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: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$ $$K_1 = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$ For Reaction 2: $2SO_3(g) \rightarrow 2SO_2(g) + O_2(g)$ $$K_2 = \frac{[SO_2]^2[O_2]}{[SO_3]^2}$$ **Step 2: Identify the relationship between reactions** Reaction 2 is the exact reverse of Reaction 1. **Step 3: Apply the reverse reaction rule** When a reaction is reversed, its equilibrium constant becomes the reciprocal of the original: $$K_2 = \frac{1}{K_1}$$ Or equivalently: $$K_1 \cdot K_2 = 1$$ **Answer:** $K_1 \cdot K_2 = 1$ (or $K_2 = \frac{1}{K_1}$) This is correct because the two equilibria are exact opposites—Reaction 2 consumes the products of Reaction 1. At the same temperature, the equilibrium constants must be reciprocals of each other.