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

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

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

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

Explanation / Answer

Kp = Kc(RT)delta-n, delta-n = 2 in this case.

0.120 = {[NH3][H2S]/[NH4HS]}(8.314*298)2

Since NH4HS is solid [NH4HS] = PNH4HS = 1

Therefore, [NH3][H2S] = 1.955x10-8

Initial concentration of H2S = 0.3/34*1/5 = 1.765x10-3

Let at equilibrium, [NH3] is x and therefore [H2S] is 1.765x10-3 -x

Substituting and simplifying, we get x(1.765x10-3 -x) = 1.955x10-8 and assuming x << 1.765x10-3 and solving for x,

x = 1.955x10-8/1.765x10-3 = 1.108x10-5 M

Total pressure of flask is P = nRT/V = (5*(1.765x10-3 +1.108x10-5)* 8.314*298/0.005 = 4400.4 Pa

PNH3 = 1.108x10-5/(1.765x10-3 +1.108x10-5)*4400.4 = 27.4 Pa

and PH2S = 4372.9 Pa

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