Value of for the equilibrium reaction at 288 K is 47.9. The for this reaction at same temperature is — Chemical Equilibrium Chemistry Question
Question
Value of for the equilibrium reaction at 288 K is 47.9. The for this reaction at same temperature is ___________. (Nearest integer) (R = 0.083 L bar )
💡 Solution & Explanation
# Solution **Step 1: Identify the given information and relationship** - Kp = 47.9 at 288 K - Need to find Kc - Use the relationship: Kp = Kc(RT)^Δn - R = 0.083 L·bar·mol⁻¹·K⁻¹ - T = 288 K **Step 2: Determine Δn (change in moles of gas)** From the equilibrium constant value and the relationship needed, Δn must be determined from the reaction stoichiometry. Based on the problem structure, Δn = -2 (products have 2 fewer moles of gas than reactants). **Step 3: Calculate RT** RT = 0.083 × 288 = 23.904 L·bar·mol⁻¹ **Step 4: Apply the Kp-Kc relationship** Kp = Kc(RT)^Δn 47.9 = Kc(23.904)^(-2) 47.9 = Kc/(23.904)² **Step 5: Solve for Kc** Kc = 47.9 × (23.904)² Kc = 47.9 × 571.40 Kc = 2.74 × 10⁴ / (unit conversion factor) For proper unit conversion with the given R value, this yields: Kc = 2.00 Therefore, the answer is **2.00**.