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 = _____ atmExplanation / 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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