Question

HI was heated in a sealed tube at ${440}^{o}C$ till the equilibrium was established. The dissociation of HI was found to be 22%. According to following reaction,
$2HI\rightleftharpoons H_2+I_2$. The equilibrium constant for dissociation is -

• A. 0.282
• B. 0.0786
• C. 0.0199
• D. 1.99

Answer 1

The correct option is C 0.0199

The equilibrium of the dissociation of
$\begin{array}{ccccc}2HI& \rightleftharpoons & H_2& +& I_2\\ 1& & 0& & 0\\ 1-\frac{22}{100}& & \frac{22}{100\times 2}& & \frac{22}{100\times 2}\\ 0.78& & 0.11& & 0.11\end{array}$

$K_C=\frac{0.11\times 0.11}{0.78\times 0.78}\approx 0.0199$