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Gre4nikov [31]
3 years ago
8

Ammonium hydrogen sulfide decomposes according to the following reaction, for which Kp = 0.11 at 250°C: NH4HS(s) ⇌ H2S(g) + NH3(

g) If 46.5 g of NH4HS(s) is placed in a sealed 5.0−L container, what is the partial pressure of NH3(g) at equilibrium? PNH3 = atm
Chemistry
1 answer:
Scrat [10]3 years ago
5 0

Answer:

0,33atm

Explanation:

For the reaction:

NH₄HS(s) ⇌ H₂S(g) + NH₃(g)

kp is defined as:

kp = 0.11 = P(H₂S) P(NH₃) <em>(1)</em>

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

<em>-Where I is an initial concentration that is not relevant for the problem and x is the </em>NH₄HS<em> that reacts-</em>

H₂S(g): x

NH₃(g): x

Replacing in (1):

0.11 = X×X

0.11 = X²

<em>0.33 = X</em>

That means P(NH₃) is <em>0.33 atm</em>

<em></em>

I hope it helps!

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