The value of ΔS° for reaction is - 22.2 J/K.mol
[tex]SO_{2}(s)+NO_{2} (g)[/tex] → [tex]SO_{3}(g)+NO(g)[/tex]
Calculation,
Given value of S°(J/K.mol) for
[tex]SO_{2}(s)[/tex] = 248.5
[tex]NO_{2} (g)[/tex] = 240.5
[tex]NO(g)[/tex] = 210.6
[tex]SO_{3}(g)[/tex] = 256.2
Formula used:
ΔS° (Reaction) = ∑S°(Product) - ∑S°(Reactant)
ΔS° = (256.2 + 210.6 ) - ( 248.5 + 240.5) = 466.8 - 489 = - 22.2 J/K.mol
The change in stander entropy of reaction is - 22.2 J/K.mol. The negative sign indicates the that entropy of reaction is decreases when reactant converted into product.
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