Answer : The equilibrium constant
for the reaction is, 0.1133
Explanation :
First we have to calculate the concentration of
.


Now we have to calculate the dissociated concentration of
.
The balanced equilibrium reaction is,

Initial conc. 1.731 M 0
At eqm. conc. (1.731-x) (2x) M
As we are given,
The percent of dissociation of
=
= 1.2 %
So, the dissociate concentration of
= 
The value of x = 0.2077 M
Now we have to calculate the concentration of
at equilibrium.
Concentration of
= 1.731 - x = 1.731 - 0.2077 = 1.5233 M
Concentration of
= 2x = 2 × 0.2077 = 0.4154 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be :
![K_c=\frac{[Br]^2}{[Br_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBr%5D%5E2%7D%7B%5BBr_2%5D%7D)
Now put all the values in this expression, we get :

Therefore, the equilibrium constant
for the reaction is, 0.1133