Suppose the gas phase isomerization reactions: A ⇌ B, A ⇌ C and B ⇌ C achieve equilibrium simultaneo — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the three equilibrium reactions and their constants The three isomerization reactions all involve 1 mole on each side: \[A \rightleftharpoons B \quad (K_1), \quad A \rightleftharpoons C \quad (K_2), \quad B \rightleftharpoons C \quad (K_3)\] Step 2 - Express [B] and [C] in terms of [A] using K1 and K2 \[K_1 = \frac{[B]}{[A]} \implies [B] = K_1[A]\] \[K_2 = \frac{[C]}{[A]} \implies [C] = K_2[A]\] Note: The three equilibria are not independent; K3 = [C]/[B] = K2/K1, so K3 is determined by K1 and K2. Step 3 - Find the total moles and mole fraction of A \[\text{Total} = [A] + [B] + [C] = [A] + K_1[A] + K_2[A] = [A](1 + K_1 + K_2)\] \[x_A = \frac{[A]}{[A](1 + K_1 + K_2)} = \boxed{\frac{1}{1 + K_1 + K_2}}\] Step 4 - Evaluate all options - **Option (A) K1/(K1+K2+K3)**: Incorrect. This would be the mole fraction of B in a different formulation. - **Option (B) 1/(1+K1+K2)**: Correct. From the two independent equilibria K1=[B]/[A] and K2=[C]/[A], total moles = [A](1+K1+K2), giving x_A = 1/(1+K1+K2). - **Option (C) 1/(K1+K2+K3)**: Incorrect. Missing the "1" term for [A] in the denominator; K3 is also redundant. - **Option (D) 1/(K1+K1*K2)**: Incorrect. Incorrect algebraic manipulation.