Question

What is the equilibrium concentration of ICl if 0.45 mol of I2 and 0.45 mol of Cl2 are initially mixed in a 2.0-L flask

Answer 1

`[ICI]_(eq)=0.271M`

Step-by-step explanation:

Hello,

In this case, considering that the equilibrium constant for the following reaction:

`I_2+Cl_2\rightleftharpoons2ICl`

Is Kc=9.09, we can compute the equilibrium concentrations by the ICE procedure, so we first compute the initial concentrations:

`[I_2]_0=[Cl_2]_0=(0.45mol)/(2.0L)= 0.225M`

Next, we write the law of mass action in terms of the change
`x` due to the reaction extent:

`Kc=([ICI]^2)/(([I_2])([CI_2])) \\\\Kc=((2x)^2)/(([I_2]_0-x)([CI_2]_0-x)) \\\\9.09=((2x)^2)/((0.225-x)(0.225-x))`

Thus, solving by using solver we have:

`x=0.135M`

Therefore, equilibrium concentration of ICl is:

`[ICI]_(eq)=2*0.135M`

`[ICI]_(eq)=0.271M`

Regards.