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