Question

25 mL of $H_2$ and 18 mL of $I_2$ vapours were heated in a sealed tube at ${456}^{o}C$, when, at equilibrium 30.8 mL of $HI$ was formed. Calculate the degree of dissociation of pure $HI$ at ${456}^{o}C$.

Answer 1

$H_2+I_2=2HI$
$25$ $18$ $0$
$(25-154)(18-15.4)30.8$
$9.6$ $2.6$
$K_c=\frac{\left[HI\right]}{\left[H_2\right]\left[I_2\right]}=\frac{(30.8{)}^2}{9.6\times 2.6}=38$
$2HI\rightleftharpoons H_2+I_2$
$2$ $0$ $0$
$K_c=\frac{1}{38}=\frac{x-x}{(2-2x)}={\left(\frac{x}{2-2x}\right)}^2$
$\frac{x}{2-2x}=0.162$
$x=0.245$
$=\frac{2x}{2}\times 100$
$=100x$
$=100\times 0.245$
$=24.5$
$H_2+I_2=2HI$
Initially assume that $18mL$ of $I_2$ will combine with $18mL$ of $H_2$ to form $2\times 18=36mL$ of $HI$. Now assume that of $236$mLof$HI$,$x$ mL dissociate
$2HI\rightleftharpoons H_2+I_2$
$(36-x)$ mL $HI$ will be present after dissociation. But at equilibrium $30.8mL$ of $HI$ was formed.
$(36-x)=30.8$
$x=36-30.8=5.2mL$
Percent degree of dissociating of $HI$
$=\frac{x\times 100}{36}=\frac{5.2\times 100}{36}$
$=14.4$
$\therefore$ Percent degree of dissociation of $HI$ at ${465}^{o}C$ is $14.4$.