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Question:

5.1 g NH4SH is introduced in a 3.0 L evacuated flask at 327°C.

1×10⁻⁸ atm²

0.242×10⁻⁸ atm²

0.242 atm²

4.9×10⁻⁷ atm²

Solution:

NH4SH(s)⇌NH3(g) + H2S(g)

n = 5.1/51 = 0.1 mole

Let α be the degree of dissociation.

Initially: 0.1 0 0
At equilibrium: 0.1(1-α) 0.1α 0.1α

Total moles at equilibrium = 0.1(1-α) + 0.1α + 0.1α = 0.1(1+α)

Total pressure, P = (nRT)/V = [0.1(1+α)RT]/V

Partial pressure of NH3 = (0.1αRT)/V
Partial pressure of H2S = (0.1αRT)/V

Kp = PNH3 × PH2S = [(0.1αRT)/V]²

Assuming α to be small, total moles ≈ 0.1 moles
P = (0.1 × 0.0821 × 600)/3 = 1.642 atm

Kp = (1.642α)² = 2.695α²

Assuming ideal gas behavior, the equilibrium constant Kp can be related to the degree of dissociation (α):
Kp = (P_NH3)(P_H2S) = (αP/2)² = α²P²/4

Solving for α and substituting the values provided in the problem, we can find Kp. Note that the problem statement is incomplete and does not provide a value for Kp or additional information needed to solve for α and subsequently Kp. More information is needed to complete the calculation.