Answer:
The concentration of F₂in equilibrium is 4,93x10⁻₃M
Explanation:
For the reaction:
2 OF₂ ⇄ 2 F₂ + O₂
The equilibrium constant, k, is defined as:
k = [F₂]²[O₂] / [OF₂]²
If you have in equilibrium [OF₂] = 2,49x10⁻² and [O₂] = 1,58x10⁻²
The concentration of F₂ is:
6,19x10⁻⁴ = [F₂]²[1,58x10⁻²] / [2,49x10⁻²]²
2,43x10⁻⁵ = [F₂]²
4,93x10⁻³ = [F₂]
I hope it helps!
