To determine the enthalpy of reaction (ΔH_reaction) for the decomposition of calcium carbonate (CaCO₃) to calcium oxide (CaO) and carbon dioxide (CO₂), we can use the given enthalpy changes of formation (ΔH_f) for each compound involved in the reaction. The reaction is:
[tex]\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \][/tex]
The enthalpy change of the reaction can be calculated using the equation:
[tex]\[ \Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants} \][/tex]
For our reaction, the products are CaO and CO₂, and the reactant is CaCO₃. Therefore, the equation becomes:
[tex]\[ \Delta H_{reaction} = [\Delta H_f(\text{CaO}) + \Delta H_f(\text{CO}_2)] - \Delta H_f(\text{CaCO}_3) \][/tex]
Let's plug in the given enthalpy values:
[tex]\[ \Delta H_f(\text{CaO}) = -157.3 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f(\text{CO}_2) = -393.5 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f(\text{CaCO}_3) = -1207.1 \, \text{kJ/mol} \][/tex]
Now, substitute these values into the equation:
[tex]\[ \Delta H_{reaction} = [(-157.3) + (-393.5)] - (-1207.1) \][/tex]
Simplify within the brackets first:
[tex]\[ \Delta H_{reaction} = -157.3 + -393.5 = -550.8 \][/tex]
Now, subtract the reactant term:
[tex]\[ \Delta H_{reaction} = -550.8 - (-1207.1) \][/tex]
We can rewrite the subtraction of a negative as an addition:
[tex]\[ \Delta H_{reaction} = -550.8 + 1207.1 \][/tex]
Finally, perform the addition:
[tex]\[ \Delta H_{reaction} = 656.3 \, \text{kJ} \][/tex]
Therefore, the enthalpy of reaction for the decomposition of calcium carbonate is:
[tex]\[ 656.3 \, \text{kJ} \][/tex]
