Question

At a certain temperature, 0.3211 of N2N2 and 1.701 mol of H2H2 are placed in a 2.00 L container.

  

N2(g)+3H2(g)↽−−⇀2NH3(g)

At equilibrium, 0.1801 mol of N2N2 is present. Calculate the equilibrium constant, ?c

Answer 1

N2 + 3H2 <-------------> 2NH3

initially

[N2] = 0.3211 / 2.0 = 0.160 M

[H2] = 1.701 / 2.0 = 0.850 M

at equilibrium

[N2] = 0.1801 / 2.0 = 0.09 M

N2 consumed = 0.160 - 0.09 = 0.07 M

H2 must consume = 3 x 0.07 = 0.21

at equilibrium [NH3] = 2 x 0.07 = 0.14 M

[H2] = 0.850 - 0.21 = 0.64 M

Kc = [NH3]² / [N2] [H2]³

Kc = [[0.14]² / [0.09] [0.64]³

Kc = 0.0196 / 0.0236

Kc = 0.830