Respuesta :
Answer: The value of [tex]\Delta S^o[/tex] for the given reaction is -620.1 J/Kmol.
Explanation:
Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.
Mathematically,
[tex]\Delta S_{rxn}=\sum [n\times S^o_{products}]-\sum [n\times S^o_{reactants}][/tex]
For the given chemical equation:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)[/tex]
Taking the standard entropy of formation for the following:
[tex]S^o_{C_2H_6}=229.6Jmol^{-1}K^{-1}\\S^o_{CO_2}=213.6Jmol^{-1}K^{-1}\\S^o_{H_2O}=69.9Jmol^{-1}K^{-1}\\S^o_{O_2}=205Jmol^{-1}K^{-1}[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(4\times S^o_{CO_2})+(6\times S^o_{H_2O})]-[(2\times S^o_{C_2H_6})+(7\times S^o_{O_2})][/tex]
[tex]\Delta S^o=[(4\times 213.6)+(6\times 69.9)]-[(2\times 229.6)+(7\times 205)]=-620.1Jmol^{-1}K^{-1}[/tex]
Hence, the [tex]\Delta S^o[/tex] of the reaction is [tex]-620.1Jmol^{-1}K^{-1}[/tex]
The standard entropy change for the complete combustion of ethane is -620.8 J/K.mol.
What is standard entropy change?
The standard entropy change is equal to the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants.
Let's consider the combustion of ethane.
2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l)
The standard entropy change (ΔS°) is:
ΔS° = n(products) × S°products - n(reactants) × S°reactants
where,
- n is the stoichiometric coefficient.
- S° is the standard entropy.
ΔS° = 4 × 213.74 J/mol.K + 6 × 69.91 J/mol.K - 2 × 229.60 J/mol.K - 7 × 205.14 J/mol.K = -620.8 J/K.mol
The standard entropy change for the complete combustion of ethane is -620.8 J/K.mol.
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