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Why can an equlibirum only be reached when the reaction takes place in a closed.. Watch

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    Why is it that an equilibrium only be reached when the reverisble reaction takes place in a closed system?

    Also, what does 'dynamic' mean in the context of 'dynamic equilibrium'?
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    (Original post by SANTR)
    Why is it that an equilibrium only be reached when the reverisble reaction takes place in a closed system?

    Also, what does 'dynamic' mean in the context of 'dynamic equilibrium'?
    Think about what a 'closed system' really means...
    It's a system that is isolated from any other systems or environments. So the only thing happening in this system is whatever I've put in it, e.g. reactants. Those reactants are limited by the size of the system.

    Conversely, if I had an open system, then equilibrium wouldn't be reached because products could be escaping, or new reactants would be introduced that would affect the rate of both the forward and backward reaction.

    Speaking of which, that is what 'dynamic' refers to. In an equilibrium, you have a constant concentration of products and reactants, but that's not because the reaction suddenly stops. It's because the forward reaction is producing a product at the same rate, the backward reaction is consuming that product (to reproduce the reactant).
    In other words, the rate of the forward and the backward reactions are equal.
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    (Original post by RMNDK)
    Think about what a 'closed system' really means...
    It's a system that is isolated from any other systems or environments. So the only thing happening in this system is whatever I've put in it, e.g. reactants. Those reactants are limited by the size of the system.

    Conversely, if I had an open system, then equilibrium wouldn't be reached because products could be escaping, or new reactants would be introduced that would affect the rate of both the forward and backward reaction.

    Speaking of which, that is what 'dynamic' refers to. In an equilibrium, you have a constant concentration of products and reactants, but that's not because the reaction suddenly stops. It's because the forward reaction is producing a product at the same rate, the backward reaction is consuming that product (to reproduce the reactant).
    In other words, the rate of the forward and the backward reactions are equal.
    Thanks for the response.
    I have a few question further questions.
    What causes the rate of the forward and backward reactions to be the SAME eventually?
    Also, I would appreciate it if you explain why the rate of the forward or backward reaction is affected if the products formed are removed as the reaction happens?
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    (Original post by SANTR)
    Thanks for the response.
    I have a few question further questions.
    What causes the rate of the forward and backward reactions to be the SAME eventually?
    Also, I would appreciate it if you explain why the rate of the forward or backward reaction is affected if the products formed are removed as the reaction happens?
    I think a good way (not the best way) to explain it would be to consider what happens to the reactants and products in a reaction.

    If you can't see the picture,
    http://www.examstutor.com/chemistry/...figure_7_2.gif

    Imagine I had this reaction A + B <-> C + D
    And I put some A and B into a closed system. They are going to react and the rate of the reaction is going to be at it's maximum because that's the maximum concentration. But as the reaction progresses, the concentration of A and B fall. Thus, kinetic theory dictates that the rate of reaction should fall as well.

    Now look at C and D. We had no concentration of it before so the rate of reaction is going to be zero as well at time 0. But as the reaction progresses, the concentration of C and D increase so the rate of reaction begins to increase.

    But the rates of both reactions will eventually settle at a steady rate because something is limiting the reaction: Each other.
    A and B react to produce C and D. Eventually A and B become really low in supply, so less C and D is formed. But because C and D increase in their rate of reaction, A and B are regenerated again which causes the supply of C and D to become low. But because A and B re-increase in their rate of reaction, they refrom C and D.... etc etc
    Do you see how we're flickering from one side of the reaction to the other all the time? And eventually, this amount of 'flickering' decreases until the point that every time some A and B go to C and D, the same amount of C and D go back to A and B. It's because of the fact that we're in a closed system that the rates of reaction have to eventually reach an equilibrium given enough time.

    Which leads nicely into your second question. I'm producing C and D, which are going to go back into A and B. But if you're asking to remove C and D, there's no chance for any regeneration of A and B. i.e. you just want a reaction where A and B goes forward but doesn't 'flicker' back to A and B. In other words, this is a one way reaction. The rate of reaction just simply decreases, not as a straight line but fast then slow.
    Spoiler:
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    Don't get this confused with le Chatelier's principle which is orientated around yield rather than rate of reaction
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    (Original post by RMNDK)
    I think a good way (not the best way) to explain it would be to consider what happens to the reactants and products in a reaction.

    If you can't see the picture,
    http://www.examstutor.com/chemistry/...figure_7_2.gif

    Imagine I had this reaction A + B <-> C + D
    And I put some A and B into a closed system. They are going to react and the rate of the reaction is going to be at it's maximum because that's the maximum concentration. But as the reaction progresses, the concentration of A and B fall. Thus, kinetic theory dictates that the rate of reaction should fall as well.

    Now look at C and D. We had no concentration of it before so the rate of reaction is going to be zero as well at time 0. But as the reaction progresses, the concentration of C and D increase so the rate of reaction begins to increase.

    But the rates of both reactions will eventually settle at a steady rate because something is limiting the reaction: Each other.
    A and B react to produce C and D. Eventually A and B become really low in supply, so less C and D is formed. But because C and D increase in their rate of reaction, A and B are regenerated again which causes the supply of C and D to become low. But because A and B re-increase in their rate of reaction, they refrom C and D.... etc etc
    Do you see how we're flickering from one side of the reaction to the other all the time? And eventually, this amount of 'flickering' decreases until the point that every time some A and B go to C and D, the same amount of C and D go back to A and B. It's because of the fact that we're in a closed system that the rates of reaction have to eventually reach an equilibrium given enough time.

    Which leads nicely into your second question. I'm producing C and D, which are going to go back into A and B. But if you're asking to remove C and D, there's no chance for any regeneration of A and B. i.e. you just want a reaction where A and B goes forward but doesn't 'flicker' back to A and B. In other words, this is a one way reaction. The rate of reaction just simply decreases, not as a straight line but fast then slow.
    Spoiler:
    Show
    Don't get this confused with le Chatelier's principle which is orientated around yield rather than rate of reaction
    Thank you very much for the thorough explanation.
    Could you also say that if the system was not closed, products would escape and this would mean that the equilibrium would shift to the right to oppose this change and therefore the forward reaction would be favoured i.e. to shift the equilibrium to the right, the rate at which the forward reaction occurs would increase. And all of this would mean that an equilibrium would be prevented from being reached,
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    (Original post by SANTR)
    Thank you very much for the thorough explanation.
    Could you also say that if the system was not closed, products would escape and this would mean that the equilibrium would shift to the right to oppose this change and therefore the forward reaction would be favoured i.e. to shift the equilibrium to the right, the rate at which the forward reaction occurs would increase. And all of this would mean that an equilibrium would be prevented from being reached,
    You're right. An open system is an example of your idea of removing the products and the equilibrium would indeed shift to the right to re-increase the concentration of the products.

    Ah, here's the problem. A lot of people equate equilibrium shift and rate of reaction together but sometimes in the wrong way.

    The equilibrium would shift to the right because the forward reaction is being favoured (in fact its the only one really happening in this open system) but it doesn't mean the rate of reaction is increasing.

    Imagine a closed system.
    Say the rate of the forward reaction is at amoldm-3s-1. This occurs at the start of the reaction, time = 0s because that's when concentration is highest. It then drops to bmoldm-3s-1 eventually when equilibrium is reached and that is the rate it's at continuously.
    Then you turn the closed system into an open system to allow the products to escape.
    You're right, now the forward reaction is more favoured because there are less collisions happening between the products to reform the reactants.
    But how can the rate of reaction of the forward reaction increase from bmoldm-3s-1 ? The only way the rate of a reaction begins to increase again is if we increase the concentration of the reactants, or the pressure, or increase the temperature.
    The only reason the rate is staying constant at b is because of the equilbrium set up. The products are turning back into the reactants as much as the reactants go to the products. Now the products aren't going back into the reactants, but that doesn't mean the reactants turn into the products any faster.
    Thus, the rate can only drop to some lower rate, cmoldm-3s-1

    Another way to visualize this would be to think that removing the products once a reaction is at equilibrium is like creating the same reaction all over again.
    i.e.
    Stage 1: Only A + B. Rate of reaction starts high (a) then low (b).
    Stage 2: A + B and C + D, rate of reaction remains low (b).
    Stage 3: Remove products.
    Stage 4: Only A + B. Rate of reaction starts low (b) then even lower (c)
    See how stage 4 is the same stage as stage 1, only the concentrations are different.

    The rest of your understanding is correct. It's just that technical detail of understanding how rate of reaction is integrated into this equilibrium topic.
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    (Original post by RMNDK)
    You're right. An open system is an example of your idea of removing the products and the equilibrium would indeed shift to the right to re-increase the concentration of the products.

    Ah, here's the problem. A lot of people equate equilibrium shift and rate of reaction together but sometimes in the wrong way.

    The equilibrium would shift to the right because the forward reaction is being favoured (in fact its the only one really happening in this open system) but it doesn't mean the rate of reaction is increasing.

    Imagine a closed system.
    Say the rate of the forward reaction is at amoldm-3s-1. This occurs at the start of the reaction, time = 0s because that's when concentration is highest. It then drops to bmoldm-3s-1 eventually when equilibrium is reached and that is the rate it's at continuously.
    Then you turn the closed system into an open system to allow the products to escape.
    You're right, now the forward reaction is more favoured because there are less collisions happening between the products to reform the reactants.
    But how can the rate of reaction of the forward reaction increase from bmoldm-3s-1 ? The only way the rate of a reaction begins to increase again is if we increase the concentration of the reactants, or the pressure, or increase the temperature.
    The only reason the rate is staying constant at b is because of the equilbrium set up. The products are turning back into the reactants as much as the reactants go to the products. Now the products aren't going back into the reactants, but that doesn't mean the reactants turn into the products any faster.
    Thus, the rate can only drop to some lower rate, cmoldm-3s-1

    Another way to visualize this would be to think that removing the products once a reaction is at equilibrium is like creating the same reaction all over again.
    i.e.
    Stage 1: Only A + B. Rate of reaction starts high (a) then low (b).
    Stage 2: A + B and C + D, rate of reaction remains low (b).
    Stage 3: Remove products.
    Stage 4: Only A + B. Rate of reaction starts low (b) then even lower (c)
    See how stage 4 is the same stage as stage 1, only the concentrations are different.

    The rest of your understanding is correct. It's just that technical detail of understanding how rate of reaction is integrated into this equilibrium topic.
    Would I be correct in saying that, removing some of the product formed i.e. lowering the concentration of the product would lower the rate of the backwards reaction (this is because there are fewer product molecules, lower frequency of collisions etc) but the forwards reaction is still continuing at the same rate as it was at equilibrium and this therefore means that the rate of the forwards reaction is > the backwards reaction (and so the rates of the respective reactions are no longer the same). Consequently, more product will be formed.
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    (Original post by SANTR)
    Would I be correct in saying that, removing some of the product formed i.e. lowering the concentration of the product would lower the rate of the backwards reaction (this is because there are fewer product molecules, lower frequency of collisions etc) but the forwards reaction is still continuing at the same rate as it was at equilibrium and this therefore means that the rate of the forwards reaction is > the backwards reaction (and so the rates of the respective reactions are no longer the same). Consequently, more product will be formed.
    Absolutely correct.
 
 
 
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