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 **Given Equilibria:** - Reaction 1: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ with $K_1$ - Reaction 2: $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$ with $K_2$ **Step 1: Identify the relationship between reactions** Reaction 2 is the **exact reverse** of Reaction 1. **Step 2: Apply the reverse reaction rule** When a reaction is reversed, the equilibrium constant becomes the reciprocal: $$K_{reverse} = \frac{1}{K_{forward}}$$ **Step 3: Write the mathematical relationship** Since Reaction 2 is the reverse of Reaction 1: $$K_2 = \frac{1}{K_1}$$ Or equivalently: $$K_1 \cdot K_2 = 1$$ **Conclusion:** The correct answer is **C**, which states that $K_1 \cdot K_2 = 1$ (or $K_2 = \frac{1}{K_1}$). This fundamental principle applies because the equilibrium constant depends on the direction of the reaction—reversing the reaction inverts the constant.