can i just take a minute to say how much ur amazing? cuz yeah my beloved teacher did not tell us ANY of this and had it not been for u, guess who would have failed yet another chemistry quiz?
thank you. I understand if the differences are same it would confirm the order. However, plotting the graph and finding the R^2 value compares directly and whichever is closer to one shows that it is that order. In this case, when I plotted the graph using your data, I get the R^2 value of 0.9999 for Time Vs ln[A], so it is First order.
Can't we just plot the [A], ln[A] or 1/[A] vs time to determine the order? If any of these plots yield a straight line, we can determine the order. Also, the negative of slope yields the rate constant. Correct me if I am wrong.
Basically you calculated everything to plot a graph then what is the point of trick..... You plot the graph or calculate difference whichever gives constant difference is the linear graph, which is the funda of linear graph
A= Aoe^-kt is the first order reaction. my question is it would be the same to say LnA = -kt +LnAo and thus since the slope is -k then half time might be found by 0.693/K?
Once you have found the order of the reaction, eg if its first order, there will be a specific formula you can use. For first order its ln ([A]/[A]o) = - kt. therefore if you plot ln conc on the y and time on the x axis, you get -k for the gradient of the straight line on the graph.
this doesnt make any sense , the ln[A] decreases at about the same rate as [A] , nothing stays ''about the same''' , how do we find what order the reaction is scientifically? (i.e with maths)
Im having problems crunching my numbers. My data is given into in mols, and I get each value into Molarity and I'm using those numbers. However when I'm doing it for ln[A] all of my values are negative????
Dennis Chiu Thanks Dennis, my values for for ln[A] are going up and down up and down, so does that mean I can omit the first order since the values aren't going downwards (-k)? Sorry if my terminology isn't making any sense.
Jereme leBourgeois Check your values for [A]. As [A] decreases over time (and it should do so during a chemical reaction), ln[A] should decrease as well. If your values for ln[A] are fluctuating as you say, that probably means the original values for [A] are incorrect.
The overall order is the sum of the order of each reactant so just say A was first order and B was second order, you add 1 and 2 together and it's third order. I think this is correct anyway
Your inputs on the calculator are hilarious. Thanks for the laugh while studying :)
Yeah 😂😂😂
Was just thinking that 🤣
I actually freaked out😂😂😂
@@epicmoments1964 HE BASHED IN RANDOM SHIT
Might be good to note that the difference in time should also remain constant when calculating differences in concentrations.
Love the way he uses the calc! It's on fire. 😂 The video was really helpful. Thanks!
let me just take the lawn of 5
How do you use the calculator like that, make a tutorial on that please.
lol why is he tapping random buttons, we get that you are calculating something
can i just take a minute to say how much ur amazing? cuz yeah my beloved teacher did not tell us ANY of this and had it not been for u, guess who would have failed yet another chemistry quiz?
MAKES SO MUCH SENSE. God Bless you.
your calculator typing skills are out of this world
thank you. I understand if the differences are same it would confirm the order. However, plotting the graph and finding the R^2 value compares directly and whichever is closer to one shows that it is that order. In this case, when I plotted the graph using your data, I get the R^2 value of 0.9999 for Time Vs ln[A], so it is First order.
My pre-lab makes sense now. Thank you, thank you, thank you!
Took all day trying to understand this thank you sir🎉
lol! you are the only person i know who can actually make me crack up while doing chemistry...unnnbelieve able!
fam really? LAWN?
thanks for the video though
That's what it's really called!
I YAWN instead
thank you for being the professor that mine could never be
Can't we just plot the [A], ln[A] or 1/[A] vs time to determine the order? If any of these plots yield a straight line, we can determine the order. Also, the negative of slope yields the rate constant.
Correct me if I am wrong.
yes, you can and I prefer that too. thank you for adding your points to the comments.
You saved the day bro. Many gratitude !!!
This vedio is titled how to solve your homework in 5 minutes.
Super thanks
So nice of you
Basically you calculated everything to plot a graph then what is the point of trick..... You plot the graph or calculate difference whichever gives constant difference is the linear graph, which is the funda of linear graph
what if none of the 3 work out?
Third order
yeah i was actually stuck with a problem whose answer key is 1.5 order. This clip is amazing but still doesn't cover that.
It took me forever to find this thank you
Do you use the same method of you are given Pressure instead of Concentration?
what if the time intervals arent constant?
Great tips by the way :)
rainman doing calculations in his mind
This was a very helpful video, thank you!
LMFAOOOOO when he hits random buttons instead of just saying he already calculated them
Thank you so much for this vid, it helped me a lot!!
Now, this is how u explain a topic! Thanks! It helps a lot :D
A= Aoe^-kt is the first order reaction. my question is it would be the same to say
LnA = -kt +LnAo and thus since the slope is -k then half time might be found by
0.693/K?
Awesome video, it was very helpful and saved me time on my first Pchem exam!
i legitimately thought he was really doing the calculations that fast guys
Does this apply to [B] as well?
Taught me better than my college professor
but when time intervalls are not steady?
Wow thanks man you are super skilled
lol. I had no clue what you were talking about until half way in the video.
So helpful! Thank you!
How to calculate for 3rd order reaction?
i wish i could give you 1000 more likes, thanks a bunch !!! :D
Why in the world my professor did not tell me this tip? Ugh... thank you so much kind sir
So what happens if none of the first differences are linear?????
Emmanuel Makume choose best fit or redo experiment
How do you determine the rate constant after your calculations?
Once you have found the order of the reaction, eg if its first order, there will be a specific formula you can use. For first order its ln ([A]/[A]o) = - kt. therefore if you plot ln conc on the y and time on the x axis, you get -k for the gradient of the straight line on the graph.
Thank You! :)
Mohammad Sukkar no problem. Message me if you have any more questions
Will do!
That dude got a mean trigger finger
Thank you very much for this straight forward explanation..... :)
how will you know if it has 3 or more order of reactions
excellent calculator work! skillful..
what does approximately constant mean what is your definition of that
best explanation
You are so good thank you so much
How to graph method sir this calculation
What if your time is not going by 20 seconds? The time given to me is not constant, damn this problem really wants me to graph it doesn't it :/
thanks teacher god keep u
damn how do you calculate like that breh, strong fingers
thanks aloooot man u r awesome
. soo helpful
but how do you get the k from that?
that means ,in calculation time doesn't take part??
Very Helpful indeed **Thanks a lot
hey thanks for this video....really helped me to understand some concepts :)
vervy very very very very very very well!!
Could you show how to do it the proper way?
this doesnt make any sense , the ln[A] decreases at about the same rate as [A] , nothing stays ''about the same''' , how do we find what order the reaction is scientifically? (i.e with maths)
You know how to use a calculator so well that it doesn't make sense to untrained eyes.
Thank you so much
Best video🎥
not me using my own calculator to check the values cause he punched in the values waaayyyy to fast lol
The problem is calculator is not allowed in entrance exams.
Thank you so much!!
Im having problems crunching my numbers. My data is given into in mols, and I get each value into Molarity and I'm using those numbers. However when I'm doing it for ln[A] all of my values are negative????
ln[A] will come out negative if [A] is between 0 and 1. I think the technique shown in the video should still apply for negative ln[A] values as well.
Dennis Chiu Thanks Dennis, my values for for ln[A] are going up and down up and down, so does that mean I can omit the first order since the values aren't going downwards (-k)? Sorry if my terminology isn't making any sense.
Jereme leBourgeois Check your values for [A]. As [A] decreases over time (and it should do so during a chemical reaction), ln[A] should decrease as well. If your values for ln[A] are fluctuating as you say, that probably means the original values for [A] are incorrect.
hahaha! so cool:) i love you man....thanks for the extremely clear presentation and explanation!
3rd order??
jefbord: I think you have to graph them.
You have to graph it
very interesting
REALLY?!?! I don't think anyone else noticed that!
this is straight up calculator abuse lmao thanks for the quick info and good laugh!
Are you Canadian?
"Turns Ouuut to be..... "
hahahahaha you're great! love your inputs on the calc so funny
thank you
Thank you!!
Ok so really people that others were fools 9 years ago!?
what if you have more than 1 product :/
The overall order is the sum of the order of each reactant so just say A was first order and B was second order, you add 1 and 2 together and it's third order. I think this is correct anyway
You seemed to be much fast😂
how do you do for 3 order
It only goes up to 2
I'd assume that if none of these fit then it is the third order
Okay but why does he pretend to press the calculator?💀
nice one bro!😄😄😄😄😄
whoops not product i meant to say reactants
The lawn? You mean natural log? Lol
Your difference explanation isn't clear...
jizuz you saved my ass bro
thanks :)
বিদেশী......পাগল 💂💂💂💂💂💂💂💂💂👲👲👲👲👲👲👲👲👲👲👲👲👲👲👲👲👲👲👲👲👲
Wow this was bam😂😊
Am I the only one getting Harlem shake and Kpop?
noy good very tedious snd time taking. m an mbbs aspirant from india we r not allowed any calculator there...
cool man,
love this! haha
اعجبني
ahaahhahaha amazing
thanks teacher god keep u
thank you!