Answer : The equilibrium constant
for the reaction is, 0.1133
Explanation :
First we have to calculate the concentration of
.
![\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DBr_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DBr_2%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
![\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DBr_2%3D%5Cfrac%7B1.35moles%7D%7B0.780L%7D%3D1.731M)
Now we have to calculate the dissociated concentration of
.
The balanced equilibrium reaction is,
![Br_2(g)\rightleftharpoons 2Br(aq)](https://tex.z-dn.net/?f=Br_2%28g%29%5Crightleftharpoons%202Br%28aq%29)
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
= ![C\alpha=1.731M\times \frac{1.2}{100}=0.2077M](https://tex.z-dn.net/?f=C%5Calpha%3D1.731M%5Ctimes%20%5Cfrac%7B1.2%7D%7B100%7D%3D0.2077M)
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 :
![K_c=\frac{(0.4154)^2}{1.5233}=0.1133](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%280.4154%29%5E2%7D%7B1.5233%7D%3D0.1133)
Therefore, the equilibrium constant
for the reaction is, 0.1133