Which is the hardest topic in A2 chemistry?

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Mina_
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Is it just me that finds carboxylic acids, amines, esters and organic synthesis to be the most difficult? Or the concepts in NMR
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Not_a_horse
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I find organic OK, I don't particularly like electrode potentials.
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alow
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None of them.
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EierVonSatan
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I used to find the concept of equilibrium being shifted, yet equilibrium constants remaining the same pretty confusing.
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helpme:)
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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
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Clark_282
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Depends on the person, I don't find the ones you listed too bad, but anything inorganic chemistry is definitely the worst for me...
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cajj112
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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
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alow
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(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
What don't you understand exactly?
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cajj112
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(Original post by alow)
What don't you understand exactly?
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. It just seems a little contradictory to me that kc tells you where the position of equilibrium is but in this instance when the position changes, kc doesn't.

Hope that makes sense
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Pigster
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Graphite is really hard.

No wait.
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alow
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(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.
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.
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cajj112
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(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.
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
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alow
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(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
There is no equilibrium after you add more reactant. The position of the equilibrium doesn't move, there is no equilibrium.
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cajj112
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(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.
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?
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EierVonSatan
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Ah and so the nightmare of this concept returns :p: 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
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alow
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(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?
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.
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cajj112
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(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.
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
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Wsupermain2
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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.
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LauraMacleod579
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I'd say NMR with carbon 13 and proton NMR. I just hate it; I get it but hate it
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alow
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(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
The definition of the equilibrium is:

the condition existing when a chemical reaction and its reverse reaction proceed at equal rates.
A system in which the forward and reverse reactions are not proceeding at the same rate (i.e. you are making more of B from A than vice versa) is by definition not in equilibrium.
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