Question

For the reaction:

$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$

The initial concentrations of $A$ and $B$ are equal. The equilibrium concentration of $C$ is two times the equilibrium concentration of $A$. What is the value of the equilibrium constant?

Answer 1

For the reaction:
$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$

The initial concentrations of $A$ and $B$ are equal.
$[A{]}_0=[B{]}_0=x$ M (say)

The equilibrium concentrations of $A$ and $B$ are-
$[A{]}_{eq}=[B{]}_{eq}=x-y$ M.

The equilibrium concentrations of $C$ and $D$ are-
$[C{]}_{eq}=[D{]}_{eq}=y$ M.
The equilibrium concentration of $C$ is two times the equilibrium concentration of $A$.
$[C{]}_{eq}=2[A{]}_{eq}$
$y=2(x-y)$
$y=2x-2y$
$3y=2x$
$x=1.5y$
The equilibrium concentrations of $A$ and $B$ are-
$[A{]}_{eq}=[B{]}_{eq}=1.5y-y=0.5y$ M.
The equilibrium concentrations of $C$ and $D$ are-
$[C{]}_{eq}=[D{]}_{eq}=y$ M.
The value of equilibrium constant is $\frac{\left[C\right]\times \left[D\right]}{\left[A\right]\times \left[B\right]}=\frac{y\times y}{0.5y\times 0.5y}=4$.