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: Finding Equilibrium Constant **Setting up the ICE table:** Let initial concentrations of A and B = $a$ mol/L each. $$\begin{array}{c|cccc} & A & B & C & D \\ \hline \text{Initial} & a & a & 0 & 0 \\ \text{Change} & -x & -x & +x & +x \\ \text{Equilibrium} & a-x & a-x & x & x \end{array}$$ **Using the given condition:** "Final concentration of A and B are **equal**" ✓ (Both are $a-x$) "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:** - $[A]_{eq} = a - \frac{2a}{3} = \frac{a}{3}$ - $[B]_{eq} = \frac{a}{3}$ - $[C]_{eq} = \frac{2a}{3}$ - $[D]_{eq} = \frac{2a}{3}$ ✓ (Check: $\frac{2a}{3} = 2 \times \frac{a}{3}$ ✓) **Calculating $K_c$:** $$K_c = \frac{[C][D]}{[A][B]} = \frac{\frac{2a}{3} \times \frac{2a}{3}}{\frac{a}{3} \times \frac{a}{3}} = \frac{\frac{4a^2}{9}}{\frac{a^2}{9}} = \boxed{4}$$ The equilibrium constant is **4** (Option D).