Question

What is δs° for the reaction so2(s) no2(g) → so3(g) no(g)? substance s°(j/k • mol) so2(g) 248.5 so3(g) 256.2 no(g) 210.6 no2(g) 240.5

Answer 1

The value of ΔS° for reaction  is - 22.2 J/K.mol

$SO_{2}(s)+NO_{2} (g)$ → $SO_{3}(g)+NO(g)$

Calculation,

Given value of S°(J/K.mol) for

$SO_{2}(s)$ = 248.5

$NO_{2} (g)$ = 240.5

$NO(g)$ = 210.6

$SO_{3}(g)$ = 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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