Calculate the reaction entropy (S°rxn) for the following reaction: NaCl + AgNO3 >
AgCl + NaNO3
S° (NaCl) = 72.13 J/mol•K
S° (AgNO3) = 140.9 J/mol•K
S° (AgCl) = 96 J/mol•K
S° (NaNO3) = 116.5 J/mol•K
Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.
NaCl = AgNO3 = AgCl + NaNO3
∆S° = S°product – S°reactant
∆S° = (96+116.5) - (72.35+141.04) = -0.88 J/mol•K
Answer: ∆S = -0.88 J/mol•K
Comments
Leave a comment