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    When you say increasing the concentration of a species shifts equilibrium position to the other side of the equation is this a temporary shift in the sense that once the concentration of that particular species has reduced and kc been restores the position of equilibrium will be in its original position

    thanks
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    It doesn't return to its original position, it goes to a position such that the ratio of concentrations once again equals Kc.
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    (Original post by Pigster)
    It doesn't return to its original position, it goes to a position such that the ratio of concentrations once again equals Kc.
    I don't really understand, say you had the equilibrium

    a <---> b

    and you increase the concentration of b Kc at this point will not have its real value so the equilibrium position shifts to the left to create more a and use up b. When Kc has finally been restored the position of equilibrium will be in the same position relative to a and b,. This is because I see the position of equilibrium as the ratio of reactants produced to products.

    I thought that temperature is the only factor which changes equilibrium permanantly because it changes the ratio of reactants to products when equilibrium is once again achieved. (after the initial change in temperature of a particular equilibrium).

    Have I got this wrong?
    Thanks
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    For an A <-> B equilibrium, adding B make the ratio of [B] / [A] no longer = to Kc, so will favour the backwards reaction until the ratio once again returns to Kc. The position of the equilibrium will be unchanged eventually.

    For an A + B <-> C reaction, adding B will make more A convert into C, the position of the equilibrium has shifted to the right.
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    (Original post by Pigster)
    For an A <-> B equilibrium, adding B make the ratio of [B] / [A] no longer = to Kc, so will favour the backwards reaction until the ratio once again returns to Kc. The position of the equilibrium will be unchanged eventually.

    For an A + B <-> C reaction, adding B will make more A convert into C, the position of the equilibrium has shifted to the right.
    Thanks for the reply,

    i understand if you add B the equilibrium will shift to the right. My understanding is that B will react with A to produce C. So B decreases (but is higher than the initial B before any was added) A decreases and C increases. What results is a restored value of Kc.

    Even though A is lower than initially, B is higher than initially which cancels any increase in C and so would i be wrong if i said eventually the equilibrium would return to its initial position, when Kc has been restored?

    Thanks
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    (Original post by 111davey1)
    ..so would i be wrong if i said eventually the equilibrium would return to its initial position, when Kc has been restored?
    If at equilibrium you had 1 mol of A, B & C all in 1 dm3, then Kc = 1 / 1x1

    Adding 1 mol of B -> 1 / 1x2 i.e not Kc

    B + A -> C is favoured until...

    [A] = 0.732
    [B] = 1.732
    [C] = 1.268

    Now, Kc = 1.268 / (0.732 x 1.732)

    The position isn't the same, but the ratio of terms is. There is, as you say, more B and C and less A - that isn't the same position.
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    (Original post by Pigster)
    If at equilibrium you had 1 mol of A, B & C all in 1 dm3, then Kc = 1 / 1x1

    Adding 1 mol of B -> 1 / 1x2 i.e not Kc

    B + A -> C is favoured until...

    [A] = 0.732
    [B] = 1.732
    [C] = 1.268

    Now, Kc = 1.268 / (0.732 x 1.732)

    The position isn't the same, but the ratio of terms is. There is, as you say, more B and C and less A - that isn't the same position.
    Thanks, so at equilibrium has the position of equilibrium shifted to the right or left?
    Im confused as to what the position actually means, i thought Kc was like a quantitative measure of the position.
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    (Original post by 111davey1)
    Thanks, so at equilibrium has the position of equilibrium shifted to the right or left?
    Im confused as to what the position actually means.
    Normally, you'd be interested in the yield of the product (that which is quoted on the RHS), so you'd say the equilibrium has shifted to the right.
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    (Original post by Pigster)
    Normally, you'd be interested in the yield of the product (that which is quoted on the RHS), so you'd say the equilibrium has shifted to the right.
    Thanks, but the yield of B has also increased, the net increase is the same on both sides so why has the equilibrium position shifted?
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    You don't really talk about the yield of a reactant.
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    (Original post by Pigster)
    You don't really talk about the yield of a reactant.
    so for the a <--> b equilibrium mentioned earlier if you add a and wait until equilibrium is established does the position has actually lie to the right because more b has formed? Oh other way round but same idea
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    (Original post by Pigster)
    You don't really talk about the yield of a reactant.
    My textbook states that 'a Kc Value of 1 would indicate that the position of equilibrium is halfway between the reactants and products. The reason i am confused is because in the initial

    a+b<--> c

    reaction a b and c was given 1 mold-3 concentrations. hence Kc value of 1 and according to my textbook the position of equilibrium is halfway between reactants and products. After some B was added yes more C formed but Kc reformed its value of 1 and my textbook would suggest that the position of equilibrium also reformed to halfway between the reactants and the products and so the position did not change in the end...
 
 
 
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