To calculate the enthalpy change, \(\Delta H_{\text{rxn}}\), for the reaction:
[tex]\[ \text{CaO(s)} + \text{CO}_2(\text{g}) \rightarrow \text{CaCO}_3(\text{s}) \][/tex]
we can use the given enthalpies of two related reactions:
1. \( \text{Ca(s)} + \text{CO}_2(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{CaCO}_3(\text{s}) \), \( \Delta H = -812.8 \, \text{kJ} \)
2. \( 2 \, \text{Ca(s)} + \text{O}_2(\text{g}) \rightarrow 2 \, \text{CaO(s)} \), \( \Delta H = -1269.8 \, \text{kJ} \)
First, we note that reaction 2 involves 2 moles of \(\text{Ca(s)}\) producing 2 moles of \(\text{CaO(s)}\). So, per mole of \(\text{CaO(s)}\), the enthalpy change is:
[tex]\[ \Delta H = \frac{-1269.8 \, \text{kJ}}{2} = -634.9 \, \text{kJ} \][/tex]
Now, we need to reverse this reaction to express the decomposition of \(\text{CaO(s)}\) back to \(\text{Ca(s)}\) and \(\text{O}_2\):
[tex]\[ \text{CaO(s)} \rightarrow \text{Ca(s)} + \frac{1}{2} \text{O}_2(\text{g}) \][/tex]
which changes the sign of \(\Delta H\):
[tex]\[ \Delta H = 634.9 \, \text{kJ} \][/tex]
Next, we need to combine these reactions to match the target reaction:
1. \(\text{Ca(s)} + \text{CO}_2(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{CaCO}_3(\text{s})\), \( \Delta H = -812.8 \, \text{kJ} \)
2. \(\text{CaO(s)} \rightarrow \text{Ca(s)} + \frac{1}{2} \text{O}_2(\text{g})\), \( \Delta H = 634.9 \, \text{kJ} \)
We add these reactions together to achieve the target reaction:
[tex]\[
\text{CaO(s)} + \text{CO}_2(\text{g}) \rightarrow \text{CaCO}_3(\text{s})
\][/tex]
Combining their enthalpy changes:
[tex]\[
\Delta H_{\text{rxn}} = -812.8 \, \text{kJ} + 634.9 \, \text{kJ} = -177.9 \, \text{kJ}
\][/tex]
Thus, the enthalpy change for the reaction \( \text{CaO(s)} + \text{CO}_2(\text{g}) \rightarrow \text{CaCO}_3(\text{s}) \) is:
[tex]\[
\Delta H_{\text{rxn}} = -177.9 \, \text{kJ}
\][/tex]
The enthalpy change expressed to four significant figures is:
[tex]\[
\Delta H_{\text{rxn}} = -177.9 \, \text{kJ}
\][/tex]
