TIP FOR ANYONE THAT DIDNT KNOW, CLICK ON SETTING ON THE VIDEO BAR AND SPEED UP THE VIDEO IN ORDER TO SAVE TIME FOR ALL YOU LAST MINUTE CRAMMERS LIKE ME :)
I wish I'd known about these last year when sitting AS Chem. I like textbooks and all, but having someone explain the topics really helps :) especially since I'm resitting.
IMPORTANT NOTE: u have to mentiion "more succesful collisions per second" or something to that affect (per seconds is accepted in the mark schemes) because otherwise if u omit the mention of successful collions in a given time frame you may drop marks
Hey father Rintoul, i'm not sure if this is a dumb question or not, but at 9:09 you said that no particles have zero energy, so how come at the beginning of the curve the line is touching the x-axis? Just wondering if you could explain that please, cheers.
I think I found that last question you were on about at the end [June 2010 CHEM2 Question 1 (b) (ii)]. How do you know whether or not the question was an equilibrium question or a kinetics question? Is it because both sides of the reaction have the same number of moles, so a change in pressure will make no difference to the yield in terms of equilibria? So in that case do you just answer as if its a kinetics question? Thanks
Thank you! It's been 2 years since I last studied this. I'm an A2 student for Biology and Physics, but I'm doing an AS in Chemistry as I dropped maths from last year. So some of the work from GCSE chemistry mentioned has left my memory.
Hey, just wanted to say I understood the topic in the textbook, but I ALWAYS prefer it when someone explains it and reads it. You explain it well! But I do have a question. At around 15:10 you were explaining about the effect of decrease in temperature, by which you said decrease temperature= less particles and you said >/ Ea (Therefore less successful collisions) However, shouldn't it be
what he meant was, when there is a decrease in temp., less particles have energy equal or higher than activation energy, therefore chances of successful collisions to occur is less.
after having done my alevels 2 years ago I remembered how great you are that I came back and these videos are actually still relevant to my uni course in the biosciences xD
at around 13:00, with the boltzman distribution curve, why is the height of the second curve (temp B) lower (before Ea) than that of curve A?? [EDIT] Is it to accommodate for the total number of molecules in the system being the same in both curves?
Hi please reply: At 7:31 have you made a mistake? I think the 2.0 moldm^-3 shouldnt level off that left of the graph in relation to R. The rate of reaction has doubled which is the number of successful collisions in a give time. If the curve was drawn to level off at the same time as curve R, but twice as high, the rate of reaction is doubled because it is steeper. No need to move it left, if i am not mistaken?
I think your maxwell boltzmann distribution may be wrong? I was told at that start it should never be curve out like a "s"? Is this correct or... Also, I don't think "alternate" would be accepted for what you said about catalysts around 17:11, it's "alternative".... Like I said, it's what I've been taught
Hiya. I have a question about the effect of concentration on collisions . Basically, in class we were told that an increase in concentration/pressure increases the number of particles with the energy higher than the activation energy. So there are more frequent collisions but not more successful because the proportion of partices is still the same. You said in the video however that the number of successful collisions also increases, but we were told you would loose marks for putting that. Please could you clear this up for me. Thanks
+Amy If a reaction only occurs because the collisions that occur are successful, for the rate to increase, the number of successful collisions must increase. It's all down to statistics; whilst the energy does not increase, as you say, there will be more molecules that have an energy equal to or greater than the activation energy, thus more successful collisions.
+Armita Azar I believe it refers to the mass dissolved into one litre. I may be wrong but I think that's the case! I know that the technician at my school often talks about 40 vol. and 100 vol.
Thankyou so so much, for some strange reason my school did equilibrium before kinetics, and so some of my concepts were really confusing to me - but this video was really helpful!! ☺☺
Thanks for the helpful videos but I also have a querie regarding the increase in temperature in the boltzman distribution curve; so when you shifted the curve to the right I noticed that the area of the curve was less than temperature a which would evince that less molecules are present- but I have read that the molecules present are the same so should the area of the curve been roughly the same?
+BhadMan316 Because more molecules have a higher energy. However, the area under the graph is the total number of molecules and so this must remain the same.
SharpSpoons no because there is a fixed volume, therefore if the concentration is doubled you will have double the number of moles/particles. Therefore if there is more particles, more oxygen is produced
E Rintoul Perfect. Honestly, your videos are absolutely amazing. I'm struggling with A2 so if you could do a few videos on that spec, I would be so so happy :D thanks :)
no because the curves are changing due to a change in reactant (h2o2), but the y axis is the product (o2). curve q has half the concentration but the same volume as curve r, which means it has half as many moles of h2o2, which means it can only produce half as many moles of o2. the o2 is measured under a constant concentration, therefore its volume is halved :)
No it’d be higher up as the y-axis is of volume. When you increase your concentration of reactant you’re increasing your volume of products, hence it will show a curve higher up to show this. With what your saying it’s for the Q line, where you’ve halved the concentration hence lower volume of product.
THNAKKKKKKKSSSSSSSSSSSS ALOTTTTTTTT MATEEEE BUT IF UR EVER FREE IN LIFE U SHOULD MAKE A VEDIO ON WHOLE AS LVL SYLLABUS REVSISON IT WOULD MEAN ALOT TO ALL THE STUDENTS SUFFERING IN THIS WRLD COZ OUR SCHOOL TEACHERS DONT N NVR WILL BE ABLE TO TEACH AS GOOD N SPECTACULAR AS U DO
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It may have been 5 years since these videos were published, but I want you to know students still use these and they re a life saver you are a hero
Morgan Roper totally agree xx
fr, I have my last Chem exam tomorrow, his videos are saving me
Periodt
coming up to 8 years now, this bloke is the goat
@@lewisbrerernumber1hater181 still goated
7 years have passed and your videos are still helpful
8 years gone and these still rock!
TIP FOR ANYONE THAT DIDNT KNOW, CLICK ON SETTING ON THE VIDEO BAR AND SPEED UP THE VIDEO IN ORDER TO SAVE TIME FOR ALL YOU LAST MINUTE CRAMMERS LIKE ME :)
DynamicArts Ramp it up to double-time.
I DO THAT EXACT SAME THING OMG I thought it was only me XDDDD :D
What grade did you get?
why are you so last minute that you cant even watch a 20 minute video? are you right outside the exam hall?
@@joelcarter9264 Finished my A-Level Chem with an A lol
I wish I'd known about these last year when sitting AS Chem. I like textbooks and all, but having someone explain the topics really helps :) especially since I'm resitting.
Superheroes come in all sizes
IMPORTANT NOTE: u have to mentiion "more succesful collisions per second" or something to that affect (per seconds is accepted in the mark schemes) because otherwise if u omit the mention of successful collions in a given time frame you may drop marks
This was such a good way to explain kinetics, I learn so much better with visual representation. Thank you!!
Freesciencelessons: My true successor
Cute pfp tho 🤧
Wheres it from???
its 2020 and you are still a hero sir.
You are the only reason I'm going to get a chemistry A level.
Great job..... Thanks alot
Nahid H You are the reason, not me!
Any updates on your career bud?
@@hhendi9636 any updates on your career bud
damnnnn 7 years ago, what r u doing now?
Hey father Rintoul, i'm not sure if this is a dumb question or not, but at 9:09 you said that no particles have zero energy, so how come at the beginning of the curve the line is touching the x-axis? Just wondering if you could explain that please, cheers.
Hi, the beginning of the curve has the coordinates (0,0) meaning that zero particles have zero energy, if that makes sense.
I will definitely post my As level grade on your channel nice I got them. If I do well, it's all thanks to you.
@@Abcdefghujklmao loooool
We demand a response
he got an a* in chem i was the paper btw
@@Cornyyylieus yeah bro i was the pen he used
I think I found that last question you were on about at the end [June 2010 CHEM2 Question 1 (b) (ii)]. How do you know whether or not the question was an equilibrium question or a kinetics question? Is it because both sides of the reaction have the same number of moles, so a change in pressure will make no difference to the yield in terms of equilibria? So in that case do you just answer as if its a kinetics question?
Thanks
jeesus i wish you were my teacher , your videos have got me an A in my mock, thanks so much do you accept donations?
+Gavin Sanghera I do indeed, there's a link on my dashboard thing. You really don't have to, though!
Thank you! It's been 2 years since I last studied this. I'm an A2 student for Biology and Physics, but I'm doing an AS in Chemistry as I dropped maths from last year. So some of the work from GCSE chemistry mentioned has left my memory.
Glad it helped!
so kind of you post videos, defo a thumbs up and I've defo recommend to any and everyone keep it up, my favourite science teacher by far
Hey, just wanted to say I understood the topic in the textbook, but I ALWAYS prefer it when someone explains it and reads it. You explain it well!
But I do have a question. At around 15:10 you were explaining about the effect of decrease in temperature, by which you said decrease temperature= less particles and you said >/ Ea (Therefore less successful collisions) However, shouldn't it be
what he meant was, when there is a decrease in temp., less particles have energy equal or higher than activation energy, therefore chances of successful collisions to occur is less.
'A + B -> C + D' one of my favourites too...
Only just found your channel but these lessons are going to really help me scrape the B I need in chemistry!!
did u get it
after having done my alevels 2 years ago I remembered how great you are that I came back and these videos are actually still relevant to my uni course in the biosciences xD
at around 13:00, with the boltzman distribution curve, why is the height of the second curve (temp B) lower (before Ea) than that of curve A?? [EDIT] Is it to accommodate for the total number of molecules in the system being the same in both curves?
Yeah that exactly the reason
What India said.
E Rintoul Thanks, while your here, do you have anything on polymerization of alkenes? ( im not with aqa)
it's 2021 ad u are still a HERO
Everytime i play eliot Rintoul video. first thing i hear,
arrighttt!!
love it. so useful thanksss
Hi please reply: At 7:31 have you made a mistake? I think the 2.0 moldm^-3 shouldnt level off that left of the graph in relation to R. The rate of reaction has doubled which is the number of successful collisions in a give time. If the curve was drawn to level off at the same time as curve R, but twice as high, the rate of reaction is doubled because it is steeper. No need to move it left, if i am not mistaken?
I have no idea though I am confused so please help
You didnt move the 0.5moldm^-3 to the right of the graph which is right, therefore you shouldnt have moved left i think for the other curve
You are correct, the graph would still have a steeper gradient so rate would be increased :) hope this isn't too late for you!
Charlie Hargrave thanks for the reply bro. And yeah i think thats what it looks like i guess. Not too late at all im in year 12 still hby?
Year 13 :) mocks start Monday... Shit starting to get real but this channel helps a lot, guy is a legend. What you studying?
I think your maxwell boltzmann distribution may be wrong? I was told at that start it should never be curve out like a "s"? Is this correct or...
Also, I don't think "alternate" would be accepted for what you said about catalysts around 17:11, it's "alternative".... Like I said, it's what I've been taught
Thank you so much for this helpful video! A life saver even in 2022!
Hi, do you have any videos on the Arrhenius equation for AQA A level chemistry?
Hello Sir, I just want to ask are you going to do a video of a past paper on 2014???
Bartholomew Pang I'm planning on doing CHEM2 June 14 tomorrow morning...
THANK YOU SO MUCH THESE VIDEOS ARE SAVING MY GRADES!!!!!!
sophie mocton That's OK :)
Great teaching, as always!
Keep working hard, I used to watch this nd I’m now in Uni 👍🏼
Thank you so much for your hard work! I have hope that i can get an A now in chem
no words bruh.. you are a ledge!!
Lots of love and respect for you man. Can't express them in words!!
Hiya. I have a question about the effect of concentration on collisions . Basically, in class we were told that an increase in concentration/pressure increases the number of particles with the energy higher than the activation energy. So there are more frequent collisions but not more successful because the proportion of partices is still the same. You said in the video however that the number of successful collisions also increases, but we were told you would loose marks for putting that. Please could you clear this up for me. Thanks
+Amy If a reaction only occurs because the collisions that occur are successful, for the rate to increase, the number of successful collisions must increase. It's all down to statistics; whilst the energy does not increase, as you say, there will be more molecules that have an energy equal to or greater than the activation energy, thus more successful collisions.
excellent videos makes revision so much more efficient and easy
Good work Man , Favourite teacher :)
+live life That's very kind of you to say :)
Eliot Rintoul, the hero we all need but don't deserve
I got a question which said 1dm^3 of '20 volume' hydrogen peroxide (it was a decomposition question) and i was wondering what this '20 volume' meant?
+Armita Azar I believe it refers to the mass dissolved into one litre. I may be wrong but I think that's the case! I know that the technician at my school often talks about 40 vol. and 100 vol.
E Rintoul you have saved my A levels
YOU ARE A LIFE SAVER!!!!!!!!!
pretty sure people in 10 years will commenting about how great this video is
Thankyou so so much, for some strange reason my school did equilibrium before kinetics, and so some of my concepts were really confusing to me - but this video was really helpful!! ☺☺
2024 AND STILL USEFULL U R THE GOATTT
Thanks for the helpful videos but I also have a querie regarding the increase in temperature in the boltzman distribution curve; so when you shifted the curve to the right I noticed that the area of the curve was less than temperature a which would evince that less molecules are present- but I have read that the molecules present are the same so should the area of the curve been roughly the same?
You can blame my wonky drawing for that!
For sure the number of molecules remains the same, it's just the energy that changes.
I have revised the book again, but it's to late to be reading at 00:28, so i shall sit back and watch as many videos as I can today x
Boro Jen xX Do it!
On the distribution graph, why was higher temp red curve lower than initial one?
the amount of particles are the same aka they're equal areas
+BhadMan316 Because more molecules have a higher energy. However, the area under the graph is the total number of molecules and so this must remain the same.
E Rintoul Lol yea I've revised and understand this topic now but thanks keep up the good vids.
Is the most probable energy always the peak of the curve?
+Dom yep
NOT ALL HERO'S WEAR CAPES AND YOU CERTAINLY ARE A HERO 🙏🏻🙏🏻🙏🏻
6:30 surely changing the concentration would only affect the time and not the volume produced?
SharpSpoons no because there is a fixed volume, therefore if the concentration is doubled you will have double the number of moles/particles. Therefore if there is more particles, more oxygen is produced
Hello. Are you going to do a video on kinetics for A2? Thanks:)
Theresaa Packk I shall be at some point, yep!
E Rintoul Perfect. Honestly, your videos are absolutely amazing. I'm struggling with A2 so if you could do a few videos on that spec, I would be so so happy :D thanks :)
At 5:12 shouldn't the new curve reach the same volume as curve R?
because the volume is the same
No because if the concentration is lower, then a lower volume of oxygen can be produced
oh lol there were so many curves I meant curve Q
no because the curves are changing due to a change in reactant (h2o2), but the y axis is the product (o2). curve q has half the concentration but the same volume as curve r, which means it has half as many moles of h2o2, which means it can only produce half as many moles of o2. the o2 is measured under a constant concentration, therefore its volume is halved :)
Thanks so much guys now I get. Good luck for the exams if you're doing it
Thank you for this video, understood everything. Very grateful
Can you make a video looking at past papers and going through how you would answer the questions please?
last minute revision, watching all the videos!
Mellisa H Go for it!
aint that the story of my life
my eyes are so red right now they are glowing
one of the best explanations
What is the equilibrium?
+naveed ehsan Can you be a little more specific...?
Sorry I watched he ur video on equilibrium so it's ok thanks for ur help
6:53 for the line at 2.0 moldm-3 wouldn't it be halfway between the line below it?
No it’d be higher up as the y-axis is of volume. When you increase your concentration of reactant you’re increasing your volume of products, hence it will show a curve higher up to show this. With what your saying it’s for the Q line, where you’ve halved the concentration hence lower volume of product.
Thank you so much! =D For some reason this and mechanism are my worst topics...
That is strange! However, we are all different! Hopefully this has helped to alleviate some stress and build on some understanding!
great video again
can you make a video on the unit 2 paper for 2014 like you done for unit 1 as that helped a lot
thanks
M Hussain That's the plan for tomorrow morning!
Could you please make a video for the topic of " Transition Metals " ?
It's huge topic so I think he's breaking them up
mayynn you are superherooo
Thank you soooo much! This helped me 👌🏻
great video
nice
keep it up bro
That is indeed quite the delightful line good sir --_--
lovely
thanks for the great video =D
Dineshkumar .Kanesan No worries!
Thankyou Very Much :)
Thank you!
Sian Royle No problem.
This really helped me
Love your explanations! Haha why are you so damn smart? Are you a professor?
Haha thanks! Nope, just a teacher!
Good video :)
love you
Here 2020, teaching myself 😐
Feel this 👌
Got my exam in 20 minutes fml
5:01 Mosquito 0_o
Exam in 10 mins 💀
Uhhhh u2 is after tomorrow
EXAMS TOMORROW LAST MINUTE REVISON GANG
How did u do?
Squid says you have a funny voice.
Haha did u see when I posted it ?
My slime
:D
THNAKKKKKKKSSSSSSSSSSSS ALOTTTTTTTT MATEEEE BUT IF UR EVER FREE IN LIFE U SHOULD MAKE A VEDIO ON WHOLE AS LVL SYLLABUS REVSISON IT WOULD MEAN ALOT TO ALL THE STUDENTS SUFFERING IN THIS WRLD COZ OUR SCHOOL TEACHERS DONT N NVR WILL BE ABLE TO TEACH AS GOOD N SPECTACULAR AS U DO