Question

Equilibrium constants are given for the following two equilibria: \[ (i)\; A_2(g) + B_2(g) \rightleftharpoons 2AB(g); \quad K_1 = 2 \times 10^{-4} \] \[ (ii)\; 2AB(g) + C_2(g) \rightleftharpoons 2ABC(g); \quad K_2 = 2 \times 10^{-2}\, L\, mol^{-1} \] Calculate the equilibrium constant for: \[ ABC(g) \rightleftharpoons \frac{1}{2}A_2(g) + \frac{1}{2}B_2(g) + \frac{1}{2}C_2(g) \]

• A. $500$ mol$^{1/2}$ L$^{1/2}$
• B. $4 \times 10^{-6}$ mol$^{1/2}$ L$^{1/2}$
• C. $500$ mol$^{-1/2}$ L$^{1/2}$
• D. $200$ mol$^{1/2}$ L$^{-1/2}$
• E. $500$ mol$^{1/2}$ L$^{-1/2}$

Hint:

Always track three transformations: reversing (reciprocal), multiplying (power), and adding (product of constants).

Answer 1

Answer: (E)

Concept: When reactions are:
• Added → multiply equilibrium constants
• Reversed → take reciprocal
• Multiplied/divided → raise power accordingly

Step 1: Add given equations. (i) $A_2 + B_2 \rightarrow 2AB$ (ii) $2AB + C_2 \rightarrow 2ABC$ Adding: \[ A_2 + B_2 + C_2 \rightarrow 2ABC \]

Step 2: Multiply constants. \[ K = K_1 \times K_2 = (2 \times 10^{-4})(2 \times 10^{-2}) = 4 \times 10^{-6} \]

Step 3: Required reaction is reverse and halved. Reverse: \[ 2ABC \rightarrow A_2 + B_2 + C_2 \Rightarrow K = \frac{1}{4 \times 10^{-6}} = 2.5 \times 10^5 \] Now divide by 2: \[ ABC \rightarrow \frac{1}{2}A_2 + \frac{1}{2}B_2 + \frac{1}{2}C_2 \]

Step 4: Take square root of K. \[ K = \sqrt{2.5 \times 10^5} = 500 \]

Step 5: Units handling. Original units: \[ K_2 = L\, mol^{-1} \Rightarrow final = mol^{1/2} L^{-1/2} \] \[ \boxed{500 \text{ mol}^{1/2} L^{-1/2}} \]