Question

2IBr <-- --> I2 + Br2. (The arrows are overlapping each other not sure how to do that)

the equilibrium constant Kc at 100 degrees C is 0.026. If 2.40 x 10^-2 mol IBr is placed in a .800-L vessels at 100 degrees C, what is the molarities at equilibrium in the vapor? A) IBr____M. B) I2____M. C) Br2____M.

Answer 1

no of moles of IBr =  2.40 x 10^-2 mol

volume = 0.800L

concentration of IBr - moles / volume =  2.40 x 10^-2 / 0.8 = 0.03M

now construct the ICE table

2IBr <-- --> I2 + Br2

I 0.03 0 0

C -2x +x +x

E 0.03-x +x +x

Kc = [I2][Br2] / [IBr]²

0.026 = x² / [0.03-x]²

0.974x² + 0.00156x - 0.0000234 = 0

solve the x value using quadratic equation

x = 0.00416 M

molarities at equilibrium is

[IBr] = 0.03 -x = 0.03 - 0.00416 = 0.0258 M

[I2] = x = 0.00416 M

[Br2] = x 0.00416 M