For a reversible reaction: A <=(K1/K2)=> B, the initial molar concentration of A and B are a M and b — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Construct the ICE Table For the reversible reaction: $\ce{A <=>[K_1][K_2] B}$ Initial concentrations: $[\ce{A}]_0 = a\text{ M}$, $[\ce{B}]_0 = b\text{ M}$. Let $x\text{ M}$ of A react to reach equilibrium. \[\begin{array}{lccc} \text{Species} & \ce{A} & \ce{<=>} & \ce{B} \ \hline \text{Initial (M)} & a & & b \ \text{Change (M)} & -x & & +x \ \text{Equilibrium (M)} & a-x & & b+x \ \end{array}\] Step 2 - Apply the Kinetic Equilibrium Condition At equilibrium, the forward rate equals the backward rate: \[r_f = r_b \implies K_1[\ce{A}] = K_2[\ce{B}]\] Substituting equilibrium concentrations: \[K_1(a - x) = K_2(b + x)\] Step 3 - Solve for x Expanding: \[K_1 a - K_1 x = K_2 b + K_2 x\] \[K_1 a - K_2 b = K_1 x + K_2 x = x(K_1 + K_2)\] \[x = \frac{K_1 a - K_2 b}{K_1 + K_2}\] Step 4 - Evaluate the Options * **(A) $\frac{K_1 a - K_2 b}{K_1 + K_2}$**: Correct. Derived by equating forward and backward rates at equilibrium. * **(B) $\frac{K_1 a - K_2 b}{K_1 - K_2}$**: Incorrect. Wrong denominator — algebraic error in grouping x terms. * **(C) $\frac{K_2 a - K_1 b}{K_1 + K_2}$**: Incorrect. Numerator has reversed rate constants. * **(D) $\frac{K_1 a + K_2 b}{K_1 + K_2}$**: Incorrect. Plus sign in numerator — incorrect sign manipulation. \[\boxed{\text{A}}\]