Question

An amount of solid NH₄HS is placed in a flask already
containing ammonia gas at a certain temperature and 0.50
atm pressure. Ammonium hydrogen sulphide decomposes
to yield NH₃ and H₂S gases in the flask. When the
decomposition reaction reaches equilibrium, the total
pressure in the flask rises to 0.84 atm? The equilibrium
constant for NH₄HS decomposition at this temperature is
(a) 0.11 (b) 0.17 (c) 0.18 (d) 0.30

Answer 1

Answer:

please forgive me I don't know this answer....

Answer 2

The equilibrium constant for NH₄HS decomposition at a particular temperature is 0.11 atm². Hence, option (a) is the correct answer.

• Given,

Pressure of ammonia gas already present in the flask = 0.50 atm

• The decomposition of ammonium sulphide is represented by the following reaction :

NH₄HS → NH₃ + H₂S

Let the initial pressure of NH₄HS be x atm.

The initial pressure of NH₃, as per question, is 0.5 atm.

The initial pressure of H₂S is 0 atm.

• Since the reaction is in equilibrium, the products exert the same pressure as reactants on the flask.

Therefore, pressure exerted by NH₃ (PNH₃) at equilibrium = 0.5 atm + x

Pressure exerted by H₂S at equilibrium = x atm

• Given, total pressure in the flask = 0.84 atm

According to the question,

(0.5 atm + x) + x  = 0.84 atm

Or, 0.5 atm + x + x  = 0.84 atm

Or, 0.5 atm + 2x = 0.84 atm

Or, 2x = 0.84 atm - 0.5 atm

Or, 2x = 0.34 atm

Or, x = 0.34 atm / 2

Or, x = 0.17 atm

• Therefore, the pressure exerted by NH₃ = 0.5 atm + 0.17 atm = 0.67 atm

Pressure of H₂S in the flask = 0.17 atm

• Now, the equilibrium constant for for NH₄HS decomposition is given as :

k = PNH₃ × PH₂S

Or, k = 0.67 atm × 0.17 atm

Or, k = 0.113 atm²

Or, k = 0.11 atm²

Therefore, the required answer is 0.11 atm².