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

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

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?

Part E.

What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.250 g of pure H2S(g), at 25 C to achieve equilibrium?

Explanation / Answer

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

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

a) No. of moles of H2S = 0.250 g/34g/mol = 0.00735 mole

b) Pressure of H2S in 5.00 L vessel at 25 C is given by

PV = nRT

PH2S = nRT/V = 0.00735*0.0821*298/5 = 0.036 atm

c) Partial pressures of NH3 and H2S at equilibrium can be calculated as

Kp = PNH3*PH2S
at equilibrium, PNH3 = x and PH2S = (0.00735+x)

0.120 = x*(0.00735+x)

x^2 + 0.00735x - 0.120 = 0
Use the quadratic formula to solve for x.
One of the roots to this equation is: x = 0.342 atm

So, PNH3 = 0.342 atm and PH2S = 0.350 atm

These are the partial pressures of each gas at equilibrium

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

Mole fractions of gases are related to the partial pressures, so
mole fraction of H2S = 0.350 atm / 0.692 atm = 0.505

e) AMount of NH4HS can be calculated as

No. of moles of H2S initially added = 0.00735 moles

PH2S = nRT/V = 0.00735*0.0821*298/5 = 0.036 atm

PNH3 can be found as

0.120 = x*(0.00735+x)

X = 0.342 atm

No. of moles of NH3 = PV = nRT ; 0.342(5.0 liters) = n(0.0821)298K
Moles of NH3 = n = 0.0698

Moles of NH3 = moles of NH4HS = 0.0698 moles

AMount of NH4HS = moles*molar mass = 0.0698moles*51 g NH4HS/1mole = 2.86 g of NH4HS

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