Question

Calculate the equilibrium concentration of $H_2,I_2$ and $HI$ at $300K$ if initially $2mol$ of $H_2$ and $I_2$ are taken in a closed container of having volume $10$ liter.

[Given: $H_2+I_2\rightleftharpoons 2HI;K=100$ at $300K$]

Answer 1

Balanced reaction :- $H_2+I_2=2HI$

Initial concentration :- $0.2$ $0.2$ $0$

Equilibrium concentration $0.2x$ $0.2-x$ $2x$

We know that

Equilibrium constant $K=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{{\left[2x\right]}^2}{[0.2-x][0.2-x]}=100$ (given)

Solve for $x$

${4x}^2={[0.2-x]}^{2}100$

${4x}^2=[x^2-0.4x+0.04]100$

${4x}^2={100x}^2-40x+4$

${96x}^2-40x+4=0$

On solving we get $x=0.25$ and $x=0.1667$

As $0.25$ is larger value that the given concentration of reactants it can be neglected.

So, the value of $x=0.1667$

$\therefore$ concentration of $H_2,I_2$ and $HI$ at equilibrium respectively

$=0.2-x,0.2-x,2x$

$=0.2-1.667,0.2-1.667,2\left(1.667\right)$

$=0.333,0.333,3.334$ $\frac{moles}{liters}$