Question 14.1: For a chemical reaction H2 + I2 → 2HI, Find out the equilibr......

Given, [HI] = 0.0276  mol/L,  [H_2] = 0.0037  mol/L and [I_2] = 0.0037  mol/L.
(a)   Equilibrium constant from the mass action law is given by:

K= \frac{\left[HI\right]^2 }{\left[H_2\right]^1 \left[I_2\right]^1 } = \frac{0.0276^2}{0.0037 \times 0.0037} = 55.6

(b)   For the reverse equation:

2HI \longrightarrow H_2 + I_2
K^\prime = \frac{\left[H_2\right]^1 \left[I_2\right]^1 }{\left[HI\right] } = \frac{0.0037 \times 0.0037}{0.0276^2} = \frac{1}{55.6}