Answer: The equilibrium constant is [tex]3.3\times 10^{-4}[/tex]
Explanation:
Initial concentration of [tex]I_2[/tex] = 0.095 M
The given balanced equilibrium reaction is,
[tex]I_2(g)\rightleftharpoons 2I(g)[/tex]
Initial conc. 0.095 M 0 M
At eqm. conc. (0.095-x) M (2x) M
Given : 2x = 0.0055
x = 0.00275
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[l]^2}{[I_2]}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(0.0055)^2}{(0.095-0.00275)}[/tex]
[tex]K_c=\frac{(0.0055)^2}{0.09225}=0.00033[/tex]
Thus the equilibrium constant is [tex]3.3\times 10^{-4}[/tex]