OCR A 5.1.1 How Fast? REVISION

[@sabthesarcastic]

this legend is getting me through chemistry I've been off sick so have missed the last few lessons where i was told that i have a test i found out today and you're helping me so much you sir will be the reason why I will pass a level chem!!

[@tapkoang]

I learned chemistry without a teacher during this covid-19 holiday. Thank you sir

[@AlleryChemistry]

Happy to help

[@evaaa5100]

Super helpful video! just wish there was a cursor so when you are talking about something you can see exactly what you mean, like when you were rearranging the equation it would have been helpful because I found it bit confusing where each thing goes. Otherwise great :)

[@jamilahmz9949]

Your videos are amazing thanks a lot helped me with no teachers during these few months

[@AlleryChemistry]

Happy to help!

[@Michelle_boops]

55:33 - Spoke with my teachers. According to them C shouldn't be in the rate equation because it's an INTERMEDIATE, not a catalyst. But aside from that, so far I can tell, it's a great video, thank you!

[@khan99ish]

It doesnt mention on the spec for the how fast section specifically to know about the iodine clock experiment should we still be aware of it?

[@thatperson4410]

Thank you soo much. You truly are a legend!

[@AlleryChemistry]

You're welcome!

[@ScottGoesFishing]

just to point out at 44:29 there is a mistake, the rate isn't trebled, 0.63 x 3 = 1.89 not 1.92

[@mnabdvy3180]

NO2 is also 2nd order..not 1st

[@jsuuzhe]

@@mnabdvy3180 no because the NO was multiplied by 4 and so was initial rate, had the initial rate multiplied by 8 it would be a second order

[@amber_x2521]

@@jsuuzhe the initial rate x4 is actually 2.52 not 2.58 so its not first order im guessing?

[@marco3391]

@@amber_x2521 it doesn't have to be exact, just roughly the same multiplication. So neither of those are mistakes, they are both first order

[@jallmo2]

I don’t understand why the C is there in the rates equation

[@duderandom7023]

Hi, I love you content ❤️👍🏼 is it possible for you to go through year 13 chemistry Edexcel PowerPoint presentation like you did for year 12 please

[@AlleryChemistry]

Yes, definitely. Follow me on twitter - twitter.com/allerytutors or @allerytutors and I set polls for people to vote for the next exam board to do.

[@duderandom7023]

Allery Chemistry kl thanks

[@SPRING-nd2rm]

Hi, can you please make similar kind of videos for ocr as level chemistry? Thanks

[@AlleryChemistry]

Hi. I already have these videos here - kzbin.info/aero/PLX4e2DxFRGQK2hBRarrpaEdnPqMj21xeD

[@SPRING-nd2rm]

@@AlleryChemistry thank you very much for the reply.

[@elliesmith7518]

Thank you! Very helpful for my mock on Monday

[@AlleryChemistry]

You're very welcome!

[@jalajarora2331]

Hi sir, I love your videos so thank you very much for them! Also I am curious to know which A-level subjects did you do?

[@AlleryChemistry]

Heh. It was while since I studied A Levels but I took Economics, Chemistry, Physics and Geography.

[@zarakeles1371]

SIR YOU CARRIED ME IN OCR CHEM PAPER 1 TODAY, THANK YOU🙏🏼

[@AlleryChemistry]

Sounds positive, keep the momentum up 👍

[@sophiekrise]

i don’t understand how you rearranged the arrhenius equation to have activation energy at the start? 1:03:54 surely you divide by InK?

[@akikabanu5885]

they +Ea/KT to the other side of the equation and then took away lnK to the other side, become Ea/KT = lnA - lnK

[@beagematgcsechem3505]

at 45:18 why did you use experiment 1 and not 2 or 3? I worked out the rate constants for experiments 2 and 3 and they are 32 and 32.25 respectively so they're not the same as the 31.5 in experiment 1. You said they're the same but they're not so please can you clarify this? Thanks!

[@HamzaKhan0711]

I also have the same question. Have you found the answer yet?

[@neithonosmani9783]

I also have the same question. My guess is that they all round to 32? Is that what is meant by them being them same?

[@relaxurself2960]

44:29 NO is quadrupled so how is it first order it should be second order shouldnt it ?

[@dasola6533]

Thats what I thought as well

[@akikabanu5885]

he means they are both quadrupled (times by 4) so it's 1st order, if one was x4 and the other was x16 then it would be second order

[@akikabanu5885]

@@dasola6533 he means they are both quadrupled (times by 4) so it's 1st order, if one was x4 and the other was x16 then it would be second order

[@jalajarora2331]

Sir at 55:09 how can you tell just from the equation that A is first order, B is first order and C is second order? I thought orders of reagents could only be determined from experimental data and not the stoicheometic coefficients in the balanced equation? Please explain this. Thank you!

[@AlleryChemistry]

You're right however we can use this rule when looking at multistep mechanisms like the example you refer to.

[@emmanuel4257]

Thanks for this video sir! I just wanted to know why E isnt in the rate equation at 55:35

[@xrzkidult5225]

C is a derivative of A and B which is the rate determining step meaning it is in the rate equation. E is neither in the rate determining step or the rate equation. A B and C are all basically versions of the themselves (rate determining step)

[@queenbless7787]

at 1:09:14 why is it times 3.1 *10^3? surely it should be times 1/3.1 *10^3? I dont get it

[@faizakabir7862]

The value of 3.1 is already 1/T that’s why he hasn’t changed it

[@muktarmohamud8094]

Please sir may I get Chemistry 1 for OCR. Thank you 🙏

[@khloex2173]

Any chance you can section this video so don’t have to search through one hour for one topic 😅 great video doh!

[@MustafaAhmed-db2kq]

can we study for the cie specification with these videos

[@BasimaSbenaty-lh6fp]

Can i please ask about the half life, where did the 2.6 come from half of 5 is 2.5 and I thought you half from where the graph starts

[@xxxmusicxxx1406]

For these questions you don't need to start at the concentration at which the graph also starts at. As long as you choose an initial concentration from the graph which you can halve and read off the time at the halved concentration to determine the half life (e.g. you could read off the half life between 2.6 and 1.3 as the graph extends to these values) then you can start from any concentration. For example you could alternatively use concentrations of 2 and 1 since the graph extends to those concentrations and so the half life can still be determined.

[@Habiba.786]

Does the half life of second order decrease?

[@rahanlol]

35:00 for the first order graph, why is the graph of rate/time a y=mx graph and not a y=-mx graph, as the rate of the reaction will only decreases as time goes on

[@Ojisgay]

There is no rate/time graph?

[@naja7610]

20:10 Iodine clock exp

[@queenbless7787]

how do you work out the units of A

[@hlovestxt8726]

It’s concentration so i think it’s just moldm-3

[@neymariousjr]

can someone explain how he got 15.1x10-3moldm-3s-1 from @49:54

[@playlive9107]

no clue

[@ewanmessenger8377]

So already worked out it’s 2nd order, yes? Then he’s now multiplying that value of 3.78 by 4 (because conc of A has doubled) which gives 15.1. This is worked out as if B isn’t involved at all, which obviously it shows that it is because this value isn’t what the rate turns out to be.