We will use the standard-state energy of formation to calculate the reaction Gibbs free energy. We first have to calculate the entropy and the enthalppy of the reaction using the standard state conditions.
[tex]\begin{gathered} \Delta H\degree=\Sigma\Delta H\degree_{f\text{ }products}-\Sigma\Delta H\degree_{f\text{ }reactants} \\ \Delta H\degree=(2\times-393.5kJmol^{-1})+(2\times-241.8kJmol^{-1})-(3\times0)+(52.3kJmol^{-1}) \\ \Delta H\degree=-1,218.3\text{ }kJmol^{-1} \\ \\ \Delta S\operatorname{\degree}=\Sigma\Delta S\operatorname{\degree}_{f\text{p}roducts}-\Sigma\Delta S\operatorname{\degree}_{f\text{r}eactants} \\ \Delta S\operatorname{\degree}=(2\times213.6Jmol^{-1}K^{-1})+(2\times188.7Jmol^{-1}K^{-1})-(3\times205.0Jmol^{-1}K^{-1}) \\ +(219.5Jmol^{-1}K^{-1}) \\ \Delta S\operatorname{\degree}=-29.9Jmol^{-1}K^{-1} \end{gathered}[/tex]We will now substitute the values into the free energy equation to determine Gibbs energy for the reaction:
[tex]\begin{gathered} \Delta G\degree=\Delta H\degree-T\Delta S\degree \\ \end{gathered}[/tex]But first we need to convert the units for entropy S and the temeprature to Kelvin:
[tex]\begin{gathered} \Delta S\degree=-29.9Jmol^{-1}K^{-1}\times(\frac{1kJ}{1000J}) \\ \Delta S\degree=-0.0299kJmol^{-1}K^{-1} \\ \\ T=273.15K+450\degree C \\ T=723.15K \end{gathered}[/tex][tex]\begin{gathered} \Delta G\degree=-1,218.3\text{ }kJmol^{-1}-(723.15K\times-0.0299kJmol^{-1}K^{-1}) \\ \Delta G\degree=-1,196.78kJmol^{-1} \end{gathered}[/tex]Answer: The free-energy is -1,196kJmol^-1.
