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.

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

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