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 mole of $H_2$ and 9 mole of $I_2$ at the same temperature, X mole of $HI$ will be formed. Also degree of dissociation of $HI$ at this temperature will be Y. 100(X+Y) is

Answer 1

Assume total volume to be 1 L.
(1) For first equilibrium

	$H_2$	$I_2$	$HI$
Initial concentration (M)	4	6	0
Equilibrium concentration (M)	$4-1=3$	$6-1=5$	2

$K=\frac{2^2}{3\times 5}=\frac{4}{15}$
(2) For second equilibrium

	$H_2$	$I_2$	$HI$
Initial concentration (M)	4	9	0
Equilibrium concentration (M)	$4-0.5x=3$	$9-0.5x=5$	x

$K=\frac{x^2}{(4-0.5x)(9-0.5x)}=\frac{4}{15}$
$3.5x^2+6.5x-36=0$
$x=2.41$
(3) Degree of dissociation of HI

	$HI$	$I_2$	$H_2$
Initial concentration (M)	2	6	0
Equilibrium concentration (M)	$2-2y=3$	$y$	y

$K=\frac{y^2}{\left(2\right(1-y){)}^2}=\frac{15}{4}$
$Y=0.8$
Hence, $100(x+y)=100(241+0.8)=321$