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Equilibrium constant Kp,eq issue

  1. Aug 29, 2016 #1
    1. The problem statement, all variables and given/known data
    question1-sentence
    View attachment 105257
    question1-answer
    View attachment 105258

    question2-sentence
    View attachment 105259
    question2-answer
    View attachment 105260
    2. Relevant equations
    Kp,eq=Pproduct^stoichiometric/Preactant^stoichiometric=yproduct^stoichiometric/yreactant^stoichiometric

    3. The attempt at a solution
    I did not understand why in question 1 why Kp,eq is done like that and why Psat(P) is underneath.
    And why in question 2 Psat(P) is at the top in Kp,eq formula.

    Why one time P is below and 1 time P is at the top in Kp,eq formula?
     
  2. jcsd
  3. Aug 29, 2016 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Please post HW questions in the proper HW forum.
     
  4. Aug 29, 2016 #3
    What is the change in the number of moles in the balanced equation for the first reaction?
    What is the change in the number of moles in the balanced equation for the second reaction?
     
  5. Aug 29, 2016 #4
    Thank you for your quick response.
    I would really appreciate it if you could answer why they changed the P position in the formula.

    I am in kind of a rush to understand this because I will have an exam soon on those things, and there is not much time to focus on.

    I need to be fast, and understand when to put P on top and when to put it on the bottom.

    I don't really get what you want to mean with "the change in number of moles in the balanced equation"
    I don't get what has the change of number of moles to do with the kp,eq.
    I want to understand why they changed P , why 1 time is underneath in formula and why 1 time is on the top.

    For the 1st reaction extent of reaction ξ=0.873 mol
    For the 2nd reaction extent of reaction ξ=0.877 mol

    Q1 Capture.JPG
    Q1-Answer Capture2.JPG


    Q2- Capture3.JPG

    Q2-Answer Capture4.JPG
     
  6. Aug 29, 2016 #5

    Borek

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    Staff: Mentor

    What is Δv in these formulas? Do you understand where it comes from?

    (Actually it is exactly the same question Chestermiller asked).
     
  7. Aug 29, 2016 #6
    In problem 1, the give you the mole fractions of the 3 species in the final state. What are the partial pressures of these three species algebraically in terms of the total pressure P?
     
  8. Aug 30, 2016 #7
    Thank you for your answer, I do not know what it is and where it comes from.
    I really need help to understand this as soon as possible.
    Thank you
     
  9. Aug 30, 2016 #8
    We can interpret the partial pressure P1 let's say as
    P1=x1*P1,sat=y1*P
     
  10. Aug 30, 2016 #9

    Borek

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    Staff: Mentor

    Sorry then, go back to your book and notes.

    Look for a ways to convert between Kc and Kp.
     
  11. Aug 30, 2016 #10
    Thank you, it does not say anywhere what that is, I looked everywhere,and tomorrow I have exam on this, please help me with this.
    It is much appreciated
     
  12. Aug 30, 2016 #11
    I know partial pressure P1=y1*P
     
  13. Aug 30, 2016 #12

    Borek

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    Staff: Mentor

    Have you tried googling for "converting Kc to Kp"?
     
  14. Aug 30, 2016 #13
    Thanks for the answer ,I know that Kc=Product (P/RT)^delta v /Reactant (P/RT)^delta v
     
  15. Aug 30, 2016 #14
    I need to understand when to know when delta v= -1 and when delta v=1
     
  16. Aug 30, 2016 #15

    Borek

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    Staff: Mentor

    Come on, look at any page where the conversion is described and check what is the definition of the Δv and how it is calculated. It is really trivial.
     
  17. Aug 30, 2016 #16
    Okay. Now substitute this into the equation for ##K_p## (i.e., with ##K_p## is expressed in terms of partial pressures) for your A+B=C problem. What do you get? (And please use LaTex).
     
  18. Sep 1, 2016 #17
    This applies to vapor-liquid equilibrium, not to chemical reactions. But, the y part is correct. What do you get when you substitute this for A, B, and C into the equilibrium relationship.
     
  19. Sep 12, 2016 #18

    Dr Uma Sharma

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    Gold Member

    delta v = total number of moles of gases on product side - total number of moles of gases on the reactant side
    like 2A(g)+B(g)---> 3C (g)
    in this case products moles = 3
    reactants = 3
    delta v= 0
    similarly it can be negative or positive
    Remember mole is equal to volume if gases are at the same temp and pressure.
     
  20. Sep 12, 2016 #19
    Did you really want to say this?
     
  21. Sep 12, 2016 #20

    Dr Uma Sharma

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    Gold Member

    Sorry they are not equal but proportional
    Moles ratio of gases and volume ratio of the gases are same if temp and pressure are kept constant.
    Avogadro's law
    ...that equal volumes of gases at the same temperature and pressure contain equal numbers of moles.
     
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