Question

A gaseous mixture having $4$ mole of $H_2$ and $6$ mole of $I_2$ on heating shows equilibrium concentration of $HI$ as $2$ mole. If we start with $4$ moles of $H_2$ and $9$ mole of $I_2$ at the same temperature, how many mole of $HI$ will be formed? Also calculate degree of dissociation of $HI$ at this temperature.

Answer 1

$H_2+I_2\leftrightharpoons 2HI$

$\left(a\right)$ initial $4$ $6$ $0$

Eqb. $(4-x)$ $(6-x)$ $2x$

Put, $x=1$

$3$ $5$ $2$ $⟹K_c=\frac{2^2}{3\times 5}$

$\left(b\right)$ initial $4$ $9$ $0$

Eqb. $4x$ $9x$ $2x$ $⟹K_c^{\prime }=\frac{(2x{)}^2}{(4-x)(9-x)}$

At same temp$⟹K_c=K_c^{\prime }$

$\frac{4}{3\times 5}=\frac{4x^2}{(4-x)(9-x)}⟹{15x}^2=x^2+36-13x⟹14x^2+13x-36=0$

$x>0⟹x=1.205$

number of moles of $HI=2x=2-41$

Degree of dissociation $⟹2HI⟶H_2+I_2$

$K^{\prime }=\frac{1}{K_c}$

$K^{\prime }=\frac{x^2}{2(1-x)}$

$⟶\frac{x^2}{2(1-x)}=\frac{3\times 5}{4}$

$⟹2x^2=15-15x$

$⟹2x^2+15x-15=0$

$x>0$ $⟶x=0.8935$

Degree of dissociation of $HI=0.8935$