The reaction is studied in a one-litre vessel at . The initial concentration of was and that of was — Chemical Equilibrium Chemistry Question
Question
The reaction $A(g) + B(g) \rightleftharpoons C(g) + D(g)$ is studied in a one-litre vessel at $250^\circ\text{C}$. The initial concentration of $A$ was $3n$ and that of $B$ was $n$. When equilibrium was attained, the equilibrium concentration of $C$ was found to be strictly equal to the equilibrium concentration of $B$. What is the equilibrium constant $K_c$ for the reaction?
💡 Solution & Explanation
Initial concentrations: $[A]_0 = 3n$, $[B]_0 = n$, $[C]_0 = 0$, $[D]_0 = 0$. At equilibrium, let $x$ moles react. $[A] = 3n - x$, $[B] = n - x$, $[C] = x$, $[D] = x$. We are given $[C] = [B] \Rightarrow x = n - x \Rightarrow 2x = n \Rightarrow x = 0.5n$. Equilibrium concentrations: $[A] = 3n - 0.5n = 2.5n$, $[B] = 0.5n$, $[C] = 0.5n$, $[D] = 0.5n$. $K_c = \frac{[C][D]}{[A][B]} = \frac{(0.5n)(0.5n)}{(2.5n)(0.5n)} = \frac{0.5}{2.5} = \frac{1}{5} = 0.2$.