Question

28 g of N2 and 6 gm of H2 were kept at 400 in 1L vessel , the equillibrium mixture contained 27.54g of NH3 . The approximate value of Kc for the above reaction can be

Answer 1

initial number of mole of
H2 = 6/2 = 3 moles
N2 = 28/28 = 1 moles
  
   initial number of moles:                      1        3          0
                                                         N2 + 3H2 ⇔  2NH3
at equilibrium number of moles:          1-x      3-3x      2x  

at equilibrium number of mole of NH3 = 27.54/17 = 1.62
                 2x = 1.62
                   x = 0.81
at equilibrium number of mole of N2 = 1 - 0.81 = 0.19
at equilibrium number of mole of H2 = 3(1 - 0.81) = 0.57
equilibrium concentration of
N2 = 0.19/1 = 0.19M
H2 = 0.57/1 = 0.57M
NH3 = 1.62/1 = 1.62M
equilibrium constant Kc = [NH3]² / {[N2]*[H2]³}
                             Kc = 1.62² / (0.19*0.57³)
                             Kc = 2.6244 / 0.0352
                             Kc ≈ 74.6