Half-life of a first-order reaction | Kinetics | AP Chemistry | Khan Academy
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- Опубликовано: 18 окт 2014
- The half-life of a reaction is the time required for a reactant to reach one-half its initial concentration or pressure. For a first-order reaction, the half-life is independent of concentration and constant over time. View more lessons or practice this subject at www.khanacademy.org/science/a...
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Thank you so much. This video fills a BIG gap in youtube chemistry.
Thank you for these amazing lessons!!
beautifully explained. thank you so much
You should put number to the concentration.
Great information! Really helpful! But please slow down a little!
This is the only teacher on Khan that I don't really like and he is in all the videos I need :/
He’s the AP chem teacher at my high school
very good videos
Thank you! My textbook was so confusing and the derivation of the equation was not clearly explained anywhere! This made everything click
I. Love. This.
i loved the teaching though the video is of low quality. Peace ,from Kenya.
thank you
It is good but make your speed moderate
had to find the rate constant for an equation using that half life formula but I got a resonance cascade instead. any idea on what I did wrong?
also -ln 0.5 = ln 2
bro i did type in the search ''how to use the jetpack in half life'' and this what I got.
i was looking for the video game half life not this
I pressed [log] then .5 and i got this answer -0.301029995 and he got -.693147... what am i doing wrong? do I need to press other buttons to get that answer?
+Kerwyn204 it's a natural logarithm, log is usually interpretend as decimal logarithm (or lg) by default, find a button that says ln instead of log
perfect thanks!
K3K it is natural log of 2(and not the half) upon the rate constant (as written in all the text books) which gives the half life ..n u r right idk how he went wrong bout that concept
There is this one question I don't quite understand : "Calculate half-life for first-order reaction if 68% of a substance is reacted within 66 s."
So, how do i calculate this. And I wonder why must we use what is left of substance in the equation and not what is used up?
ln [A] = -kt + ln [A]o
ln 0.32 = - k (66 s) + ln 1
k = 0.0172642 s-1
-Why must i use 32% in the equation and not 68%?
-And Why use ln (1) for ln [A]o in the equation?
1) You're calculating the current concentration percentage of [A], 68% has reacted, so you're left with 32%
2) 1 = 100/100 (100%), your initial concetration
Omg I have this exact same question on one of my tests so weird.
I have exactly the same question in one of my chemistry exercises !!!
Naweed Babakarkhail could you contact me please! As soon as possible
Am I the only dumb one here
I was looking for half life videos i have no clue wtf any of this means
Why were the initial concentrations cancelled out ?
Are you really living a life if you're not free? You're alive, but you're only living a Half-Life.
he sounds like remy from ratatouille
I genuinely hate school