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