At a certain temperature, 0.3811mol of N2 and 1.521 mol of H2 are placed in a 4 L container.

Question

N2 (g) + 3H2 (g) ⇔ 2NH3 (g)

At equilibrium, 0.1201 mol of N2 is present. Calculate the equilibrium constant Kc.

(please be as detailed as possible, thank you so much)

Answer 1

Set up an ICE table to help visualize what is happening.

N₂(g) + 3H₂(g) <==> 2NH₃(g)

0.3811....1.521..............0............Initial

-x..........-3x.................+2x...........Change

0.3811-x...1.521-3x..........2x...........Equilibrium

Since we are told that at equilibrium, there are 0.1201 mol N₂, we can now determine the value of x.

0.3811 - x = 0.1201

x = 0.261

So, at equilibrium (see equilibrium line on the ICE table), we have

moles N₂ = 0.1201 mols N₂

moles H₂ = 1.521 - 3 * 0.261 = 1.521 - 0.783 = 0.738 mols H₂

moles NH₃ = 2x = 2 * 0.261 = 0.522 mols NH₃

Keq = [NH3]² / [N2][H2]³

Since the volume of the reaction is 4 L, we can convert these to concentrations:

[N₂] = 0.1201 mol / 4 L = 0.030 M

[H₂] = 0.738 mol / 4 L = 0.1845 M

[NH₃] = 0.522 mol / 4 L = 0.1305 M

Keq = [0.1305]² / [0.030][0.1845]³ = 0.0170 / 0.000188

Keq = 90.2