Answer:
[tex][I_2]=[Br]=0.31M[/tex]
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:
[tex]I_2+Br_2\rightleftharpoons 2IBr[/tex]
Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:
[tex]K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}[/tex]
Thus, we solve for x as show below:
[tex]\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M[/tex]
Therefore, the concentrations of both bromine and iodine are:
[tex][I_2]=[Br]=2.0M-1.69M=0.31M[/tex]
Regards!
