# 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?

## $\text{NH"_4"HS (s)"\rightleftharpoons"NH"_3"(g)"+"H"_2"S(g)}$ ${\text{K}}_{p} = 0.108$ at 25°C

Jul 16, 2018

The total pressure is $\text{1.15 atm}$. (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 ${\text{NH}}_{3} \left(g\right)$.

Assuming it is ideal,

$P V = n R T$

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

$\implies {P}_{N {H}_{3}} = \frac{{n}_{N {H}_{3}} R T}{V} _ \text{flask}$

The ${K}_{p}$ is implied to be using units of $\text{atm}$, and I am going by Tro's text. Therefore, $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$.

${P}_{N {H}_{3}} = \left(\text{0.100 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "298.15 K")/("2.58 L}\right)$

$=$ $\text{0.9483 atm}$

This becomes its initial pressure in the decomposition, because that was in the flask with the $\text{NH"_4"HS} \left(s\right)$ in the first place.

$\text{NH"_4"HS"(s) rightleftharpoons "NH"_3(g) " "+" " "H"_2"S} \left(g\right)$

$\text{I"" "-" "" "" "" "0.9483" "" "" "" } 0$
$\text{C"" "-" "" "" "+P_i" "" "" "" } + {P}_{i}$
$\text{E"" "-" "" "" "0.9483+P_i" "" "" } {P}_{i}$

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.

${K}_{p} = 0.108 = {P}_{N {H}_{3}} {P}_{{H}_{2} S}$

$= \left(0.9483 + {P}_{i}\right) {P}_{i}$

$= 0.9483 {P}_{i} + {P}_{i}^{2}$

${P}_{i}^{2} + 0.9483 {P}_{i} - 0.108 = 0$

And using the quadratic formula, this has the physical solution of ${P}_{i} = \text{0.1028 atm}$, since pressure can never be negative.

Just to check,

${K}_{p} = \left(0.9483 + 0.1028\right) \left(0.1028\right) \approx 0.108$ 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:

$\textcolor{b l u e}{{P}_{t o t}} = {P}_{N {H}_{3}} + {P}_{{H}_{2} S}$

$= \left(0.9483 + {P}_{i}\right) + \left({P}_{i}\right)$

$= \text{0.9483 atm} + 2 {P}_{i}$

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

$=$ $\textcolor{b l u e}{\text{1.154 atm}}$