Go to first unread
Back
to top
Which is the hardest topic in A2 chemistry?
Watch
Announcements
Is it just me that finds carboxylic acids, amines, esters and organic synthesis to be the most difficult? Or the concepts in NMR
0
reply
Report
#4
I used to find the concept of equilibrium being shifted, yet equilibrium constants remaining the same pretty confusing.
3
reply
Report
#5
Organic synthesis is hard due to the amount you have to learn. Electode potentials as a concept r ok but i always seem to mess up the questions. And also equilibrium constants not changing is weird, like if pressure changes surely Kc would change apparently not
3
reply
Report
#6
Depends on the person, I don't find the ones you listed too bad, but anything inorganic chemistry is definitely the worst for me...
1
reply
Report
#7
I'm another one who can't get my head around position of equilibrium changing but not kp or kc. I can answer the questions and stuff but just can't understand why
1
reply
Report
#8
(Original post by chemistryguy123)
I'm another one who can't get my head around position of equilibrium changing but not kp or kc. I can answer the questions and stuff but just can't understand why
I'm another one who can't get my head around position of equilibrium changing but not kp or kc. I can answer the questions and stuff but just can't understand why
0
reply
Report
#9
(Original post by alow)
What don't you understand exactly?
What don't you understand exactly?
Hope that makes sense
0
reply
Report
#11
(Original post by chemistryguy123)
I understand it well enough o be able to answer questions on it but the fact that kp and kc increase and decrease depending on position of equilibrium in general but I don't quite get why you change the concentration, the position of equilibrium moves but kc doesn't.
I understand it well enough o be able to answer questions on it but the fact that kp and kc increase and decrease depending on position of equilibrium in general but I don't quite get why you change the concentration, the position of equilibrium moves but kc doesn't.
0
reply
Report
#12
(Original post by alow)
That's a contradiction. The position of the equilibrium does not move, on the addition of more reaction the system is no longer in equilibrium. Once equilibrium is reestablished it will have the same Kc.
That's a contradiction. The position of the equilibrium does not move, on the addition of more reaction the system is no longer in equilibrium. Once equilibrium is reestablished it will have the same Kc.
0
reply
Report
#13
(Original post by chemistryguy123)
Correct me if I'm wrong but isn't that also a contradiction since you've said the position of equilibrium does not move but then it's not in equilibrium, which surely means the position of equilibrium does have to move in order to establish equilibrium again
Correct me if I'm wrong but isn't that also a contradiction since you've said the position of equilibrium does not move but then it's not in equilibrium, which surely means the position of equilibrium does have to move in order to establish equilibrium again
0
reply
Report
#14
(Original post by alow)
There is no equilibrium when you add more reactant. The position of the equilibrium doesn't move, there is no equilibrium.
There is no equilibrium when you add more reactant. The position of the equilibrium doesn't move, there is no equilibrium.
If there is equilibrium before, then you add more reactant the position of equilibrium has to move in order to maintain equilibrium, doesn't it? That's what I've always been told.
How can there suddenly be no equilibrium?
0
reply
Report
#15
Ah and so the nightmare of this concept returns 
The value of Kc does not alter with changes in concentration. However, the composition of equilibrium does change - i.e. the amounts of each species.
Take a reaction A + B --> C + D where Kc = 4.1. If we have 1 mol of A and B initially, then at equilibrium we would have (ignoring rounding). We can ignore volume here for simplicity as it cancels.
[A] = 0.33
[B] = 0.33
[C] = 0.67
[D] = 0.67
Increase the concentration of [A] by adding 1 mol to the mixture, which disturbs the equilibrium.
[A] = 1.33
[B] = 0.33
[C] = 0.67
[D] = 0.67
As with Le Chatelier's principle, the equilibrium shifts to oppose the change - so it makes more C and D to restore the position (i.e. Kc)
[A] = 1.15
[B] = 0.15
[C] = 0.85
[D] = 0.85
The value of Kc does not alter with changes in concentration. However, the composition of equilibrium does change - i.e. the amounts of each species.Take a reaction A + B --> C + D where Kc = 4.1. If we have 1 mol of A and B initially, then at equilibrium we would have (ignoring rounding). We can ignore volume here for simplicity as it cancels.
[A] = 0.33
[B] = 0.33
[C] = 0.67
[D] = 0.67
Increase the concentration of [A] by adding 1 mol to the mixture, which disturbs the equilibrium.
[A] = 1.33
[B] = 0.33
[C] = 0.67
[D] = 0.67
As with Le Chatelier's principle, the equilibrium shifts to oppose the change - so it makes more C and D to restore the position (i.e. Kc)
[A] = 1.15
[B] = 0.15
[C] = 0.85
[D] = 0.85
1
reply
Report
#16
(Original post by chemistryguy123)
Sorry but this has confused me even more.
If there is equilibrium before, then you add more reactant the position of equilibrium has to move in order to maintain equilibrium, doesn't it? That's what I've always been told.
How can there suddenly be no equilibrium?
Sorry but this has confused me even more.
If there is equilibrium before, then you add more reactant the position of equilibrium has to move in order to maintain equilibrium, doesn't it? That's what I've always been told.
How can there suddenly be no equilibrium?
I add another 0.5mol of A to the reaction mixture, therefore increasing [A]. This means that the forward reaction will now be proceeding faster than the reverse (as there is more A being converted to B than vice versa).
The system is therefore no longer in equilibrium as the forward and reverse reactions are not proceeding at the same rate.
After some time equilibrium is reestablished as [B] "catches up" to [A], with the same ratio of [B] to [A] as before the addition of extra A.
1
reply
Report
#17
(Original post by alow)
Let's say I have a reversible reaction in solution which is in equilibrium: A ⇌ B
I add another 0.5mol of A to the reaction mixture, therefore increasing [A]. This means that the forward reaction will now be proceeding faster than the reverse (as there is more A being converted to B than vice versa).
The system is therefore no longer in equilibrium as the forward and reverse reactions are not proceeding at the same rate.
After some time equilibrium is reestablished as [B] "catches up" to [A], with the same ratio of [B] to [A] as before the addition of extra A.
Let's say I have a reversible reaction in solution which is in equilibrium: A ⇌ B
I add another 0.5mol of A to the reaction mixture, therefore increasing [A]. This means that the forward reaction will now be proceeding faster than the reverse (as there is more A being converted to B than vice versa).
The system is therefore no longer in equilibrium as the forward and reverse reactions are not proceeding at the same rate.
After some time equilibrium is reestablished as [B] "catches up" to [A], with the same ratio of [B] to [A] as before the addition of extra A.
0
reply
Report
#18
When you add more reactant, the equilibrium is disrupted, over a period of time the concentrations of the species in the chemical equation will return to a state such that the value of the expression for Kc is once again Kc. Ie. in real life after adding species to mixture, it is of course not in equilibrium immediately afterwards.
The reason that the the relative concentrations can change such that the proportions are not the same as in the original state is that Kc is non linear and often dependent on more than just a single product and reactant.
The reason that the the relative concentrations can change such that the proportions are not the same as in the original state is that Kc is non linear and often dependent on more than just a single product and reactant.
0
reply
Report
#20
(Original post by chemistryguy123)
Does this not mean that the position of equilibrium has moved to the right in order to make more of B? This is what I've been taught
Does this not mean that the position of equilibrium has moved to the right in order to make more of B? This is what I've been taught
the condition existing when a chemical reaction and its reverse reaction proceed at equal rates.
0
reply
X
Quick Reply
Today on TSR
-
Ofqual consultationTeachers to use 'mini-exams' and other work to decide grades
- Is it too late to do well in my A-levels?
- What Harry Potter house are you in?
- 2021 wellbeing blog launch
- Student tips on lockdown mental health