chemistry

What is the pressure change for the reaction below at 25 C in the process of reaching equilibrium if there is initially enough ammonium hydrogen sulfide to reach equilibrium?

Kp= 4.0*(10^4)

NH3(g)+H2S(g) ---><--- NH4HS(s)

Answer: 1.0*[10(-2)]

Not sure what to do here; I know the the Qc expression is:

Qc= 1/([NH3][H2S])

and Kp=Kc[(RT)^(delta n)]

but I'm not sure how to advance.

Is "delta n" -1 (1 mole product - 2 mole reactant)

OR

-2 (0 mole gaseous product - 2 mole gaseous reactant)

And how do I factor in the change? What is changing?

Thank you.

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asked by l
  1. Call pNH3 = p, then at equilibrium, partial pressure H2S must be p also, then
    Kp = 4E4 = (1/pNH3*pH2S)
    4E4 = (1/p2)
    Solve for p which will give you partial pressure H2S and partial pressure NH3. Add them together to find the total pressure change. 1E-2 is the correct answer.

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