# position of equilibrium

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

So i realised that there is a contradiction in my book. It says that Kc measures the extent of a chemical reaction and its magnitude can give you an indication of where the equilibrium lies. I understand this.. if it is very large or very small it means equilibrium lies to the right or left respectively and if it is moderate it would suggest equilibrium lies somewhere in the middle. Say its very large it would mean you have little reactant and lots of product at equilibrium.

It also says that it is not changed by concentration or pressure.

so if we had this equilibrium: A + B <---> c

and increase the concentration of A we would say the equilibrium position has shifted to the right and more c would be produced. I understand how Kc stays constant i just don't fully understand the terminology.

Basically, i am wondering how the equilbrium position moves to the right to produce more c yet according to Kc the extent of the reaction (or the position of equilibrium) hasn't actually changed.

It also says that it is not changed by concentration or pressure.

so if we had this equilibrium: A + B <---> c

and increase the concentration of A we would say the equilibrium position has shifted to the right and more c would be produced. I understand how Kc stays constant i just don't fully understand the terminology.

Basically, i am wondering how the equilbrium position moves to the right to produce more c yet according to Kc the extent of the reaction (or the position of equilibrium) hasn't actually changed.

0

reply

Report

#2

(Original post by

So i realised that there is a contradiction in my book. It says that Kc measures the extent of a chemical reaction and its magnitude can give you an indication of where the equilibrium lies. I understand this.. if it is very large or very small it means equilibrium lies to the right or left respectively and if it is moderate it would suggest equilibrium lies somewhere in the middle. Say its very large it would mean you have little reactant and lots of product at equilibrium.

It also says that it is not changed by concentration or pressure.

so if we had this equilibrium: A + B <---> c

and increase the concentration of A we would say the equilibrium position has shifted to the right and more c would be produced. I understand how Kc stays constant i just don't fully understand the terminology.

Basically, i am wondering how the equilbrium position moves to the right to produce more c yet according to Kc the extent of the reaction (or the position of equilibrium) hasn't actually changed.

**111davey1**)So i realised that there is a contradiction in my book. It says that Kc measures the extent of a chemical reaction and its magnitude can give you an indication of where the equilibrium lies. I understand this.. if it is very large or very small it means equilibrium lies to the right or left respectively and if it is moderate it would suggest equilibrium lies somewhere in the middle. Say its very large it would mean you have little reactant and lots of product at equilibrium.

It also says that it is not changed by concentration or pressure.

so if we had this equilibrium: A + B <---> c

and increase the concentration of A we would say the equilibrium position has shifted to the right and more c would be produced. I understand how Kc stays constant i just don't fully understand the terminology.

Basically, i am wondering how the equilbrium position moves to the right to produce more c yet according to Kc the extent of the reaction (or the position of equilibrium) hasn't actually changed.

The equilibrium constant is a number equal to the quotient of the reaction when the reaction is at equilibrium. This only changes with temperature

The position of equilibrium is the relative amount of product compared to reactant. This

*can*change with pressure or by removing one component from the reaction.

Take an imaginary case of X(g) <==> 2Y(g) in a 1 litre vessel

The value of kc = [Y]

^{2}/[X] = 4 (remember this is imaginary)

The position of equilibrium is the moles of Y compared to the moles of X, which in this case must be equal to 2 mol Y for each 1 mol X

If we now double the pressure by halving the volume of the flask, then the new concentrations are: [Y] = 4 mol / litre and [X] = 2 mol / litre

The quotient is now equal to 4

^{2}/2 = 8.

This is clearly not the same as 4 and so, as kc must be restored, the system responds by moving the position of the equilibrium towards the reactants to reduce the concentration of [Y] and increase the concentration of [X] until the quotient equals 4 again.

0

reply

Report

#3

**111davey1**)

So i realised that there is a contradiction in my book. It says that Kc measures the extent of a chemical reaction and its magnitude can give you an indication of where the equilibrium lies. I understand this.. if it is very large or very small it means equilibrium lies to the right or left respectively and if it is moderate it would suggest equilibrium lies somewhere in the middle. Say its very large it would mean you have little reactant and lots of product at equilibrium.

It also says that it is not changed by concentration or pressure.

so if we had this equilibrium: A + B <---> c

and increase the concentration of A we would say the equilibrium position has shifted to the right and more c would be produced. I understand how Kc stays constant i just don't fully understand the terminology.

Basically, i am wondering how the equilbrium position moves to the right to produce more c yet according to Kc the extent of the reaction (or the position of equilibrium) hasn't actually changed.

So, the equilibrium position will move, but, due to Le Chatelier's principle, it will move back to the normal place (so normal Kc) with time, as the rate of formation of C increases

0

reply

(Original post by

Yup, can be confusing.

The equilibrium constant is a number equal to the quotient of the reaction when the reaction is at equilibrium. This only changes with temperature

The position of equilibrium is the relative amount of product compared to reactant. This

Take an imaginary case of X(g) <==> 2Y(g) in a 1 litre vessel

The value of kc = [Y]

The position of equilibrium is the moles of Y compared to the moles of X, which in this case must be equal to 2 mol Y for each 1 mol X

If we now double the pressure by halving the volume of the flask, then the new concentrations are: [Y] = 4 mol / litre and [X] = 2 mol / litre

The quotient is now equal to 4

This is clearly not the same as 4 and so, as kc must be restored, the system responds by moving the position of the equilibrium towards the reactants to reduce the concentration of [Y] and increase the concentration of [X] until the quotient equals 4 again.

**charco**)Yup, can be confusing.

The equilibrium constant is a number equal to the quotient of the reaction when the reaction is at equilibrium. This only changes with temperature

The position of equilibrium is the relative amount of product compared to reactant. This

*can*change with pressure or by removing one component from the reaction.Take an imaginary case of X(g) <==> 2Y(g) in a 1 litre vessel

The value of kc = [Y]

^{2}/[X] = 4 (remember this is imaginary)The position of equilibrium is the moles of Y compared to the moles of X, which in this case must be equal to 2 mol Y for each 1 mol X

If we now double the pressure by halving the volume of the flask, then the new concentrations are: [Y] = 4 mol / litre and [X] = 2 mol / litre

The quotient is now equal to 4

^{2}/2 = 8.This is clearly not the same as 4 and so, as kc must be restored, the system responds by moving the position of the equilibrium towards the reactants to reduce the concentration of [Y] and increase the concentration of [X] until the quotient equals 4 again.

Ok so you double the pressure which doubles the concentration but the amount in mol stays the same. Equilibrium shifts to the left to produce more mol of X and reduce Y. Kc is restored in the process... but you said the position of equilibrium is the moles of y compared to the moles of x which would suggest that the position of equilibrium has in fact changed eventhough Kc has returned to its original position. So the position of equilibrium has changed in the end? i think i was confusing the position with the constant.

But still does Kc give you an indication of where the position lies?

So take the equilibrium

A+B<--> C

and you increase the concentration of A by adding more of it, equilibrium will shift to the right -- now this is saying the forward reaction is going quicker than the backwards reaction -- and results in more mol of c and less of A and B but how would you determine if the position had shifted because wouldn't the net increase in mol would be the same

0

reply

Report

#5

(Original post by

Thanks,

Ok so you double the pressure which doubles the concentration but the amount in mol stays the same. Equilibrium shifts to the left to produce more mol of X and reduce Y. Kc is restored in the process... but you said the position of equilibrium is the moles of y compared to the moles of x which would suggest that the position of equilibrium has in fact changed eventhough Kc has returned to its original position. So the position of equilibrium has changed in the end? i think i was confusing the position with the constant.

But still does Kc give you an indication of where the position lies?

So take the equilibrium

A+B<--> C

and you increase the concentration of A by adding more of it, equilibrium will shift to the right -- now this is saying the forward reaction is going quicker than the backwards reaction -- and results in more mol of c and less of A and B but how would you determine if the position had shifted because wouldn't the net increase in mol would be the same

**111davey1**)Thanks,

Ok so you double the pressure which doubles the concentration but the amount in mol stays the same. Equilibrium shifts to the left to produce more mol of X and reduce Y. Kc is restored in the process... but you said the position of equilibrium is the moles of y compared to the moles of x which would suggest that the position of equilibrium has in fact changed eventhough Kc has returned to its original position. So the position of equilibrium has changed in the end? i think i was confusing the position with the constant.

But still does Kc give you an indication of where the position lies?

So take the equilibrium

A+B<--> C

and you increase the concentration of A by adding more of it, equilibrium will shift to the right -- now this is saying the forward reaction is going quicker than the backwards reaction -- and results in more mol of c and less of A and B but how would you determine if the position had shifted because wouldn't the net increase in mol would be the same

__Initial equilibrium situation__:

X(g) <==> 2Y(g)

2 dm

^{3}flask containing 4 mol of Y(g) and 2 mol of X(g)

Initial concentrations: [X] = 2/2 = 1 mol dm

^{-3}

[Y] = 4/2 = 2 mol dm

^{-3}

DEFINE: The position of equilibrium is the relative amounts of product/reactant = 4/2 = 2

Reaction quotient = Kc = [Y]

^{2}/[X] = 4/1 = 4 (this is equal to the equilibrium constant when the system is at equilibrium)

__NOW Change the conditions__:

New volume of the flask = 1dm

^{3}

New concentrations: [X] = 2/1 = 2 mol dm

^{-3}

[Y] = 4/1 = 4 mol dm

^{-3}

New reaction quotient = [Y]

^{2}/[X] = 16/2 = 8 (

__this is now not equal to kc__)

The position of equilibrium

__must__change to restore the quotient to the value of kc.

This can only happen by the reverse reaction occurring: 2Y --> X

Let 2n moles of Y react to re-establish equilibrium:

New moles at equilibrium: Y mol = (4 – 2n)

X mol = (2 + n)

New concentrations at equilibrium: [Y] = (4 - 2n)/1 = (4 – 2n)

[X] = (2 + n)/1 = (2 + n)

And as the equilibrium constant is re-established then:

Kc = (4 – 2n)

^{2}/(2 + n) = 4

Rearrange: (16 – 16n + 4n

^{2}) = 4(2 + n)

Collect terms: (8 – 20n + 4n

^{2}) = 0

Divide through by 4: (2 – 5n + n

^{2}) = 0

Solve quadratic for n: n = 4.56 or 0.438.

The first answer is impossible, therefore n= 0.438 mol

Hence moles at equilibrium: Y = 4 – (2 x 0.438) = 4 – 0.877 = 3.123 mol

X = 2 + 0.438 = 2.438 mol

New concentrations at equilibrium: [Y] = 3.123 mol dm

^{-3}

[X] = 2.438 mol dm

^{-3}

Quotient = [Y]

^{2}/[X] = (3.123)

^{2}/2.438 = 9.753/2.438 = 4.00 (quotient now equals kc again)

The position of equilibrium = relative moles of Y to X = 3.123/2.438 = 1.28

Originally this was equal to 2.00 and now it has changed to 1.28.

Hence, increasing the pressure has changed the position of equilibrium towards the side of the fewer moles of gas.

0

reply

Thanks,

Im just going to outline what i think about equilibrium:

Kc can give you an idea about the extent of a reaction:

If Kc<<1 equilibrium position lies to the left

if Kc is around 1 equilibrium position is near centre

If Kc is>>1 equilibrium position lies to the right - To say this is to say the amount of product is much greater than the amount of reactant at equilibrium.

If you take the equilibrium

A (g) <--> B (g) + C (g)

And double the pressure the concentration of all species will double. But the top of the reaction quotient will increase by more and so its value would increase above Kc. Therefore, the position of equilibrium shifts to the left increasing A and using up B and C to restore the quotient to the value of Kc.

The equilibrium amount in mol of B and C will decrease and the equilibrium amount a A will increase until equilibrium is once again established. Kc doesn't change but equilibrium has been shifted to the left.

now if you had

A <--> B

And you increase the concentration of A (by adding more) The reaction quotient for the reaction would increase to a value higher than Kc. Therefore, the equilibrium position would shift to the right so the quotient once again equals Kc. when it does the position of equilibrium will lie in its original position. If these were gases after a change in pressure the quotient would remain the same as Kc.

If you had

A + B<---> C

and you increase the concentration of A (by adding more) The quotient will have a lower value than Kc. The equilibrium will shift to the right to produce more and use up A and B. the quotient will once again be the same as Kc.

Has equilibrium shifted:

If 1 mol of A, B and C was in 1 dm3 (at equilibrium), then Kc = 1 / 1x1 Adding 1 mol of B -> 1 / 1x2 so not Kc B + A -> C is favoured until... [A] = 0.732 [B] = 1.732 [C] = 1.268

The net increase in moles is the same on both sides so unsure...

And Lastly (sorry for the length),

http://pmt.physicsandmathstutor.com/...%20A-level.pdf

Question 2bii) Obviously increasing the temperature will shift equilibrium to the left and increasing the pressure would shift equilibrium to the right. Now im just checking you can look at this two ways. Firstly, as they have not added anything like in my previous examples which makes it dodgy and just changed temp and pressure you can see that when the system is in equilibrium after these changes The position may be to the left, in the original position (if they cancel each other) or to the right. Secondly, you could also look directly after the changes have been made to see if the position has shifted in one direction or the other and therefore would eventually, when at equilibrium be shifted to the left or right.

And can you just confirm that a shift in equilibrium does not necessarily mean a change in the position permanently as in example A<-->B. and that shifting equilibrium means one of either the forward reaction or backwards reaction is favoured.

Thank you for your time

Im just going to outline what i think about equilibrium:

Kc can give you an idea about the extent of a reaction:

If Kc<<1 equilibrium position lies to the left

if Kc is around 1 equilibrium position is near centre

If Kc is>>1 equilibrium position lies to the right - To say this is to say the amount of product is much greater than the amount of reactant at equilibrium.

If you take the equilibrium

A (g) <--> B (g) + C (g)

And double the pressure the concentration of all species will double. But the top of the reaction quotient will increase by more and so its value would increase above Kc. Therefore, the position of equilibrium shifts to the left increasing A and using up B and C to restore the quotient to the value of Kc.

The equilibrium amount in mol of B and C will decrease and the equilibrium amount a A will increase until equilibrium is once again established. Kc doesn't change but equilibrium has been shifted to the left.

now if you had

A <--> B

And you increase the concentration of A (by adding more) The reaction quotient for the reaction would increase to a value higher than Kc. Therefore, the equilibrium position would shift to the right so the quotient once again equals Kc. when it does the position of equilibrium will lie in its original position. If these were gases after a change in pressure the quotient would remain the same as Kc.

If you had

A + B<---> C

and you increase the concentration of A (by adding more) The quotient will have a lower value than Kc. The equilibrium will shift to the right to produce more and use up A and B. the quotient will once again be the same as Kc.

Has equilibrium shifted:

If 1 mol of A, B and C was in 1 dm3 (at equilibrium), then Kc = 1 / 1x1 Adding 1 mol of B -> 1 / 1x2 so not Kc B + A -> C is favoured until... [A] = 0.732 [B] = 1.732 [C] = 1.268

The net increase in moles is the same on both sides so unsure...

And Lastly (sorry for the length),

http://pmt.physicsandmathstutor.com/...%20A-level.pdf

Question 2bii) Obviously increasing the temperature will shift equilibrium to the left and increasing the pressure would shift equilibrium to the right. Now im just checking you can look at this two ways. Firstly, as they have not added anything like in my previous examples which makes it dodgy and just changed temp and pressure you can see that when the system is in equilibrium after these changes The position may be to the left, in the original position (if they cancel each other) or to the right. Secondly, you could also look directly after the changes have been made to see if the position has shifted in one direction or the other and therefore would eventually, when at equilibrium be shifted to the left or right.

And can you just confirm that a shift in equilibrium does not necessarily mean a change in the position permanently as in example A<-->B. and that shifting equilibrium means one of either the forward reaction or backwards reaction is favoured.

Thank you for your time

0

reply

Report

#7

(Original post by

Thanks,

Im just going to outline what i think about equilibrium:

Kc can give you an idea about the extent of a reaction:

If Kc<<1 equilibrium position lies to the left

if Kc is around 1 equilibrium position is near centre

If Kc is>>1 equilibrium position lies to the right - To say this is to say the amount of product is much greater than the amount of reactant at equilibrium.

If you take the equilibrium

A (g) <--> B (g) + C (g)

And double the pressure the concentration of all species will double. But the top of Kc, the numerator, would increase by more and thus you would find that Kc increases.

**111davey1**)Thanks,

Im just going to outline what i think about equilibrium:

Kc can give you an idea about the extent of a reaction:

If Kc<<1 equilibrium position lies to the left

if Kc is around 1 equilibrium position is near centre

If Kc is>>1 equilibrium position lies to the right - To say this is to say the amount of product is much greater than the amount of reactant at equilibrium.

If you take the equilibrium

A (g) <--> B (g) + C (g)

And double the pressure the concentration of all species will double. But the top of Kc, the numerator, would increase by more and thus you would find that Kc increases.

Therefore, the position of equilibrium shifts to the left increasing A and using up B and C to restore Kc.

The equilibrium amount in mol of B and C will decrease and the equilibrium amount a A will increase until equilibrium is once again established. Kc is restores but equilibrium has been shifted to the left.

now if you had

A <--> B

And you increase the concentration of A Kc for the reaction would be disturbed. Therefore, the equilibrium position would shift to the right to restore Kc. Kc will be restored but when it is restored the position of equilibrium will lie in its original position... (This might be a special case or i might be getting it wrong)

The equilibrium amount in mol of B and C will decrease and the equilibrium amount a A will increase until equilibrium is once again established. Kc is restores but equilibrium has been shifted to the left.

now if you had

A <--> B

And you increase the concentration of A Kc for the reaction would be disturbed. Therefore, the equilibrium position would shift to the right to restore Kc. Kc will be restored but when it is restored the position of equilibrium will lie in its original position... (This might be a special case or i might be getting it wrong)

If you had

A + B <---> C

and you increase the concentration of A (by adding more) Kc will be disturbed.

What is disturbed is the equilibrium, The system is no longer at equilibrium and must respond in such a way as to restore the equality between Kc and the reaction quotient.

The equilibrium will shift to the right to produce more C and use up A and B. Kc will be restored.

I dont think the position at equilibrium has changed but if it has, it has as long as i'm getting the idea about position.

And Lastly (sorry for the length),

http://pmt.physicsandmathstutor.com/...%20A-level.pdf

Question 2bii) Obviously increasing the temperature will shift equilibrium to the left and increasing the pressure would shift equilibrium to the right. Now im just checking you can look at this two ways. Firstly, as they have not added anything like in my previous examples which makes it dodgy and just changed temp and pressure you can see that when the system is in equilibrium after these changes The position may be to the left, in the original position (if they cancel each other) or to the right. Secondly, you could also look directly after the changes have been made to see if the position has shifted in one direction or the other and therefore would eventually, when at equilibrium be shifted to the left or right.

0

reply

(Original post by

Thanks,

Im just going to outline what i think about equilibrium:

Kc can give you an idea about the extent of a reaction:

If Kc<<1 equilibrium position lies to the left

if Kc is around 1 equilibrium position is near centre

If Kc is>>1 equilibrium position lies to the right - To say this is to say the amount of product is much greater than the amount of reactant at equilibrium.

If you take the equilibrium

A (g) <--> B (g) + C (g)

And double the pressure the concentration of all species will double. But the top of the reaction quotient will increase by more and so its value would increase above Kc. Therefore, the position of equilibrium shifts to the left increasing A and using up B and C to restore the quotient to the value of Kc.

The equilibrium amount in mol of B and C will decrease and the equilibrium amount a A will increase until equilibrium is once again established. Kc doesn't change but equilibrium has been shifted to the left.

now if you had

A <--> B

And you increase the concentration of A (by adding more) The reaction quotient for the reaction would decrease to a value lower than Kc. Therefore, the equilibrium position would shift to the right so the quotient once again equals Kc. when it does the position of equilibrium will lie in its original position. If these were gases after a change in pressure the quotient would remain the same as Kc.

If you had

A + B<---> C

and you increase the concentration of A (by adding more) The quotient will have a lower value than Kc. The equilibrium will shift to the right to produce more C and use up A and B. the quotient will once again be the same as Kc.

Has equilibrium shifted:

If 1 mol of A, B and C was in 1 dm3 (at equilibrium), then Kc = 1 / 1x1 Adding 1 mol of B -> 1 / 1x2 so not Kc B + A -> C is favoured until... [A] = 0.732 [B] = 1.732 [C] = 1.268

The net increase in moles is the same on both sides so unsure...

And Lastly (sorry for the length),

http://pmt.physicsandmathstutor.com/...%20A-level.pdf

Question 2bii) Obviously increasing the temperature will shift equilibrium to the left and increasing the pressure would shift equilibrium to the right. Now im just checking you can look at this two ways. Firstly, as they have not added anything like in my previous examples which makes it dodgy and just changed temp and pressure you can see that when the system is in equilibrium after these changes The position may be to the left, in the original position (if they cancel each other) or to the right. Secondly, you could also look directly after the changes have been made to see if the position has shifted in one direction or the other and therefore would eventually, when at equilibrium be shifted to the left or right.

And can you just confirm that a shift in equilibrium does not necessarily mean a change in the position permanently as in example A<-->B. and that shifting equilibrium means one of either the forward reaction or backwards reaction is favoured.

Thank you for your time

**111davey1**)Thanks,

Im just going to outline what i think about equilibrium:

Kc can give you an idea about the extent of a reaction:

If Kc<<1 equilibrium position lies to the left

if Kc is around 1 equilibrium position is near centre

If Kc is>>1 equilibrium position lies to the right - To say this is to say the amount of product is much greater than the amount of reactant at equilibrium.

If you take the equilibrium

A (g) <--> B (g) + C (g)

And double the pressure the concentration of all species will double. But the top of the reaction quotient will increase by more and so its value would increase above Kc. Therefore, the position of equilibrium shifts to the left increasing A and using up B and C to restore the quotient to the value of Kc.

The equilibrium amount in mol of B and C will decrease and the equilibrium amount a A will increase until equilibrium is once again established. Kc doesn't change but equilibrium has been shifted to the left.

now if you had

A <--> B

And you increase the concentration of A (by adding more) The reaction quotient for the reaction would decrease to a value lower than Kc. Therefore, the equilibrium position would shift to the right so the quotient once again equals Kc. when it does the position of equilibrium will lie in its original position. If these were gases after a change in pressure the quotient would remain the same as Kc.

If you had

A + B<---> C

and you increase the concentration of A (by adding more) The quotient will have a lower value than Kc. The equilibrium will shift to the right to produce more C and use up A and B. the quotient will once again be the same as Kc.

Has equilibrium shifted:

If 1 mol of A, B and C was in 1 dm3 (at equilibrium), then Kc = 1 / 1x1 Adding 1 mol of B -> 1 / 1x2 so not Kc B + A -> C is favoured until... [A] = 0.732 [B] = 1.732 [C] = 1.268

The net increase in moles is the same on both sides so unsure...

And Lastly (sorry for the length),

http://pmt.physicsandmathstutor.com/...%20A-level.pdf

Question 2bii) Obviously increasing the temperature will shift equilibrium to the left and increasing the pressure would shift equilibrium to the right. Now im just checking you can look at this two ways. Firstly, as they have not added anything like in my previous examples which makes it dodgy and just changed temp and pressure you can see that when the system is in equilibrium after these changes The position may be to the left, in the original position (if they cancel each other) or to the right. Secondly, you could also look directly after the changes have been made to see if the position has shifted in one direction or the other and therefore would eventually, when at equilibrium be shifted to the left or right.

And can you just confirm that a shift in equilibrium does not necessarily mean a change in the position permanently as in example A<-->B. and that shifting equilibrium means one of either the forward reaction or backwards reaction is favoured.

Thank you for your time

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top