Standard entropy

Question

Calculate the standard entropy change ∆S° for the reaction of liquid I so octane C₈H₁₈ with O₂(g) to give CO₂(g) and H₂O(g) at 298K

C₈H₁₈ (1) + O₂(g) -----> CO₂(g) + 9H₂O(g) Given that standard entropy (S°) for CO₂, H₂0, C₈H₁₈ and O₂ are 213.8J/mol/K, 188.8J/mol/K, and 205.2J/mol respectively.

Answer 1

C₈ H₁₈ (l) + 25/2 O₂ (g) → 8CO₂ + 9H₂ O(g)

Standard entropy change 

ΔS° = (Standard entropy of product) - (standard entropy of reactant)

Given, 

standard entropy (S°) of 

S° Co₂ = 213.8 J/mol/k

S° H₂° = 188.8 J/mol/K

S°C₈H₁₈(l) =  360 J/mol/K

S° O₂ = 205.2 J/mol/K

∴ ΔS° = (8 x 213.8) + (9x 188.8) - (360 + 25/2 x 205.2)

= (1710.4 + 1699.2) - (360 + 2565)

= 3409 .6 - 2925

 ΔS° = 484.6 J/K