# Equilibrium constant/yield question.

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#1
I understand that a value of Kc greater than 1 means that the equilibrium position lies towards the right and so the reaction is product favoured. But does this mean that, at equilibrium, the concentrations of the products are greater than the reactants?

also, i know the only way to change Kc is by changing the temperature. Does this result in an increase in the equilibrium yield of either the reactants or products and a decrease in the equilibrium yield of the other?

And lastly is it true that an increase in concentration or pressure of an equilibirum mixture will increase the equilibrium yield of both the reactants and the products?

Thanks
0
4 years ago
#2
(Original post by 111davey1)
I understand that a value of Kc greater than 1 means that the equilibrium position lies towards the right and so the reaction is product favoured. But does this mean that, at equilibrium, the concentrations of the products are greater than the reactants?

also, i know the only way to change Kc is by changing the temperature. Does this result in an increase in the equilibrium yield of either the reactants or products and a decrease in the equilibrium yield of the other?

And lastly is it true that an increase in concentration or pressure of an equilibirum mixture will increase the equilibrium yield of both the reactants and the products?

Thanks
1) Think about the equation mathematically. The product of the concentrations of the products must be greater than the product of the concentrations of the reactants.

e.g for the equilibrium A B

where [B] is the concentration of B, and [A] is the concentration of A.

Therefore to be greater than 1 [B] must be > [A].

e.g2 for the equilibrium A B + C

Therefore now to be greater than 1, the product [B][C] must be greater than [A].

2) Yes it does. Again think of the equation for Kc. If Kc is changing, then the equilibrium concentrations must change to reach the new value of Kc. Whether the concentrations or reactants/products increase/decrease is determined whether the reaction is exothermic or endothermic.

3) Mols must be conserved. You could use Le Chatlier's principle e.g if you increase concentration of one reactant/product, the system will respond to try and minimize the change imposed on it, by decreasing the concentration of reactant/product. The key is that Kc/Kp does not change with concentration/pressure, the system adapts with changes of pressure/concentration so that Kc/Kp overall remain the same.

Edit: MexicanKeith gives a nice example!
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#3

So am i wrong to say if you have an equilibrium

a goes to b b goes to a

and you increase the concenration of a the equilibrium will shift to b to produce more of it and thus to restore the concentraion ratios (Kc) and what you are left with is an equilibrium mixture with a greater concentration of both a and b
0
4 years ago
#4
(Original post by 111davey1)

So am i wrong to say if you have an equilibrium

a goes to b b goes to a

and you increase the concenration of a the equilibrium will shift to b to produce more of it and thus to restore the concentraion ratios (Kc) and what you are left with is an equilibrium mixture with a greater concentration of both a and b
The only thing the system can do to maintain Kc is convert some of the excess A to B. If you left the mixture for long enough, enough of the excess A would be converted to B in order for Kc to stay the same and the equilibrium to be restored.

Of course the time this takes depends on the rate constant governing the forward/reverse rates of the reaction, which is a whole other story!
0
4 years ago
#5
(Original post by 111davey1)

So am i wrong to say if you have an equilibrium

a goes to b b goes to a

and you increase the concenration of a the equilibrium will shift to b to produce more of it and thus to restore the concentraion ratios (Kc) and what you are left with is an equilibrium mixture with a greater concentration of both a and b
You're right, in that you would have more of both A and B at the end than at the beginning, take a very simple example of your equilibrium and assuming an equilibrium constant of 1, meaning concentrations of A and B are the same at equilibrium.

if we started with 0.5moldm^-3 of A and 0.5 moldm^-3 of B we would be at equilibrium.

Now say we added enough A that the concentration of A would increase to 0.6 moldm^-3.

The equilibirum is disturbed, and to maintain the value of Kc some of the added A must be turned into B.

So when the system reaches equilibrium you would now have 0.55 moldm^-3 of A and 0.55moldm^3 of B. ie more of both reactant and product.

This makes sense, equilibrium says that [A] and [B] must have a given ratio, if you add more stuff (and don't change the volume) then, to maintain the ratio, both [A] and [B] must end up larger than they were to begin with
2
#6
Thanks,
Just for further understanding if you have the equilibrium again

a to b b to a

and you increase the concentraion of a, equilibrium will shift to b using up a and producing more b thus restoring Kc. My question is that is this a temporary shift in the position of equilibrium in the sense that when it once again returns to equilibrium the position of equilibrium will be where it started.

From this is it true that the value of Kc gives you the true position of the equilibrium and thus tells you whether the reaction is producing more reactant or product at equilibrium.

If this is true, surely it follows that the only way to shift equilibrium permanantly is to change the temperature as this will alter the relative proportions of reactant and product once it returns to equilibrium.
Thanks
0
4 years ago
#7
(Original post by 111davey1)
Thanks,
Just for further understanding if you have the equilibrium again

a to b b to a

and you increase the concentraion of a, equilibrium will shift to b using up a and producing more b thus restoring Kc. My question is that is this a temporary shift in the position of equilibrium in the sense that when it once again returns to equilibrium the position of equilibrium will be where it started.

From this is it true that the value of Kc gives you the true position of the equilibrium and thus tells you whether the reaction is producing more reactant or product at equilibrium.

If this is true, surely it follows that the only way to shift equilibrium permanantly is to change the temperature as this will alter the relative proportions of reactant and product once it returns to equilibrium.
Thanks
Yep exactly - temperature is the only way to change Kc. Or say you want a particular product, if you continuously remove a product (e.g H2O or a gas) you will continuously be driving the reactants over to products (although the value of Kc at that particular temperature is not changing - just the reaction is changing its relative concentrations of A/B in order to maintain Kc)
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#8
(Original post by MexicanKeith)
You're right, in that you would have more of both A and B at the end than at the beginning, take a very simple example of your equilibrium and assuming an equilibrium constant of 1, meaning concentrations of A and B are the same at equilibrium.

if we started with 0.5moldm^-3 of A and 0.5 moldm^-3 of B we would be at equilibrium.

Now say we added enough A that the concentration of A would increase to 0.6 moldm^-3.

The equilibirum is disturbed, and to maintain the value of Kc some of the added A must be turned into B.

So when the system reaches equilibrium you would now have 0.55 moldm^-3 of A and 0.55moldm^3 of B. ie more of both reactant and product.

This makes sense, equilibrium says that [A] and [B] must have a given ratio, if you add more stuff (and don't change the volume) then, to maintain the ratio, both [A] and [B] must end up larger than they were to begin with
sorry to come back to this but i understand that if you add more stuff to the mixture you will end up will a greater concentraion of both and a greater amount in mol at equilibrium.

Just to check if they were gases and you increase the pressure of the system you would be increasing the concentration of all but the amount in mol would stay the same (if same amount of gas molecules on each side). if however there is more moles on one side and you increase the pressure equilibrium would shift to the side with fewer gas molecules and the amount (and therefore concentration of this species) would increase to restore Kc.

My question is in this example does the concentration of all increase at equilibrium but the amount in mol of the side with fewer gas molecules increase and the amount in mol of the side with the most gas molecules decrease. And also in your example is it the concentration AND amount in mol of all species which increase.

Thanks
0
4 years ago
#9
(Original post by 111davey1)
sorry to come back to this but i understand that if you add more stuff to the mixture you will end up will a greater concentraion of both and a greater amount in mol at equilibrium.

Just to check if they were gases and you increase the pressure of the system you would be increasing the concentration of all but the amount in mol would stay the same (if same amount of gas molecules on each side). if however there is more moles on one side and you increase the pressure equilibrium would shift to the side with fewer gas molecules and the amount (and therefore concentration of this species) would increase to restore Kc.

My question is in this example does the concentration of all increase at equilibrium but the amount in mol of the side with fewer gas molecules increase and the amount in mol of the side with the most gas molecules decrease. And also in your example is it the concentration AND amount in mol of all species which increase.

Thanks
Check out this interactive on the effect of pressure on equilibrium concentrations and Kc
0
4 years ago
#10
(Original post by 111davey1)
sorry to come back to this but i understand that if you add more stuff to the mixture you will end up will a greater concentraion of both and a greater amount in mol at equilibrium.

Just to check if they were gases and you increase the pressure of the system you would be increasing the concentration of all but the amount in mol would stay the same (if same amount of gas molecules on each side). if however there is more moles on one side and you increase the pressure equilibrium would shift to the side with fewer gas molecules and the amount (and therefore concentration of this species) would increase to restore Kc.

My question is in this example does the concentration of all increase at equilibrium but the amount in mol of the side with fewer gas molecules increase and the amount in mol of the side with the most gas molecules decrease. And also in your example is it the concentration AND amount in mol of all species which increase.

Thanks
Imagine an equilibrium

A(g) <=> 2B(g)

lets say the equilibrium constant is equal to 1 for simplicity

Kc = [B]^2 / [A] = 1

It's worth pointing out that for gases you should actually use Kp (not Kc) but as some a level course don't really cover Kp i'll work with Kc for now.

if we initially have an equilibrium where
[A] = [A]o and
[B] = [B]o (the subscript o just means that we are considering initial conditions)

if you half the volume of the container then you double the pressures of all the gases (Boyle's law).

now
[A] = 2[A]o
[B] = 2[B]o ie each one is double its initial value

so now think about the expression for Kc
(2[B]o)^2 / (2[A]o)
= 4[B]o^2 / 2[A]o
= 2 [B]o^2 / [A]o
= 2Kc

Ie the concentrations aren't at equilibrium so the fractions will begin to shift to make the fraction smaller ie [B] will decrease and [A] will increase.

So at this point the moles of A will increase to greater than than initially and the moles of b will decrease to less than initially.

However, these new numbers of moles are still in half the volume that we initially had, so the concentration of A and B are BOTH higher than initially.

The important thing is that, if you double the pressure, then in the end

[A] > 2[A]o

[B] < 2[B]o

Because they both initially double, then the equilibrium shifts as predicted by Le Chatelier.

So, for A in this example:
• volume halves
• number of moles increases
• concentration more than doubles

for B:
• volume halves
• number of moles decreases
• concentration increases, but less than doubles
0
#11
(Original post by MexicanKeith)
Imagine an equilibrium

A(g) <=> 2B(g)

lets say the equilibrium constant is equal to 1 for simplicity

Kc = [B]^2 / [A] = 1

It's worth pointing out that for gases you should actually use Kp (not Kc) but as some a level course don't really cover Kp i'll work with Kc for now.

if we initially have an equilibrium where
[A] = [A]o and
[B] = [B]o (the subscript o just means that we are considering initial conditions)

if you half the volume of the container then you double the pressures of all the gases (Boyle's law).

now
[A] = 2[A]o
[B] = 2[B]o ie each one is double its initial value

so now think about the expression for Kc
(2[B]o)^2 / (2[A]o)
= 4[B]o^2 / 2[A]o
= 2 [B]o^2 / [A]o
= 2Kc

Ie the concentrations aren't at equilibrium so the fractions will begin to shift to make the fraction smaller ie [B] will decrease and [A] will increase.

So at this point the moles of A will increase to greater than than initially and the moles of b will decrease to less than initially.

However, these new numbers of moles are still in half the volume that we initially had, so the concentration of A and B are BOTH higher than initially.

The important thing is that, if you double the pressure, then in the end

[A] > 2[A]o

[B] < 2[B]o

Because they both initially double, then the equilibrium shifts as predicted by Le Chatelier.

So, for A in this example:
• volume halves
• number of moles increases
• concentration more than doubles

for B:
• volume halves
• number of moles decreases
• concentration increases, but less than doubles
Ah yes thank you very much. very well explained
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