Question

HI was heated in a closed tube at ${440}^{\circ }C$ till equilibrium is reached. At this temperature 22%of HI was dissociated. The equilibrium constant for this dissociation is;

• A. 0.282
• B. 0.0796
• C. 0.0199
• D. 1.99

Answer 1

The correct option is C 0.0199

The equilibrium for the dissociation of HI is $2HI\left(g\right)\iff H_2\left(g\right)+I_2\left(g\right)$.
22% of HI dissociates. The concentrations of various species are given blow.

	$2HI$	$H_2$	$I_2$
Initial	100	0	0
Change	-22	11	11
Equilibrium	78	11	11

The expression for the equilibrium constant is $K_c=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$.
Substitute values in the above expression.
$K_c=\frac{11\times 11}{(78{)}^2}=0.0199$.