At a certain temperature 05011 mol of N2 and 1781 mol of H2

At a certain temperature, 0.5011 mol of N2 and 1.781 mol of H2 are placed in a 5.00-L container N,(g) + 3H2(g) 2NH3(g) At equilibrium, 0.1201 mol of N2 is present. Calculate the equilibrium constant, Kc. Number e146.7

Solution

          N2(g)   +   3H2(g) <---->    2NH3(g)

initial 0.5011/5      1.781/5             0

           0.1 M      0.356 M             0

change    -0.076      -0.228            +0.152

equilib   0.024        0.128            0.152 M

at equilibrium, N2 = 0.1201/5 = 0.024 M

   Kc = [NH3]^2/[N2][H2]^3

   Kc = 0.152^2/(0.128^3*0.024)

      = 459.03