At a certain temperature, 0.4811 mol of N2 and 1.761 mol of H2 are placed in a 4.50 L container. N2(g)+3H2(g)↽−−⇀2NH3(g) At equilibrium, 0.1201 mol of N2 is present.

Question

Calculate the equilibrium constant, 𝐾c .

Answer 1

We need to find K_c, which we can set up as K_c = [NH₃]²/[H₂]³[N₂]. The brackets indicate concentrations (usually molarities = mol/L) of each species at equilibrium. This equation is given in a more generic form on the AP Chemistry equations sheet.

To find equilibrium concentrations, I'd set up a RICE Table: reaction, initial, change, and equilibrium!

R: N_2(g) + 3H_2(g) ⇆ 2NH_3(g)

I: 0.4811 mol/4.50 L 1.761 mol / 4.50 L 0M

C: - x -3x +2x

E: 0.1069 - x M 0.3913 - 3x M 2x = 0.1201 mol / 4.50 L

From the RICE table, the equilibrium concentration of NH₃ is 2x, which is given to be 0.1201 mol / 4.5 L:

2x = 0.1201 mol / 4.5 L = 0.026689 M → x = .013344 M

We can use this x for the other equilibrium concentrations from the RICE table and plug them into K_c:

K_c = [NH₃]²/[H₂]³[N₂] = (2x)²/[(0.3913 - 3x)³(0.1069 - x)] = (0.1201)²/(.3513)³(0.09356) = 0.176