1 mol H2  & 2 mol I2  are taken in a 10 lit flask & heated. Now, at equilibrium, if 0.4  mole HI is present, then

H2+I2⇌2HI           1         2             0             x        x            2x_ (1−x)    (2−x)     2x                         Given  2x=0.4                                   ⇒x=0.2                       ⇒[I2]>[H2]>[HI]