± Heterogeneous Equilibrium of Ammonium Bisulfide Ammonium bisulfide, NH4HS , fo

Question

± Heterogeneous Equilibrium of Ammonium Bisulfide

Ammonium bisulfide, NH4HS , forms ammonia, NH3 , and hydrogen sulfide, H2S , through the reaction

NH4HS(s)NH3(g)+H2S(g)

This reaction has a Kp value of 0.120 at 25 C .

An empty 5.00-L flask is charged with 0.350 g of pure H2S(g) , at 25 C .

Part B

What is the initial pressure of H2S(g) in the flask?

Express your answer numerically in atmospheres.

P= 5.02 * 10 ^-2 atm

I was able to get this far on my own, but now im completely lost for part C-E

Addition of ammonium bisulfate

In addition to the H2S already present in the flask, solid NH4HS is added until there is excess unreacted solid remaining.

Part C

What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S , respectively?

Enter the partial pressure of ammonia followed by the partial pressure of hydrogen sulfide numerically in atmospheres separated by a comma

Part D

What is the mole fraction, , of H2S in the gas mixture at equilibrium?

Express your answer numerically.

Part E

What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.350 g of pure H2S(g) , at 25 C to achieve equilibrium?

Express your answer numerically in grams.

P= 5.02 * 10 ^-2 atm

I was able to get this far on my own, but now im completely lost for part C-E

Addition of ammonium bisulfate

In addition to the H2S already present in the flask, solid NH4HS is added until there is excess unreacted solid remaining.

Part C

What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S , respectively?

Enter the partial pressure of ammonia followed by the partial pressure of hydrogen sulfide numerically in atmospheres separated by a comma

Part D

What is the mole fraction, , of H2S in the gas mixture at equilibrium?

Express your answer numerically.

Part E

What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.350 g of pure H2S(g) , at 25 C to achieve equilibrium?

Express your answer numerically in grams.

Explanation / Answer

Initial pressure of H2S

i.e pH2S = 0.0502 atm [from question]

c]

From this last equilibrium, you know that

Kp = 0.120 = PNH3 X PH2S

You also know, as you stated, that at equilibrium, PNH3 = x and PH2S = 0.0502 + x. Pllugging these into the expression for Kp gives
0.120 = (x) (0.0502 + x)
Multiply through and rearrange to get the equation:
x^2 + 0.0502x - 0.120 = 0
Use the quadratic formula to solve for x.
One of the roots to this equation is: x = 0.322

So, PNH3 = 0.322 atm and PH2S = 0.3722 atm (These are the partial pressures of each gas at equilibrium).

d]

Mole fractions are related to partial pressures,

Partial pressure = Mole fraction * Total pressure

mole fraction H2S = 0.3722 atm / 0.3722+0.322 atm = 0.536

Mole fraction of NH3 = 0.464

e]

Moles of NH3 at equilibrium = ?

PV = nRT

P = 0.322 atm

n = PV /RT = 0.0658

Moles of NH3 = Moles of NH4HS to be added

Mass of NH4HS = 0.0658*35 = 2.303 gms

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