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.
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.