In: Chemistry
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, x, 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?
I think there is a mistake in the question you cannot put pure H2S in the flask and expect the reaction
NH4HS(s)NH3(g)+H2S(g) to go since you have only the product
I will do the calculation with 0.3 g of NH4HS(s) since that will be correct
C)
0.30 g of NH4HS(s) will be
0.30g/51.11 g/mol = 0.0058 moles
Ammonium bisulfide will react to form NH3 and H2S in 1:1 molar ratios
Kp = pNH3 x pH2S
0.12 = X x X
X2 = 0.12
X = 0.12 = 0.346 atm
So 0.346 atm each of H2S and NH3 will be formed.
D) Mole fraction of H2S can be calculated using the formula
PV=nRT
0.346 x 5L = n x 0.08205 x 298
n = 0.346 x 5/0.08205 x 298
n = 0.0708
since n will be 0.0708 for both gases. Mole fraction of H2S will be 0.5
E)
Calculate number of moles of H2S from MW of 34.08 g /mol
moles H2S = 0.30g/34.08 = 0.0088 moles
Then P = nRT/V.
P = 0.0088 x 0.08205 x 298/5
P = 0.043 atm
NH4HS NH3 +
H2S
Initial solid 0 0.043
Change x x
Equilibrium solid x 0.043+x
Kp = pNH3 x pH2S
0.12 = x * (0.043+x)
0.12 = x2 + 0.043x
x2 + 0.043x -0.12 = 0
solve the quadratic equation
x = 0.325 and-0.368
Since negative values are not possible the partial pressure of NH3 would be 0.325
again from PV=nRT calculate n
n = 0.325 x 5/0.08205 x 298
n = 0.066 moles
As per the equation 1 mole NH4HS gives 1 mole of NH3 so to get 0.066 moes of NH3 you will need minimum 0.066 moles of NH4HS
That is 0.066 mole x 51.11 g/mole = 3.39 g of NH4HS is required