Question

For the reaction
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
The value of equilibrium constant is 9.0. The degree of dissociation of HI will be-

• A. 2
• B. $\frac{2}{5}$
• C. $\frac{5}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{2}{5}$

Equilibrium constant of the reaction $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is 9.0.
So the equilibrium constant for the dissociation of HI i.e. $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$ will be $\frac{1}{9}$.
$2HI\rightleftharpoons H_2+I_2$
$100$
$1-x\frac{x}{2}\frac{x}{2}$
$K_C=\frac{x}{2}\times \frac{x}{2}\frac{1}{(1-x)}\times \frac{1}{(1-x)}$
$\frac{1}{9}=\frac{x^2}{2\times 2(1-x{)}^2};$
$\frac{1}{3}=\frac{x}{2(1-x)}$
$or2-2x=3x$
$5x=2$
$x=\frac{2}{5}$