Given the following equilibrium constants: (i) (ii) (iii) What is the equilibrium constant for the r — Chemical Equilibrium Chemistry Question
Question
Given the following equilibrium constants: (i) $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad K_1$ (ii) $N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \quad K_2$ (iii) $H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g) \quad K_3$ What is the equilibrium constant $K_4$ for the reaction $2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$?
💡 Solution & Explanation
The target reaction can be constructed by applying transformations. Reversing reaction (i) gives $2NH_3 \rightleftharpoons N_2 + 3H_2$ with constant $1/K_1$. Adding reaction (ii) directly gives $N_2 + O_2 \rightleftharpoons 2NO$ with constant $K_2$. Multiplying reaction (iii) by $3$ gives $3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O$ with constant $K_3^3$. Adding these three manipulated reactions yields the target reaction. Thus, $K_4 = (1/K_1) \cdot K_2 \cdot K_3^3 = \frac{K_2 K_3^3}{K_1}$.