Kc: I get it but I don't understand it
I accept that Kc does not change with pressure because that's what I've been taught but I don't understand WHY!
If pressure/concentration changes in an equilibrium, according to Le Chatelier's, the equilibrium will shift to compensate for this change. Therefore the concentrations of the reactants will change, and since Kc is worked out from concentration ratios... surely it must change!!
I've researched it and the internet tells me that it's because Kc must be constant so the concentrations re-adjust to keep it the same... but why??
If anyone can help me understand, I'd really appreciate it! Thanks in advance
If pressure/concentration changes in an equilibrium, according to Le Chatelier's, the equilibrium will shift to compensate for this change. Therefore the concentrations of the reactants will change, and since Kc is worked out from concentration ratios... surely it must change!!
I've researched it and the internet tells me that it's because Kc must be constant so the concentrations re-adjust to keep it the same... but why??
If anyone can help me understand, I'd really appreciate it! Thanks in advance
Original post by HannaRose
I accept that Kc does not change with pressure because that's what I've been taught but I don't understand WHY!
If pressure/concentration changes in an equilibrium, according to Le Chatelier's, the equilibrium will shift to compensate for this change. Therefore the concentrations of the reactants will change, and since Kc is worked out from concentration ratios... surely it must change!!
I've researched it and the internet tells me that it's because Kc must be constant so the concentrations re-adjust to keep it the same... but why??
If anyone can help me understand, I'd really appreciate it! Thanks in advance
If pressure/concentration changes in an equilibrium, according to Le Chatelier's, the equilibrium will shift to compensate for this change. Therefore the concentrations of the reactants will change, and since Kc is worked out from concentration ratios... surely it must change!!
I've researched it and the internet tells me that it's because Kc must be constant so the concentrations re-adjust to keep it the same... but why??
If anyone can help me understand, I'd really appreciate it! Thanks in advance
Here's an interactive that I prepared just for this question.
The effect of pressure changes on equilibria.
Check it out ...
The fact that the equilibrium shifts is what makes the Kc stay the same.
Picture a reaction
N2O4(g) --> 2NO2(g)
If you increase the pressure of this reaction, the value for NO2 is going to increase as there are twice as many gas moles as N2O4, okay?
Our Kc expression for this reaction will be:
Kc = [NO2]^2
.........N2O4
As the value of NO2 increases more than N2O4 the fraction becomes more top heavy as NO2 increases, so Kc will appear larger.
However, due to le Chatelier's principle, we know that increasing the pressure will favor the side with the fewer gas moles to minimise the increase in pressure.
Therefore, the equilibrium shifts to the left, lowering the value of NO2 and increasing the value of N2O4. The point of Kc is to tell us 'How far' the position of equilibrium is between the reactants and the products. Therefore the change in equilibrium will shift to the left until the Kc value is the same as the Kc value before the pressure was increased.
Does that make any sense?
Picture a reaction
N2O4(g) --> 2NO2(g)
If you increase the pressure of this reaction, the value for NO2 is going to increase as there are twice as many gas moles as N2O4, okay?
Our Kc expression for this reaction will be:
Kc = [NO2]^2
.........N2O4
As the value of NO2 increases more than N2O4 the fraction becomes more top heavy as NO2 increases, so Kc will appear larger.
However, due to le Chatelier's principle, we know that increasing the pressure will favor the side with the fewer gas moles to minimise the increase in pressure.
Therefore, the equilibrium shifts to the left, lowering the value of NO2 and increasing the value of N2O4. The point of Kc is to tell us 'How far' the position of equilibrium is between the reactants and the products. Therefore the change in equilibrium will shift to the left until the Kc value is the same as the Kc value before the pressure was increased.
Does that make any sense?
Original post by charco
Here's an interactive that I prepared just for this question.
The effect of pressure changes on equilibria.
Check it out ...
The effect of pressure changes on equilibria.
Check it out ...
Thank you!! that's really helpful because it explains both situations
Original post by herothing11
The fact that the equilibrium shifts is what makes the Kc stay the same.
Picture a reaction
N2O4(g) --> 2NO2(g)
If you increase the pressure of this reaction, the value for NO2 is going to increase as there are twice as many gas moles as N2O4, okay?
Our Kc expression for this reaction will be:
Kc = [NO2]^2
.........N2O4
As the value of NO2 increases more than N2O4 the fraction becomes more top heavy as NO2 increases, so Kc will appear larger.
However, due to le Chatelier's principle, we know that increasing the pressure will favor the side with the fewer gas moles to minimise the increase in pressure.
Therefore, the equilibrium shifts to the left, lowering the value of NO2 and increasing the value of N2O4. The point of Kc is to tell us 'How far' the position of equilibrium is between the reactants and the products. Therefore the change in equilibrium will shift to the left until the Kc value is the same as the Kc value before the pressure was increased.
Does that make any sense?
Picture a reaction
N2O4(g) --> 2NO2(g)
If you increase the pressure of this reaction, the value for NO2 is going to increase as there are twice as many gas moles as N2O4, okay?
Our Kc expression for this reaction will be:
Kc = [NO2]^2
.........N2O4
As the value of NO2 increases more than N2O4 the fraction becomes more top heavy as NO2 increases, so Kc will appear larger.
However, due to le Chatelier's principle, we know that increasing the pressure will favor the side with the fewer gas moles to minimise the increase in pressure.
Therefore, the equilibrium shifts to the left, lowering the value of NO2 and increasing the value of N2O4. The point of Kc is to tell us 'How far' the position of equilibrium is between the reactants and the products. Therefore the change in equilibrium will shift to the left until the Kc value is the same as the Kc value before the pressure was increased.
Does that make any sense?
That makes perfect sense, thank you so much!
I think I understand it now
Original post by herothing11
The fact that the equilibrium shifts is what makes the Kc stay the same.
Picture a reaction
N2O4(g) --> 2NO2(g)
If you increase the pressure of this reaction, the value for NO2 is going to increase as there are twice as many gas moles as N2O4, okay?
Picture a reaction
N2O4(g) --> 2NO2(g)
If you increase the pressure of this reaction, the value for NO2 is going to increase as there are twice as many gas moles as N2O4, okay?
I wonder if you could help me, I'm getting really confused with this!
I thought that if you increase the pressure, the equilibrium shifts to the side with fewer gaseous moles, so the LHS, and therefore the value of N2O4 increases?
Original post by 1901
I wonder if you could help me, I'm getting really confused with this!
I thought that if you increase the pressure, the equilibrium shifts to the side with fewer gaseous moles, so the LHS, and therefore the value of N2O4 increases?
I thought that if you increase the pressure, the equilibrium shifts to the side with fewer gaseous moles, so the LHS, and therefore the value of N2O4 increases?
That's right, yes. If you increase the pressure, initially the NO2 will increase as there are more moles.
Le Chatelier's principle states that the position of equilibrium will shift to minimise this change, to do this it will therefore shift towards the left to decrease NO2 and increase N2O4.
Hope that clears it up for you.
Original post by herothing11
That's right, yes. If you increase the pressure, initially the NO2 will increase as there are more moles.
Le Chatelier's principle states that the position of equilibrium will shift to minimise this change, to do this it will therefore shift towards the left to decrease NO2 and increase N2O4.
Hope that clears it up for you.
Le Chatelier's principle states that the position of equilibrium will shift to minimise this change, to do this it will therefore shift towards the left to decrease NO2 and increase N2O4.
Hope that clears it up for you.
Ohhhhh okay I see, thanks alot!
Original post by shepan6
So, for example with N2 + 3H2 <-> 2NH3. If you increase the pressure, and [N2] will increase as that side of the reaction has the highest number of moles?
Yes. The left hand side will increase because it has more gas moles, this causes the equilibrium to shift to the right however, as the change must be minimised.
Original post by herothing11
Yes. The left hand side will increase because it has more gas moles, this causes the equilibrium to shift to the right however, as the change must be minimised.
I have no idea from where you are getting your logic or understanding ...
If you increase the pressure of a system at equilibrium it moves towards the side of fewer moles to restore the value of kc
The initial increase in pressure disturbs the equilibrium.
The system responds by making more of the side which has fewer moles.
At no time is there 'more moles' of the other side!!!!
Ok, so take the reaction A(g) --> 2B(g).
Kc is given by p(B)^2/p(A) where p(B) denotes the partial pressure of B.
If you increase the pressure, the partial pressures of B and A both change - A will increase, where B will decrease (according to Le C. - equilibrium shifts left, more A is produced, so its partial pressure increases, where B's decreases.)
But the effects of the changes of B and A cancel each other out, leaving Kc the same. So even though the equilibrium does shift, the value of the equilibrium constant remains the same.
Hope this is helps :-)
Kc is given by p(B)^2/p(A) where p(B) denotes the partial pressure of B.
If you increase the pressure, the partial pressures of B and A both change - A will increase, where B will decrease (according to Le C. - equilibrium shifts left, more A is produced, so its partial pressure increases, where B's decreases.)
But the effects of the changes of B and A cancel each other out, leaving Kc the same. So even though the equilibrium does shift, the value of the equilibrium constant remains the same.
Hope this is helps :-)
Original post by charco
I have no idea from where you are getting your logic or understanding ...
If you increase the pressure of a system at equilibrium it moves towards the side of fewer moles to restore the value of kc
The initial increase in pressure disturbs the equilibrium.
The system responds by making more of the side which has fewer moles.
At no time is there 'more moles' of the other side!!!!
If you increase the pressure of a system at equilibrium it moves towards the side of fewer moles to restore the value of kc
The initial increase in pressure disturbs the equilibrium.
The system responds by making more of the side which has fewer moles.
At no time is there 'more moles' of the other side!!!!
There is a change in concentration values actually.
The position of equilibrium doesn't just move towards the side with fewer moles for the sake of it.
In a reaction, for example
Kc = [NO2]^2
.........N2O4
Increasing the pressure increases the concentration terms for NO2 more than that of N2O4. The fraction therefore becomes more top heavy. That's why it shifts to the left hand side, to minimize this change.
This is standard knowledge. It's even found in the mark schemes of the OCR papers.
(edited 10 years ago)
Original post by herothing11
There is a change in concentration values actually.
The position of equilibrium doesn't just move towards the side with fewer moles for the sake of it.
In a reaction, for example
Kc = [NO2]^2
.........N2O4
Increasing the pressure increases the concentration terms for NO2 more than that of N2O4. The fraction therefore becomes more top heavy. That's why it shifts to the left hand side, to minimize this change.
This is standard knowledge. It's even found in the mark schemes of the OCR papers.
The position of equilibrium doesn't just move towards the side with fewer moles for the sake of it.
In a reaction, for example
Kc = [NO2]^2
.........N2O4
Increasing the pressure increases the concentration terms for NO2 more than that of N2O4. The fraction therefore becomes more top heavy. That's why it shifts to the left hand side, to minimize this change.
This is standard knowledge. It's even found in the mark schemes of the OCR papers.
Your wording makes it sound as though the reaction first goes backwards to then go forwards!
Initially the actual amount (moles) of the side with a greater number of moles of gas does not increase, the concentration of the gas (moles per unit volume) increases more than that of the other side.
Original post by charco
Your wording makes it sound as though the reaction first goes backwards to then go forwards!
Initially the actual amount (moles) of the side with a greater number of moles of gas does not increase, the concentration of the gas (moles per unit volume) increases more than that of the other side.
Initially the actual amount (moles) of the side with a greater number of moles of gas does not increase, the concentration of the gas (moles per unit volume) increases more than that of the other side.
That's what I said. You must have misunderstood.
I assumed we were talking about concentrations as Kc is expressed in terms of concentration values.
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