A + B C + D. If finally the concentration of A and B are both equal but at equilibrium concentration — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution Let me set up the equilibrium problem systematically. **Initial Setup:** - Reaction: $A + B \rightarrow C + D$ - Let initial concentration of A = $a$ and initial concentration of B = $b$ - Let $x$ = amount of A reacted at equilibrium **At Equilibrium:** $$[A] = a - x$$ $$[B] = b - x$$ $$[D] = x$$ **Using Given Conditions:** **Condition 1:** Final concentrations of A and B are equal $$a - x = b - x$$ $$a = b$$ So A and B started with equal concentrations. **Condition 2:** Concentration of D is twice that of A $$[D] = 2[A]$$ $$x = 2(a-x)$$ $$x = 2a - 2x$$ $$3x = 2a$$ $$x = \frac{2a}{3}$$ **Finding Equilibrium Concentrations:** $$[A] = a - \frac{2a}{3} = \frac{a}{3}$$ $$[B] = a - \frac{2a}{3} = \frac{a}{3}$$ $$[D] = \frac{2a}{3}$$ $$[C] = \frac{2a}{3}$$ (by stoichiometry) **Equilibrium Constant:** $$K_c = \frac{[C][D]}{[A][B]} = \frac{\frac{2a}{3} \times \frac{2a}{3}}{\frac{a}{3} \times \frac{a}{3}} = \frac{\frac{4a^2}{9}}{\frac{a^2}{9}} = 4$$ **Answer: $K_c = 4$** (Option D)