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Sagot :
To determine the enthalpy change (ΔH) for the decomposition of calcium carbonate ([tex]\( \text{CaCO}_3 \)[/tex]), we need to compute the difference between the total enthalpy of the products and the total enthalpy of the reactants.
The chemical reaction given is:
[tex]\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \][/tex]
Given the following enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) for each compound involved:
- For [tex]\( \text{CaO} (s) \)[/tex]: [tex]\( \Delta H_{CaO} = -635.5 \, \text{kJ/mol} \)[/tex]
- For [tex]\( \text{CaCO}_3 (s) \)[/tex]: [tex]\( \Delta H_{CaCO_3} = -1207.0 \, \text{kJ/mol} \)[/tex]
- For [tex]\( \text{CO}_2 (g) \)[/tex]: [tex]\( \Delta H_{CO_2} = -393.5 \, \text{kJ/mol} \)[/tex]
1. Determine the enthalpy of the reactants:
Since [tex]\( \text{CaCO}_3 \)[/tex] is the only reactant in this decomposition reaction:
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_{CaCO_3} = -1207.0 \, \text{kJ/mol} \][/tex]
2. Determine the enthalpy of the products:
The products are [tex]\( \text{CaO} \)[/tex] and [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[ \Delta H_{\text{products}} = \Delta H_{CaO} + \Delta H_{CO_2} = -635.5 \, \text{kJ/mol} + (-393.5 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1029.0 \, \text{kJ/mol} \][/tex]
3. Calculate the enthalpy change of the reaction ([tex]\( \Delta H_{\text{reaction}} \)[/tex]):
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1029.0 \, \text{kJ/mol} - (-1207.0 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1029.0 \, \text{kJ/mol} + 1207.0 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = 178.0 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change for the decomposition of calcium carbonate ([tex]\( \text{CaCO}_3 \)[/tex]) into calcium oxide ([tex]\( \text{CaO} \)[/tex]) and carbon dioxide ([tex]\( \text{CO}_2 \)[/tex]) is [tex]\( \Delta H_{\text{reaction}} = 178.0 \, \text{kJ/mol} \)[/tex].
The final tabular data would be:
\begin{tabular}{|r|r|}
\hline
Compound & [tex]$\Delta H_f (\text{kJ/mol})$[/tex] \\
\hline
[tex]$\text{CaO} (s)$[/tex] & -635.5 \\
\hline
[tex]$\text{CaCO}_3 (s)$[/tex] & -1207.0 \\
\hline
[tex]$\text{CO}_2 (g)$[/tex] & -393.5 \\
\hline
\end{tabular}
This result signifies that the reaction is endothermic with an enthalpy change of [tex]\( 178.0 \, \text{kJ/mol} \)[/tex].
The chemical reaction given is:
[tex]\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \][/tex]
Given the following enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) for each compound involved:
- For [tex]\( \text{CaO} (s) \)[/tex]: [tex]\( \Delta H_{CaO} = -635.5 \, \text{kJ/mol} \)[/tex]
- For [tex]\( \text{CaCO}_3 (s) \)[/tex]: [tex]\( \Delta H_{CaCO_3} = -1207.0 \, \text{kJ/mol} \)[/tex]
- For [tex]\( \text{CO}_2 (g) \)[/tex]: [tex]\( \Delta H_{CO_2} = -393.5 \, \text{kJ/mol} \)[/tex]
1. Determine the enthalpy of the reactants:
Since [tex]\( \text{CaCO}_3 \)[/tex] is the only reactant in this decomposition reaction:
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_{CaCO_3} = -1207.0 \, \text{kJ/mol} \][/tex]
2. Determine the enthalpy of the products:
The products are [tex]\( \text{CaO} \)[/tex] and [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[ \Delta H_{\text{products}} = \Delta H_{CaO} + \Delta H_{CO_2} = -635.5 \, \text{kJ/mol} + (-393.5 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1029.0 \, \text{kJ/mol} \][/tex]
3. Calculate the enthalpy change of the reaction ([tex]\( \Delta H_{\text{reaction}} \)[/tex]):
[tex]\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1029.0 \, \text{kJ/mol} - (-1207.0 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1029.0 \, \text{kJ/mol} + 1207.0 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = 178.0 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change for the decomposition of calcium carbonate ([tex]\( \text{CaCO}_3 \)[/tex]) into calcium oxide ([tex]\( \text{CaO} \)[/tex]) and carbon dioxide ([tex]\( \text{CO}_2 \)[/tex]) is [tex]\( \Delta H_{\text{reaction}} = 178.0 \, \text{kJ/mol} \)[/tex].
The final tabular data would be:
\begin{tabular}{|r|r|}
\hline
Compound & [tex]$\Delta H_f (\text{kJ/mol})$[/tex] \\
\hline
[tex]$\text{CaO} (s)$[/tex] & -635.5 \\
\hline
[tex]$\text{CaCO}_3 (s)$[/tex] & -1207.0 \\
\hline
[tex]$\text{CO}_2 (g)$[/tex] & -393.5 \\
\hline
\end{tabular}
This result signifies that the reaction is endothermic with an enthalpy change of [tex]\( 178.0 \, \text{kJ/mol} \)[/tex].
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