Given the following equation, N₂O(g) + NO₂(g) → 3NO(g) Δ_rG° = -23.0 kJ mol^-1
Calculate Δ_rG° for the following reaction:
3N₂O(g) + 3NO₂(g) → 9NO(g)

Question

Given the following equation, N₂O(g) + NO₂(g) → 3NO(g) Δ_rG° = -23.0 kJ mol^-1 Calculate Δ_rG° for the following reaction: 3N₂O(g) + 3NO₂(g) → 9NO(g)

Quizplus · Accepted Answer

The given equation has a Δ_rG° of -23.0 kJ mol^-1, and the new equation has three times the amount of reactants and products. According to the equation Δ_rG° = Σ_rG°(products) - Σ_rG°(reactants), the entropy change will also be tripled for the new equation. Therefore, Δ_rG° for the new equation is -23.0 kJ mol^-1 x 3 = -69.0 kJ mol^-1.

Given the Following Equation, N2O(g) + NO2(g) → 3NO(g) ΔrG°

Given the Following Equation, N₂O(g) + NO₂(g) → 3NO(g) Δ_rG°