$60mL\text{\hspace{0.17em}of }{H}_2\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}42mL\text{\hspace{0.17em}of\hspace{0.17em}}{I}_2$ are heated in a closed vessel. At equilibrium the vessel contains $28mLHI.$ Calculate degree of dissociation of $HI?$

$H_2+I_2\rightleftharpoons 2HI$

Volume at $t=0$                         60          42           0

Volume at equilibrium       (60 – x)  (42 – x)     2x

Given, 2x = 28

$\therefore x=14$   

                                            (60 – 14)    (42 – 14) 28

Since at constant p and T, mole $\alpha$ 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 $\Delta n=0$

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

Now for dissociation of HI

                                       $2HI\rightleftharpoons H_2+I_2$

Mole at $t=0$                  1       0      0
Mole at equilibrium       $(1-\alpha )\frac{\alpha }{2}\frac{\alpha }{2}$       
Where $\alpha$ is degree of dissociation

$K_{C_1}=\frac{{\alpha }^2}{4{(1-\alpha )}^2}=\frac{1}{K_C}$

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

$\therefore \alpha =0.719or71.9\%$