At 600K, 2 mol of NO are mixed with 1 mol of O . 2NO + O 2NO The reaction occurring as above comes t — Chemical Equilibrium Chemistry Question
Question
At 600K, 2 mol of NO are mixed with 1 mol of O . 2NO + O 2NO The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is _______. (Nearest integer). 2 (g) 2(g) 2(g)
💡 Solution & Explanation
**Step 1: Set up an ICE table** Reaction: 2NO(g) + O₂(g) ⇌ 2NO₂(g) | | 2NO | O₂ | 2NO₂ | |---|---|---|---| | Initial (mol) | 2 | 1 | 0 | | Change (mol) | -2x | -x | +2x | | Equilibrium (mol) | 2-2x | 1-x | 2x | **Step 2: Find x using the given equilibrium data** At equilibrium, O₂ = 0.6 mol 1 - x = 0.6 x = 0.4 mol **Step 3: Calculate equilibrium moles** - NO: 2 - 2(0.4) = 1.2 mol - O₂: 0.6 mol - NO₂: 2(0.4) = 0.8 mol - Total: 2.6 mol **Step 4: Calculate mole fractions and partial pressures** At 1 atm total pressure: - P(NO) = (1.2/2.6) × 1 = 0.462 atm - P(O₂) = (0.6/2.6) × 1 = 0.231 atm - P(NO₂) = (0.8/2.6) × 1 = 0.308 atm **Step 5: Apply the equilibrium constant expression** Kₚ = [P(NO₂)]² / ([P(NO)]² × P(O₂)) Kₚ = (0.308)² / [(0.462)² × 0.231] Kₚ = 0.0949 / (0.214 × 0.231) = 0.0949 / 0.0494 = 1.92 ≈ 2.00 Therefore, the answer is 2.00.