A + B C + D. If finally the concentration of A and B are both equal but at equilibrium concentration — Chemical Equilibrium Chemistry Question
Question
A + B C + D. If finally the concentration of A and B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.
💡 Solution & Explanation
# Solution Let me set up the equilibrium problem systematically. **Initial Setup:** Let initial concentrations of A and B both be $a$ mol/L. **At Equilibrium:** - Concentration of A = Concentration of B (given) - Concentration of D = 2 × Concentration of A (given) **From Stoichiometry:** $$A + B \rightarrow C + D$$ Let $x$ = amount of A that reacted. Then: - $[A]_{eq} = a - x$ - $[B]_{eq} = a - x$ - $[D]_{eq} = x$ - $[C]_{eq} = x$ **Using Given Condition:** Since $[D]_{eq} = 2[A]_{eq}$: $$x = 2(a-x)$$ $$x = 2a - 2x$$ $$3x = 2a$$ $$x = \frac{2a}{3}$$ **Finding Equilibrium Concentrations:** - $[A]_{eq} = a - \frac{2a}{3} = \frac{a}{3}$ - $[B]_{eq} = \frac{a}{3}$ - $[C]_{eq} = \frac{2a}{3}$ - $[D]_{eq} = \frac{2a}{3}$ **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: The equilibrium constant is 4** (Option D)