Question

$32$ moles of $HI\left(g\right)$ were heated in a sealed bulb at ${444}^{o}C$ till the equilibrium was reached. Its degree of dissociation was found to be $20$%. Calculate the number of moles of hydrogen iodide, hydrogen and iodine present at the equilibrium point and determine the value of the equilibrium constant for the reaction $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$. Considering the volume of the container $1L$.

Answer 1

Equilibrium process is:

$2HI\to H_2+I_2$

Initial concentration (mol/L):

$3200$

At equilibrium:

$32-2xxx$

Since degree of dissociation, $\alpha =0.2$ (20%) = fraction of reactant dissociated

$\therefore \alpha =2x/32$

Given $0.2=2x/32$

or, $x=3.2$

Therefore equilibrium concentrations (in 1L volume) are:

$\left[HI\right]=32-2\times 3.2=25.6M$, $\left[I_2\right]=x=3.2M=\left[H_2\right]$

Using the relation for equilibrium constant as:

$K_C=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$

Upon substitution we get:

$K_C=\frac{\left[3.2\right]\left[3.2\right]}{(25.6{)}^2}$

$K_C=0.0156$ and number of moles of reactant and product present at equilibrium are:

$\left[HI\right]=32-2\times 3.2=25.6mol$, $\left[I_2\right]=x=3.2mol=\left[H_2\right]$