This is one of my more harder problems: Calculate K from Inital + One Final Concentration

Question

A student ran the following reaction in the laboratory at 668 K:

2NH₃(g) N₂(g) + 3H₂(g)

When she introduced 7.53×10^-2 moles of NH₃(g) into a 1.00 liter container, she found the equilibrium concentration of NH₃(g) to be 5.07×10^-3 M.

Calculate the equilibrium constant, K_c, she obtained for this reaction.

Answer 1

K_C = [N₂][H₂]³/[NH₃]² Let x be N₂ produced or "extent of reaction"

[N₂] at equilibrium: x

[NH₃] at eq is 5.07 x 10^-3

[H₂] = 3x

We can find x from the NH₃ reacted: ( 7.53x10^-2 - 5.07x10^-3)/2

Now plug in.