To find the enthalpy change of the reaction, we use the formula:
[tex]\[
\Delta H_{\text{rxn}} = \sum\left(\Delta H_{\text{f,products}}\right) - \sum\left(\Delta H_{\text{f,reactants}}\right)
\][/tex]
Given the enthalpy values in the table, let’s identify the values for the products and reactants:
Products:
- [tex]\(CO_2 (g) \rightarrow -393.509 \, \text{kJ/mol} \)[/tex]
- [tex]\(H_2O (g) \rightarrow -241.818 \, \text{kJ/mol} \)[/tex]
Reactants:
- [tex]\(CaO (s) \rightarrow -635.09 \, \text{kJ/mol} \)[/tex]
- [tex]\(CO (g) \rightarrow -110.525 \, \text{kJ/mol} \)[/tex]
Step-by-Step Solution:
1. Calculate the total enthalpy of formation for the products:
[tex]\[
\Delta H_{\text{f, products}} = (-393.509) + (-241.818)
\][/tex]
[tex]\[
\Delta H_{\text{f, products}} = -635.327 \, \text{kJ/mol}
\][/tex]
2. Calculate the total enthalpy of formation for the reactants:
[tex]\[
\Delta H_{\text{f, reactants}} = (-635.09) + (-110.525)
\][/tex]
[tex]\[
\Delta H_{\text{f, reactants}} = -745.615 \, \text{kJ/mol}
\][/tex]
3. Use the formula to find the enthalpy change of the reaction:
[tex]\[
\Delta H_{\text{rxn}} = \Delta H_{\text{f, products}} - \Delta H_{\text{f, reactants}}
\][/tex]
[tex]\[
\Delta H_{\text{rxn}} = (-635.327) - (-745.615)
\][/tex]
[tex]\[
\Delta H_{\text{rxn}} = -635.327 + 745.615
\][/tex]
[tex]\[
\Delta H_{\text{rxn}} = 110.288 \, \text{kJ/mol}
\][/tex]
Therefore, the enthalpy of the reaction is [tex]\(110.288 \, \text{kJ/mol} \)[/tex].
Looking at the multiple-choice options provided:
- [tex]\(-453.46 \, \text{kJ}\)[/tex]
- [tex]\(-226.73 \, \text{kJ}\)[/tex]
- [tex]\(226.73 \, \text{kJ}\)[/tex]
- [tex]\(453.46 \, \text{kJ}\)[/tex]
Our calculated value ([tex]\(110.288 \, \text{kJ/mol}\)[/tex]) does not exactly match any of the given multiple-choice options. This might indicate a potential error in the provided options or a specific context where certain conventions or approximations were used. However, we stand by the calculated enthalpy change based on the given enthalpy values from the table.
