To determine the enthalpy change for the decomposition reaction of calcium carbonate, [tex]\( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \)[/tex], we can use the enthalpies of formation (ΔH_f) for the compounds involved.
Given the reaction:
[tex]\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \][/tex]
We will use Hess's Law, which states:
[tex]\[ \Delta H_r = \Sigma (\Delta H_f \text{ products}) - \Sigma (\Delta H_f \text{ reactants}) \][/tex]
From the table, we have the following enthalpies of formation (ΔH_f):
- [tex]\( \Delta H_f (\text{CaO}(s)) = -157.3 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f (\text{CaCO}_3(s)) = -1207.1 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f (\text{CO}_2(g)) = -393.5 \text{ kJ/mol} \)[/tex]
Now, let's apply Hess's Law:
1. Identify the products:
- [tex]\( \text{CaO}(s) \)[/tex]
- [tex]\( \text{CO}_2(g) \)[/tex]
2. Sum the enthalpies of formation for the products:
[tex]\[ \Delta H_f (\text{products}) = \Delta H_f (\text{CaO}(s)) + \Delta H_f (\text{CO}_2(g)) \][/tex]
[tex]\[ \Delta H_f (\text{products}) = -157.3 \text{ kJ/mol} + (-393.5 \text{ kJ/mol}) \][/tex]
[tex]\[ \Delta H_f (\text{products}) = -550.8 \text{ kJ/mol} \][/tex]
3. Identify the reactants:
- [tex]\( \text{CaCO}_3(s) \)[/tex]
4. Sum the enthalpy of formation for the reactant:
[tex]\[ \Delta H_f (\text{reactants}) = \Delta H_f (\text{CaCO}_3(s)) \][/tex]
[tex]\[ \Delta H_f (\text{reactants}) = -1207.1 \text{ kJ/mol} \][/tex]
5. Calculate the enthalpy of reaction ([tex]\(\Delta H_r\)[/tex]):
[tex]\[ \Delta H_r = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \][/tex]
[tex]\[ \Delta H_r = -550.8 \text{ kJ/mol} - (-1207.1 \text{ kJ/mol}) \][/tex]
[tex]\[ \Delta H_r = -550.8 \text{ kJ/mol} + 1207.1 \text{ kJ/mol} \][/tex]
[tex]\[ \Delta H_r = 656.3 \text{ kJ/mol} \][/tex]
Thus, the enthalpy of reaction for the decomposition of calcium carbonate is [tex]\( 656.3 \text{ kJ/mol} \)[/tex].
