Question

In: Chemistry

Ammonium bisulfide, NH4HS, forms ammonia, NH3, and hydrogen sulfide, H2S, through the reaction NH4HS(s)?NH3(g)+H2S(g) This reaction...

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?

Solutions

Expert Solution

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



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