Br2(g) + 3 F2(g) ⇄ 2 BrF3(g) Kp = 5.4 × 108 0.30 atm of Br2 and 0.60 atm of F2 are placed in a 3.0 L container and the system is allowed to reach equilibrium. Calculate the pressureof Br2 at equilibrium.

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:

= 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

i do not get it   at all