Question

Consider the following reaction: A2 (g) + B2 (g) ⇄ 2AB (g) Kp = 25.6 The three species are mixed in a 5.000 L flask and have the following partial pressures: PAB = 0.520 atm, PA₂ = 0.020 atm, PB₂ = 0.450 atm. What is the equilibrium partial pressure of AB in atm? Report your answer with four decimals. Enter numbers only; do not enter units.

Answer 1

To find the equilibrium partial pressure of AB in the reaction A2 (g) + B2 (g) ⇄ 2AB (g), we can use the given equilibrium constant (Kp) and the partial pressures of A2, B2, and AB.

The equilibrium constant expression for this reaction is:

Kp = (PAB)^2 / (PA₂ * PB₂)

Given:

Kp = 25.6

PAB = 0.520 atm

PA₂ = 0.020 atm

PB₂ = 0.450 atm

Substituting the given values into the equilibrium constant expression, we have:

25.6 = (0.520)^2 / (0.020 * 0.450)

Simplifying the equation:

25.6 = 0.2704 / 0.009

To find the equilibrium partial pressure of AB, we rearrange the equation:

(PAB)^2 = 25.6 * (PA₂ * PB₂)

Substituting the given values:

(PAB)^2 = 25.6 * (0.020 * 0.450)

Simplifying:

(PAB)^2 = 0.2304

Taking the square root of both sides:

PAB = √0.2304

PAB ≈ 0.4800 atm

Therefore, the equilibrium partial pressure of AB in atm is approximately 0.4800, with four decimal places.