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Sagot :
Answer:
ΔS°reaction = 100.9 J K⁻¹ (mol C₃H₈)⁻¹
Explanation:
The equation for the reaction is given as;
C₃H₈(g) + 5O₂(g) → 4H₂O(g) + 3CO₂(g)
In order to determine the entropy change, we have to use the entropy valuues for the species in the reaction. This is given as;
S°[C₃H₈(g)] = 269.9 J K⁻¹ mol⁻¹
S°[O₂(g)] = 205.1 J K⁻¹ mol⁻¹
S°[H₂O(g)] = 188.8 J K⁻¹ mol⁻¹
S°[CO₂(g)] = 213.7 J K⁻¹ mol⁻¹
The unit of entropy is J K⁻¹ mol⁻¹
Entropy change for the reaction is given as;
ΔS°reaction = ΔS°product - ΔS°reactant
ΔS°reaction = [(4 * 188.8) + (3 * 213.7)] - [269.9 + (5 * 205.1)]
ΔS°reaction = 100.9 J K⁻¹ (mol C₃H₈)⁻¹
The entropy change for the combustion of gaseous propane, under standard state conditions 100.9 J K⁻¹.
The equation for the combustion of gaseous propane is given as follows:-
C₃H₈(g) + 5O₂(g) → 4H₂O(g) + 3CO₂(g)
In order to determine the entropy change, we have to use the entropy values for the species in the reaction. This is as follows:-
S°[C₃H₈(g)] = 269.9 J K⁻¹ mol⁻¹
S°[O₂(g)] = 205.1 J K⁻¹ mol⁻¹
S°[H₂O(g)] = 188.8 J K⁻¹ mol⁻¹
S°[CO₂(g)] = 213.7 J K⁻¹ mol⁻¹
The unit of entropy is J K⁻¹ mol⁻¹
Entropy change for the reaction is given as;
ΔS°reaction =nproduct ΔS°product - nreactantΔS°reactant
ΔS°reaction = [(4mol * 188.8 J K⁻¹ mol⁻¹) + (3 mol * 213.7 J K⁻¹ mol⁻¹)] - [(1 mol*269.9J K⁻¹ mol⁻¹ )+ (5 mol* 205.1 J K⁻¹ mol⁻¹)]
ΔS°reaction = 100.9 J K⁻¹
So, the entropy change for the combustion of gaseous propane is 100.9 J K⁻¹ .
To know more about:-
brainly.com/question/14818671
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