For the following reaction, the equilibrium constant Kc is 0.80 at a certain temperature. If the concentration of NO (g) and NOBr (g) are both 0.80 M, at equilibrium, what is the concentration of Br2 (g) ? 2NO (g) + Br2 (g) ⇌ 2NOBr (g)

[Br₂] = 1.25M

Explanation:

2NO (g) + Br₂ (g) ⇄ 2NOBr (g)

Eq 0.80M? 0.80M

That's the situation told, in the statement.

Let's make the expression for Kc

Kc = [NOBr]² / [Br₂]. [NO]²

Kc = 0.80² / [Br₂]. [0.80]²

0.80 = 1 / [Br₂]

[Br₂] = 1 / 0.80 → 1.25