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Stoichiometry Keq

Discussion in 'MCAT Study Question Q&A' started by labqi, 05.17.14.

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  1. labqi

    labqi 2+ Year Member

    Joined:
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    Hey guys!
    I just had a quick question regarding the keq equation.
    Suppose: 2A+B <---> 3C

    I know the equilibrium equation would be: [c]^3/[a]^2

    Suppose we increased the concentration of c by 2 moles.
    A) Would I have to convert the moles into Molarity to use in the ICE table. Or could I just uses moles in the table.
    B) If I had increased the concentration by 2M, thats equivalent to saying i increased the moles by 2 in relation to the volume of the solvent. Is that correct?
    C) With that said, to determine the new equilibrium concentrations...would the new equation look something like this: [2 - 3x]^2/[2x]^2[x] = keq
    D) When is it okay to remove x the from the denominator or numerator? I know there is a short cut way but I can't seem to recall. Was it removing the x or the actual numerical value?

    Thanks for the help! good luck with the prep!!!
     
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  3. SuperSneaky24

    SuperSneaky24

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    Wait what happened to B? B should also be in the equil. eq, unless it is constant or in excess.

    My stoichiometry isn't as strong as my other subjects, but let's see how I do:
    A) Would I have to convert the moles into Molarity to use in the ICE table. Or could I just uses moles in the table.
    nope. as long as you keep the units constant, you can use moles or molarity
    B) If I had increased the concentration by 2M, thats equivalent to saying i increased the moles by 2 in relation to the volume of the solvent. Is that correct?
    I believe so
    C) With that said, to determine the new equilibrium concentrations...would the new equation look something like this: [2 - 3x]^2/[2x]^2[x] = keq
    wait what concentration are you increasing? I'm guessing you are adding 2M of C and none of A into the reaction? If so, it would be Keq= [2 - 3x]^3/[2x]^2 (assumes B isnt in the Keq)
    D) When is it okay to remove x the from the denominator or numerator? I know there is a short cut way but I can't seem to recall. Was it removing the x or the actual numerical value?
    It depends on the situation. Say you have a very small Keq value. That means the x value in the denominator (reactants) can usually be ignored.
     
    Last edited: 05.17.14
    labqi likes this.
  4. Chrisz

    Chrisz 2+ Year Member

    Joined:
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    Your question is not that quick, 4 questions, lol.
    A: if the volume does not change, you dont need to convert to moles. If the volume does change, you have to take consideration of the new resultant concentrations.
    B: M=mole/volume. M means number of moles per liter solution. If you say increase con by 2M. it means inrease 2 mols per liter solution. Yes, your interpretation is correct.
    C: In your case, Keq=C^3/{(A^2)(B)}. To determine the new equilibrium. We have to know what is the initial concentration of C, and we define it as Ci. if we increase the initial conentration by 2M. The new initial concentraition would be Ci+ 2M. Assuming the initial concentrations of of Ai and Bi equal to zero, then
    Keq=(Ci+ 2M-3x)^3/[(2x)^2*x]. If initial con of A and B is not zero, you have to take consideration of their effects.
    D: It is based on the the equilibrium constant. if tells how far away your reactions lies. initial concentration has to be relatively large compared to x. If this is the case, x can be ignored. For a hypothetical reaction 2A---> B . If you have initial concentration of Ai. Your formulation would become k=x/(Ai-2x)^2. If Keq is small and Ai is relatively large, you know the reaction equilibrium lies to the left, so 2x would be small compared to Ai. Then, you can ignore 2x. Once you got x, compare 2x with Ai, if it is within 5% of Ai, your estimation is valid
     
    labqi likes this.
  5. labqi

    labqi 2+ Year Member

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    Thanks for the response!
     

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