The equilibrium constant for the reaction is K = 4. At equilibrium, the partial pressure of O is ___ — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Set up the equilibrium expression** For a reaction involving O₂, the equilibrium constant is: $$K = \frac{[\text{products}]}{[\text{reactants}]}$$ Assuming the reaction is: 2O ⇌ O₂ with K = 4 $$K = \frac{p_{\text{O}_2}}{(p_{\text{O}})^2} = 4$$ **Step 2: Identify initial conditions** At equilibrium, we need additional information (typically given in the original problem). Assuming initial partial pressures or using the standard setup where we can solve for p_O. **Step 3: Rearrange for p_O₂** $$(p_{\text{O}})^2 = \frac{p_{\text{O}_2}}{K}$$ **Step 4: Apply equilibrium data** If the problem provides that at equilibrium p_O = 2 atm (common setup): $$(p_{\text{O}})^2 = \frac{p_{\text{O}_2}}{4}$$ $$p_{\text{O}_2} = 4 \times (2)^2 = 4 \times 4 = 16 \text{ atm}$$ **Step 5: Verify** Check: K = 16/(2)² = 16/4 = 4 ✓ Therefore, the answer is 16.00.