Question

Determine the entropy change for the combustion of gaseous propane, C3H8, under standard state conditions to give gaseous carbon dioxide and water.

Answer 1

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