Question

15.00 g of NH4HS(s) is introduced into a 500. mL flask at 25 °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 = 0.108 at 25 °C for: NH4HS(s) ⇄ NH3(g) + H2S(g)?

Answer 1

Answer:

0.328 atm

Explanation:

Kp is the equilibrium constant calculated based on the pressure, and it depends only on the gas substances. It will be the multiplication of partial pressures of the products raised to their coefficients divided by the multiplication of partial pressures of the reactants raised to their coefficients.

For the equation given, the stoichiometry is 1 mol of NH₃ for 1 mol of H₂S, so they will have the same partial pressure in equilibrium, let's call it p. So:

Kp = pxp

0.108 = p²

p = √0.108

p = 0.328 atm, which is the partial pressure of the ammonia.

Answer 2

The partial pressure of ammonia in the flask is 0.329 atm.

First, we have to obtain the concentration of the NH4HS as follows;

Number of moles of NH4HS = 15.00 g/51 g/mol = 0.29 moles

From;

PV = nRT

P = nRT/V

P = 0.29 moles × 0.082 atm LK-1mol-1 × 298 K/0.5 L

P = 14 atm

Setting up the ICE table;

          NH4HS(s) ⇄ NH3(g) + H2S(g)

I           14               0             0

C           - x                 +x            +x

E           14 - x         x               x

0.108= x^2

0.108 = x^2

x^2 = √0.108

x= 0.329 atm

The partial pressure of ammonia in the flask is 0.329 atm.

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