The standard free energy change ( ) for 50% dissociation of N O into NO at 27 C and 1 atm pressure i — Chemical Equilibrium Chemistry Question
Question
The standard free energy change ( ) for 50% dissociation of N O into NO at 27 C and 1 atm pressure is –x J mol . The value of x is _____. (Nearest Integer) [Given: R = 8.31 J K mol , log 1.33 = 0.1239 , ln 10 = 2.3] 2 4 2 o –1 –1 –1
💡 Solution & Explanation
# Solution: ΔG° for 50% Dissociation of N₂O₄ into NO₂ **Step 1: Set up the equilibrium** For N₂O₄ ⇌ 2NO₂ with 50% dissociation: - Initial: 1 mole N₂O₄ - At equilibrium: 0.5 mole N₂O₄ and 1 mole NO₂ - Total moles = 1.5 **Step 2: Calculate partial pressures** - P(N₂O₄) = (0.5/1.5) × 1 = 1/3 atm - P(NO₂) = (1/1.5) × 1 = 2/3 atm **Step 3: Calculate Kₚ** $$K_p = \frac{[P(NO_2)]^2}{P(N_2O_4)} = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} = 1.33 \text{ atm}$$ **Step 4: Use the ΔG° formula** $$\Delta G° = -RT \ln K_p$$ Convert Kₚ to natural logarithm: $$\ln K_p = \ln(1.33) = \ln(10) × \log(1.33)$$ $$= 2.3 × 0.1239 = 0.285$$ **Step 5: Calculate ΔG°** $$\Delta G° = -8.31 × 300 × 0.285$$ $$= -2493 × 0.285 = -710.5 \text{ J/mol}$$ Therefore, the answer is **710**.