Determine $\Delta$S° for the reaction SO₃(g)+ H₂O(l)$\rarr$H₂SO₄(l). A)169.2 J/K·mol B)1343.2 J/K·mol C)-169.2 J/K·mol D)-29.4 J/K·mol E)29.4 J/K·mol

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the formula ∆G = ∆H - T∆S, where ∆G is the change in Gibbs free energy, ∆H is the change in enthalpy, ∆S is the change in entropy, and T is the temperature in Kelvin. We can find ∆H and ∆S from the balanced equation: SO₃(g) + H₂O(l) → H₂SO₄(l) The coefficients in the balanced equation give the stoichiometry of the reaction, so we can use them to find the change in enthalpy and entropy per mole of reaction. ∆H°f(SO₃,g) = 45.05 kJ/mol ∆H°f(H₂O,l) = -469.15 kJ/mol ∆H°f(H₂SO₄,l) = -186.17 kJ/mol ∆H°r = Σ∆H°f(products) - Σ∆H°f(reactants) ∆H°r = (-186.17 kJ/mol) - [(1 mol)(45.05 kJ/mol) + (1 mol)(-469.15 kJ/mol)] ∆H°r = -169.27 kJ/mol ∆S°rxn = ΣS°(products) - ΣS°(reactants) ∆S°rxn = [(1 mol)(205.97 J/K·mol) + (1 mol)(69.91 J/K·mol)] - [(1 mol)(207.21 J/K·mol) + (1 mol)(130.59 J/K·mol)] ∆S°rxn = -29.38 J/K·mol Substituting these values into the ∆G equation gives: ∆G° = -169.27 kJ/mol - (298 K)(-29.38 J/K·mol/1000 J/kJ) ∆G° = -169.27 kJ/mol + 8.78 kJ/mol ∆G° = -160.49 kJ/mol Finally, we convert ∆G° from kJ/mol to J/K·mol by multiplying by 1000 and dividing by the temperature in Kelvin: ∆G°/T = (-160.49 kJ/mol)(1000 J/kJ)/(298 K) = -539.5 J/K·mol Therefore, the answer is (C) -169.2 J/K·mol.

Determine Δ\DeltaΔ S° for the Reaction SO3(g)+ H2O(l) →\rarr→ H2SO4(l)

Determine $\Delta$ S° for the Reaction SO₃(g)+ H₂O(l) $\rarr$ H₂SO₄(l)