Arrhenius Plots
ฝัง
- เผยแพร่เมื่อ 15 ต.ค. 2016
- Arrhenius plots are just another addition this Nobel prize winning Swedish scientist brought to Chemistry. In this video you will find out how to calculate activation energy and the Arrhenius constant using an Arrhenius plot. Yes.. y=mx+c makes an appearance.
Just wanted to say, I'm an A2 student (OCR) and need an A for uni. Thank you so much for doing these videos and putting in such effort. Honestly you are an amazing teacher, you make things simple to understand whilst making sure we understand why and how it works as well as just being able to do it which is really important for remembering in exams. So thank you so much. You make this so understandable in 14 minutes when at college I was struggling for 2 hours! Amazing teaching skills.
Ah thanks so much! Means a lot to get feedback, I really never know how useful they are until people tell me. I hope you get your Amazon what a result that will be!! What are planning on studying at uni?
Allery Chemistry yes they are very helpful, please keep making them!! Haha. I'm going to study Veterinary Medicine so need AAA. I'm studying chemistry, biology and geography :)
Seeing as your exam has now passed how did it go? are you doing veterinary medicine?
What did you get if you don’t mind me asking
@@HappyHorseJumping I'm curious did you get there? I'm in the same position now. Year 13, I need a AAB for Pharmacy! I'm studying Biology Chemistry and Computer science, though I wish I did geography instead of Computer Science
so much clearer than the text books/revision guides. THANKS!!!!!!
Great! So pleased they are helping you.
Was so confused on this in class, you really have a great way of explaining tricky topics in a short amount of time very clearly and concisely :) This channel is a lifesaver, thank you so much!
watching on the day of my exam. I wish I had found this sooner 😖😫
just warching 43 min before my exam. thank you
Thank you so so much this really helped me - I'm not a maths student and struggled with this concept. This clearly explained it to me and I finally get it!
i didn't think I will ever understand this graph :) but your video is making a MASSIVE difference- thank you so much!! :'))))
Thank you so much for this video you don't understand how easy you made this to understand I am so glad I found ur channel and I cannot wait to donate and listen to your podcasts
Thank you so much for these videos; you honestly help out so many people including myself.
I'm doing the A2 specification for Edexcel and you have clearly explained all of the points we need to know for our unit 4 exam.
THANK YOU SO MUCH!
Cannot thank you enough for this video you’ve explained it so well!!!
Definitely the hardest part of A level Chemistry imo
Yes I think you may be right! Hope the vid made it a little bit easier for you.
I think you are brilliant Allery. I have to teach this and now I feel more confident explaining this to my students. Thank you
Mans got such a cute smile 😊
Thank you so much, your videos are such a help!!
Okay this was AMAZING!! U were born to teach
Thank you very much for this!! very well explained and taught!! my understanding of this is so much better!! thanks once again
Thank you so so much. My friend told me about u. I understnd this so much now and it looked really complicated so i left it to the last min.
You are brilliant, you make much more sense than my teacher
paper 3 tomorrow chemistry A I was shocked how it was on paper 1 8 marks in total just on this split in 2 sections. Everything I didn't want coming up all came u- on a positive note I really appreciate your videos n I truly feel u don't get enough credit!!!! thank you
Thank you so much for your help! You're amazing.
So beyond useful!!
Glad it was helpful!
Amazing explanation 👏 thanks
You are a life saver !!!!! Thank you so much
No worries!
Couldn't have come at a better time! First test on Thursday
Fab! Good luck for your test! Please share the video, the more people that benefit from it the better!
Great video.Thank you so much
when i got that big number for lnA i thought my calculator was just broken and was trying to fix it, until i unpaused the video and realised that was the correct answer
This video is excellent!
I have got full marks on my exam question, eventhough I'm not a maths student, I just followed the steps.
Excellent!
THANK YOU SO MUCH SUCH A GREAT HELP
Really helpful thanks so much!
wow this is great thank you so much
We are doing these next! Starting it thursday woo thank you xx
A bit of a head start then! Better to do that than to be behind and struggling. Keep up the hard work!
Really helpful, thank you 😊
You’re welcome 😊
thank you for being a hero
are the units for A dm^3mol^-1s^-1 (time stop in video is 13:49) or moldm^-3s^1? This was such a great video! I am currently revising for my year 13 mocks, so thank you!!!
Thank you sooooo much!!!!!!!
You're welcome!
just wanted to say that the units for A are the same as the units for K
Thank you. May I ask how to best convert the Arrhenius equation to solve for (or predict) 't' from the increase of carbon? Similar to Arrhenius' original predictions.
Thank you very much..🖤
No worries!
I have a question. When calculating A doesn't the section in brackets have to be multiplied by -1 as it is -(Ea/RT) ? As before, when you calculated Ea you did - (-11,000 x 8.31).
Hi
I was just wondering why in the part for the calculation of the activation energy you put..........
- (-11,300x8.31) whereas you didn’t put a minus in front of the bracket in the calculation for A?
Why wouldn’t it be -4= -(-1.1/0.1x10^3)?
Sorry if it’s a silly question!
Thanks for the really helpful video
Well said. It is confusing.
how do you work out the units for A?
Just wondering that when you rearrange for Ea from -Ea/R= gradient why does the minus in front of the Ea not become positive so Ea = +(-11,000x8.31)?
Ur are the best ❤
Hi, was wondering if you could cover the specification point for AQA on defining the terms order of a reaction and defining the rate constant. Just comment this back to me pls. I understand all the calculations and how to deduce the rate equation, but I’m not sure how to actually define these if asked.
Hi Chris, you explained this really well, thank you? To work out the arrhenius constant if the straight line touched the y-axis could you just get lnA like that?
Theoretically yes
there is another way to calculate A because its the y intercept if the line is extrapolated u can get a figure of A on the y-axis however the risk is if you are a decimal off, the result is a complete different answer
Thanks
How did you get those units for A? (or is that a stupid question..)
It's complex so you just need to remember them unfortunately.
Hello, once we know the value of A and the gradient, we could technically plot the graph, right?
Hiya! I was just wondering.. You know the last example in this video? It's so much more complex than the only one in my textbook.. And another textbook for the same spec doesn't have arrhenius at all! So where did you find this/do you have anymore/ did you make it up? And do you know if the AQA spec this year requires this difficulty of question because we can't find any hard examples more complex than rearranging. No gradient working out etc. 😕 thank you so much if you can help xx
Yes you will need to know both. The spec requires you to know how to use Arrhenius Equation. This leaves it open for them to ask questions using an equation or graph. They are just using 2 different ways. I just made up the questions as there are very few questions available, it is very new to majority of boards.
Allery Chemistry It's awkward with no questions to do! We have looked so long for some 😭Thank you so much for your help and thank you for your examples you gem!
No problem at all. If you look through the old edexcel/OCR salters past papers you may find some on there. Edexcel/OCR Salters were the only board that had Arrhenius to my knowledge.
Why is it -(-1100*8.31)
Why is there two minuses
@@fatimaa8770 It was -Ea/R initially, so to save some time in rearranging the equation later on, he minused a minus. -Ea = -11000*8.31. Hence Ea=-(-11000*8.31). Multiplying both sides by -1 to cancel the minus on Ea.
Hi :) If we calculate A from a point on the graph, A is dependent on temperature, so wouldn't we get a different value of A from each of the three points (because of the different temperatures)? And when I use the actual y intercept to get A, I get a different value altogether. If we extrapolate your line back to the y intercept, it would be something like -3, so e^(-3) = 0.050 I'm probably missing something obvious. Help, I'm confused haha
Why is the unit dm3mol-s- at the end?
Just wanted to ask is this for AQA
Legend
you're the best
Was just wondering, do the exams give us both the original and the simplified Arrhenius Equations?
I would assume they don't give you the equation. But definitely check with your specification. The simplifying of it is definitely something you need to do though.
So it does not matter which constant and temperature you choose when calculating the Arrhenius equation?
Don,t we need to determine the value of k constant first to move forward to plote ln(k) on Y-axis and 1/T on x-axis. How to determine the value of ln k, when we don,t have value of k?
Also how did you know the units for A? Thank you
am a bit late but apparently its complex so u just need to know them
Can u zoom into the board a bit please
I want know how to convert from Differential scanning calorimetery graph into Arrhenius graph
Not on A Level spec so not needed.
Stupid question- is 1/t time or temperature? If it’s time, where does the temperature part fit into all of this?
Absolute temperature (K)
THANKS just saying!!!!!
is the formula given in the exam?
Hi I think I get it mostly but when you worked out Ea you did -Ea/R but what happened to the T I thought it should be -Ea/RT... I think the T was the 3.1?
If y = mx + c, and lnk = lnA - Ea/RT, y = lnk, c = lnA, and what's left is -Ea/RT. -Ea/RT can be re-written in the format -Ea/R * 1/T, as when they're multiplied, they form -Ea/RT. The mx, can therefore be assigned as x = 1/T (which is why it forms the x axis), making m (or the gradient) be equal to -Ea/R.
Louis Greenhalgh thanks for replying 1 day after my chem exam 😂😅it's okay it never came up
Depending on your spec, it still could. It's on my next paper (edexcel paper 2) and could also be on paper 3
Louis Greenhalgh it can only come up in paper 3 now, I hope it does I was kind of upset it didn't come up in paper 1
the gradient is equal to -Ea/r
1/T is your x coordinate , so as we worked oout gradient we equate it to -Ea/r
how was ln(K) value found? I'm guessing experimentally and then ln( 1/time)
Lnk is the y axis
Assuming the rate constant can be found from the rate equation? however the initial rates needed to find the rate constant (k) are found experimentally.
good timing for this video lol
I must be psychic!
Can you do a video on directing groups?
why is k proportional to 1/t.
Anyone watching just before 2019 exams???
This video for not clear
In what way?
Allery Chemistry are you sure your LnA/A calculations are right sir? I’m even more confused