AP Chemistry Exam Question About Equilibrium?

Question

NH4HS(s)  NH3(g) + H2S(g) Hº = +93 kilojoules
The equilibrium above is established by placing solid NH4HS in an evacuated container at 25ºC. At equilibrium, some solid NH4HS remains in the container. Predict and explain each of the following.
(a) The effect on the equilibrium partial pressure of NH3 gas when additional solid NH4HS is introduced into the container
(b) The effect on the equilibrium partial pressure of NH3 gas when additional solid H2S is introduced into the container
(c) The effect on the mass of solid NH4HS present when the volume of the container is decreased
(d) The effect on the mass of solid NH4HS present when the temperature is increased.


Thank you!

Answer 1

Equilibrium is a balancing act. A stable system in equilibrium will try to stay in equilibrium. Equlibrium is usually measured as having a particular set of concentrations (which are temperature and pressure dependent).

If you apply stress to the system, the system responds in a way to relieve the stress (after some period of time) and restore the equilibrium.

In this system, you have a solid that turns into a gas and some energy.

If you increase the pressure, the gases are driven back towards the solid. If you add more solid, the system responds by converting some of it into gases to re-establish the equilibrium.

I hope this helps you figure out the answer for yourself. Good luck.

Answer 2

1. Equilibrium is for the gas phase components only. Adding more H-ammonium sulfide in the solid phase does nothing.
The equilibrium term is Kp=[NH3][H2S]
2. Something is screwed up. Solid H2S doesn't exist at 25 degC. Adding H2S(g) would drop the partial pressure of NH3 as more NH4HS is ppt out.
3. Shrinking the tank would cause NH4HS to ppt from the gas phase. The conc of the gases is increased by the shrinkage, and the reaction must reverse to re-establish the equilibrium.
4. Increasing temp increases partial pressure and conc of both gases. To compensate, NH4HS solid mass would increase. [however, Kp is a function of temp, so the quanitative increase would be hard to compute]

Answer 3

i be attentive to a skill to try this, yet i'm not sure whether you be attentive to of this. If my answer is authentic, you could nicely, settle for this equipment. H2(g) + I2(g) <-> 2 hi(g) At t = 0, concentrations: 0.4 0.4 0.4 At t = eq., concentrations: 0.4(a million-x) 0.4(a million-x) 0.4(a million+2x) Given Kc = 9.0 --> ( 0.4(a million+2x) )^2 / ( (0.4(a million-x)) )^2 (utilising the expression for equilibrium consistent) fixing the above, x = 8/11. So, [hi] at equilibrium = 0.4(a million+2x) = 0.4(a million+(sixteen/11)) fixing, [hi] = 0.ninety 8 M and since the quantity of field is a million L, type of moles of hi = 0.ninety 8