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Ammonium hydrogen sulfide decomposes according to the following reaction, for which Kp = 0.11 at 250ºC: NH4HS(s) H2S(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

Answer :

ajeigbeibraheem

Answer:

0.33 atm

Explanation:

Equation for the reaction is expressed as follows:

[tex]NH_4HS_{(s)}[/tex]      ⇄     [tex]H_2S_{((g)} + NH _{3(g)}[/tex]

[tex]K_p[/tex] = 0.11

The equilibrium constant for the reaction can be represented as:

[tex]K_p[/tex] = [tex]\frac{P_{H_2S_{((g)}} P_{NH _{3(g)}}}{P_{NH_4HS_{(s)}}}[/tex]

[tex]{P_{NH_4HS_{(s)}}}[/tex] = 1   since it is solid and solid and it is known that any solid has uniformity at equilibrium.

So; [tex]K_p[/tex] = [tex]{P_{H_2S_{((g)}} P_{NH _{3(g)}}}[/tex]

At equilibrium ;

[tex]{P_{H_2S_{((g)}} =P_{NH _{3(g)}}}[/tex]

Let x = [tex]{P_{H_2S_{((g)}} =P_{NH _{3(g)}}}[/tex]

So;

[tex]K_p[/tex] = [tex][x][x][/tex]

[tex]K_p[/tex] = [tex]x^2[/tex]

[tex]x= \sqrt{K_p}[/tex]

[tex]x=\sqrt{0.11}[/tex]

[tex]x = 0.33[/tex] atm

Therefore,  the partial pressure of NH3(g) at equilibrium [tex]{P_{H_2S_{((g)}} =P_{NH _{3(g)}}}[/tex] = 0.33 atm

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