Question

Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equilibrium with 9.56x102 MNH3(g) and 7.62x102 M H2S(g), what is the value of the equilbrium constant at 302 K? K =

Answer 1

Kc = 3.72 × 10⁶

Step-by-step explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

`Kc=([NH_(3)] * [H_(2)S] )/([NH_(4)HS]) =(9.56 * 10^(2)  * 7.62  * 10^(2))/(0.196) =3.72 * 10^(6)`