An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 670 K in a % litre flask contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide. Calculate equilibrium constant.

Reaction will be given as:
H₂ (g) + I₂ (g) = 2 HI (g)
Therefore, the equilibrium constant K will be given as:
K = [HI] ² / [H₂] [I₂]
The molar concentrations of the species at equilibrium are:
[H₂] = 0.4 / 5 = 0.08 mol L^-1
[I₂] = 0.4 / 5 = 0.08 mol L^-1
[HI] = 2.4 / 5 = 0.48 mol L^-1
Substituting these values in the expression of K, we will get:
K = (0.48) ² / (0.08) (0.08)
K = 36.0
Hence the equilibrium constant is equal to 36.0