[RAYMONDg]

Two separate one litre vessels both at the same pressure (100kPa) and same temperature. One contains NH3 vapour and the other contains Hydrogen Vapour:

The complete contents of Hydrogen is injected into the NH3 vessel is the resulting sum pressure 200kPa or is the pressure greater because molecules have volume?

Is the resulting mixture homogeneous or stratified?

[fledarmus]

And your reasoning would be...

[Bryan Sanctuary]

Mixing two or more non-reactive gases always results in a homogeneous phase.  The gases mix and in the absence of any reactions, they simply diffuse together. Liquids and solids can be heterogeneous but gases are always in a homogeneous phase.

If the homogeneous mixture acts as an Ideal Gas then the volume will double to 200 kPa. That is Dalton's law of partial pressures.  When gases become non-ideal, think of the van deer Waals equation, then there is a volume change upon mixing.

Let us assume the mixture acts like an ideal gas.

However in this case, adding H₂ to NH₃ causes a reaction.  In fact a well established way of making ammonia is the Haber process http://en.wikipedia.org/wiki/Haber_process: catalysts and raising the temperature helps, but they will still react under standard conditions--think of le chatelier principle.

The way to increase the production of ammonia is to maintain the hydrogen and nitrogen concentrations high and bleed off the product: ammonia.  In this case we have no nitrogen present so ammonia will decompose.

Actually ammonia would not be pure in the original vessel, but rather decompose to an equilibrium mixture:

N₂ (g) + 3 H₂ (g) ⇌ 2 NH₃ (g)

However adding a lot, equal pressure, of hydrogen to ammonia will shift the equilibrium.

I will not do the calculation, but with 100 kPa of H₂ and 100 kPa of NH₃ you can find the number of moles of each, say n_H2 and n_NH3.  Then we have:

                N₂ (g) + 3 H₂ (g) ⇌ 2 NH₃ (g)

                   0            n_H2              n_NH3
 equil:          α         n_H2 + 3α      n_NH3 - 2α

So at equilibrium the total number of moles is:

n_Total = n_H2  + n_NH3+ 2α

Find the partial pressures from Dalton's Law in terms of the mole fractions:

P_i/P_Total=X_i

So using P=nRT/V,

P_Total=(n_H2  + n_NH3+ 2α)RT/V

So you can see that the total pressure of the mixture is indeed the sum of the nitrogen and ammonia pressure, except for the amount of nitrogen formed, α.

To find α use the equilibrium expression:

K_P =P_NH3²/[P_H2³N_NH3]

Since the equilibrium constant is quite small at, say 300 K, I would expect that α will be quite small and so likely the total final pressure will be about 200 kPa, but to be sure one needs to determine α from the equilibrium constant at the temperature given.

[Caustikola]

Well, for a start, you  may need to specify the temperature. Ammonia will not be decomposing at 300K so you won't have any Nitrogen, and no reaction as well. The pressure will roughly double, as you suggest