Answer: [tex]\Delta S^{0}[/tex] for the reaction is -186.75 J/K
Explanation:
Change in entropy ([tex]\Delta S^{0}[/tex]) for the given reaction under standard condition is given by-
[tex]\Delta S^{0}[/tex]= [tex][3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}][/tex]
So [tex]\Delta S^{0}[/tex] = [tex][3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol][/tex] = -186.75 J/K
