28 g of N2 and 6 gm of H2 were kept at400 in 1L vessel , the equillibrium mixturecontained 27.54g of NH3 . The approximate value of Kc forthe above reaction can be

28 g of N2 and 6 gm of H2 were kept at400 in 1L vessel , the equillib

1.	28 g of N2 and 6 gm of H2 were kept at400 in 1L vessel , the equillibrium mixturecontained 27.54g of NH3 . The approximate value of Kc forthe above reaction can be
Answer» 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 / (0.190.57³) Kc = 2.6244 / 0.0352 Kc ≈ 74.6