Calculate the equilibrium concentration for HIg for the reac

Calculate the equilibrium concentration for HI_(g) for the reaction (H_2(g) + I_2(g)<-> HI_(g)) whose K_eq is 6.22 x 10² present in a 4.00 L reaction vessel when the equilibrium moles at 25^degrees C were found to be; moles of H₂= 0.0505; moles of I₂= 0.0498.

Solution

[H2]    = no of moles/volume in L

          = 0.0505/4   = 0.012625M

[I2]     =   no of moles/volume in L

          = 0.0498/4   = 0.01245M

           H_2(g) + I_2(g)<-> 2HI_(g)

          Kc   = [HI]^2/[H2][I2]

        6.22*10^2   = [HI]^2/0.012625* 0.01245

      [HI]^2             = 6.22*10^2* 0.012625* 0.01245

     [HI]^2            = 0.0977

         [HI]              = 0.3125M