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

A) What are the partial pressures of

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

B) What is the mole fraction,

, of H2S in the gas mixture at equilibrium?

C) 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

SOLUTION:

(A) Kp is given as

Kp = [PNH3][PH2S]

OR 0.12 = (x)2

x = 0.35 atm

Hence parial pressure of NH3 and H2S is 0.35 atm each.

(B) mole fraction of H2S (xH2S) = partial pressure of H2S / Total pressure

Total presure = 0.35 + 0.35 = 0.7

Therefore  xH2S = 0.35 / 0.7 = 0.5

(C) Concentration of H2S = n/V

n = number of moles of H2S = given mass / molar mass = 0.25g / 34 = 0.007

V = volume in liters = 5

Concentration = 0.007 / 5 = 0.0015M

Kc = [H2S][NH3] / [NH4HS]

Minimum conc. of NH3 should be equal to H2S

0.12 = (0.0015 X 0.0015) / [NH4HS]

[NH4HS] = 1.9 X 10-5M

Number of moles in above concentration = C X V = 1.9 X 10-5M X 5L = 9.5 X 10-5 moles

Mass of NH4HS = molar mass X number of moles = 51 X 9.5 X 10-5 = 0.0048g

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