Question

${\text{3}}{\text{.2}}$mole of HI were heated in a sealed bulb at ${\text{44}}{{\text{4}}^{\text{o}}}{\text{C}}$ till the equilibrium state was reached. Its degree of dissociation was found to be ${\text{20}}$%. Calculate the number of moles of hydrogen iodide present at equilibrium point and determine the equilibrium constant.

Answer 1

First we will determine the moles of HI dissociated and then the equilibrium concentrations for each of the reactant and product of the reaction. Then we will write equilibrium constant expressions. By substituting the equilibrium concentration we will determine the equilibrium constant.

Complete step-by-step answer:
The formula to determine the degree of dissociation is as follows:
Degree of dissociation = moles dissociated / moles present at initial
On substituting ${\text{3}}{\text{.2}}$mole for initial concentration of HI and ${\text{20}}$% for Degree of dissociation,
$\dfrac{{{\text{20}}}}{{{\text{100}}}}\, = \,\dfrac{{{\text{moles}}\,{\text{dissociated}}}}{{{\text{3}}{\text{.2}}}}$
${\text{moles}}\,{\text{dissociated}}\, = \,\dfrac{{{\text{20}}}}{{{\text{100}}}}\,\, \times 3.2$
${\text{moles}}\,{\text{dissociated}}\, = \,0.64$
So, the initial amount of the HI was $3.2$ moles out of which $0.64$ moles dissociates.
The balanced equation for the dissociation of HI is as follows:
${\text{2HI}}\, \to \,{{\text{H}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,\,{{\text{I}}_{\text{2}}}{\text{(g)}}$

Now we will determine the change in concentration of each reactant and product and their equilibrium concentration as follows:
We know from the above balanced equation that two moles of HI giving one mole of hydrogen and one mole of iodine so, $0.64$ moles will give,
${\text{2}}\,{\text{mol}}\,{\text{HI}}\,\,{\text{ = }}\,\,{\text{1}}\,{\text{mol}}\,\,{{\text{H}}_{\text{2}}}$
${\text{0}}{\text{.64}}\,\,{\text{mol}}\,{\text{HI}}\,\,{\text{ = }}\,\,0.32\,\,{\text{mol}}\,\,{{\text{H}}_{\text{2}}}$
And
${\text{2}}\,{\text{mol}}\,{\text{HI}}\,\,{\text{ = }}\,\,{\text{1}}\,{\text{mol}}\,\,{{\text{I}}_{\text{2}}}$
${\text{0}}{\text{.64}}\,\,{\text{mol}}\,{\text{HI}}\,\,{\text{ = }}\,\,0.32\,\,{\text{mol}}\,\,{{\text{I}}_{\text{2}}}$
So, the concentration of HI is changing from $3.2$ moles to $0.64$ moles. The concentration of products is changing by $0.32$ moles so, now we can prepare the ICE chart as follows:

                 ${\text{2HI}}\, \to \,{{\text{H}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,\,{{\text{I}}_{\text{2}}}{\text{(g)}}$  

Initial conc. $3.2\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0$
Change $0.64\,\,\,\,\,\,\,\,0.32\,\,\,\,\,\,\,0.32$
Equi. conc.$\left( {3.2 - 0.64} \right)\,\,\,\,\,\,\,\,0.32\,\,\,\,\,\,\,0.32$
So, the equilibrium concentration of HI is,
$3.2 - 0.64\, = \,2.56$
So, the equilibrium concentration of HI is $2.56$moles.

Now we will write the equilibrium constant expression for the reaction as follows:
${\text{2HI}}\, \to \,{{\text{H}}_{\text{2}}}{\text{(g)}}\,{\text{ + }}\,\,{{\text{I}}_{\text{2}}}{\text{(g)}}$
The equilibrium constant expression for the above reaction is as follows:
${{\text{k}}_{{\text{equi}}}}{\text{ = }}\,\dfrac{{\left[ {{{\text{H}}_2}} \right]\,\,\left[ {{{\text{I}}_2}} \right]}}{{{{\left[ {{\text{HI}}} \right]}^{\text{2}}}}}\,$
On substituting equilibrium concentration of each,
${{\text{k}}_{{\text{equi}}}}{\text{ = }}\,\dfrac{{\left[ {0.32} \right]\,\,\left[ {0.32} \right]}}{{{{\left[ {2.56} \right]}^{\text{2}}}}}\,$
${{\text{k}}_{{\text{equi}}}}{\text{ = }}\,\dfrac{{\left[ {0.1024} \right]\,\,}}{{\left[ {6.5536} \right]}}\,$
${{\text{k}}_{{\text{equi}}}}{\text{ = }}\,0.0156\,$
So, the equilibrium constant is $0.0156$.
Therefore, the number of moles of hydrogen iodide present at equilibrium point is $2.56$and the equilibrium constant is $0.0156$.

Note: When we determine the change in concentration then we multiply the change with the stoichiometry of the product and divide the change by the stoichiometry of reactant whereas the change in concentration of reactant is multiplied with stoichiometry of reactant only. Degree of dissociation tells how much amount from initial is undergoing dissociation.