Question

# 23.4758 g of solid ammonium hydrogen sulfide is introduced into a 318.242-mL flask at 18.228 °C;...

23.4758 g of solid ammonium hydrogen sulfide is introduced into a 318.242-mL flask at 18.228 °C; the flask is sealed, and the system is allowed to reach equilibrium. What is the partial pressure of ammonia in this flask if Kp = 0.803 at 18.228 °C for the reaction

NH4HS(s) ⇌ NH3(g) + H2S(g)?

Report your answer to three significant figures.

#### Homework Answers

Answer #1

Given equation shows that equal numbers of moles of the two gases ammonia and hydrogen sulfide are produced during the reaction, so the partial pressures of two gases must be the same when the equilibrium has been established.

For the given heterogeneous equilibrium, Kp = pNH3(g) * pH2S(g )

(Note: Vapour pressure of a solid in a heterogeneous equilibrium is incorporated into Kp value and so pNH4HS(s) does not appear in the equilibrium expression)

Given Kp = 0.803
So, 0.803 = (pNH3)2                 (As both the partial pressures are same)

So, pNH3(g) = Square root of 0.803

= 0.896

The partial pressure of ammonia is 0.896

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