Answer :
To determine the enthalpy of reaction for the decomposition of calcium carbonate ([tex]\(CaCO_3(s)\)[/tex]), we utilize the enthalpy of formation values for the reactants and products. The enthalpy of reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) can be calculated using Hess's Law, which states that the change in enthalpy for a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
Let's break down the calculation step-by-step.
1. Identify the given enthalpy of formation values:
- Enthalpy of formation of [tex]\(CaCO_3(s)\)[/tex]: [tex]\(\Delta H_f^\circ(CaCO_3(s)) = -1206.9 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of [tex]\(CaO(s)\)[/tex]: [tex]\(\Delta H_f^\circ(CaO(s)) = -635.5 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of [tex]\(CO_2(g)\)[/tex]: [tex]\(\Delta H_f^\circ(CO_2(g)) = -393.5 \, \text{kJ/mol}\)[/tex]
2. Write the balanced chemical equation:
[tex]\[ CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \][/tex]
3. Apply Hess's Law to calculate the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \left[ \Delta H_f^\circ(CaO(s)) + \Delta H_f^\circ(CO_2(g)) \right] - \Delta H_f^\circ(CaCO_3(s)) \][/tex]
4. Substitute the values into the equation:
[tex]\[ \Delta H_{\text{reaction}} = \left[ -635.5 \, \text{kJ/mol} + (-393.5 \, \text{kJ/mol}) \right] - (-1206.9 \, \text{kJ/mol}) \][/tex]
5. Simplify the equation:
[tex]\[ \Delta H_{\text{reaction}} = \left( -635.5 - 393.5 \right) + 1206.9 \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1029.0 + 1206.9 \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = 177.9 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy of reaction for the decomposition of calcium carbonate ([tex]\(CaCO_3\)[/tex]) is [tex]\(177.9 \, \text{kJ/mol}\)[/tex].
Let's break down the calculation step-by-step.
1. Identify the given enthalpy of formation values:
- Enthalpy of formation of [tex]\(CaCO_3(s)\)[/tex]: [tex]\(\Delta H_f^\circ(CaCO_3(s)) = -1206.9 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of [tex]\(CaO(s)\)[/tex]: [tex]\(\Delta H_f^\circ(CaO(s)) = -635.5 \, \text{kJ/mol}\)[/tex]
- Enthalpy of formation of [tex]\(CO_2(g)\)[/tex]: [tex]\(\Delta H_f^\circ(CO_2(g)) = -393.5 \, \text{kJ/mol}\)[/tex]
2. Write the balanced chemical equation:
[tex]\[ CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \][/tex]
3. Apply Hess's Law to calculate the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{reaction}} = \left[ \Delta H_f^\circ(CaO(s)) + \Delta H_f^\circ(CO_2(g)) \right] - \Delta H_f^\circ(CaCO_3(s)) \][/tex]
4. Substitute the values into the equation:
[tex]\[ \Delta H_{\text{reaction}} = \left[ -635.5 \, \text{kJ/mol} + (-393.5 \, \text{kJ/mol}) \right] - (-1206.9 \, \text{kJ/mol}) \][/tex]
5. Simplify the equation:
[tex]\[ \Delta H_{\text{reaction}} = \left( -635.5 - 393.5 \right) + 1206.9 \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -1029.0 + 1206.9 \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = 177.9 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy of reaction for the decomposition of calcium carbonate ([tex]\(CaCO_3\)[/tex]) is [tex]\(177.9 \, \text{kJ/mol}\)[/tex].
