Question

$0.96$ g of $HI$ was heated to attain equilibrium: $2HI\rightleftharpoons H_2+I_2$. The reaction mixture on titration requires $15.7$ ml of $N/10$ hypo solution. Calculate the degree of dissociation of $HI$.

• A. ${\alpha }_{HI}=0.633$
• B. ${\alpha }_{HI}=0.381$
• C. ${\alpha }_{HI}=0.121$
• D. ${\alpha }_{HI}=0.209$

Answer 1

The correct option is D ${\alpha }_{HI}=0.209$

$2HI\rightleftharpoons H_2+I_2$

$\frac{0.96}{128}$ $0$ $0$ Moles at $t=0$
$=7.5\times {10}^{-3}$
$(7.5\times {10}^{-3}-x)$ $\frac{x}{2}$ $\frac{x}{2}$ Moles at equilibrium
mEq. of $I_2$ formed at equilibrium $=$ mEq. of hypo used for reaction mixture
$\therefore \frac{w_{I_2}}{E}\times 1000=15.7\times \frac{1}{10}$
$\therefore \left(\frac{w}{E}\right)of{I}_2=1.57\times {10}^{-3}$
Moles of $I_2$ formed $=\frac{1.57\times {10}^{-3}}{2}=0.785\times {10}^{-3}$
$\frac{x}{2}=0.785\times {10}^{-3}\Rightarrow x=1.57\times {10}^{-3}$
Degree of dissociation of $HI$ $=\frac{\text{Moles dissociated}}{\text{Moles taken}}=\frac{15.7\times {10}^{-3}}{7.5\times {10}^{-3}}$
${\alpha }_{HI}=0.209or20.9\%$