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Ammonium bisulfide, NH4HS , forms ammonia, NH3 , and hydrogen sulfide, H2S , thr

ID: 880319 • Letter: A

Question

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 0.300 g of pure H2S(g) , at 25 C .

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PART A

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.

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PART B

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

Express your answer numerically.

H2S = ?

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PART C

What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.300 g of pure H2S(g) , at 25 C to achieve equilibrium?

PNH3 , PH2S = _____ atm

Explanation / Answer

Temperature = 25 oC = 298 K

Volume = 5 L

Mass of H2S = 0.30 g

Moles of H2S = 0.30 / 34.08

= 0.0088

Using PV = nRT

Initial pressure of H2S = 0.0088 * 0.0821 * 298 / 5

= 0.043 atm

Kp = 0.120

(a).                          NH4HS(s)           NH3(g)      +       H2S(g)
Initial                             -                         0.0                     0.043
Change                         -                          +x                       +x
Equilibrium                   -                           x                    0.043 + x

Kp = P(NH3) * P(H2S)

0.120 = x (0.043 + x)

x = 0.325 atm

Hence, at equilibrium:

Partial pressure of NH3 = 0.325 atm

Partial pressure of H2S = 0.043 + 0.325 = 0.368 atm

(b). Total pressure at equilibrium = 0.325 + 0.368

= 0.693 atm

We know that:

Partial pressure = mole fraction * total pressure

0.368 = X * 0.693

X = 0.368 / 0.693

X = 0.531

(c).   Moles of H2S at equilibrium = PV /RT

= 0.368 * 5 / 0.0821 * 298

= 0.075

Moles of H2S produced from NH4HS = 0.075 - 0.0088

= 0.0662

Moles of NH4HS required = 0.0662

Minimum mass of NH4HS that must be added = 0.0662 * 51.11

= 3.38 g

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