Question

$60mL$ of $H_2$ and $42mL$ of $I_2$ are heated in a closed vessel. At equilibrium the vessel contains $28mL$ of $HI$. Calculate degree of dissociation of $HI$.

Answer 1

Given that,

$H_2+I_2\rightleftharpoons 2HI$
$60$ $42$ $0$ --- At time, $t=0$
$(60-x)(42-x)$ $2x:2x=28\therefore x=14$ --- At equilibrium
$(60-14)(42-14)$ $28$

Since at constant $P$ and $T$, $moles\propto volume$ of gas (by $PV=nRT)$. Thus, volume of gases given can be directly used as concentration. This can be done only for reactions having $△n=0$.

$\therefore K_c=\frac{28\times 28}{46\times 28}=\frac{28}{46}$

Now for dissociation of $HI;2HI\rightleftharpoons H_2+I_2$
Moles at $t=0$ $1$ $0$ $0$
Moles at equilibrium $(1-\alpha )$ $\alpha /2\alpha /2$
Where $\alpha$ is the degree of dissociation

$K_{c_1}=\frac{{\alpha }^2}{4(1-\alpha {)}^2}=\frac{1}{K_c}$

$\therefore \frac{\alpha }{2(1-\alpha )}=\sqrt{\left(\frac{46}{28}\right)}$

$\therefore \alpha =0.719$ or $71.9\%$.