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

The entropy change for the combustion of gaseous propane, C3H8, can be determined using the equation: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g).

Step-by-step explanation:

The entropy change for the combustion of gaseous propane, C3H8, under standard state conditions to give gaseous carbon dioxide and water can be determined using the equation:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Entropy change (ΔS) can be calculated using the equation:

ΔS = ΣnS(products) - ΣnS(reactants)

Where n represents the stoichiometric coefficients and S represents the molar entropy of each substance.

Using the molar entropies of the substances involved in the equation, you can calculate the entropy change for the combustion reaction of propane.

Answer 2

ΔS°reaction = 100.9 J K⁻¹ (mol C₃H₈)⁻¹

Step-by-step 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