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Chemical Equilibrium Problem

  1. Jan 9, 2005 #1
    Hi everyone

    Here's the problem:

    The decomposition of ammonium hydrogen sulfide,

    [tex]NH_{4}HS(s) ----> NH_{3}(g) + H_{2}(g)[/tex] (this rxn is reversible)

    is an endothermic process. A 6.1589 gm sample of solid is placed in an evacuated 4 litre vessel at exactly 24 degrees C. After equilibrium has been established, the total pressure inside 0.709 atm., some solid ammonium hydrogen sulfide remains in the vessel.

    (a) what is [tex]K_{p}[/tex] for the reaction?
    (b) what percentage of the solid has decomposed?
    (c) If the volume of the vessel were doubled at constant temperature, what would be the amount of solid (in moles) in the vessel?

    I have been able to do parts (a) and (b). The answers I get are

    (a) [tex]K_{p} = 0.1256 atm^2[/tex]
    (b) percentage associated = 9.4

    For part (c) I figured that at constant temperature, the equilibrium constant would be the same as in part (a). Now if volume were doubled, the pressure would be halved. But I can't solve this further. I do not know the answer but I would be grateful if someone could help me do part (c).


  2. jcsd
  3. Jan 9, 2005 #2


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    I get a different answer for (b). One of us must be wrong :grumpy:

    MW (NH4HS) = 51g/mol
    So 6.1589 g = 0.1208 mol of NH4HS

    At equilibrium, there are n moles of NH3 and n moles of H2S, a total of 2n moles.

    From the ideal gas equation : 0.709 * 4.0 = (2n)*0.0821*297 , which gives

    2n = 0.1164, (or n = 0.05815)

    Now Kp = (0.709/2)^2 = 0.1256 atm^2, which seems okay.

    But I get % decomposed = 0.1164/0.1208 = 96.36%

    Who's making the mistake ?
  4. Jan 9, 2005 #3


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    As for part (c) :

    Do not assume that doubling the volume will halve the pressure. This implies that the number of moles remains constant, which is not true. By increasing the volume, you instantaneously decrease the partial pressures of the products. As a result, Q < Kp. So, according to le Chatelier, more (moles of) product will be formed to try and equalize Q with Kp. However, in this case, there is an upper limit on Q at any volume V, given by

    Q(max) = (RT*0.0604/V)^2, which results from complete (100%) decomposition. So, as V increases, Q(max) decreases, until it falls below Kp at some volume V(max). At this volume, you will have 100% decomposition. Increasing V beyond this V(max) has no effect on the reaction, because it has already achieved complete decomposition, and there's nothing more it can do.

    So, in this particular case, if you double the volume, have you exceeded V(max), or not ?
  5. Jan 9, 2005 #4
    Thanks Gokul. Yes, I think I made the calculation mistake though I used a totally similar method to compute the number of moles.

    I think what I need to do is to use the expression for Q(max) that you have given in your post and compute the critical value of V(max) beyond which Q = K and equilibrium would be reattained right?

    So this gives

    [tex]V_{max} = \frac{0.0604RT}{\sqrt{K_{p}}}[/tex]

    Now I understand that Vmax is never actually attained and the actual volume is less than Vmax. Here V = 2 litres for part(c) so if Vmax < 2L nothing should happen whereas if Vmax > 2L then the reaction should proceed forward to form more of the product, right? In case the latter possibility holds, how do we proceed to find the equilbrium concentrations?

    Thanks again for your help...you've given me a new perspective of this (mathematically, Qmax is a piecewise function of volume as you have pointed out and it varies as the inverse square of volume at a given temperature so the point of intersection of Qmax(V) with the horizontal constant line representing the equilibrium constant gives the value Vmax.I hope I am right with this one :smile:)

    Last edited: Jan 9, 2005
  6. Jan 9, 2005 #5
    a) I got the same thing for Kp....

    However, i don't have 96.36% decomposed....

    There was a mole ratio of 1:1 with ammonium hydrogen sulfide and 1 of its product...

    so n = 0.0583225 moles of ammonium hydrogen sulfide decomposed...

    That's = to 2.974 grams of ammonium hydrogen sulfide decomposed...

    So 2.974/6.1589 = 48.3% decomposed...
  7. Jan 9, 2005 #6
    Goku.... Where did you find that formula with Q(max)....
  8. Jan 9, 2005 #7


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    apchem :

    there can not be a 1:1 ratio between a reactant and a product. Such a ratio can only exist between products. For every molecule of NH4HS that decomposes, 1 molecule of NH3 and 1 molecule of H2S will be formed. So there will always be a 1:1 molar ratio between NH3 and H2S.

    As, for the formula for Q(max) : I did not find it anywhere. It comes directly from definition.
    Q = p(NH3)*p(H2S) = RTn(NH3)/V * RTn(H2S)/V, but n(NH3) = n(H2S).
    So, Q = (nRT/V)^2. But n(max) = 0.5*n(NH4HS-initial). This determines Q(max).

    By definition, Q = Kp, when the system has reached equilibrium. If Q <> Kp, the equilibrium shifts to the right/left in order to attain equality between them.

    maverick :
    I think you seen to have understood it. I created the idea of Q(max) - it's not something I've come across before - specifically for this problem, because I realized the following :

    1. Increasing the volume shifts the equilibrium to the right requiring more product formation, the number of moles of product being a linear function of the volume

    2. You can not indefinitely increase the number of moles of product formed, so beyond a point increasing the volume doesn't do anything to shift the equilibrium.

    I thought it useful to visualize this in terms of a Q(max). Your thinking of Q(max) as a function that falls off as V^-2 intersecting the Kp line at some value of V(max) is just what I have in my head too.

    In this particular problem, since we are so close to 100% dissociation in a 4L container, there is very little more increase (3.6% increase in volume) that is required to actually reach full decomposition. So really, any volume greater than about 4.15 L makes no difference to the compositions...though, in real life, a large increase will give rise to a larger reaction rate, so the only difference is a matter of the time it takes for complete decomposition.

    PS : In (c) it says V = 8L (not 2L)
    Last edited: Jan 9, 2005
  9. Jan 9, 2005 #8
    Goku... just look at the reaction

    NH4HS <-> NH3 + H2S

    It's balanced... and thus there is a mole ratio of 1:1 with NH4HS and H2S...

    You're only getting H2S from decompostion of NH4HS so why won't it be 1:1... With the H2S or NH3 created... you can determine the number of grams that has been decomposed by NH4HS... It's in equilibrium so it's safe to assume their moles will not change...
  10. Jan 9, 2005 #9


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    What the balnced chemical reaction tells you is that if you start with 1 mole of NH4HS and it decomposes completely, you will get one mole each of H2S and NH3. However, in general, you do not have complete decomposition (or reaction). In general you will reach a steady state with some fraction of the reactants in equilibrium with the products. In many of the popular reactions, (neutralization of strong acid with strong base, reaction of metal with acid, etc.) the equilibrium constant is so large that the reaction does go (virtually) to completion.

    Here's a better way to understand this :

    Let's say that you put 1 mole of NH4HS in a closed container, and allow it to slowly decompose over time, till after a very long time T, it reaches some equilibrium composition. In each row below, I state the amounts of each component present in the reaction vessel.

    Code (Text):

                 NH4HS <---------------> NH3    +    H2S

    @ t=0 :     1 mole                0 moles       0 moles

    @ t=T :  (1- f) moles             f moles       f moles
    The argument is the one I made in my previous post. Every molecule of NH4HS that decomposes gives rise to 1 molecule of NH3 and 1 molecule of H2S (this is what the balanced equation tells us). So, if f moles of NH4HS are lost, f moles each of NH3 and H2S are formed.

    The balanced equation tells you that 1 mole of reactant gives rise to 1mole of each product. It does NOT say that 1 mole of reactant co-exists with 1 mole of each product.
  11. Jan 9, 2005 #10
    Just think about it... 1 mole of NH3 and 1 mole of H2S is formed.... The only place it came from is the decomposition of NH4HS. Conservation of mass...

    Mass consumed from the reactant is = to the mass gained by the product... So you know the mass gained by the product... you can calculate the % decomposition...

    You get the same answer if you calculate the mass of the product formed divide it by initial mass of reactant... or the mass of the reactant consumed from the mole ratio 1:1 divide it by the initial mass of the reactant... try it...

  12. Jan 9, 2005 #11


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    apchem, maverick :

    I did make a calculation mistake in my first post.

    0.0583 moles of each product is formed.

    So, the moles of NH4HS remaining is 0.1208 - 0.0583 = 0.0625 moles. (I subtracted twice the number, by mistake)

    So, the percentage decomposition is 48.3%

    apchem : If this is what you've been trying to point out to me...thanks, I misunderstood what you were trying to say. I thought you were trying to say that there would be 0.0583 moles of NH4HS remaining. Guess I didn't read your first post carefully. My bad.

    Notice that the corrected version is what I've been saying in all the subsequent posts, so I still hold they (the subsequent posts) are accurate (only the first post with the calculation has this error). Do you disagree with this ?

    maverick : since less than 50% has decomposed, a doubling of the volume will not exceed the limit.
    Last edited: Jan 9, 2005
  13. Jan 9, 2005 #12
    With the increase of Volume... it increases the number of moles of the products right?...

    I get the answer as .11645 moles for H2S and .11645 moles NH3... so almost all of NH4H2S is consumed...
    Last edited: Jan 9, 2005
  14. Jan 9, 2005 #13


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    That looks right. The numbers will double.
  15. Jan 9, 2005 #14
    Hi Gokul

    Thanks for the help. Could you please summarize your approach for part (c)? That would be very helpful. And anyway calculation mistakes aren't a problem because I would've worked this out all by myself again.

  16. Jan 10, 2005 #15


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    Will do it tomorrow.
  17. Jan 12, 2005 #16


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    If there are n moles each, of H2S and NH3, then
    [tex]K_p = p(NH_3)~p(H_2S) = p^2 = \frac{(nRT)^2}{V^2} [/tex]

    Or, [tex] n = \frac {V}{RT} \sqrt{K_p} [/tex]

    So, as V increases, n will increase proportionally, till [itex] n = n_{max} = 0.1208/2 [/itex]

    This value determines the upper limit of V, beyond which any increase will make no difference to the equilibrium, but a possible difference to the reaction rate. Thus :

    [tex]V_{max} = n_{max}RT \frac{1}{\sqrt{K_p}} [/tex]

    If [itex]V > V_{max}[/itex], then [tex]Q = \frac{(n_{max}RT)^2}{V^2} < K_p [/tex]

    So, the system will never reach equilibrium under this condition.
  18. Jan 14, 2005 #17
    Thanks so much Gokul :-)
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