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Chem Equilibrium Q

  1. May 4, 2005 #1
    A(g) <---
    <--- 2B(g) + C(g)

    When 1.00mol of A is placed in a 4.00L container at temperature t, the concentration of C at equilibrium is 0.050mol/L. What is the equilibrium constant for the reaction at temperature t?

    Thanks for the help.
     
  2. jcsd
  3. May 5, 2005 #2
    A(g)-> 2B(g)+ C(g)
    1mol
    (1-x)/4 2x/4 x/4
    they have given that x=0.05mol/L
    kc= (0.05/4*0.01/16)/0.95/4
    =1/19*100mol/L
     
  4. May 5, 2005 #3
    OK lets look at the problem. First what is the initial concentration of A. Second given the stoichiometry of the problem and the value of C at equilibrium what are the concentrations of A and B ? Finally how due you write the equilibrium constant (and don't forget powers). Due the math ;)
     
  5. May 5, 2005 #4
    can u explain to me how u set that up? and how do you solve it? (do the calculations)
     
  6. May 5, 2005 #5

    GCT

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    [tex]Ka= [2x]^{2}[x]/[initial conc.-x][/tex]

    Can you explain to us why it'll be in such a form?
     
  7. May 5, 2005 #6
    the equilibrium constant for the reaction at temperature t should be Kc so considering
    the moles used to start this reaction which is 1mol of A(g) you can determine the moles used to get the right side of the equation at equilibrium. So the moles used from A(g) should be (1-x)moles and they have given that at equilibrium the concentration of C is 0.05mol(x=0.05) and it equals to the concentration of B(g) but there are two moles of B so it should be 2*0.05
    and by deviding these moles by the volume of the container you get mol/L of each gas
    kc= [C(g)] *[B(g)]*[B(g)]/[A(g)]

    I hope this helps and my guess is that it should be 0.05mol and not 0.05mol/L
     
    Last edited: May 5, 2005
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