COF₂(g) passed over catalyst at 1000°C comes to equilibrium: 2COF₂(g) ⇌ CO₂(g) + CF₄(g). Analysis of — Chemical Equilibrium Chemistry Question
Question
COF₂(g) passed over catalyst at 1000°C comes to equilibrium: 2COF₂(g) ⇌ CO₂(g) + CF₄(g). Analysis of the equilibrium mixture (after quick cooling to freeze the equilibrium) shows that 500 ml of the equilibrium mixture (STP) contains 300 ml (STP) of (COF₂ + CO₂) taking the total pressure to be 10 atm. The value of K_p for the reaction is
💡 Solution & Explanation
Total volume = 500 ml. Volume of (COF₂ + CO₂) = 300 ml. Therefore, Volume of CF₄ = 500 - 300 = 200 ml. From the stoichiometry, 1 mole of CO₂ is formed for every 1 mole of CF₄. Thus, Volume of CO₂ = Volume of CF₄ = 200 ml. Volume of COF₂ = 300 - 200 = 100 ml. Since volume is proportional to moles at STP, the mole ratio COF₂ : CO₂ : CF₄ is 100 : 200 : 200 = 1 : 2 : 2. K_p = (P_CO₂ × P_CF₄) / (P_COF₂)² = (X_CO₂ × X_CF₄) / (X_COF₂)² = (2 × 2) / (1)² = 4. Therefore, correct answer is 4.