A sample of ${NH}_{4} HS$ $"NH"_4"HS"$ (s) is placed in a 2.58 L flask containing 0.100 mol ${NH}_{3}$ $"NH"_3$ (g). What will be the total gas pressure when equilibrium is established at 25°C?

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A sample of ${NH}_{4} HS$ $"NH"_4"HS"$ (s) is placed in a 2.58 L flask containing 0.100 mol ${NH}_{3}$ $"NH"_3$ (g). What will be the total gas pressure when equilibrium is established at 25°C?

${NH}_{4} HS (s) ⇌ {NH}_{3} (g) + H_{2} S(g)$ $"NH"_4"HS (s)"\rightleftharpoons"NH"_3"(g)"+"H"_2"S(g)"$

$K_{p} = 0.108$ $"K"_p=0.108$ at 25°C

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Answer 1 · 2018-07-16T07:27:44

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

Here you just have to recognize that you have been given a concentration but also a $K_{p}$ $K_p$ instead of a $K_{c}$ $K_c$ . You'll first have to determine the partial pressure of ${NH}_{3} (g)$ $"NH"_3(g)$ .

Assuming it is ideal,

$P V = n R T$ $PV = nRT$

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

$\Rightarrow P_{N H_{3}} = \frac{n_{N H_{3}} R T}{V_{flask}}$ $=> P_(NH_3) = (n_(NH_3)RT)/V_"flask"$

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

$P_{N H_{3}} = \frac{0.100 mols \cdot 0.082057 L \cdot atm/mol \cdot K \cdot 298.15 K}{2.58 L}$ $P_(NH_3) = ("0.100 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "298.15 K")/("2.58 L")$

$=$ $=$ $0.9483 atm$ $"0.9483 atm"$

This becomes its initial pressure in the decomposition, because that was in the flask with the ${NH}_{4} HS (s)$ $"NH"_4"HS"(s)$ in the first place.

${NH}_{4} HS (s) ⇌ {NH}_{3} (g) + H_{2} S (g)$ $"NH"_4"HS"(s) rightleftharpoons "NH"_3(g) " "+" " "H"_2"S"(g)$

$I - 0.9483 0$ $"I"" "-" "" "" "" "0.9483" "" "" "" "0$
$C - + P_{i} + P_{i}$ $"C"" "-" "" "" "+P_i" "" "" "" "+P_i$
$E - 0.9483 + P_{i} P_{i}$ $"E"" "-" "" "" "0.9483+P_i" "" "" "P_i$

Each gas gains pressure $P_{i}$ $P_i$ , and from the $K_{p}$ $K_p$ we can then find the total pressure after finding $P_{i}$ $P_i$ . The coefficients are all $1$ $1$ , so the gains $P_{i}$ $P_i$ are equal.

$K_{p} = 0.108 = P_{N H_{3}} P_{H_{2} S}$ $K_p = 0.108 = P_(NH_3)P_(H_2S)$

$= (0.9483 + P_{i}) P_{i}$ $= (0.9483 + P_i)P_i$

$= 0.9483 P_{i} + P_{i}^{2}$ $= 0.9483P_i + P_i^2$

This becomes the quadratic:

$P_{i}^{2} + 0.9483 P_{i} - 0.108 = 0$ $P_i^2 + 0.9483P_i - 0.108 = 0$

And using the quadratic formula, this has the physical solution of $P_{i} = 0.1028 atm$ $P_i = "0.1028 atm"$ , since pressure can never be negative.

Just to check,

$K_{p} = (0.9483 + 0.1028) (0.1028) \approx 0.108$ $K_p = (0.9483 + 0.1028)(0.1028) ~~ 0.108$ $\sqrt{}$ $color(blue)(sqrt"")$

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:

$P_{t o t} = P_{N H_{3}} + P_{H_{2} S}$ $color(blue)(P_(t ot)) = P_(NH_3) + P_(H_2S)$

$= (0.9483 + P_{i}) + (P_{i})$ $= (0.9483 + P_i) + (P_i)$

$= 0.9483 atm + 2 P_{i}$ $= "0.9483 atm" + 2P_i$

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

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

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A sample of ${NH}_{4} HS$ $"NH"_4"HS"$ (s) is placed in a 2.58 L flask containing 0.100 mol ${NH}_{3}$ $"NH"_3$ (g). What will be the total gas pressure when equilibrium is established at 25°C?

${NH}_{4} HS (s) ⇌ {NH}_{3} (g) + H_{2} S(g)$ $"NH"_4"HS (s)"\rightleftharpoons"NH"_3"(g)"+"H"_2"S(g)"$

$K_{p} = 0.108$ $"K"_p=0.108$ at 25°C

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A sample of "NH"_4"HS"NH4HS"NH"_4"HS" (s) is placed in a 2.58 L flask containing 0.100 mol "NH"_3NH3"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)"NH4HS (s)⇌NH3(g)+H2S(g)"NH"_4"HS (s)"\rightleftharpoons"NH"_3"(g)"+"H"_2"S(g)" "K"_p=0.108Kp=0.108"K"_p=0.108 at 25°C

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A sample of ${NH}_{4} HS$ $"NH"_4"HS"$ (s) is placed in a 2.58 L flask containing 0.100 mol ${NH}_{3}$ $"NH"_3$ (g). What will be the total gas pressure when equilibrium is established at 25°C?

${NH}_{4} HS (s) ⇌ {NH}_{3} (g) + H_{2} S(g)$ $"NH"_4"HS (s)"\rightleftharpoons"NH"_3"(g)"+"H"_2"S(g)"$

$K_{p} = 0.108$ $"K"_p=0.108$ at 25°C