Soaking times are presupposed on the type, amount and time of fix used and if a wash aid is incorporated in a post fix bath as well as volume of water used.
The reason fiber paper needs to be washed so long is to remove to final few percent of fixer that setttles in the paper base. Most fixer that is in the emulsion is cleared with a wash agent and in the first few minutes of a rinse in running water or a print washer. That is why the original Ilford archival sequence used a rapid fixer (ammonium thiosulfate) for only 30 seconds at a 1-3 dillution. the purpose being to prevent as little fix as possible to penetrate the paper base. Ilford at one time used a 5 minute final wash with constant water flow over the surface. I think the current time is more. Ilford later modified its fix times to 60 without changing the amount of fix taken up by the base.
As far as soaking vs constant flow of water over the prints I have read several articles in Photo Techniques, in books and on line about the subject. Here is what the consensus seems to be.
Removing fix from a print is a diffusion process. Once fix is leeched from the paper it does not re-enter it. therefore the only purpose of keeping prints apart during the wash is so they don't stick together. it doesn't matter is you add new prints, the old ones will not abosrb the additional fixer.
However, the more fix is in the water, the less efficient the process becomes. The key is the interface between the water and the paper surface. I won't bore you with the chemsitry because I don't remember it all but simply put the less fix in the water at the interface, the more fresh water for fix in the print to leech into.
That leads to the third point about agitiation. If there is no agitation or water flow, to constantly move fresh water over the surface, the longer the procedure takes.
Finally, you don't need to constantly add fresh water and remove the old water. You do need to have a sufficient supply of water to keep it fairly clear of fix for that volume. However, large water volume is of no use in a system that exchanges water since you are constantly replenishing the water at the interface. The most efficient system would be to have a constant sheet of water flowing over both sides of the paper without a resevior.
So for the original question it depends on how big a tray, and if you agitate it any and use an HCA. I don't think one change is enough. If I wash a big print I run it under a shower head front and back for about 10min after HCA and then soak for 45 min with occasional agitiation and 15 min intervals in 20x24 inch tray.
I also have a 20x24 tay which I drilled holes in one end to allow very slow drainage and a very slow water flow to sort of exchange the water volume every 15min. Don't know if it is any better, but it is what I use nowfor large prints.