A sample of #"NH"_4"HS"# (s) is placed in a 2.58 L flask containing 0.100 mol #"NH"_3# (g). What will be the total gas pressure when equilibrium is established at 25°C?

The total pressure is #"1.15 atm"#. (to two sig figs.)

Here you just have to recognize that you have been given a concentration but also a #K_p# instead of a #K_c#. You'll first have to determine the partial pressure of #"NH"_3(g)#.

Assuming it is ideal,

#PV = nRT#

the partial pressure can be found with the ideal gas law:

#=> P_(NH_3) = (n_(NH_3)RT)/V_"flask"#

The #K_p# is implied to be using units of #"atm"#, and I am going by Tro's text. Therefore, #R = "0.082057 L"cdot"atm/mol"cdot"K"#.

#P_(NH_3) = ("0.100 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "298.15 K")/("2.58 L")#

#=# #"0.9483 atm"#

This becomes its initial pressure in the decomposition, because that was in the flask with the #"NH"_4"HS"(s)# in the first place.

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

#"I"" "-" "" "" "" "0.9483" "" "" "" "0#
#"C"" "-" "" "" "+P_i" "" "" "" "+P_i#
#"E"" "-" "" "" "0.9483+P_i" "" "" "P_i#

Each gas gains pressure #P_i#, and from the #K_p# we can then find the total pressure after finding #P_i#. The coefficients are all #1#, so the gains #P_i# are equal.

#K_p = 0.108 = P_(NH_3)P_(H_2S)#

#= (0.9483 + P_i)P_i#

#= 0.9483P_i + P_i^2#

This becomes the quadratic:

#P_i^2 + 0.9483P_i - 0.108 = 0#

And using the quadratic formula, this has the physical solution of #P_i = "0.1028 atm"#, since pressure can never be negative.

Just to check,

#K_p = (0.9483 + 0.1028)(0.1028) ~~ 0.108# #color(blue)(sqrt"")#

Therefore, since we assumed ideal gases earlier, just as is assumed in Dalton's law, the total pressure is given from Dalton's law of partial pressures:

#color(blue)(P_(t ot)) = P_(NH_3) + P_(H_2S)#

#= (0.9483 + P_i) + (P_i)#

#= "0.9483 atm" + 2P_i#

#= "0.9483 atm" + 2("0.1028 atm")#

#=# #color(blue)("1.154 atm")#