Well first some key points.
First, your pressure would not be 2.5 atm, it would be higher. Your world, assuming a distribution equal to that of the Earth, should have a gravity of approximately 3.3 g since you have a world 4 times more massive, although somewhat larger than the Earth (I do not want to do the algebra to calculate the gravity of your world exactly, if you are interested, use Newton's equation a = GM / R^2).
Now we need to know exactly the height, density and composition of your atmosphere, let's say your atmosphere is proportional to the earth's with your modification and the composition of the first layer is 50% diatomic nitrogen, 12% nitrogen oxides, 8% water, 4% CO2 and 26% other gases (warning here, I'm about to put up a wall of text with numbers and data for whoever cares, feel free to skip it) (note this is to one atmosphere what a spherical cow in a vacuum is to a cow) here the densities:
N2: 1.2506 g/ml
NOx: 1.58 g/ml
H2O: approx 1 g/ml
CO2: 0.001976 g/ml
(Note that ml is equivalent 1 to 1 to cm3)
75% of the atmosphere is in the first 11 km on Earth, let's say that in your world it is in the first 20 km (to round off) and the diameter of the Earth is approximately 13,000 km (INTA data) then in your world it would be 29,900 km then the volume of your atmosphere would be 4/3pir^3 (I'm not using the integral for a hollow sphere, instead I subtracted the volume of the rock from the total volume) 1.12195E14 km3 - 1.1197E14 km3 = 225000000000 km3 = 2.25e+26 ml which after calculating gives 2.0137E23 kg on average (I'm calculating quickly) then the average pressure in your world (I'm taking 1g=10m/s2 then in your world it would be 33m/s2) (2.0137E23* 33)/( 29900000^24pi)= 591502343.02954 Pa or 5837.67424652889167 atm (this number is very inflated by many simplifications I made, it's probably between 1000 to 1200 atm because neither the density is the same, the atmosphere may be thinner or the mass may be much smaller etc. But as I said your world is very small and very massive).
If I took the time to calculate all this it is for an important reason and that is that playing with the proportions of a planet is a problem, that value makes sense for the conditions of your world, but because your world is in a very uncomfortable place where it is very massive but very small. In your case, either you calculate everything very well or you only use one pressure. I will say that your world has 2.5 atm and I will ignore all this, but let it be clear that based on the data you gave, this would not be the case.
(Gosh, I did a lot of quick calculations, I apologize in advance for any mistakes, if I find any I'll correct them)
So second question, partial vapor pressures, well pure ammonia NH3 is liquid at -15 °C and 2.5 atm, that means 2 things, first, ammonia is not a strong base, it is a weak base (if you ever have a doubt about whether or not it is a strong base, see if it has an OH, if it doesn't have one and it is not an alkaline metal, it is almost certainly a weak base), the second thing it means is that it is very cold, under those conditions many natural rocks would not be affected, here is a list of the types of rocks that would be predominant:
Silicon oxides (quartz, obsidian, etc.)
Iron oxides (almost all rocks contain iron oxides or sulfates)
Basalts (these are almost all minerals formed by high pressure at the bedrock
level and are usually exposed in areas of high volcanic activity)
Carbonates (water is much worse than ammonia)
Essentially, water is worse than ammonia when it comes to rocks, and the low temperature is a problem when it comes to dissolving things. At those temperatures, things like methanol, CO2, and even oxygen would be found as liquid or solid.
On the other hand, with those pressures and temperatures you would have very interesting rocks, such as stable nitrates (things like N3Fe or N2Ca could exist)
My best advice: Look at Titan, that moon is very similar to what you want, and it defines your planet very well because I suspect that even with my truncated mathematics I found a serious flaw in the approach.