Angelica Watkins yeah because rate law is high school chem and ur teacher assumes that you know that as highschool chem is a pre req to higher lvl chem lol
She is making it fast so it isn't like a 15 minute video. It is for the person who wants to get to the point without lots of time being spent. Teachers teach long lessons, youtube is for the quick review of what you learn in class
And, for all the people who are saying that she is talking too fast: if you notice, all of her videos are under 10 minutes in length, so she has to compress half an hour of lecture to within ten minutes meaning that she had to rehearse this video and talk faster than usual in order to fit it into a ten-minute timeframe. You might have to go back and watch it a few times, but that is how this format works.
hi dr... please answer my question :: a solution of adrug contained 500unit/ml when prepared it was analyzed after 40 days and was found to contain 300unit/ml assuming the decompostion is first order at what time will the drug have decomposed to one half of its original concentration?? how to be the right solution???
Thankyou so much! You're videos are the only studying i look forward to after school . I love your teaching method and you make everything seem so easy.Go Brightstorm- and this chemistry teacher!
Thank you sooo much. I wish you were my actual teacher, because my teacher doesn't really explain and then expects us to be able to take a quiz on it the next day..
Can't believe those people asking you to slow down, im so happy that I found someone like you who doesn't blabble for 20 minutes and that gets straight to the point and fast. Absolutely love it !! Thank you
if you think she's talking too fast you can change the speed of the video by going to the bottom right corner of the video, click on settings and change from "normal" to "0.5".
what you're referring to is the order of the reaction. this is not the rate of the reaction itself. that 1/2 that she calculated was the ratio of the first concentration of a to the second concentration of a raised to m which was equal to the ratio of the first rate of reaction to the second rate of reaction. since m is an exponent we needed to prove that 1/2 raised to m was equal to 1/2 hence finding m. since anything raised to one is that number we say that m was equal to 1.
Great video... One thing though. When you were finding what m and n were, you skips the steps that shows how to transpose the equation for [ B ]2. For newbie, that may have been a bit confusing to figure out how you got the equation : (.100/.200)m =(2x 10-3/4x10-3) Really appreciate the video, its pretty detailed... Please keep making more.
I swear you are an angel...i struggle with these problems last week ended up not finishing my HW...after i watched your videos, i sailed through those problems!...Thank YOU
Hey! At 5:30 how do you get from 1/2 and 1/2 to 1, and 1/2 nd 1/4 to 2? Sorry if I'm being stupid but I can't seem to get a whole number with my example question
If you were wondering how she got to second order: Think of n being the subject of the equation, change the fractions into decimals simply for clarity. You are then left with, n = 0.5/0.25 n= 2 Simple as that :)
If you know how natural log works then you just take ln of both sides of the eqn. m*ln(0.1/0.2)=ln(1/2) m = [ln(1/2)]/[ln(0.1/0.2)] m = 1 Her examples were easy enough that you can do it intuitively but for more complex problems you gotta use ln.
guys just flip the numbers and it would fit perfectly without using ln just divide the big number over the small one and you would have the correct number
My professor would probably be offended by me saying this, but I watch your videos to learn chemistry because he may as well be speaking Latin when he talks. Thanks for uploading these videos
Exponents and logarithms helped me here. The first question was what is her ratio of concentrations. it was 1:2 (.1M of A, compared to .2M of A) and then looked at the difference in her rates of change 1:2 as well (2e-3 compared to 4e-3). 2 to 4 is the same as 1 to 2... so we ask ourselves, what number do we raise 1/2 to, to give us 1/2 and that number is 1. Yet when we looked at [B] we saw the rates went from 4, to 16! and we had only doubled our [B]. So we need to compare the 2 values again.
because you have 1/2 on each side and to isolate the variable "m" you must take the reciprocal and mult. each side by 2/1 which gets rid of the 1/2 on the "m" and turns 1/2 on the left to 1 because 1/2 x 2/1 = 1
You use log rules. If you take the natural log of both sides, you can bring the exponent down... (1/2)^m=1/2 take natural log of both sides ln(1/2)^m=ln(1/2) because of log rules, the exponent "m" becomes coefficient m x ln(1/2)=ln(1/2) divide to get m by itself m=ln(1/2)/ln(1/2) m=1
Where have you been my entire life in college? I am studying for the OAT and you are soooooo awesome. I love how simple and easy you explain everything and with great depth to where anyone can understand.
does it matter which value is placed above in the fraction/ calculation. cause i was taught that Eg if using experiments 1 and 2 then the values of experiment 2 must be the numerator and values of one must be the denom.
Can anyone reading this please help me out. Trying everywhere but can't find the answer. The question is, "Given rate of reaction = k [A]^m[B]^n, What are the units for k when m=0 and n=1? What is the overall order of the reaction?" Thanks! :)
She was solving for “m” in the equation. So the problem was: (½)^m =½ You can probably see intuitively that for ½ to equal ½, then “m” must equal 1 (Any number to the first power equals itself). Alternatively, you can prove that m=1 by solving with logarithms. :)
could you please explain to me how I go about choosing which trials I'm going to use ? if I was looking for the order of a reaction with respect to A , same for B
I was taught the same steps but that it was final over initial (with both rates and concentrations. So my professor taught me that it was Rate 2 over Rate 1 = k [A]final^m [B]final^n all over k [A]initial^m [B]initial^n. Help me please
u are a life saver:) can u please explain how to get a value of a or b or even the rate of reactiion when they arent given, beacuse sometimes u are asked to figure out some missing values when u have been given some already?
Doing Rate 2/Rate 1 will most likely give you whole #s for the base instead of fractions so it will be easier and faster for some people to find the order since integers are easier to work with than fractions. It will give you the same answer either way.
Mandy Pannu nope it doesn't matter, just remember to be consistent. if you do rate 2/rate 1, then it has to equal concentration 2/concentration 1; and vice versa :D
@brightstorm2 I think it's necessary to correct the wrong explanation about how concentration affect the rate of a reaction. You said "the higher more concentrated the reactants are the faster they're going to collide". However, increasing concentration does not change the speed of the molecules, it only increases number of collision. It is temperature which affects the speed (kinetic energy) of the molecules which lead to higher ratio of successful to unsuccessful collisions.
we're going to find 2 ratios again. The concentration in the 2nd test, to the concentration in the 3rd. well that is 1:2 (or 1/2, 1 to 2, whichever way you prefer to read the ratio). How about in the rates? That went from 4e-3 to 16e-3! thats 4 times as much, or a ratio of 1:4. The letter 'n' represents the power or amount, that you have to change 1/2 to give you 1/4. That number is 2. If you are familiar with logs, they will directly give you this number. I didn't understand what happened,
What does the definition of the reaction rate law state? The writing isnt legible.. when it says between the rate of a _______ and concentration of reactants???
I'm taking the ACS General Chemistry exam in about an hour, and after 4 months of sitting in a classroom not understanding anything, I finally understand reaction rates.
Y i still don't get it! how did she know that m is 1 when 1/2^m = 1/2? plus 1/2^m = 1/2 just doesn't make sense to me... and also the N part too T.T can someone explain to me plz?
You are my lifesaver! Thanks so much! I've watched a lot of your videos and you cover the material in ways I can understand! I never need to go to class again because i know i can be taught the material better on TH-cam. :)
This is a much easier method than the one my teacher taught me. She wants us to make a graph to using the integrated rate laws every time to see if what order of reaction it is.
Forgive me, but I still don't get why 1/2n = 1/4! And when calculating the rate (on bottom left at 7:52) why are both sides divided by 0.001 instead of 0.100?
Thnx alot and btw the ppl who are bitching about the vid quality (it was uploaded in 2010 where 240p was the standard) and if you find it fast paced just change the playback to 0.75 as I did and it helped me alot.
You might want to differentiate next time between Reaction Rate and Rate Law. I think what a lot of people are confused on is that their professors usually lecture separately on reaction rates and then on rate laws. I understand it's all incorporated into one equation in the end; however, it might be easier for your viewers. Also, do talk a little slowly if possible. I am the same way when talking about biology. It's the passion that gets the best of us :)
This video was certainly helpful! Thank you so much for taking the time to make this review to have clearer understanding for us college students. General Chemistry does get harder!
This was much more helpful than the lecture class I paid $1500 for,
+Angelica Watkins this is so so sos osososososososo true...like what are we paying these people for you know???
especially Pogil ??
use rate my professor
Angelica Watkins yeah because rate law is high school chem and ur teacher assumes that you know that as highschool chem is a pre req to higher lvl chem lol
no its not. its more so AP chem...
I ask the same thing.
She is making it fast so it isn't like a 15 minute video. It is for the person who wants to get to the point without lots of time being spent. Teachers teach long lessons, youtube is for the quick review of what you learn in class
And, for all the people who are saying that she is talking too fast: if you notice, all of her videos are under 10 minutes in length, so she has to compress half an hour of lecture to within ten minutes meaning that she had to rehearse this video and talk faster than usual in order to fit it into a ten-minute timeframe. You might have to go back and watch it a few times, but that is how this format works.
For anyone confused on how she derived "n", she uses:
natural log (ln) 1/4 divided by: natural log (ln) 1/2 = 2
The Observer thankyou😭
Hira Mazhar and how’d she get “m”
Hina Chan log(1/2) / log(1/2)
Hina Chan you just have to make the bases same!
@@hinachan6298 by division of trial 1and2ofA=rate of trial A
hi dr... please answer my question :: a solution of adrug contained 500unit/ml when prepared it was analyzed after 40 days and was found to contain 300unit/ml assuming the decompostion is first order at what time will the drug have decomposed to one half of its original concentration?? how to be the right solution???
You just explained 50 minutes of lecture in less that 10 minutes.
Thank you.
:D
Yea this video helped more than a year of chemistry with Ms Trela.
What up yi
@@nateewing600 hello Nathaniel Ewing
It took just over 9 minutes to help me understand something that my professor has yet to be able to do in a week! Awesome video!
put 3 days of work into a 9:16 minute video.
Man, thank god for TH-cam. I understood bits and pieces in my teachers lecture, but this made it much more clear.
Thankyou so much! You're videos are the only studying i look forward to after school .
I love your teaching method and you make everything seem so easy.Go Brightstorm- and this chemistry teacher!
Thank you sooo much.
I wish you were my actual teacher, because my teacher doesn't really explain and then expects us to be able to take a quiz on it the next day..
I watched 7 videos... and this is the BEST!!.... She is SOO much better than all 5 of my chem teachers...
Can't believe those people asking you to slow down, im so happy that I found someone like you who doesn't blabble for 20 minutes and that gets straight to the point and fast. Absolutely love it !! Thank you
if you think she's talking too fast you can change the speed of the video by going to the bottom right corner of the video, click on settings and change from "normal" to "0.5".
It works great😀😀 I change it into 0.75
Thankyou....
what you're referring to is the order of the reaction. this is not the rate of the reaction itself. that 1/2 that she calculated was the ratio of the first concentration of a to the second concentration of a raised to m which was equal to the ratio of the first rate of reaction to the second rate of reaction. since m is an exponent we needed to prove that 1/2 raised to m was equal to 1/2 hence finding m. since anything raised to one is that number we say that m was equal to 1.
Great video... One thing though. When you were finding what m and n were, you skips the steps that shows how to transpose the equation for [ B ]2.
For newbie, that may have been a bit confusing to figure out how you got the equation : (.100/.200)m =(2x 10-3/4x10-3)
Really appreciate the video, its pretty detailed... Please keep making more.
I swear you are an angel...i struggle with these problems last week ended up not finishing my HW...after i watched your videos, i sailed through those problems!...Thank YOU
Hey! At 5:30 how do you get from 1/2 and 1/2 to 1, and 1/2 nd 1/4 to 2? Sorry if I'm being stupid but I can't seem to get a whole number with my example question
haha dude thanks for asking cause i'm wondering too
If you were wondering how she got to second order:
Think of n being the subject of the equation, change the fractions into decimals simply for clarity.
You are then left with,
n = 0.5/0.25
n= 2
Simple as that :)
If you know how natural log works then you just take ln of both sides of the eqn.
m*ln(0.1/0.2)=ln(1/2)
m = [ln(1/2)]/[ln(0.1/0.2)]
m = 1
Her examples were easy enough that you can do it intuitively but for more complex problems you gotta use ln.
guys just flip the numbers and it would fit perfectly without using ln just divide the big number over the small one and you would have the correct number
it's ln(1/4)/ln(1/2) to get x=2
this is very educational! thank you very much. i have a test in 30 minutes and for 9 minutes i learned more than i did for 3 weeks
My professor would probably be offended by me saying this, but I watch your videos to learn chemistry because he may as well be speaking Latin when he talks. Thanks for uploading these videos
do you go to university
Exponents and logarithms helped me here. The first question was what is her ratio of concentrations. it was 1:2 (.1M of A, compared to .2M of A) and then looked at the difference in her rates of change 1:2 as well (2e-3 compared to 4e-3). 2 to 4 is the same as 1 to 2... so we ask ourselves, what number do we raise 1/2 to, to give us 1/2 and that number is 1. Yet when we looked at [B] we saw the rates went from 4, to 16! and we had only doubled our [B]. So we need to compare the 2 values again.
because you have 1/2 on each side and to isolate the variable "m" you must take the reciprocal and mult. each side by 2/1 which gets rid of the 1/2 on the "m" and turns 1/2 on the left to 1 because 1/2 x 2/1 = 1
You use log rules. If you take the natural log of both sides, you can bring the exponent down...
(1/2)^m=1/2
take natural log of both sides
ln(1/2)^m=ln(1/2)
because of log rules, the exponent "m" becomes coefficient
m x ln(1/2)=ln(1/2)
divide to get m by itself
m=ln(1/2)/ln(1/2)
m=1
Thanx for clear it simply...
From INDIA
at 4:43, you said the rate of reaction for [a] is 1 although the answer was 1/2 ^m = 1/2.
Why is the rate 1 when you calculated it as 1/2 ?
Oh wonderful thanks
so is that how you would work out the rate constant for the reaction and is it really necessary to include unit for k?
Perfect explanations! Thank you so much!
Where have you been my entire life in college? I am studying for the OAT and you are soooooo awesome. I love how simple and easy you explain everything and with great depth to where anyone can understand.
Whaaat you were doing this in college? I’m doing this in high school right now
very helpful. thanks. I have a test tomorrow and i didn't know what i was going to do because rate laws was one think i couldn't understand.
Thank god for this. Needed a refresher as this just popped up for the first time in 2 years.
was that an average rate which was measured for the reaction? or the initial rate?
Nice...clear, concise and to the point. Thank you clearing up the units. I was just gonna skip that part on the test. Got it now.
Can you explain how you determined the order for [B]^n? I 'm not understanding what you're doing to get [B] is 2?
does it matter which value is placed above in the fraction/ calculation. cause i was taught that Eg if using experiments 1 and 2 then the values of experiment 2 must be the numerator and values of one must be the denom.
Great video, helped me lots! Thanks!
at 7:00, why is [B] squared? i thought it was first order in trial 1
Can anyone reading this please help me out. Trying everywhere but can't find the answer. The question is, "Given rate of reaction = k [A]^m[B]^n, What are the units for k when m=0 and n=1? What is the overall order of the reaction?" Thanks! :)
She was solving for “m” in the equation. So the problem was: (½)^m =½
You can probably see intuitively that for ½ to equal ½, then “m” must equal 1 (Any number to the first power equals itself). Alternatively, you can prove that m=1 by solving with logarithms. :)
could you please explain to me how I go about choosing which trials I'm going to use ? if I was looking for the order of a reaction with respect to A , same for B
I was taught the same steps but that it was final over initial (with both rates and concentrations. So my professor taught me that it was Rate 2 over Rate 1 = k [A]final^m [B]final^n all over k [A]initial^m [B]initial^n.
Help me please
Wow, this was really clear and helpful.
Absolutely 👍👌
How do u find the order of each reactant if there isn't any constant concentration for A or B when either of the two varies
u are a life saver:)
can u please explain how to get a value of a or b or even the rate of reactiion when they arent given, beacuse sometimes u are asked to figure out some missing values when u have been given some already?
Doing Rate 2/Rate 1 will most likely give you whole #s for the base instead of fractions so it will be easier and faster for some people to find the order since integers are easier to work with than fractions. It will give you the same answer either way.
does it matter if the rate2/rate 1 or rate1/rate2 ? because in a lot of places i see rate2/rate1 where as in this video you/she did rate 1/rate 2.
Mandy Pannu nope it doesn't matter, just remember to be consistent. if you do rate 2/rate 1, then it has to equal concentration 2/concentration 1; and vice versa :D
Khánh Hạ H.T Thankyou so much ! :D
Could you have compared experiment 1 and 3? Or can you only compare experiments that are in numerical order?
@brightstorm2 I think it's necessary to correct the wrong explanation about how concentration affect the rate of a reaction. You said "the higher more concentrated the reactants are the faster they're going to collide". However, increasing concentration does not change the speed of the molecules, it only increases number of collision. It is temperature which affects the speed (kinetic energy) of the molecules which lead to higher ratio of successful to unsuccessful collisions.
for getting the k can i pick any trial to get it?
thanks for this, im just confused as to where you got the "0.001/0.001" ?
Hello professor,
For the reaction A-->B and 2A-->2B, do we get 2 different rate constants (irrespective of the reaction order)?
we're going to find 2 ratios again. The concentration in the 2nd test, to the concentration in the 3rd. well that is 1:2 (or 1/2, 1 to 2, whichever way you prefer to read the ratio). How about in the rates? That went from 4e-3 to 16e-3! thats 4 times as much, or a ratio of 1:4. The letter 'n' represents the power or amount, that you have to change 1/2 to give you 1/4. That number is 2. If you are familiar with logs, they will directly give you this number. I didn't understand what happened,
What does the definition of the reaction rate law state? The writing isnt legible.. when it says between the rate of a _______ and concentration of reactants???
You are the best , because you are so clear and you don't leave a single thing out. Thank you!
thank you for all your help.
this comes might handy the day before finals!
Extremely helpful, I have an upcoming quiz and didn't understand anything from lecture. The explanation here is much clearer, thanks a lot!
Hi, how can does this work when non of the concentrations stay constant and changing at the same time
superb mam..! i understood it very well.:)
can u plz give the defination of k..?
I'm taking the ACS General Chemistry exam in about an hour, and after 4 months of sitting in a classroom not understanding anything, I finally understand reaction rates.
Fast, Clear and Direct to the point. Perfect for Reviewing and Last Minute Studying.
Thanks Mam it really helped me out
for the first practice problem, why is m = 1 for the concentration of A ?
Thank you! Struggled with this for awhile as I couldn't find anyone who would break it down sufficiently.
In the end when you said "no it's not divided by", you could actually divide it by s because, 1/M/s=1/(M*s).
How do we decide which ratio that we should take ???
Private tutions available?
You are a great teacher, Thanks for all your vedios it is great help for student like me please do not stop this work. God bless you
thanks
it,s helpful video
She thought order in a great way, previously I saw many vedioes but I little bit of confusion is get pending by this she had cleared all my doubts
Amazing how simply you put this. Great job!
i was so lost when my teacher did this but she explained it like a pro and now i understand. THANK YOU
Great video.....You made it easy with detail explanations
thanks but do you have a video on how to work out the initial rate plsssss
its was so confusing when I first learn in school.
but now i understand it
thank you so much!
words can't express how happy i am to have found this channel! omg thank you sooo much!
Y i still don't get it! how did she know that m is 1 when 1/2^m = 1/2?
plus 1/2^m = 1/2 just doesn't make sense to me...
and also the N part too T.T can someone explain to me plz?
How did they come up with them?
You are my lifesaver! Thanks so much! I've watched a lot of your videos and you cover the material in ways I can understand! I never need to go to class again because i know i can be taught the material better on TH-cam. :)
This is a much easier method than the one my teacher taught me. She wants us to make a graph to using the integrated rate laws every time to see if what order of reaction it is.
How did u get 1 after both sides equal one-half
Thank you for making this video; it's helping me prepare to teach HS chemistry. Your obvious intelligence is impressive!
very helpful and to the point, thank you
Great presentation! Really helpful... I need to watch more of your videos.
you are a great teacher. thank you.
Is that overall order -1 for the units for k trick
Forgive me, but I still don't get why 1/2n = 1/4! And when calculating the rate (on bottom left at 7:52) why are both sides divided by 0.001 instead of 0.100?
just want to say thank you very much. that was a very very helpful video.
I could have save a whole semester of time and hundreds of dollars had I watched this before registering for Gen Chem. Great video. Thank you!!!
I can learn more in less than ten minutes for each chemistry subject with these videos, than I can in my actual chemistry class.
why is the max resolution 240p??
thanks for this video this video is really helpful for me it make my concepts more clear
Thnx alot and btw the ppl who are bitching about the vid quality (it was uploaded in 2010 where 240p was the standard) and if you find it fast paced just change the playback to 0.75 as I did and it helped me alot.
quick and insightful...... why aren't all teachers like that? short and to the point.....
You might want to differentiate next time between Reaction Rate and Rate Law.
I think what a lot of people are confused on is that their professors usually lecture separately on reaction rates and then on rate laws. I understand it's all incorporated into one equation in the end; however, it might be easier for your viewers.
Also, do talk a little slowly if possible. I am the same way when talking about biology. It's the passion that gets the best of us :)
This video was certainly helpful! Thank you so much for taking the time to make this review to have clearer understanding for us college students. General Chemistry does get harder!
1/2 to power of 1 will equal to 1/2 which we need....becoz it says 1/2 =1/2...... so itz like 1/2 power to something gives 1/2..get it????
Calculate the rate of words/secs of this instructor
answer:39/sec
Just wanted to thank you guys for explaining it so well!!
lady you are a life saver! Thank you
Very direct explanation..soo helpful thnks