A mixture of 0.600 mol bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br2(g) + I2(g) 2 IBr(g). When the mixture has come to equilibrium, the concentration of iodine monobromide is 1.190 M. What is the equilibrium constant for this reaction at 350°C?

The answer to your question is   Ke = 1.4751

Explanation:

Data

bromine = 0.6 mol

iodine = 1.6 mol

volume = 1 L

Ke = ?

Iodine monobromide concentration = 1.190 M

Process

1.- Write the balanced chemical equation

                    Br₂  +    I₂   ⇒   2IBr

2.- Write the formula for Ke

     Ke = [IBr]²/[Br₂][I₂]

3.- Substitution

      Ke = [1.19]² / [0.6][1.6]

4.- Simplification

      Ke = 1.4161 / 0.96

5.- Result

      Ke = 1.4751

the answer would be a