Question

The equilibrium constant, Ky, for the following reaction is 0.110 at 298 K. NH_HS(s) NH3(g) + H2S(g) If an equilibrium mixture of the three compounds in a 5.50 L container at 298 K contains 2.94 mol of NH_HS(s) and 0.219 mol of NH3(g), the partial pressure of H2S(g) is atm.

Answer 1

Equilibrium mole of h2s= mole of Nh3

So first calculate total mole then mole fraction

Now using kp= product/reactant

Solid have no partial pressure so ignore Nh4Hs

From there we calculate total pressure

Now we know that partial pressure = mole fraction × total pressure

Using this we can get partial pressure of H2S