The equilibrium vapor pressure for benzene is 94.4 mm Hg. When liquid benzene is in equilibrium with its vapor, we must have ________

Question

Quizplus · Accepted Answer

At equilibrium, the rates of evaporation and condensation are equal, which means the system is at dynamic equilibrium. In this process, we have no change in the free energy of the system, i.e., ΔG = 0. Therefore, ∆G° = - RTlnK = 0. Since K = P / P°, where P is the vapor pressure and P° is the standard pressure, we can write: 
ΔG° = - RTln(P / P°) = 0. Solving for ln(P / P°) = 0, we get P / P° = 1, which means P = P°. Therefore, at equilibrium, the vapor pressure of the liquid benzene should be equal to its equilibrium vapor pressure, which is 94.4 mm Hg. This can only be true when the standard free energy change is positive, ∆G° > 0, indicating that the forward reaction of vaporization is not favored under standard conditions. Hence, the correct answer is B, (F1F1F1S1F1F1F10G = 0 and F1F1F1S1F1F1F10G° > 0.)

The Equilibrium Vapor Pressure for Benzene Is 94 Δ\DeltaΔ G = 0 And

The Equilibrium Vapor Pressure for Benzene Is 94 $\Delta$ G = 0 And