In a 10-liter capacity container the following system is in equilibrium at 25 celcius degrees: NH4HS (s) <-> NH3 (g) + H2S (g) K = 1.8 x 10 ^ -4 ?

Well, here's something off the top of my head.

I assume that the given #K# is #K_c = 1.8 xx 10^(-4)#, because it's quite small, and this reaction occurs moderately at room temperature.

For the reaction

#"NH"_4"HS"(s) rightleftharpoons "NH"_3(g) + "H"_2"S"(g)#,

the mols of gas increased by #2#, so:

#K_p = K_c(RT)^(Deltan_"gas")#

#= (1.8 xx 10^(-4) "M"^2)("0.08206 L"cdot"atm/mol"cdot"K"cdot "298.15 K")^(2)#

#= "0.1077 atm"^2#

From here, the mass action expression can be written in terms of the mol fractions:

#K_p = P_(NH_3)P_(H_2S)#

#= (1-chi_(H_2S))(chi_(H_2S))P_(t ot)^2#

where #chi_(H_2S)# is the mol fraction of #"H"_2"S"#, and #chi_iP_(t ot) = P_i# is the definition of the partial pressure.

#=> P_(t ot) = sqrt((K_p)/((1-chi_(H_2S))(chi_(H_2S))))#

Now, mol fractions are relative, and these need not be #50//50# because for all we know, this could be a re-established equilibrium after a disturbance.

If you know the average molar mass, you can write it in terms of the mol fraction of each gas in the container at equilibrium, assuming ideality:

#M_(avg) = (1 - chi_(H_2S))M_(NH_3) + chi_(H_2S)M_(H_2S)#

#= M_(NH_3) + chi_(H_2S)(M_(H_2S) - M_(NH_3))#

Therefore, the mol fraction of #"H"_2"S"# in the container is:

#chi_(H_2S) = (M_(avg) - M_(NH_3))/(M_(H_2S) - M_(NH_3)) = ("17.07 g/mol" - "17.031 g/mol")/("34.081 g/mol" - "17.031 g/mol")#

#= 0.002287#

It makes sense that this is so low, because the molar mass is very close to that of ammonia. Therefore, the total pressure is:

#P_(t ot) = sqrt(("0.1077 atm"^2)/((1 - 0.002287)(0.002287))#

#=# #"6.870 atm"#

From the total pressure, the ideal gas law gives the total mols:

#n_(t ot) = ("6.870 atm" cdot "10.0 L")/("0.08206 L"cdot"atm/mol"cdot"K" cdot "298.15 K")#

#=# #"2.808 mols"#

And as a result, the mols of #"H"_2"S"# are:

#color(blue)(n_(H_2S)) = chi_(H_2S) cdot n_(t ot)#

#=# #color(blue)("0.00642 mols")#