Question

At 24°C, Kp = 0.080 for the equilibrium: NH4HS (s) NH3 (g) + H2S (g) A sample of solid NH4HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH4HS remains.

Answer 1

Answer:

Partial pressure of ammonia is 0,28.

Explanation:

For the reaction:

NH₄HS(s) ⇄ NH₃(g) + H₂S(g)

kp = 0,080 is defined as:

0,080 = P[NH₃]  P[H₂S] (1)

Where P[NH₃] is partial pressure of NH₃

Asuming amount of NH₄HS is 1, equilibrium concentration for each compound is:

[NH₄HS] = 1 - x

P [NH₃] = x

P [H₂S] = x

Replacing in (1):

0,080 = X×X

0,080 = X²

X = 0,28 = P[NH₃]

I hope it helps