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