Question

$3.2$ mole of $HI$ were heated in a sealed bulb at ${444}^{o}C$ till the equilibrium state was reached. Its degree of dissociation was found to be $20$%. Calculate the number of moles of hydrogen iodide present at the equilibrium point and determine the equilibrium constant.

Answer 1

The dissociation of $HI$ is represented by the equation,

$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$

$(a-ax)$ $ax/2$ $ax/2$

Degree of dissociation, $x=0.20$ and initial concentration of $HI$, $a=3.2$ mole

At equilibrium

No. of moles of $HI=a(1-x)=3.2\times 0.8=2.56$

No. of moles of $H_2=\frac{a.x}{2}=3.2\times 0.5\times 0.2=0.32$

No. of moles of $I_2=\frac{a.x}{2}=3.2\times 0.5\times 0.2=0.32$

also, $K_c=\frac{\left[I_2\right]{\left[H_2\right]}^{}}{{\left[HI\right]}^2}=\frac{0.32\times 0.32}{2.56\times 2.56}=0.0156$