Answer: The [tex]\Delta S^o[/tex] of the reaction is [tex]179Jmol^{-1}K^{-1}[/tex]
Explanation:
Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.
Mathematically,
[tex]\Delta S_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}][/tex]
For the given chemical equation:
[tex]Al_2O_3(s)+3H_2(g)\rightarrow 2Al(s)+3H_2O(g)[/tex]
We are given:
[tex]\Delta S^o_{Al_2O_3}=51Jmol^{-1}K^{-1}\\\Delta S^o_{H_2}=131Jmol^{-1}K^{-1}\\\Delta S^o_{Al}=28Jmol^{-1}K^{-1}\\\Delta S^o_{H_2O}=189Jmol^{-1}K^{-1}[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{H_2O})+(3\times \Delta S^o_{H_2O})]-[(1\times \Delta S^o_{Al_2O_3})+(3\times \Delta S^o_{H_2})][/tex]
[tex]\Delta S^o=[(2\times 28)+(3\times 189)]-[(1\times 51)+(3\times 131)]=179Jmol^{-1}K^{-1}[/tex]
Hence, the [tex]\Delta S^o[/tex] of the reaction is [tex]179Jmol^{-1}K^{-1}[/tex]
