Consider the reaction:

\(\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)\)

A sample of pure \(\mathrm{NH}_{4} \mathrm{HS}\) is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is \(3.5^{*} 10^{-3}\). Once the reaction reaches equilibrium, what mass of \(\mathrm{NH}_{3}\) is in the container? (Assume that the sample of \(\mathrm{NH}_{4} \mathrm{HS}\) was large enough that equilibrium could be achieved.)

Equation Transcription:

NH4HS

3.5 * 10-3

NH3

Text Transcription:

NH_4HS(s) harpoon NH_3(g) + H_2S(g)

NH_4HS

3.5 times 10^-3

NH_3

Solution 104P:

Step 1:

The equilibrium constant is

Keq = = [NH3][H2S]

Thus, we know that, NH4HS(s) is a solid, so that it is not included into the equilibrium constant.

Keq= 3.5 × 10‒3.

[NH3][H2S] = 3.5 × 10‒3.

From equation, 1 mol NH4HS(s) produces 1.0 mol of NH3(g) and (H2S each.

Thus,

[NH3] = [H2S]

Therefore.

Keq = [NH3][H2S] = [NH3]2

[NH3]2 = 3.5 × 10‒3.

[NH3] = = 0.059 M