Question

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

Answer 1

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

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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")`