Question

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)

Answer 1

Answer:

[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