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Ammonium hydrogen sulfide decomposes according to the following reaction, for which Kp = 0.11 at 250°C:NH₄HS(s) ⇄ H₂(g) + NH₃(g) If 55.0 g of NH₄HS(s) is placed in a sealed 5.0-L container, what is the partial pressure of NH₃(g) at equilibrium?

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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.

To know more about partial pressure

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