equilibrium homework help

Question

Consider the reaction: N₂ (g) + 3 Br₂ (g) ⇌ 2 NBr₃ (g) At equilibrium, the concentrations of N₂ and Br₂ are 0.34 M and 0.70 M respectively, and the concentration of NBr₃ is 0.090 M. What is Kc for this equilibrium?

Answer 1

N₂(g) + 3Br₂(g) <==> 2NBr₃(g)

Kc = [NBr₃]² / [N₂] [Br₂]³

Kc = (0.09)² / (0.34)(0.70)³ = 0.0081 / 0.1166

Kc = 0.069