Question

± Heterogeneous Equilibrium of Ammonium Bisulfide - Copy

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 0.300 g of pure H2S(g), at 25 ∘C.

Part A

What is the initial pressure of H2S(g) in the flask?

Express your answer numerically in atmospheres.

Hints

P =

4.31×10−2

atm

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Correct

Addition of ammonium bisulfate

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

Part B

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.

Answer 1

NH4HS(s)   ⇌   NH3(g)+H2S(g)

4.31*10^-2       0       0       (initial)

4.31*10^-2 -x   x       x       (at equilibrium)

Kp = p(NH3)p(H2S) / p(NH4HS)

0.120 = x*x /(4.31*10^-2 - x)

5.172*10^-3 - 0.120*x = x^2

x^2 + 0.120*x - 5.172*10^-3 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 0.12

c = -5.172*10^-3

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.509*10^-2

roots are :

x = 3.366*10^-2 and x = -0.1537

since x can't be negative, the possible value of x is

x = 3.366*10^-2

so,

p(NH3) = x = 3.366*10^-2 atm

p(H2S) = x = 3.366*10^-2 atm

P(NH4HS) = 4.31*10^-2 -x = 4.31*10^-2 - 3.366*10^-2 = 9.44*10^-3 atm

Answer: 3.366*10^-2, 3.366*10^-2