A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br2(g) + I2(g) 2IBr(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 K = 1.48

Explanation:

Balanced chemical reaction

                   Br₂   +   I₂    ⇒    2IBr

Process

1.- Calculate the concentration of each of the reactants

[Br₂] = 0.6 mol  / 1  L

        = 0.6 M

[I₂] = 1.6 mol / 1 L

     = 1.6 M

[IBr] = 1.19 M

2.- Write the equation of the Equilibrium constant

      K = [IBr]² / [Br₂][I₂]

-Substitution

      K = [1.19]² / [0.6][1.6]

- Simplification

      K = 1.4161 / 0.96

- Result

      K = 1.48