Answer:
[tex][ICI]_{eq}=0.271M[/tex]
Explanation:
Hello,
In this case, considering that the equilibrium constant for the following reaction:
[tex]I_2+Cl_2\rightleftharpoons2ICl[/tex]
Is Kc=9.09, we can compute the equilibrium concentrations by the ICE procedure, so we first compute the initial concentrations:
[tex][I_2]_0=[Cl_2]_0=\frac{0.45mol}{2.0L}= 0.225M[/tex]
Next, we write the law of mass action in terms of the change [tex]x[/tex] due to the reaction extent:
[tex]Kc=\frac{[ICI]^2}{([I_2])([CI_2])} \\\\Kc=\frac{(2x)^2}{([I_2]_0-x)([CI_2]_0-x)} \\\\9.09=\frac{(2x)^2}{(0.225-x)(0.225-x)}[/tex]
Thus, solving by using solver we have:
[tex]x=0.135M[/tex]
Therefore, equilibrium concentration of ICl is:
[tex][ICI]_{eq}=2*0.135M[/tex]
[tex][ICI]_{eq}=0.271M[/tex]
Regards.
