Answer:The the [tex]\Delta S_{rxn}[/tex] for the given reaction is 179.98 J/K.
Explanation:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g),\Delta S_{rxn}=?[/tex]
[tex] =192.8 J/mol K[/tex]
[tex]\Delta S_{O_2}=205.0 J/mol K[/tex]
[tex]\Delta S_{H_2O}=188.83 J/mol K[/tex]
[tex]\Delta S_{NO}=210.8 J/mol K[/tex]
[tex]\Delta S_{rxn}=(4 mol\times \Delta S_{NO}+6 mol\times \Delta S_{H_2O})-(4 mol\times \Delta S_{NH_3}-5 mol\times \Delta S_{O_2})[/tex]
[tex]\Delta S_{rxn}=(4\times 210.8 J/mol K+6\times 188.83 J/mol K)-(4\times 192.8 J/mol K+5\times 205.0 J/mol K)[/tex]
[tex]\Delta S_{rxn}=179.98 J/K[/tex]
The the [tex]\Delta S_{rxn}[/tex] for the given reaction is 179.98 J/K.
