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Equilibrium pressure of a gaseous system

  1. Apr 5, 2009 #1

    2NO2(g) -> N2O4(g)

    • Calculate ΔG at 298.15 K - complete
    • Calculate KP at 298.15 K
    • Calculate equilibrium pressure of a system with initial conditions of 1 mol N2O4, confined to 24.46 liters fixed volume, at 298.15 K.

    2. Relevant equations

    ΔG = -RTlnK
    KP = (Pproducts)n / (Preactants)n

    3. The attempt at a solution

    ΔG = -RTlnK
    ΔG = ΔGN2O4 - 2ΔGNO2
    ΔG = (97.9 kJ/mol) - 2(51.3 kJ/mol)
    ΔG = -4.73 kJ/mol

    KP = PN2O4 / (PNO2)2

    How do you determine the partial pressures of either NO2 or N2O4?
  2. jcsd
  3. Apr 5, 2009 #2


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    Staff: Mentor

    Calculate K first.
  4. Apr 5, 2009 #3
    K = Kp ?
  5. Apr 7, 2009 #4
    Even if K = KP, I will still have 2 unknowns in KP = PN2O4 / (PNO2)2

    Please explain.
  6. Apr 7, 2009 #5


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    Staff: Mentor

    Amounts of both gases are linked through reaction stoichiometry.
  7. Apr 7, 2009 #6
    So I'm assuming by that you mean PNO2 = 2PN2O4 ?

    i.e. KP = P / (2P)2 ?
  8. Apr 7, 2009 #7


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    Staff: Mentor

    No. You started with 1 mole of N2O4. If x moles of N2O4 reacted - how many moles of N2O4 were left? How many moles of NO2 were produced?
  9. Apr 7, 2009 #8

    Here's where I am right now:

    I calculated P (x) in the above equation to be 0.0371

    Using PV = nRT, I calculated the initial pressure of N2O4 to be 101,348 Pa:

    P = [(1 mol)(8.3145J K-1 mol-1)(298.15 K)] / (0.02446 m3)
    P = 101,348 Pa

    initial: 0
    change: +2x
    equil.: 2x

    initial: 101,348
    change: -x
    equil.: 101,348 - x

    Seems wrong.
    Last edited: Apr 7, 2009
  10. Apr 7, 2009 #9


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    Staff: Mentor

    Doesn't look wrong to me. Just enter these values into reaction quotient and solve for x. Use this x to calculate pressure - and you are done.

    Note: I am not sure about K/Kp thing, black hole between my ears :(
  11. Apr 7, 2009 #10
    I already did this and got 0.0371 (see my previous post). What does 0.0371 refer to?

    Right now I am doing this, which is clearly wrong:

    initial: 0
    change: +2x
    equil.: 2x = 2(0.0371) = 0.0742

    initial: 101,348
    change: -x
    equil.: 101,348 - x = 101,348 - 0.0371 = 101,347.96

    Please clarify.

    EDIT: If it would help, just let me know and I can type out the example given in the textbook. It's long, but it might help you see exactly what I should be doing (for some reason, I can't seem to be able to translate the example into a solution for this specific problem).
    Last edited: Apr 7, 2009
  12. Apr 7, 2009 #11


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    Final pressure is sum of both pressures - that of N2O4 and that of NO2. Use x to calculate both pressures (or number of moles, then pressures - math will be identical, as pressure is directly proportional to the number of moles, just numbers will be different).
  13. Apr 7, 2009 #12
    I still do not understand what the above calculated x value represents. Is it a proportion of Ptotal? Or is it an actual pressure value as I have calculated?

    i.e. are my above numbers correct? The change in contributed pressures seems much too small to be the correct answer (with NO2 contributing only 0.00007% to the total pressure, which strikes me as ridiculously low - the K value is nowhere near high enough to have such a bias favouring the product, N2O4).
    Last edited: Apr 7, 2009
  14. Apr 7, 2009 #13


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    Staff: Mentor

    It becomes hard to follow.

    What value of Kp have you used?

    What is the reaction quotient in terms of x (the equation that you have solved)?
  15. Apr 10, 2009 #14
    = KP
    = 6.74
    = PN2O4 / (2PNO2)2
    = PN2O4 / 4P2NO2
    = x / 4x2
  16. Apr 11, 2009 #15


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    x/4x2 is just 1/4x. You have prepared ICE table, but you are not using E line. The way it is written now it looks like both gases came from nowehere.
  17. Apr 12, 2009 #16

    I calculated Pinitial, which applies only to N2O4, as its the only form of gas initially present. According to the E line, the change in N2O4 is -x (so Pinitial - x), while the change in NO2 is +2x (so 0 + 2x):

    Can you please clarify what you mean by "came from nowhere"?

    And is x = 0.037 in Pa or in bar?
    Last edited: Apr 12, 2009
  18. Apr 13, 2009 #17


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    Staff: Mentor

    2NO2 <-> N2O4
    I: 0 1
    C: +2x -x
    E: +2x 1-x

    K = (1-x)/(2x)2

    Take a look at numerator - it is 1-x, that means initial 1 minus x that reacted; denominator is 2x produced from what is missing in the numerator. When it was x/4x2 both numerator and denominator look like they contain x produced from nowhere - nothing is consumed.
  19. Apr 13, 2009 #18
    So in other words, per 1 mol of N2O4 originally present, 2 x 0.0371 mol of N2O4 is converted to NO2. This would answer my earlier question of what the x value represents. I was under the impression that x was the pressure value itself.
  20. Apr 13, 2009 #19


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    Staff: Mentor

    This is a constant volume system, so the total pressure can change. In the case of constant pressure systems, n and partial pressures are directly protportional, so you can use whichever you want.
    Last edited: Apr 13, 2009
  21. Apr 23, 2009 #20
    Having returned to this problem after a brief absence, I am at a loss as to how I originally calculated the value of K as being 6.74. The value I am coming up with now is 1.002.

    I am going ahead with the problem assuming the latter value to be the correct one, unless someone here can correct me.

    K = KP
    K = e-ΔG/RT
    K = e-(-4.74 kJ mol-1)/(8.3145 J K-1 mol-1)(298.15 K)
    K = 1.002

    * Value previously calculated in post #1

    equilibrium values (from ICE table):
    E (2NO2): 2x
    E (N2O4): 1 - x

    K = 1.002 = (1 - x)/(2x)2
    K = 1.002 = (1 - x)/4x2
    4.008x2 + x - 1 = 0

    Using quadratic formula, solve for x...

    x = 0.390, -0.640

    Discard negative value, as molar value cannot be negative.

    x = 0.390

    Thus, per 1 mol of N2O4, 2 x 0.390 = 0.780 mol of it is converted into NO2 at equilibrium.


    Per 1 unit of pressure initially exerted by N2O4, 2 x 0.390 = 0.780 units of pressure will be exerted by NO2 at equilibrium.

    Is this correct? Yes or no?

    Assuming it is correct, then, calculating Pinitial of the system (N2O4) as follows:

    Pinitial = (nRT)/V
    Pinitial = [(1 mol)(8.3145 J K-1 mol-1)(298.15 K)]/(0.02446 m3)
    Pinitial = 101,348 Pa

    And based on the initial calculations above, the equilibrium pressure would be:

    2NO2: 79,051 Pa
    N2O4: 22,297 Pa

    Is this correct? Yes or no?
    Last edited: Apr 23, 2009

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