Question

0.96g of HI were heated to attain equilibrium

$2HI\rightleftharpoons H2+I2$. The reaction mixture on titration requires 15.7ml of $\frac{N}{10}$ hypo. Calculate % dissociation of HI.

• A. 18.9%
• B. 19.9%
• C. 20.9%
• D. 21.9%

Answer 1

The correct option is C

20.9%

$2HI\rightleftharpoons H_2+l_2$

Initial $\frac{0.96}{128}00$

moles = $7.5\times {10}^{-3}$

Moles $(7.5\times {10}^{-3}-x)\frac{x}{2}\frac{x}{2}$

at equilibrium

Now Meq. of $I_2$ formed at equilibrium = M eq. of hypo used $\frac{W}{E}\times 100=15.7\times \frac{1}{10}$ or $\frac{W}{E}$ of $I_2=1.57\times {10}^{-3}$

Moles of $I_2$ formed at equilibrium

$=\frac{1.57\times {10}^{-3}}{2}=0.785\times {10}^{-3}$

or $\frac{x}{2}=0.785\times {10}^{-3}$ or $x=1.57\times {10}^{-3}$

degree of dissociation of HI

$=\frac{molesdissociated}{initialmoles}=\frac{x}{7.5\times {10}^{-3}}$

$=\frac{1.57\times {10}^{-3}}{7.5\times {10}^{-3}}=0.209=20.9$