Really Quick Equilibrium Question?

Question

Hi,
I'm sure you guys are getting tired of my equilibrium questions by now... but here's a last really quick one:
For:
NH4HS (s) <--> NH3 (g) +H2S (g)

If you add more NH4HS (s) to the container, does the partial pressure of NH3 (g):
1- Stay the same because the air is already "saturated" with the NH3 and H2S gas, so that no more of the solid can reach the gaseous state,
or
2- Increase because more NH4HS is trying to reach a gaseous state, separating into more NH3 and H2S gas particles, therefore increasing the partial pressure of NH3 gas in the air?

Thanks in advance!

Answer 1

If we consider the previous solutions of your equilibrium questions related to the same reaction which is
NH4HS (s) <--> NH3 (g) +H2S (g)
together with the general requirements of an equilibrium system which are as follows;
(a) The temperature should be constant
(b) The system should be closed,
we can easily arrive at the following conclusions:
(1) The equilibrium constant expression for this reaction in terms of partial pressures is
Kp = P(NH3) x P(H2S)
ALSO, The equilibrium constant expression for this reaction in terms of concentrations will be
Kc = [NH3][H2S]
These expressions show that the amount of solid NH4HS has no effect upon the equilibrium system already established, that is, the addition of more solid does not disturb the equilibrium.
(2) However, addition of any one of the gaseous products will disturb the equilibrium. Because, at a certain temperature, Kp has a fixed value which is obtained by the product of the partial pressures of NH3 and H2S. This means that the total pressure inside the reaction vessel has a certain value. At the same temperature, addition of any one of the gaseous substances increases the total pressure inside the vessel. This means that the product of the partial pressures of gases will give a new Kp value which is greater than the previous one. But this is impossible, because Kp changes with temperature only. Hence, to satisfy the original Kp value, the equilibrium shifts to left to decrease the total pressure in the vessel.
(3) The reverse of this situation does not occur. That is, if you add more solid, you cannot get the more gaseous product. Because, the case explained above occurs. Total pressure increases and if the temperature is constant (not mentioned in the question , but it should be), then to satisfy the original value of Kp the gaseous products are again converted to the solid. Therefore, partial pressures of both NH3 and H2S do not change.
(4) If, however, the volume of the reaction vessel is increased, since V and P are inversely proportional, total pressure decreases and more solid is converted into gaseous products. BUT..... This does not mean that the partial pressures of the gaseous products will increases. Whenever, the partial pressure reach their original values to give the original Kp, the reaction stops and the equilibrium is attained.
I don't know that this explanation is satisfactory for you. If not, please let me know.
By the way, happy new year.

Answer 2

when more solid is added then the equilibrium will shift to right.so more nh3 will form and its partial pressure will increase.

Answer 3

MathPhysChem is correct. It's a little bit like why adding more solid sodium chloride to a saturated solution doesn't give you any more Na+ and Cl- ions floating around.

Answer 4

1- Stay the same because the air is already "saturated" with the NH3 and H2S gas, so that no more of the solid can reach the gaseous state,

because NH4HS is solid , it will not matter to gas or (aq)

Answer 5

nicely possibly it helps in case you generally draw the vectors of the strain. then you certainly will see a ninety° triangle. you generally use pythagoras in case you wanna get the element of the triangle on the different element of the the best decision attitude. in case you have that section and a attitude you ought to use your sine and cosine. trouble-free trigonometry