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Posted by on Sunday, March 11, 2012 at 8:06pm.

Ammonium bisulfide, NH2HS, forms ammonia NH3, and hydrogen sulfide, H2S through the reaction:
NH4HS (s) <--> NH3(g) + H2S(g)
Kp value of .120 at 25C
in a 5L flask is charged with .300g of H2S at 25C.

A) What are the partial pressures of NH3 and H2S at equilibrium, what are the values of P(NH3) and P(H2S)?
[i have a value of 0.043 for initial pressure of H2S, am i suppose to use this?]

B) What is the mole fraction of H2S in the gas mixture at equilibrium?
[since the temps are the same and they have the same mole to mole ratio is my answer .5?]

C)What is the minimum mass of NH4HS that must be added to the 5L flask when charged with the 0.3 g of H2S at 25C to achieve equilibrium?
[no clue what to do!? any help with these questions would be great! thank you in advance]

  • chemistry - , Sunday, March 11, 2012 at 9:34pm

    Yes. pH2S = 0.043 but I wouldn't throw away the third place since you are allowed 3 significant figures.
    Then Kp = pNH3*pH2S
    You know Kp and pH2S; calculate pNH3.

    B)XH2S = pH2S/total P
    XNH3 = pNH3/total P

    C) You have the 0.300 g H2S. From the data above you can calculate the mass NH3 in the 5 L. Add mass NH3 to mass H2S to arrive at the minimum mass of NH4HS.

  • chemistry - , Sunday, March 11, 2012 at 9:48pm

    Thank you for answering me! You don't know how much it means to me. I think I'm over thinking this but I need all the help I can get.

    So for A) I just divided .120/.0431 and got 2.78. It just seems too simple to me haha.
    For letter C I tried finding the moles of H2S. Since its one to one ratio wouldn't the moles be the same? Then placing these two over Kp to give me the moles of that?

    Once again thank you for your response!

  • chemistry - , Sunday, March 11, 2012 at 10:13pm

    No, mols are not the same. They would be if you had the pure NH4HS coming to equilibrium but you don't have that. The problem has charged the vessel with 0.300 g H2S (0.0431 moles H2S); therefore, your value of 2.78 atm is the partial pressure of NH3. Knowing p NH3 you should be able to obtain n from PV = nRT and from that you can get grams NH3.

  • chemistry - , Sunday, March 11, 2012 at 10:21pm

    Oh okay that makes more sense now. Thank you for clearing that up. I tried entering in 2.78 for my partial pressure of NH3 but the program I am using is telling me that is incorrect?

  • chemistry - , Sunday, March 11, 2012 at 10:58pm

    That's my error. The vessel is charged with 0.300 g H2O (0.04315 atm) but that isn't at equilibrium. It will shift to come to equilibrium.

  • chemistry - , Sunday, March 11, 2012 at 11:14pm

    Try this.
    If the system were allowed to come to equilibrium (without the added H2S), the partial pressure of each gas would be sqrt(Kp) = about 0.35 atm. Since the added 0.300 g H2S has a pressure of 0.04315 atm initially, we note that is smaller than the H2S would have been had no H2S been added at all. Therefore, I believe the reaction will shift to the right since there is insufficient H2S.
    .........NH4HS ==> NH3 + H2S
    Solve for p.
    If I didn't make an error my number is 0.3255 for p and 0.3687 for 0.04315+p.
    That seems reasonable since
    (0.3255*0.3687) = 0.120. Remember I have too many s.f. in my numbers; only 3 s.f. should be used.

  • chemistry - , Sunday, March 11, 2012 at 11:24pm

    Ah! Yes that worked out perfectly! Thank you so much for the help. That one really had me stuck!

  • chemistry - , Monday, March 12, 2012 at 1:50am

    Idk why maybe because its 2 am and i cant do things right at this time, but am also trying to solve this problem and when i get to the part to use the quadratic formula to find p i get a different number i have it set up as this:

    and i get two answers for p obviously, 1 is .3686 and the other one is -.3255 (which is the same as your answer except is negative) I wonder what am doing wrong >< , thx for all your help

  • chemistry - , Monday, March 12, 2012 at 4:59pm

    I re-worked the problem and I get two roots for the equation; one is -0.36866 (the same as yours but mine is minus) and +0.3255 (the same as your but mine is +). If you are solving this by hand you are slipping a sign somewhere; if you are using the calculator you are using the negative of the correct formula.

  • chemistry - , Monday, March 12, 2012 at 10:23pm

    could you please explain the last question, c, again. Ive tried a few times but I am still getting it wrong.

  • chemistry - , Monday, March 12, 2012 at 11:48pm

    yep that is what happened, i forgot one sign, thank you

  • chemistry - , Friday, March 14, 2014 at 4:48pm

    Roses are red
    Violets are blue
    Sorry to be playing on this page
    But I don't have a clue.

  • chemistry - , Friday, January 27, 2017 at 10:20am


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