For the reaction i2(g)+br2(g)←−→2ibr(g), kc=280 at 150 ∘c. suppose that 0.450 mol ibr in a 2.00-l flask is allowed to reach equilibrium at 150 ∘c. what is the equilibrium concentration of 2ibr, i2, br2

I₂ (g) + Br₂ (g) ↔ 2IBr (g)

Given is Kc = 280 at 150 degree C.

Kc = [IBr]² / [I₂][Br₂]

Initially [IBr] = 0.45 / 2 = 0.225 M

The actual reaction is:

2IBr ↔ I₂ + Br₂, Kc = 1/280 = 0.00357142

In equilibrium,

[IBr] = 0.225 - 2x

[I₂] = x

[Br₂] = x

By substituting the values we get,

K = [I₂] [Br₂] / [IBr]²

0.00357142 = x×x / (0.225 - 2x)²

√(0.00357142) = x / (0.225 - 2x)

0.0597613 (0.225 - 2x) = x

0.01344 - 2 × 0.0597613x = x

(1 + 2 × 0.0597613)x = 0.01344

x = 0.01344 / (1+2 × 0.0597613)

x = 0.01344 / 1.1195226

x = 0.012005

Substituting the values we get,

IBr = 0.225 - 2 × 0.012005 = 0.17698 M

I₂ = x = 0.012005 M

Br₂ = 0.012005 M

i'm assuming that the correct answer is indeed the first one/ "a"..