Edexcel A Level Chemistry Paper 2 Tuesday 18th June 2024 9CH0/02

Looking forward to what challenge they can pull

How are you looking forward😭 I need an extra week minimum

What is the most efficient way to revise leading up to our exams?
Thank you btw

At this point, past papers should be the best bet.
I’d recommend doing the 2018 paper 2 as a number of topics that came up on it haven’t come up in a while, plus the 6-marker involved a niche area of the spec that often isn’t taught terribly well and the most recent paper 2 you have access to (probably 2022, which you can find on PMT) as these questions will probably be more similar in style to the questions you’ll get now.
If you find reagents and conditions problematic (since there are a lot of them), try looking at Sporcle.com to find relevant quizzes on them.
Also, I don’t think Grignard reagents have come up in a while and iirc they often are pretty badly explained at A level - I would recommend looking at some past paper 3’s (2017 and 2019) to find some useful questions on them.

Former Edexcel Chemist here (sat exams in 2022 and achieved an A*, mentored A level chemistry for a few years - now studying chemistry at uni).
Happy to answer any questions you may have about the material covered on papers 2 and 3.

Aiming to do really good on this paper but lots of people are saying its going to be tough because of paper 1, what is the best way to revise to get all the challenging organic question that may come up involving problem solving?

Former Edexcel Chemist here (sat exams in 2022 and achieved an A*, mentored A level chemistry for a few years - now studying chemistry at uni).
Happy to answer any questions you may have about the material covered on papers 2 and 3.

Could you please give me a walkthrough of how to answer this question? It's a three mark question from the 2020 paper (5.d.ii). I just can't seem to make sense of it! Thank you in advance

This has made me lose my faith in chem what kinda question ???? 😭😭

Could you please give me a walkthrough of how to answer this question? It's a three mark question from the 2020 paper (5.d.ii). I just can't seem to make sense of it! Thank you in advance

CO2 is an acidic gas, so essentially the reaction with NaOH is just removing the CO2 from the mixture.
Since the volume decreases by 125 cm^3 upon this change, that must be the volume of CO2.
They’ve told you that for every mole of CxHy that reacts, you make x moles of CO2 and that the volume of CxHy was 25 cm^3.
Since the volume of gas is proportional to the number of moles, you can treat the volumes like moles and treat it like any other mole problem.
Volume of CO2 = Volume of CxHy * ratio of CO2 formed to CxHy burned
125 = 25 * x => x = 5
Now they haven’t told you how much oxygen was in the mixture. Let us just call the amount of it that reacts V cm^3.
We know that the initial mixture consisted of only O2 and CxHy and that there 25 cm^3 of CxHy. This means that the volume must have been V + 25 cm^3.
But they’ve told us that in the reaction, the volume decreased by 75 cm^3, so the volume is (V + 25) - 75 cm^3 = V - 50 cm^3.
Conveniently, this is the volume of CO2 made, so you can say that V - 50 = 125 and so V = 175.
Using a similar procedure to before, for every mole of CxHy that reacts, (x + y/4) moles of O2 react and so the volume of O2 is 25(x + y/4) = 175 and we have previously found x = 5.
So we have 25(5 + y/4) = 175
(5 + y/4) = 7
y/4 = 2
y = 8
Since x is 5 and y is 8, the hydrocarbon is C5H8.
Because they’ve told you it’s cyclic, the carbons should be arranged in a ring. Fitting all the hydrogens in in such a way that all carbons have 4 bonds should reveal that you get cyclopentene.

CO2 is an acidic gas, so essentially the reaction with NaOH is just removing the CO2 from the mixture.
Since the volume decreases by 125 cm^3 upon this change, that must be the volume of CO2.
They’ve told you that for every mole of CxHy that reacts, you make x moles of CO2 and that the volume of CxHy was 25 cm^3.
Since the volume of gas is proportional to the number of moles, you can treat the volumes like moles and treat it like any other mole problem.
Volume of CO2 = Volume of CxHy * ratio of CO2 formed to CxHy burned
125 = 25 * x => x = 5
Now they haven’t told you how much oxygen was in the mixture. Let us just call the amount of it that reacts V cm^3.
We know that the initial mixture consisted of only O2 and CxHy and that there 25 cm^3 of CxHy. This means that the volume must have been V + 25 cm^3.
But they’ve told us that in the reaction, the volume decreased by 75 cm^3, so the volume is (V + 25) - 75 cm^3 = V - 50 cm^3.
Conveniently, this is the volume of CO2 made, so you can say that V - 50 = 125 and so V = 175.
Using a similar procedure to before, for every mole of CxHy that reacts, (x + y/4) moles of O2 react and so the volume of O2 is 25(x + y/4) = 175 and we have previously found x = 5.
So we have 25(5 + y/4) = 175
(5 + y/4) = 7
y/4 = 2
y = 8
Since x is 5 and y is 8, the hydrocarbon is C5H8.
Because they’ve told you it’s cyclic, the carbons should be arranged in a ring. Fitting all the hydrogens in in such a way that all carbons have 4 bonds should reveal that you get cyclopentene.