Question

The equilibrium constant K_c for the reaction below is 0.00440 at a certain temperature.

Br₂(g) ⇌ 2Br(g)

If the initial concentrations are

[Br₂] = 0.0131 M

and

[Br] = 0.0854 M,

calculate the concentrations of these species at equilibrium.

Answer 1

[Br2] = 0.0485M

[Br] = 0.0146M

Explanation

Br2(g) <------> 2Br(g)

Kc1= [Br]^2/[Br2]= 0.00440

Consider the reverse

2Br(g)<----> Br2(g)

Kc-1 = [Br2]/[Br]^2=227.27

Initial concentration

[Br] = 0.0854M

[ Br2 ] = 0.0131M

Change in concentration

[ Br] = -2x

[ Br2] = +x

Equillibrium concentration

[ Br] = 0.0854 - 2x

[ Br2 ] = 0.0131+ x

Therefore,

(0.0131+x)/(0.0854 - 2x)^2= 227.27

0.0131 + x =( 0.007293 + 4x^2 - 0.3416x )227.3

0.0131 + x = 1.6577 + 909.2x^2 - 77.65x

909.2x^2 - 78.65x + 1.6446= 0

x = 0.0353

[Br] = 0.0854 - (2 × 0.0354) = 0.0146M

[ Br2] = 0.0131 + 0.0354= 0.0485M