Question

The equilibrium constant K_c for the reaction below is 0.00173 at a certain temperature.

Br₂(g) ⇌ 2Br(g)

If the initial concentrations are

[Br₂] = 0.0393 M

and

[Br] = 0.0940 M,

calculate the concentrations of these species at equilibrium.

[Br2]	=	M
[Br]	=	M

Answer 1

Br2 --> 2 Br

using ICE table

initial conc of Br2 , Br are 0.0393 , 0.094

change in conc of Br2 , Br are -x , +2x

equilibrium conc of Br2 , Br are 0.0393 -x , 0.094 +2x

now

Kc = [Br]^2 / [Br2]

0.00173 = ( 0.094 + 2x)^2 / (0.0393 - x)

(0.094 + 2x)^2 = 0.00173 ( 0.0393 - x)

4x2 + 0.376x + 0.008836 = 0.000067989 - 0.00173x

4x2 + 0.37773 x + 0.008768011 = 0

x = -0.0411

now

at equilibrium

[Br2] = 0.0393 - x = 0.0393 + 0.0411 = 0.0804 M

[Br] = 0.094 + 2x = 0.094 + ( 2 x -0.0411) = 0.0118 M