Question

The equilibrium constant, Kc , for the following reaction is 1.80×10-4 at 298 K.

NH4HS(s)-----> NH3(g) + H2S(g)

If an equilibrium mixture of the three compounds in a 6.21 L container at 298 K contains 3.17 mol of NH4HS(s) and 0.335 mol of NH3, the number of moles of H2S present is ____ moles??

Answer 1

[NH3] = mol of NH3 / volume

= 0.335 mol / 6.21 L

= 0.0539 M

Kc is defined as concentration of product by concentration of reactant with each concentration term raised to power that is equal to its stoichiometric coefficient in balanced equation

pure liquid and solid do not appear in Kc

So, NH4HS will not appear in Kc expression

Kc = [NH3][H2S]

1.80*10^-4 = (0.0539)*[H2S]

[H2S] = 3.34*10^-3 M

now use:

[H2S] = mol of H2S/volume

3.34*10^-3 M = mol of H2S/6.21

mol of H2S = 0.0207 mol

Answer: 0.0207 mol