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 reaction 1: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ $$K_1 = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$ For reaction 2: $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$ $$K_2 = \frac{[SO_2]^2[O_2]}{[SO_3]^2}$$ **Step 2: Identify the relationship** Reaction 2 is the **reverse** of reaction 1. When a reaction is reversed, the new equilibrium constant equals the reciprocal of the original: $$K_2 = \frac{1}{K_1}$$ Or equivalently: $$K_1 \cdot K_2 = 1$$ **Step 3: Conclusion** The correct relationship is **$K_1 \cdot K_2 = 1$** (or $K_2 = \frac{1}{K_1}$) This is option C. The reverse reaction always has an equilibrium constant that is the reciprocal of the forward reaction's constant.