = \left[ -157.3 + (-393.5) \right] - (-1207.1)
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = \left[ -157.3 - 393.5 \right] + 1207.1
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -550.8 + 1207.1
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = 656.3 \, \text{kJ}
\][/tex]
Therefore, the enthalpy of reaction for the decomposition of calcium carbonate is:
[tex]\[
\boxed{656.3 \, \text{kJ}}
\][/tex]
