Question

$25mL$ of $H_2$ and $18mL$ of $I_2$ vapours were heated in a sealed glass tube at ${465}^{o}C$ and at equilibrium $30.8mL$ of $HI$ is formed. Calculate the percentage degree of dissociation of $HI$ at ${465}^{o}C$.

Answer 1

Since, the number of moles of a gas is proportional its volume (Avogadro's
Law), the volumes in $mL$ of gases may be used instead of number of moles in the equilibrium constant expression.
Let $x$ ml of hydrogen reacts with $x$ ml of iodine to form $2x$ ml of HI.
$H_2+I_2\to 2HI$
Initial volume (ml) 25 18
Equilibrium volume (ml) $25-x$ $18-x$ $2x=30.8$
Since, $2x=30.8$ mL
So, $x=15.4$ mL
Vol. of $H_2$ at equilibrium $(25-x)=(25-15.4)=9.6$ mL
Vol. of $I_2$ at equilibrium $(18-x)=(18-15.4)=2.6$ mL
Let V L be the total volume.
$K_c=\frac{{\left[HI\right]}^2}{\left[I_2\right]{\left[H_2\right]}^{}}$
$K_c=\frac{(30.8/V{)}^2}{\left(9.6/V\right)\left(2.6/V\right)}$
$K_c=38.0$
Consider the reverse reaction (decomposition of HI)
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
$K_c^{\prime }=\frac{1}{K_c}$
$K_c^{\prime }=\frac{1}{38.0}$
$K_c^{\prime }=0.0263$

Let the degree of dissociation be y.
If we start with 2 moles of HI, 0 moles of hydrogen and 0 moles of iodine, then at equilibrium, $2(1-y)$ moles of HI, y moles of hydrogen and y moles of iodine will be present.
$K_c^{\prime }=\frac{\left[H_2\right][I_2}{[HI{]}^2}$
$K_c^{\prime }=\frac{\left(y/V\right)\times \left(y/V\right)}{{\left(2\right(1-y\left)/V\right)}^2}$
$0.0263=\frac{y^2}{4(1-y{)}^2}$
$0.105=\frac{y^2}{(1-y{)}^2}$
$0.3244=\frac{y}{(1-y)}$
$1-y=\frac{y}{0.3244}$
$1-y=3.08y$
$1=4.08y$
$y=24.5$
The percentage degree of dissociation of HI at ${465}^{o}C$ is $24.5$ %.