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
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Answer 1

Answer:

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Explanation:

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