Chemistry
Chemistry
Problem
Sulfur trioxide decomposes according to 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g) ΔH° = +198 kJ At 800 K the equilibrium partial pressures in a rigid container are P(SO₃) = 0.80 atm, P(SO₂) = 0.10 atm, P(O₂) = 0.05 atm. The equilibrium constants are Kₚ(800 K) = 1.2 × 10⁻² and Kₚ(900 K) = 4.8 × 10⁻². The temperature is rapidly increased to 900 K while the volume of the container is doubled instantaneously. No reaction occurs during the brief interval of these changes. (a) Calculate the numerical value of Q immediately after the temperature increase and volume expansion, using the new partial pressures that exist at that instant. (b) Compare the value of Q obtained in part (a) with Kₚ(900 K). Predict the direction in which the system will shift to re-establish equilibrium at 900 K and justify your answer. (c) Will the equilibrium partial pressure of O₂ at 900 K be greater than, less than, or equal to 0.025 atm (its value immediately after the volume change)? Provide a detailed explanation citing both ΔH° and the relative magnitudes of Q and Kₚ.Intuitive explanation
When you suddenly change both temperature and volume, the system is no longer at equilibrium—you need to track what happens to relative to the new to predict which direction the reaction shifts. The key is recognizing that doubling the volume immediately cuts all partial pressures in half (before any reaction), giving you a new to compare against at 900 K. Since the reaction is endothermic, higher temperature favors products, so increases—but you have to see whether the pressure change or temperature change dominates in pushing the system back to equilibrium.
Intuitive explanation
When you suddenly change both temperature and volume, the system is no longer at equilibrium—you need to track what happens to relative to the new to predict which direction the reaction shifts. The key is recognizing that doubling the volume immediately cuts all partial pressures in half (before any reaction), giving you a new to compare against at 900 K. Since the reaction is endothermic, higher temperature favors products, so increases—but you have to see whether the pressure change or temperature change dominates in pushing the system back to equilibrium.