I don't see how you came to the conclusions you did about what I said. First there is no such thing as a rate of chemical equilibrium. There is a rate of approach to equilibrium, but that is a kinetics problem that depends on the free energy differences between the different phases involved in the process and the specific mass transfer mechanism involved if the process is a mass transfer process like saturating a mass of diesel fuel with water. That said, for a tank of diesel fuel to become water saturated, there must be a pathway for water to get to the fuel. No mater what the pathway looks like geometrically, the fuel will eventually become water saturated if it is in contact with a source of water at a sufficiently high concentration. However, the geometry of the pathway controls the rate at which water vapor can access the fuel and thus the time required to achieve equilibrium. In otherwords, in unit time, a batch of fuel with a large surface exposed to air that contains water vapor will absorb much more water than a volume of fuel exposed to the same air over a very small area. If the air and fuel are static, the mass transfer mechanism for moving water from the air to the fuel is diffusion across a plane interface. That process actually involves two diffusion processes, diffusion of water vapor in the air to the interface and diffusion of dissolved water in the oil away from the interface. However, since the diffusion coefficient of water vapor in air is many times larger than that in oil, the air reservoir is effectively at constant water vapor concentration relative to the diesel fuel reservoir. In that case, the rate limiting diffusivity is the diffusivity in oil.

You misconception that the rate of mass transfer doesn't depend on the area of the interface likely comes from only seeing solutions to the diffusion equations written in terms of concentration. The area term does not show up in the concentration solution because it is burried in the concentration term itself. However, if you recast the equations into terms of mass transferred rather than concentration, the area term jumps out at you. Might I suggest that you give "The Mathematic of Diffusion", by J. Crank (Oxfor University Press) a quick perusal if you don't see this.

For your information, I did not discuss condensation on tank walls. That is a process that occurs when the air in a tank becomes water saturated due to a temperature drop in the surroundings and water condenses out of the air onto the relatively cold tank walls. That water, once the droplets reach sufficient size, will run down the tank walls to the air/oil interface. The droplets will sit there for a while until they become large enough to overcome surface tension effects, then they will sink to the bottom of the tank. This process can lead to a pool of water at the bottom of the fuel tank even though the fuel oil itself is not water saturated. That said, once water pools at the bottom of the tank, the water will gradually be abosrbed into the fuel until the fuel becomes saturated (assuming there is enough water for saturation - this depends on relative volumes of water and oil in the tank). A nearly empty tank will achieve saturation through this mechanism much more quickly than a full tank, simply because there is less oil in the empty tank and less water has to be transferred into the oil to achieve saturation. Again, the rate of approach to saturation depends on the mass transfer mechanism of the water from the pooled water into the oil. For static oil in a tank, thet mass transfer process is diffusion of water vapor in the oil. The rate of approach to saturation will be increased of the fuel is stirred in any way, for example by thermal effects, i.e., convective mixing.