Answer:
The pressure at equilibrium of Br₂ is 0.10048 atm
Explanation:
Based on the reaction:
Br₂(g) + 3 F₂(g) ⇄ 2 BrF₃(g)
Kp is defined as:
[tex]Kp = \frac{P_{BrF_3}^2}{P_{Br_2}P_{F_2}^3}[/tex] = 5.4x10⁸
If initial pressures of Br₂ and F₂ are 0.30atm and 0.60atm respectively, the pressures in equilibrium are:
Br₂ = 0.30atm - X
F₂ = 0.60atm - 3X
BrF₃ = 2X
Replacing in Kp formula:
5.4x10⁸ = [2X]² / [0.30atm - X] [0.60atm - 3X]³
5.4x10⁸ = 4X² / [0.30 - X] [0.216 - 3.24 X + 16.2 X² - 27 X³]
5.4x10⁸ = 4X² / 0.0648 - 1.188 X + 8.1 X² - 24.3 X³ + 27 X⁴
Solving for X:
X = 0.3000 → False answer because produce negative concentrations.
X = 0.19952. Replacing in equation of Br₂:
Br₂ = 0.30atm - 0.19952atm = 0.10048 atm
