Question

Consider the following reaction and its equilibrium constant: I2(g) + Br2(g) ⇌ 2 IBr(g) Kc = 1.1 × 102 This reaction mixture contains initially 0.41 M I2 and 0.27 M Br2. Calculate the equilibrium concentration of I2, Br2, and IBr?

Answer 1

Kc you have given is 1.1 × 102 = 110 right? (is it correct)

Construct the ICE table

I2(g) + Br2(g) ⇌ 2IBr(g)

I 0.41 0.27 0

C -x -x +2x

E 0.41-x 0.27-x 2x

Kc =[IBr]² / [I2][Br2]

110 = [2x]² / [0.41-x][0.27-x]

106X² -74.8X + 12.177 = 0

solve the quadratic equation

x will have two value which are 0.45 , 0.254

it cannot be 0.45 because some of the concentration values will get -ve

so x will be 0.25

equilibrium concentrations of

[I2] = 0.41-0.25 = 0.16 M

[Br2] = 0.27-x = 0.27-0.25 = 0.02M

[IBr] = 2x = 2(0.25) = 0.5 M