Question

When solid NH₄⁺HS is placed in a closed flask at 28°C, the solid dissociates according to the equation: NH₄⁺HS(s) ⇌ NH₃(g) + H₂S(g). The total pressure of the equilibrium mixture is 0.766 atm. What is the partial pressure of NH₃ in the equilibrium mixture?

1) 0.383 atm
2) 0.766 atm
3) 1.532 atm
4) Cannot be determined

Answer 1

To find the partial pressure of NH₃ in the equilibrium mixture, you can use the equilibrium constant (Kp) and the partial pressures of the reactants and products. By setting up an expression for Kp and solving for x, you can calculate the partial pressure of NH₃ to be 2.753 atm. The correct Option is not given(None of the given).

Step-by-step explanation:

To determine the partial pressure of NH₃ in the equilibrium mixture, we need to first find the equilibrium constant (Kp) for the dissociation of NH₄⁺HS. The equilibrium constant can be calculated by using the partial pressures of the products and reactants at equilibrium.

First, let's express the equation in terms of partial pressures:

NH₄⁺HS(s) ⇌ NH₃(g) + H₂S(g)

At equilibrium, the partial pressures of NH₃ and H₂S are equal, so we can represent them as x.

The partial pressure of NH₄⁺HS can be calculated by subtracting x from the total pressure of the equilibrium mixture:

NH₄⁺HS = 0.766 atm - x

Since the equation has a 1:1:1 ratio, the equilibrium constant (Kp) is equal to the ratio of the partial pressure of NH₃ and the partial pressure of NH₄⁺HS:

Kp = [NH₃] / [NH₄⁺HS] = x / (0.766 - x)

Now we can solve for x:

Kp = 1.383 = x / (0.766 - x)

1.383(0.766 - x) = x

1.056 + 1.383x = x

1.383x - x = -1.056

0.383x = -1.056

x = -1.056 / 0.383 = 2.753

The partial pressure of NH₃ in the equilibrium mixture is 2.753 atm.