Answer:
0,33atm
Explanation:
For the reaction:
NH₄HS(s) ⇌ H₂S(g) + NH₃(g)
kp is defined as:
kp = 0.11 = P(H₂S) P(NH₃) (1)
Where P(H₂S) and P(NH₃) are partial pressures of each compound.
In equilibrium, if in your system the only addition is of NH₄HS(s), the partial pressures and the concentration of each compound are:
NH₄HS: I - x
-Where I is an initial concentration that is not relevant for the problem and x is the NH₄HS that reacts-
H₂S(g): x
NH₃(g): x
Replacing in (1):
0.11 = X×X
0.11 = X²
0.33 = X
That means P(NH₃) is 0.33 atm
I hope it helps!
The partial pressure of NH3(g) is 0.33 atm.
Number of moles of NH4HS = 46.5 g/51 g/mol = 0.91 moles
Given that;
PV =nRT
P = ?
V = 5.0−L
n = 0.91 moles
R = 0.082 atm LK-1mol-1
T = 250°C + 273 = 523 K
Making P the subject of the formula;
P = nRT/V
P = 0.91 moles × 0.082 atm LK-1mol-1 × 523 K / 5.0−L
P = 7.8 atm
We must now set up the ICE table;
NH4HS(s) ⇌ H2S(g) + NH3(g)
I 7.8 atm 0 0
C -x +x +x
E 7.8 - x x x
We know that;
Kp = pH2S × pNH3
Note that NH4HS is a pure solid and does not get into the equation
Kp = 0.11
0.11 = x^2
x = √0.11
x = 0.33 atm
Since partial pressure of H2S = partial pressure of NH3 = x
The partial pressure of NH3(g) = 0.33 atm.
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