Answer:
the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.
Explanation:
3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)
∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants
∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]
∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K
∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.
For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,
then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.
