Answer:
= 0.0050 M
= 0.0155 M
Explanation:
Initial moles of
= 0.072 mole
Volume of container = 3.9 L
Initial concentration of
The given balanced equilibrium reaction is,

Initial conc. 0.018 M 0
At eqm. conc. (0.018-x) M (2x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[I]^2}{[I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BI%5D%5E2%7D%7B%5BI_2%5D%7D)

we are given : 
Now put all the given values in this expression, we get :


So, the concentrations for the components at equilibrium are:
![[I]=2\times x=2\times 0.0025=0.0050](https://tex.z-dn.net/?f=%5BI%5D%3D2%5Ctimes%20x%3D2%5Ctimes%200.0025%3D0.0050)
![[I_2]=0.018-x=0.018-0.0025=0.0155](https://tex.z-dn.net/?f=%5BI_2%5D%3D0.018-x%3D0.018-0.0025%3D0.0155)
Hence, concentrations of
and
are 0.0050 M ad 0.0155 M respectively.