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