Calculate Δ_rG° at 298 K using the following information: 4HNO₃(g) + 5N₂H₄(l) → 7N₂(g) + 12H₂O(l) Δ_rG°= ?
Δ_fH° (kJ mol^-1) -133.9 50.6 -285.8
S°(J K^-1 mol^-1) 266.9 121.2 191.6 70.0

Question

Quizplus · Accepted Answer

Δ_rG° = Δ_rH° - TΔ_rS°Δ_rH° = [7Δ_fH°(N₂) + 12Δ_fH°(H₂O)] - [4Δ_fH°(HNO₃) + 5Δ_fH°(N₂H₄)]= [7(-285.8) + 12(-285.8)] - [4(-133.9) + 5(50.6)]= -3.30 × 10³ kJ mol^-1Δ_rS° = [7S°(N₂) + 12S°(H₂O)] - [4S°(HNO₃) + 5S°(N₂H₄)]= [7(191.6) + 12(70.0)] - [4(266.9) + 5(121.2)]= -1.10 × 10³ J K^-1 mol^-1Δ_rG° = -3.30 × 10³ kJ mol^-1 - (298 K)(-1.10 × 10³ J K^-1 mol^-1)= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol^-1= -3.30 × 10³ kJ mol^-1 + 3.26 × 10³ kJ mol

Calculate ΔrG° at 298 K Using the Following Information: 4HNO3(g)

Calculate Δ_rG° at 298 K Using the Following Information: 4HNO₃(g)