Question
NH4HS(s) NH3(g) + H2S(g) ∆H_ = +93 kilojoules
The equilibrium above is established by placing solid NH4HS in an evacuated container at 25°C. 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.
The equilibrium above is established by placing solid NH4HS in an evacuated container at 25°C. 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.
Answers
You need an arrow. How do you which are the products and which the reactants?
a) If there is solid NH4HS initially, adding more solid NH4HS will not cause the equilibrium to change; therefore, no effect.
b) Confusing. There is NO solid H2S present initially; therefore, how can ADDITIONAL solid H2S be added.
c)Volume is decreased, pressure is increased. Equilibrium changes to the side with the smaller number of moles; therefore, the equilibrium will shift to the left and NH3 will be decreased, H2S is decreased, mass of NH4HS is increased.
d) This is an endothermic reaction which can be written as
NH4HS(s) + heat ==> NH3(g) + H2S(g)
Adding heat will shift the equilibrium to the right; therefore, the mass of the NH4HS will decrease.
a) If there is solid NH4HS initially, adding more solid NH4HS will not cause the equilibrium to change; therefore, no effect.
b) Confusing. There is NO solid H2S present initially; therefore, how can ADDITIONAL solid H2S be added.
c)Volume is decreased, pressure is increased. Equilibrium changes to the side with the smaller number of moles; therefore, the equilibrium will shift to the left and NH3 will be decreased, H2S is decreased, mass of NH4HS is increased.
d) This is an endothermic reaction which can be written as
NH4HS(s) + heat ==> NH3(g) + H2S(g)
Adding heat will shift the equilibrium to the right; therefore, the mass of the NH4HS will decrease.
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