Question

For the reaction
$H _{2}+ I _{2} \rightleftharpoons 2 HI$
The value of equilibrium constant is $9.0 .$ The degree of dissociation of $HI$ will be

• A. 2
• B. 2/5
• C. 5/2
• D. 1/2

Answer 1

Answer: (B)

Equilibrium constant of the reaction

$H _{2}+ I _{2} \rightleftharpoons 2 HI \text { is } 9.0$

So the equilibrium constant for the dissociation of $HI$, i.e.

$2 HI \rightleftharpoons H _{2}+ I _{2}$ will be $1 / 9$



$K_{C}=\frac{x}{2}=\frac{x}{2} \frac{1}{(1-x)} \times \frac{1}{(1-x)}$

$\frac{1}{9}=\frac{x^{2}}{2 \times 2(1-x)^{2}} ; \frac{1}{3}=\frac{x}{2(1-x)}$

$\Rightarrow x=2 / 5$