Question

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?

Answer 1

The equilibrium constant for this reaction at 350°C is determined as 5.85.

Concentration of each component

concentration of bromine, C(Br) = 0.6 mol/1 = 0.6

concentration of iodine, C(I) = 1.6 mol/1 = 1.6

Create an ICE table

What is ICE table?

An ICE table is a tabular system of keeping track of changing concentrations in an equilibrium reaction.

ICE is an abbreviation that stands for initial, change, equilibrium.

Create ICE table for the reactants and products formed;

      Br2(g)   +     I2(g)    ↔     2IBr(g)

I     0.6              1.6                 0

C    -1.19            -1.19               1.19

E    0.6 - 1.19      1.6  - 1.19       1.19

E = -0.59           0.41                1.19

Equilibrium constant

The equilibrium constant is calculated as follows;

KC = [IBr]²/[Br][I]

KC = (1.19²) / (0.59 x 0.41)

KC = 5.85

Thus, the equilibrium constant for this reaction at 350°C is determined as 5.85.

Learn more about equilibrium constant here: https://brainly.com/question/19340344

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