Question

$15$ moles of hydrogen reacting with $5.2$ moles of iodine from $10g$ mole of hydrogen iodide. Calculate the equilibrium constant of the reaction.

Answer 1

$H_2+I_2\to 2HI$
Initially, 15 moles of hydrogen and 5.2 moles of iodine are present.
At equilibrium, 10 moles of hydrogen iodide are present which are obtained by the reaction of 5 moles of hydrogen with 5 moles of iodine.
$15-5=10$ moles of hydrogen and $5.2-5=0.2$ moles of iodine remain at equilibrium.
Let V L be the total volume.
The equilibrium constant
$K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
$K_c=\frac{(\frac{\text{10 mol}}{\text{V L}}{)}^2}{\frac{\text{10 mol}}{\text{V L}}\times \frac{\text{0.2 mol}}{\text{V L}}}$
$K_c=\frac{(10{)}^2}{10\times 0.2}$
$K_c=50$