Answer:
The standard entropy change for the reaction -93.12 J/K.mol
Explanation:
The Standard entropy change:
[tex]\Delta S^{0} = \Sigma nS^{0}(products) - \Sigma mS^{0}(reactants)[/tex]
ΔS° = [42.59] - [33.17 + (205.07/2)] J/K.mol
ΔS° = [42.59] - [33.17 + 102.535] J/K.mol
ΔS° = [42.59] - [135.705] J/K.mol
ΔS° = - 93.12 J/K.mol
The study of chemicals and bonds is called chemistry.
The correct answer to the question is -93.12 J/K.mol
The Standard entropy change is stated as follows:-
Δ[tex]S^o = S^o_{products} - S^o_{reactants}[/tex]
Δ[tex]S^o = [42.59] - [33.17 + \frac{205.07}{2}] J/K.mol[/tex]
Δ[tex]S^o = [42.59] - [33.17 + 102.535] J/K.mol[/tex]
Δ[tex]S^o = [42.59] - [135.705] J/K.mol[/tex]
Δ[tex]S^o = - 93.12 J/K.mol[/tex]
Hence, the correct answer is -93.12j/K.mol
For more information, about the entropy, refer to the link:-
https://brainly.com/question/13146879
