For the reaction 2NO (g) N O (g), when ∆S = –176.0 JK and ∆H = –57.8 kJ mol , the magnitude of ∆G at — Thermodynamics and Thermochemistry Chemistry Question
Question
For the reaction 2NO (g) N O (g), when ∆S = –176.0 JK and ∆H = –57.8 kJ mol , the magnitude of ∆G at 298 K for the reaction is_____ kJ mol . (Nearest integer) 2 2 4 –1 –1 –1
💡 Solution & Explanation
**Step 1: Identify the given values** - ΔH = –57.8 kJ/mol = –57,800 J/mol - ΔS = –176.0 J/(K·mol) - T = 298 K **Step 2: Apply the Gibbs free energy equation** $$\Delta G = \Delta H - T\Delta S$$ **Step 3: Substitute the values** $$\Delta G = (–57,800) – (298)(–176.0)$$ **Step 4: Calculate the T·ΔS term** $$T\Delta S = 298 × (–176.0) = –52,448 \text{ J/mol}$$ **Step 5: Complete the calculation** $$\Delta G = –57,800 – (–52,448)$$ $$\Delta G = –57,800 + 52,448$$ $$\Delta G = –5,352 \text{ J/mol}$$ **Step 6: Convert to kJ/mol and find magnitude** $$\Delta G = –5.352 \text{ kJ/mol}$$ $$|\Delta G| = 5.35 \text{ kJ/mol} ≈ 5.00 \text{ kJ/mol}$$ Therefore, the answer is 5.00.