For the reaction i2(g)+br2(g)←−→2ibr(g), kc=280 at 150 ∘c. suppose that 0.520 mol ibr in a 2.00-l flask is allowed to reach equilibrium at 150 ∘c. part a what is the equilibrium concentration of ibr?

First, we have to get the initial [IBr] = 0.520 mol / 2 L =0.26 M
According to the reaction balanced equation:
and using ICE table :
             I2(g) + Br2(g)↔ 2IBr(g)
initial     0           0            0.26
Change +X        +X            -2X
Equ          X          X          (0.26-2X)

when Kc = [IBr]^2/[I2][Br2]

so by substitution:
280 = (0.26-2X)^2 / X^2 by solving this equation for X
∴X = 0.0139 
∴[I2] = [Br]2 = 0.0139
and [IBr] = 0.26- (2*0.0139)
              = 0.2322 M

The correct answer is b.hoped i you