At 600 K, 0.250 mol N2(g) and 0.270 mol H2(g) were placed in a 2.00 L container. At equilibrium, the concentration of NH3(g) was 0.030 M. Calculate KC at 600 K for N2(g) + 3H2(g) ⇌ 2NH3(g).

Answer 1

This isn't all that different from your previous questions which I have answered. In this problem however, you aren't given the equilibrium concentrations, so you need to find them and then use the Kc expression to find Kc. Start by setting up an ICE table and convert moles to mole/L...

N₂(g) + 3H₂(g) <==> 2NH₃(g)

0.125........0.135...................0.........I

-x..............-3x....................+2x.......C

0.125-x....0.135-3x.............2x.........E

Since we know that [NH3] @ equilibrium = 0.03, we then know that 2x = 0.03 and x = 0.015.

Equilibrium concentrations are:

[N₂] = 0.125 - 0.015 = 0.11 M

[H₂] = 0.135 - 0.045 = 0.09 M

[NH₃] = 0.015 M

Kc = [NH₃]² / [N₂][H₂]³ = (0.015)² / (0.11)(0.09)³

Kc = 2.81