# 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?

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

#"K"_p=0.108# at 25°C

##### 1 Answer

The total pressure is

Here you just have to recognize that you have been given a concentration but also a

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

#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) 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**

#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

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")#