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Effect on equilibrium, addition of inert gas

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  1. Aug 27, 2015 #1

    Titan97

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    If inert gases are added at constant volume to an equilibrium mixture, the total pressure increases. But the mole fraction of the gaseous reactants or products decrease.
    Partial pressure ##p=x\cdot P## where P is total pressure and ##x## is mole fraction.
    How can you say that ##p## remains constant?
    (I read that the equilibrium is not affected since partial pressure of the participants of equilibrium remains constant)
     
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  3. Aug 27, 2015 #2

    Borek

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    Assume initial pressure of a gas to be p0. Assume 1 mole of the gas to make calculations easier. Assume volume V. Assume you have added n moles of an inert gas. What is the new total pressure? What is the mole fraction of the original gas? What is its partial pressure? Don't guess - calculate!
     
  4. Aug 27, 2015 #3
    The mole fractions decrease by the same factor that the total pressure increases, so that the partial pressures remain the same. True or false: each gas in an ideal gas mixture exhibits PVT behavior as if the other gases are not present.

    Chet
     
  5. Aug 27, 2015 #4

    Titan97

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    Initial mole fraction = ##\frac{1}{N}##
    Initial total pressure = ##P_0##
    ##P_0=\frac{NRT}{V}##
    ##p=x.P_0=RT/V##
    Let n moles of argon be added.
    New mole fraction = ##\frac{1}{n+N}##
    New total pressure = ##\frac{(n+N)RT}{V}##
    ##p'=\frac{RT}{V}##
    Hence there is no change in partial pressure.
     
  6. Aug 27, 2015 #5

    Titan97

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    @Chestermiller Let the volume occupied by one gas be ##v##. Won't ##v## change if more gases are added? Since adding more gases means lesser volume that can be occupied.
     
  7. Aug 27, 2015 #6

    Borek

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    I assume you have just answered your own question :wink:
     
  8. Aug 27, 2015 #7
    The molecules of each gas occupy the entire volume. The partial pressures of the various gases add up to the total pressure.
     
  9. Aug 28, 2015 #8

    Titan97

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    What if noble gases are added under constant pressure?
    Let the equation be A(g)↔B(g)+C(g) (let ↔ be reversibility sign)
    In this case, mole fraction decreases while total pressure remains constant. Volume is also constant.
    Let moles of A be ##n_1##,
    moles of B=##n_2##
    and moles of C=##n_3##.
    Let moles of argon be ##a## and total number of moles before addition of argon be ##n##.
    Initially, ##K=\frac{n_2.n_3}{n_1(n)}P##
    After addition of argon, ##K'=\frac{n_2.n_3}{n(n+a)}P##
    (Pressure remains constant).
    Hence K decreases which means products are favoured.
    Is this correct?
     
  10. Aug 28, 2015 #9

    Borek

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    If you add argon, pressure doesn't stay constant.
     
  11. Aug 28, 2015 #10

    Titan97

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    The vessel is maintained at constant pressure.
     
  12. Aug 28, 2015 #11

    Borek

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    Then you are just diluting reacting gases, lowering their partial pressures - and that's a completely different case than the one you started with (you have mentioned constant volume in the very first post).
     
  13. Aug 28, 2015 #12

    Titan97

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    Yes. I was just thinking about a new situation. But in post 8, I found that K changed. But K can only change if temperature changes.
     
  14. Aug 29, 2015 #13

    Borek

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    No, you found that Q (reaction quotient) has changed, not K. After you change volume, equilibrium will shift till Q=K.
     
  15. Aug 29, 2015 #14

    Titan97

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    That solves all my doubts on Le Chatelier's principles.
     
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