Question

$HI$ was heated in a sealed tube at ${400}^{o}C$ till the equilibrium was reached. $HI$ was found to be $22$% decomposed. The equilibrium constant for dissociation is:

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

Answer 1

The correct option is B $0.0199$

The reaction is $2HI\to H_2+I_2.$
The inital concentrations of HI, hydrogen and iodine are 1 M, 0 M and 0 M respectively.
The equilibrium concentrations of HI, hydrogen and iodine are 1-0.22=0.78 M, 0.11 M and 0.11 M respectively.
The expression for the equilibrium constant is $K=\frac{\left[H_2\right]\left[I_2\right]}{\left[H{I}_2\right]}.$
Substitute values in the above expression.
$K=\frac{0.11\times 0.11}{(0.78{)}^2}=\mathrm{0.0199.}$