Question

At $1000K$, the equilibrium constant, $K_c$ for the reaction,

$CO\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons CO{Cl}_2\left(g\right)$,

is equal to $3.04$. If $1$ mole of $CO$ and $1$ mole of ${Cl}_2$ are introduced into a $1$ litre box at $1000K$, what will be the final concentration of $CO{Cl}_2$ at equilibrium?

Answer 1

$CO\left(g\right)+C{l}_2\rightleftharpoons COC{l}_2$

Initial moles 1 1 0

At equilibrium $1-x$ $1-x$ $x$

The equilibrium concentrations are :

$\left[CO\right]=\frac{\text{( 1 -x)mol}}{\text{1 L}}=\text{1-x M}$
$\left[C{l}_2\right]=\frac{\text{( 1 -x )mol}}{\text{1 L}}=\text{1-x M}$
$\left[COC{l}_2\right]=\frac{\text{x mol}}{\text{1 L}}=\text{x M}$
$K_c=\frac{\left[CO\right]\left[C{l}_2\right]}{\left[COC{l}_2\right]}$
$3.04=\frac{\text{1-x M}\times \text{1-x M}}{\text{x M}}$
$3.04x=1(1-x)-x(1-x)$
$3.04x=1-x-x+x^2$
$3.04x=1-2x+x^2$
$0=1-5.04x+x^2$
$x^2-5.04x+1=0$

This is quadratic equation with solution

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-5.04)\pm \sqrt{(-5.04{)}^2-4(1\left)\right(1)}}{2\left(1\right)}$
$x=\frac{5.04\pm 4.63}{2}$
$x=\frac{5.04\pm 4.63}{2}$
$x=4.833$ or $x=0.207$
The value $x=4.833$ is discarded as it will lead to negative value of concentration.

Hence, $x=0.207$

The final concentration of $COC{l}_2$ at equilibrium is
$\text{x M}=\text{0.207 M}$