Answer: The [tex]\Delta S^o[/tex] of the reaction is [tex]-579JK^{-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.
The equation representing entropy change of the reaction follows:
[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]4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)[/tex]
We are given:
[tex]\Delta S^o_{NH_3}=192Jmol^{-1}K^{-1}\\\Delta S^o_{O_2}=205.1Jmol^{-1}K^{-1}\\\Delta S^o_{N_2}=192Jmol^{-1}K^{-1}\\\Delta S^o_{H_2O}=70Jmol^{-1}K^{-1}[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(6\times \Delta S^o_{H_2O})+(2\times \Delta S^o_{N_2})]-[(4\times \Delta S^o_{NH_3})+(3\times \Delta S^o_{O_2})][/tex]
[tex]\Delta S^o_{rxn}=[(6\times 70)+(2\times 192)]-[(4\times 192)+(3\times 205.1)]=-579JK^{-1}[/tex]
Hence, the [tex]\Delta S^o[/tex] of the reaction is [tex]-579JK^{-1}[/tex]
