To find the standard Gibbs free energy change for the reaction [tex]\( \Delta G_{ \text{rxn} } \)[/tex], we need to use the Gibbs free energies of formation of the reactants and products. The reaction given is:
[tex]\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \][/tex]
The given Gibbs free energies of formation ([tex]\( \Delta G_{f} \)[/tex]) are:
- [tex]\( \Delta G_{f,\text{CaCO}_3} = -1128.76 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta G_{f,\text{CaO}} = -604.17 \, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta G_{f,\text{CO}_2} = -394.4 \, \text{kJ/mol} \)[/tex]
The standard Gibbs free energy change for the reaction ([tex]\( \Delta G_{ \text{rxn} } \)[/tex]) can be calculated using the following formula:
[tex]\[ \Delta G_{ \text{rxn} } = \Delta G_{ \text{products} } - \Delta G_{ \text{reactants} } \][/tex]
In this case:
[tex]\[ \Delta G_{ \text{products} } = \Delta G_{f,\text{CaO}} + \Delta G_{f,\text{CO}_2} \][/tex]
[tex]\[ \Delta G_{ \text{reactants} } = \Delta G_{f,\text{CaCO}_3} \][/tex]
Substituting the given values:
[tex]\[ \Delta G_{ \text{products} } = (-604.17) + (-394.4) \][/tex]
[tex]\[ \Delta G_{ \text{products} } = -998.57 \, \text{kJ/mol} \][/tex]
Now, we calculate [tex]\( \Delta G_{ \text{rxn} } \)[/tex]:
[tex]\[ \Delta G_{ \text{rxn} } = -998.57 - (-1128.76) \][/tex]
[tex]\[ \Delta G_{ \text{rxn} } = -998.57 + 1128.76 \][/tex]
[tex]\[ \Delta G_{ \text{rxn} } = 130.19 \, \text{kJ/mol} \][/tex]
Lastly, we round the answer to the nearest whole number:
[tex]\[ \Delta G_{ \text{rxn} } \approx 130 \, \text{kJ/mol} \][/tex]
Thus, the standard Gibbs free energy change for the reaction, [tex]\( \Delta G_{ \text{rxn} } \)[/tex], is [tex]\( 130 \, \text{kJ/mol} \)[/tex].
