Question

A combination of hydrogen and iodine is carried out by heating $60mL$ of hydrogen and $42mL$ of iodine in a closed vessel. At equilibrium $28mL$ of hydroiodic acid is present in the vessel. Calculate the degree of dissociation of $HI$.

Answer 1

Number of moles is directly proportional to volume. So in equilibrium constant expression, we can replace number of moles (or molar concentration) with volume for the reaction where $\Delta n=0$
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
60mL of hydrogen and 42mL of iodine are present initially.
At equilibrium 28mL of hydroiodic acid is present which is obtained by the reaction of 14 mL of hydrogen and 14 mLof iodine.
$60-14=46$ mL of hydrogen and $42-14=28$ mL of iodine remains at equilibrium.
The equilibrium constant $K=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
$K=\frac{[28{]}^2}{46\times 28}$
$K=0.6087$
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
Let x be the degree of dissociation of HI.
For each mole of HI, when x moles of HI dissociate, x/2 moles of hydrogen and x/2 moles of iodine are formed. $1-x$ moles of HI remains at equilibrium.
The equilibrium constant $K^{\prime }=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$
$\frac{1}{0.6087}=\frac{\left(x/2\right)\times \left(x/2\right)}{[1-x{]}^2}$
$6.57=\frac{x^2}{[1-x{]}^2}$
$2.56=\frac{x}{[1-x]}$
$2.56-2.56x=x$
$2.56=3.56x$
$x=0.72$
The degree of dissociation of HI is 0.72