Four moles of hydrogen iodide were taken in a $10$ litre flask kept at $800K$. When equilibrium was attained, the mixture was found to contain $0.42$ mole of iodine. Calculate the equilibrium constant for dissociation of $HI$.

$2HI\rightleftharpoons H_2+I_2$

Four moles of hydrogen iodide were present initially. 0.42 mole of iodine were formed from $2\times 0.42=0.84$ moles of HI. 0.42 moles of hydrogen were also formed. $4-0.84=3.16$ moles of HI were present at equilibrium.

The equilibrium constant $K_c=\frac{[HI{]}^2}{[H_2][I_2]}$
 $K_c=\frac{[\frac{\text{ 3.16 mol }}{\text{ 10 L }}{]}^2}{[\frac{\text{ 0.42 mol }}{\text{ 10 L }}][\frac{\text{0.42 mol }}{\text{ 10 L }}]}$
 $K_c=\frac{[0.316{]}^2}{[0.042][0.042]}$
 $K_c=56.6$