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Homework Help: Equilibrium constant

  1. Jun 4, 2005 #1
    Could someone please help me with this problem?

    Samples of A (2.5 mol) and B (1.0 mol) are placed in a 5.0L container and the following reaction takes place:


    2A(g) <--> B(g) + C(g)

    At equilibrium, the concentration of A is 0.20M. What is the value of Kc?

    Thanks. It's probably really simple but I can't seem to get it. I'm not that strong on this stuff.

    Also, apparently the relationship between Kc and Kp is Kc = Kp. Why is that the case?

    Thanks again.
     
  2. jcsd
  3. Jun 4, 2005 #2
    [itex]
    K_p=K_c (RT)^{dn}
    [/itex]

    where [itex]dn[/itex]= No. of moles of products- No. of moles of reactants

    Kc = Kp because the number of moles of the gaseous componentson both sides are the same.

    For your first query , what is law of mass action?....How is Kc related to the products and reactants?
     
    Last edited by a moderator: Jun 5, 2005
  4. Jun 4, 2005 #3

    GCT

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    [tex]Kc= \frac{[C]}{[A]^{2}} [/tex]

    [tex]Kc= \frac{[C_b+.25mol/5.0L][.25mol/5.0L]}{[.200M]^{2}} [/tex]

    [tex]Kc= \frac{[1.0mol/5.0L+.25mol/5.0L][.25mol/5.0L]}{[.200M]^{2}} [/tex]

    solve for [tex] Kc[/tex]
     
  5. Jun 5, 2005 #4

    siddharth

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    Initially, the concentration of all the species are
    [A] = 2.5(mol) / 5.0(L) = c1 -I
    = 1.0(mol) / 5.0(L) = c2 -II
    [C] = 0

    Now, let 'a' be the extent of the reaction

    Concentration of all species at equilibirium are

    [A]=c1(1-2a) -III
    =c2+c1(a)
    [C]=c1(a)

    Now it is given that the concentration of [A] at equilibirium is 0.2 mol/L.
    From III, c1(1-2a)=0.2
    Now, the value of c1 is known.Therefore the value of a can be found.
    From this the value of Kc=([C])/([A]^2) {The concentration of species at equilibirium}

    GCT:
    Instead of 0.25, shouldn't it be 0.75? Initially, number of moles of A are 2.5. Finally, the number of moles of A 0.2*5=1 . Therefore the number of moles of A reacted = 2.5-1=1.5 . By molar ratio, the number of moles of B and C formed are 0.75.
     
  6. Jun 5, 2005 #5
    Ok, I get it. Thanks guys :smile:
     
  7. Jun 5, 2005 #6

    GCT

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    I'm not quite sure why you're trying to find the number of moles, when the equilibrium concentration of A is given. Anyways I neglected to use the appropriate initial concentration for A.

    [tex].25mol/5.0L=.5M,~mol~A~reacted~=~.5M-.20M=.3M[/tex]

    you'll need to divide by 2, to find the equilvalent moles of B and C

    [tex].3M~A~reacted(1~mol~B/2~mol~A)~=~.15M~B,C~formed[/tex]
     
  8. Jun 5, 2005 #7

    siddharth

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    It should be 2.5mol/5L=.5M
    Anyway, finding the number of moles is the same thing. If 0.75 moles of B and C are formed, then the concentration of each is 0.75/5= 0.15 M. That's why in your initial post, instead of 0.25/5, it should be 0.75/5=0.15M. (As you wrote in the last post)
     
    Last edited: Jun 5, 2005
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