Question

One mole of $H_2$, $2$ moles of $I_2$ and $3$ moles of $HI$ are injected in a one litre flask. What will be the concentration of $H_2,I_2$ and $HI$ at equilibrium when $K_c$ is $45.9$?

Answer 1

One mole of $H_2$, 2 moles of $I_2$ and 3 moles of HI are injected in a one litre flask.
x moles of $H_2$ will react with x moles of $I_2$ to form 2x moles of
HI. Total number of moles of HI present at equilibrium will be $3+2x$
$1-x$ moles of $H_2$ and $2-x$ moles
of $I_2$ will remain at equilibrium.

The equilibrium constant $K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
$45.9=\frac{[\frac{\text{3+2x mol}}{\text{1 L}}{]}^2}{\left[\frac{\text{1-x mol}}{\text{1 L}}\right]\left[\frac{\text{1-x mol}}{\text{1 L}}\right]}$
$45.9=\frac{[3+2x{]}^2}{[1-x][2-x]}$
$45.9\left(1\right[2-x]-x[2-x\left]\right)=3[3+2x]+2x[3+2x]$
$45.9(2-x-2x+x^2)=9+6x+6x+4x^2$
$91.8-137.7x+45.9x^2=9+12x+4x^2$
$41.9x^2-149.7x+82.8=0$
This is quadratic equation with solution
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-149.7)\pm \sqrt{(-149.7{)}^2-4(41.9\left)\right(82.8)}}{2\left(41.9\right)}$
$x=\frac{149.7\pm 92.4}{83.8}$
$x=\frac{149.7\pm 92.4}{83.8}$
$x=2.88$ or $x=0.684$

The value $x=2.88$ is discarded as it will lead to negative

value of number of moles.
Hence, $x=0.684$
The equilibrium concentrations are
$\left[HI\right]=3+2x=3+2\left(0.684\right)=4.368$ mol/L
$\left[H_2\right]=1-x=1-0.684=0.316$ mol/L
$\left[I_2\right]=2-x=2-0.684=1.316$ mol/L