Question

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 $HI$.

Answer 1

$2HI\rightleftharpoons H_2+I_2$

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