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Why doesn't the concentration have an effect on the equilibrium constant?
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Hi!
So we were learning about Kc today and my teacher mentioned that concentration has no effect on Kc as a fact we just need to know, but I don't particularly understand why it doesn't have an effect.
So we were learning about Kc today and my teacher mentioned that concentration has no effect on Kc as a fact we just need to know, but I don't particularly understand why it doesn't have an effect.
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I did ask my teacher and the internet but my question wasn't really answered - she did mention that Kc takes into account concentration but not temperature
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I did ask my teacher and the internet but my question wasn't really answered - she did mention that Kc takes into account concentration but not temperature
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#2
The only thing that changes the constant is temperature, otherwise it wouldn’t be a constant.
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(Original post by John Wick11)
The only thing that changes the constant is temperature, otherwise it wouldn’t be a constant.
The only thing that changes the constant is temperature, otherwise it wouldn’t be a constant.
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#4
(Original post by _Mia101)
Yes, I know that - I just don't understand why
Yes, I know that - I just don't understand why
Forward rate = reverse rate
Concentrations unchanging
If you artificially add some more reactant, the rate of the forward reaction must increase, but the rate of the reverse reaction will not change. Hence the forward reaction moves more to the RHS until equilibrium is re-established.
Rate is a function of concentration. So is you change concentration you change rate.
But the rate constant shows the relative concentrations when rate forward = rate back.
This can only happen at one particular concentration ratio, i.e. the equilibrium constant.
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(Original post by charco)
The equilibrium is attained when the rate of the forward reaction equals the rate of the reverse reaction. At this point the concentrations of the components cannot change.
Forward rate = reverse rate
Concentrations unchanging
If you artificially add some more reactant, the rate of the forward reaction must increase, but the rate of the reverse reaction will not change. Hence the forward reaction moves more to the RHS until equilibrium is re-established.
Rate is a function of concentration. So is you change concentration you change rate.
But the rate constant shows the relative concentrations when rate forward = rate back.
This can only happen at one particular concentration ratio, i.e. the equilibrium constant.
The equilibrium is attained when the rate of the forward reaction equals the rate of the reverse reaction. At this point the concentrations of the components cannot change.
Forward rate = reverse rate
Concentrations unchanging
If you artificially add some more reactant, the rate of the forward reaction must increase, but the rate of the reverse reaction will not change. Hence the forward reaction moves more to the RHS until equilibrium is re-established.
Rate is a function of concentration. So is you change concentration you change rate.
But the rate constant shows the relative concentrations when rate forward = rate back.
This can only happen at one particular concentration ratio, i.e. the equilibrium constant.
I do still have one question though: if you change the concentrations artificially will the relative concentrations not change too?
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(Original post by John Wick11)
Try google for Le Chataliers Principle
Try google for Le Chataliers Principle
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#8
Yes and the position of equilibrium has to change in order to keep Kc, the equilibrium constant, constant
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#9
(Original post by _Mia101)
Thank you that makes a bit more sense now!
I do still have one question though: if you change the concentrations artificially will the relative concentrations not change too?
Thank you that makes a bit more sense now!
I do still have one question though: if you change the concentrations artificially will the relative concentrations not change too?
Think about it. If you add more of the left hand side, the forward reaction is now faster than the reverse reaction.
So, the system moves back towards equilibrium by making more of the right hand side and less of the left hand side.
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(Original post by charco)
Temporarily, but then the system is not at equilibrium and responds to restore the equilibrium condition.
Think about it. If you add more of the left hand side, the forward reaction is now faster than the reverse reaction.
So, the system moves back towards equilibrium by making more of the right hand side and less of the left hand side.
Temporarily, but then the system is not at equilibrium and responds to restore the equilibrium condition.
Think about it. If you add more of the left hand side, the forward reaction is now faster than the reverse reaction.
So, the system moves back towards equilibrium by making more of the right hand side and less of the left hand side.
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#11
(Original post by _Mia101)
oh okay, thank you! So the reason the equilibrium position moves is to maintain the equilibrium constant (which is the comparision of the concentration of the products over the concentration of the reactants)?
oh okay, thank you! So the reason the equilibrium position moves is to maintain the equilibrium constant (which is the comparision of the concentration of the products over the concentration of the reactants)?
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(Original post by charco)
Yes. There is only one ratio of products to reactants at which the rate forward = rate backwards.
Yes. There is only one ratio of products to reactants at which the rate forward = rate backwards.
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