F₂ + H₂ ⇔ 2HF
initial : [F2] = 3/1,5 = 2mol/l [H2] = 2mol / l [HF] = 0
reactioneaza: F2 : x moli H2 : x moli se obtin 2x moli HF
la echilibru : [F2] = (2-x) mol/l [H2] = (2 - x) [HF] = 2x
[HF] /([F2]·[H2]) = 1,15·10⁻²
4x² / (2-x)² = 1,15·10^-2
2x/(2-x) = 1,072/10 ≈ 0,1
2x²- 4x + 0,1 = 0 x² - 2x + 0,0536 = 0 ⇒ x = 1,9728
[F2] = [H2] = 0,027 moli/l [HF] = 3,9456 mol/l