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