Question

One mole of $H_2$, two moles of $I_2$ and three moles of $HI$ are injected in a $1$ litre flask. What will be the concentration of $H_2,I_2$ and $HI$ at equilibrium at ${490}^{o}C$? The equilibrium constant for the reaction at ${490}^{o}C$ is $45.9$.

Answer 1

$H_2+I_2\to 2HI$
Since volume of flaskis 1 L, the number of moles is equal to molar concentration.
The initial concentrations are
$[H_2{]}_0=1$ M
$[I_2{]}_0=2$ M
$[HI{]}_0=3$ M
The equilibrium concentrations are
$\left[H_2\right]=1-x$ M
$\left[I_2\right]=2-x$ M
$\left[HI\right]=3+2x$ M
Note: x M of $H_2$ will react with x M of $I_2$ to form 2x M of $HI$ at equilibrium.
The equilibrium constant
$K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
$45.9=\frac{[3+2x{]}^2}{[1-x][2-x]}$
$45.9=\frac{3[3+2x]+2x[3+2x]}{[2-x]-x[2-x]}$
$45.9=\frac{9+6x+6x+4x^2]}{2-x-2x+x^2}$
$45.9=\frac{9+12x+4x^2}{2-3x+x^2}$
$45.9(2-3x+x^2)=9+12x+4x^2$
$91.8-137.7x+45.9x^2=9+12x+4x^2$
$82.8-149.7x+41.9x^2=0$
$41.9x^2-149.7x+82.8=0$
This is quadratic equation with solution
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-149.7)\pm \sqrt{(-149.7{)}^2-4(41.9\left)\right(82.8)}}{2\left(41.9\right)}$
$x=\frac{149.7\pm 92.4}{83.8}$
$x=2.888$ or $x=0.684$
The value $x=2.888$ is discarded as it will lead to negative value of concentration.
$x=0.684$
The equilibrium concentrations are
$\left[H_2\right]=1-x=1-0.684=0.316$ M
$\left[I_2\right]=2-x=2-0.684=1.316$ M
$\left[HI\right]=3+2x=3+2\left(0.684\right)=4.36$ M