Ammonium hydrogen sulfide (NH4SH) was detected in the atmosphere of Jupiter subsequent to its collision with the comet Shoemaker-Levy. The decomposition of NH4SH (s) and its equilibrium with ammonia and hydrogen sulfide is described by the following reaction: NH4SH (s) NH3 (g) + H2S (g); Kp = 12.0 at 25°C. If we have a 1.00-L flask which already contains gaseous NH3 at a pressure of 4.0 atm and heat up some NH4SH, what will the equilibrium pressure of NH3 be at 25°C?

6 atm

Explanation:

We know that some decomposition of the solid will occur NH₄SH according to the equilibria:

NH₄SH (s) ⇄ NH₃ (g) + H₂S (g)

and at equilibrium it the partial pressures of NH₃ and H₂S has to obey:

Kp = p NH₃ x p H₂S = 12.0

(remember NH₄SH is solid and it is not in included in the expression for Kp)

we already have 4.0 atm of NH₃ so this expression becomes:

(4 + y) y = 12

where y is the partial pressure of NH₃ and H₂S produced from the decomposition.

Solving for y:

4y + y² = 12

y² + 4y - 12 = 0

which can be factored into

(y + 6) (y - 2) = 0 (No need to solve the quadratic equation)

then y₁ = - 6 and y₂ = 2

y₁ = - 6 is physically imposible, therefore y = 2

Therefore the equilibrium pressure of y will be:

p NH₃ = (2 + 4) atm = 6 atm