Answer:
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.