Question

The equilibrium constant K_c for the following reaction is 4.59 × 10^-7 at 730^oC.

2HBr(g) ⇌ H₂(g) + Br₂(g)

Suppose 3.20 mol HBr and 1.50 mol H₂ are added to a rigid 12.0-L flask at 730^oC. What is the equilibrium concentration (in M) of Br₂?

Answer 1

Given,

Kc = 4.59 x 10^-7

for the reaction

2HBr <----> H2 + Br2

Initial Concenrations

[HBr] = 3.2 / 12 = 0.2667 M

[H2] = 1.5 / 12 = 0.125 M

2HBr <-------------> H2 + Br2

0.2667...... ............0.125...0

-2X.......................+X......+X

0.2667-2X.........0.125+X...X

Kc = [H2] [Br2] / [HBr]^2

=> 4.59 x 10^-7 = (0.125 + X) (X) / (0.2667 - 2X)^2

Neglecting X wrt 0.125 and 2X wrt 0.2667 gives,

4.59 x 10^-7 = 0.125 x X / (0.2667)^2

=> X = 2.6 x 10^-7 = [Br2] = Equilibrium Conc. of Br2