Question

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.

Answer 1

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:

$Kp = \frac{P_{BrF_3}^2}{P_{Br_2}P_{F_2}^3}$ = 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

Answer 2

Answer:

$p_{Br_2}=0.11atm$

Explanation:

Hello,

In this case, we write the equilibrium law of mass action as shown below:

$Kp=\frac{p_{BrF_3}^2}{p_{Br_2}p_{F_2}^3}$

In such a way, we can rewrite it in terms of the partial pressures at the beginning and due to change $x$:

$5.4x10^{8}=\frac{(2x)^2}{(0.30-x)(0.60-3x)^3}$

Hence, solving for $x$ we obtain:

$x_1=0.19atm\\x_2=0.30atm$

So the feasible solution is 0.19 atm since the other solution produce a negative pressure of fluorine at equilibrium. Therefore, the pressure of bromine at equilibrium is:

$p_{Br_2}=0.30atm-0.19atm\\\\p_{Br_2}=0.11atm$

Best regards.