Question

For the reaction $H_2+I_2\rightleftharpoons 2HI$
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

The correct option is B 2/5

Equilibrium constant of the reaction
$H_2+I_2\rightleftharpoons 2HI\text{is}9.0$
So the equilibrium constant for the dissociation of $HI$ i.e. $2HI\rightleftharpoons H_2+I_2$ will be 1/9.
$\begin{array}{ccccc}2HI& \rightleftharpoons & H_2& +& I_2\\ 1& & 0& & 0\\ 1-x& & \frac{x}{2}& & \frac{x}{2}\end{array}$

$K_C=\frac{x}{2}\times \frac{x}{2}\frac{1}{(1-x)\times (1-x)}$
$\frac{1}{9}=\frac{x^2}{2\times 2(1-x{)}^2};$
$\frac{1}{3}=\frac{x}{2(1-x)}$
or $2-2x=3x$
$5x=2$
$x=2/5$