Sure, let's start with the information given in the table for the enthalpies of formation (ΔHf) of different compounds:
| Compound | ΔHf (kJ/mol) |
|--------------------|--------------|
| [tex]\( C_2H_2(g) \)[/tex] | 226.73 |
| [tex]\( CaCO_3(s) \)[/tex] | -1206.92 |
| [tex]\( CaO(s) \)[/tex] | -635.09 |
| [tex]\( CO(g) \)[/tex] | -110.525 |
| [tex]\( CO_2(g) \)[/tex] | -393.509 |
| [tex]\( H_2O(l) \)[/tex] | -285.8 |
| [tex]\( H_2O(g) \)[/tex] | -241.818 |
| [tex]\( C(s) \)[/tex], diamond| 1.895 |
| [tex]\( C(s) \)[/tex], graphite| 0.0 |
We are tasked with finding the enthalpy change of a specific reaction, which we assume is:
[tex]\[ C_2H_2(g) + CaO(s) \rightarrow CaCO_3(s) + CO(g) \][/tex]
To find the enthalpy change of the reaction (ΔHrxn), 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]
First, we'll sum the enthalpies of formation of the products:
Products:
- For [tex]\( CaCO_3(s) \)[/tex]: -1206.92 kJ/mol
- For [tex]\( CO(g) \)[/tex]: -110.525 kJ/mol
Summing these:
[tex]\[ \Delta H_{\text{f, products}} = (-1206.92) + (-110.525) \][/tex]
[tex]\[ \Delta H_{\text{f, products}} = -1317.445 \, \text{kJ/mol} \][/tex]
Next, we'll sum the enthalpies of formation of the reactants:
Reactants:
- For [tex]\( C_2H_2(g) \)[/tex]: 226.73 kJ/mol
- For [tex]\( CaO(s) \)[/tex]: -635.09 kJ/mol
Summing these:
[tex]\[ \Delta H_{\text{f, reactants}} = (226.73) + (-635.09) \][/tex]
[tex]\[ \Delta H_{\text{f, reactants}} = 226.73 - 635.09 \][/tex]
[tex]\[ \Delta H_{\text{f, reactants}} = -408.36 \, \text{kJ/mol} \][/tex]
Now, we can find the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{f, products}} - \Delta H_{\text{f, reactants}} \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = (-1317.445) - (-408.36) \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -1317.445 + 408.36 \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -909.085 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change of the reaction, [tex]\(\Delta H_{\text{rxn}}\)[/tex], is [tex]\(-909.085 \, \text{kJ/mol}\)[/tex]. The options given might not match this exactly, but based on the provided calculations, this is the accurate enthalpy change of the reaction.
