Two gaseous equilibria SO (g) O (g) (g)and 2(g) 2(g) + (g) have equilibrium constants and respective — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Write the equilibrium expressions** For the first equilibrium: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ $$K_1 = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$ For the second equilibrium: $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$ $$K_2 = \frac{[SO_2]^2[O_2]}{[SO_3]^2}$$ **Step 2: Identify the relationship** Comparing the two expressions, we observe that: $$K_2 = \frac{1}{K_1}$$ or equivalently: $$K_1 \cdot K_2 = 1$$ **Step 3: Reasoning** The second equilibrium is the **reverse** of the first equilibrium. When a reaction is reversed, the equilibrium constant of the reversed reaction equals the reciprocal of the original reaction's constant. This is a fundamental principle in chemical equilibrium. **Answer:** $K_1 \cdot K_2 = 1$ (or $K_2 = \frac{1}{K_1}$)