Answer : The correct option is, [tex][I_2]<[IBr][/tex]
Solution : Given,
Moles of [tex]I_2[/tex] = Moles of [tex]Br_2[/tex]
[tex]K_{eq}=280[/tex]
The balanced equilibrium reaction is,
[tex]I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)[/tex]
The expression for equilibrium constant is,
[tex]K_{eq}=\frac{[IBr]^2}{[I_2][Br_2]}[/tex]
As per question, Moles of [tex]I_2[/tex] = Moles of [tex]Br_2[/tex]
Now put all the given values in above formula, we get the relation between the [tex][IBr][/tex] and [tex][I_2][/tex].
[tex]280=\frac{[IBr]^2}{[I_2][I_2]}[/tex]
[tex]280=\frac{[IBr]^2}{[I_2]^2}[/tex]
[tex]\sqrt{280}=\frac{[IBr]}{[I_2]}[/tex]
[tex][IBr]=16.733\times [I_2][/tex]
From this we conclude that the concentration of [tex]IBr[/tex] is greater than the concentration of [tex]I_2[/tex].
Therefore, the correct option is, [tex][I_2]<[IBr][/tex]
