Worked example: Determining a rate law using initial rates data | AP Chemistry | Khan Academy

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  • Опубликовано: 11 окт 2014
  • The rate law for a chemical reaction can be determined using the method of initial rates, which involves measuring the initial reaction rate at several different initial reactant concentrations. In this video, we'll use initial rates data to determine the rate law, overall order, and rate constant for the reaction between nitrogen dioxide and hydrogen gas. View more lessons or practice this subject at www.khanacademy.org/science/a...
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Комментарии • 52

  • @DoctorMario8
    @DoctorMario8 7 лет назад +65

    Helped so much!! Doesnt make sense that I pay thousands for tuition, only to come to Khan Academy to learn for free!

    • @nitinaravindraj6753
      @nitinaravindraj6753 5 месяцев назад

      absolutely right! going to donate a huge amount after I grow up

  • @ashleyc7277
    @ashleyc7277 6 лет назад +18

    thank god for khan academy. The tutoring service, Study Edge, at UF is crazy expensive and I was worried that i would fail my exam just because I can't afford it but Khan Academy is helping soooooo much.

    • @prachisharma8076
      @prachisharma8076 6 лет назад +2

      This is what I don't like how education has a price!

  • @RealEverythingComputers
    @RealEverythingComputers 7 месяцев назад +2

    Awesome tutorial! Way more simplified explanation than textbooks!

  • @Sir_Lloyds
    @Sir_Lloyds 3 года назад +3

    Brilliantly articulated 🤝🏅🔥

  • @Aditya_Singh11
    @Aditya_Singh11 Год назад +4

    Even after 8 years, this channel is underrated

  • @lonicemwamba5918
    @lonicemwamba5918 2 года назад +2

    Wow this was so helpful 👏

  • @shahed6010
    @shahed6010 Год назад

    this was very helpful thank you Khan

  • @july2052
    @july2052 Год назад

    Thank you for your help!!

  • @chickoomate
    @chickoomate 8 лет назад

    Thank You 😎

  • @Malika296
    @Malika296 8 лет назад +19

    fucking brilliant mate. this was a life saver

  • @SheaBaby31
    @SheaBaby31 5 лет назад +7

    Too bad I cannot use a calculator on the MCAT

  • @vanessaescalera4117
    @vanessaescalera4117 6 лет назад +2

    wow. this is what was went over in 2 lectures. saving my grade!!!

  • @charlottesandor5545
    @charlottesandor5545 2 года назад +1

    Khan academy videos are always the best! Thank you!

  • @user-te2kd9zg6u
    @user-te2kd9zg6u 8 лет назад +1

    THANK YOU!!!!!

  • @sciencenerd7639
    @sciencenerd7639 3 месяца назад

    thank you

  • @temitayookeowo3010
    @temitayookeowo3010 7 лет назад +1

    I love y'all

  • @parkerc-h4589
    @parkerc-h4589 Месяц назад

    just saved me ass, its 7am and i have a quiz on this today 😭thanks so much

  • @monikakhobragade1062
    @monikakhobragade1062 2 года назад

    Hi, if we arent given the experimental data how to determine the orders
    and can you give an example of methane and air order of reaction
    order of oxygen

  • @zekeskulski7146
    @zekeskulski7146 7 лет назад

    Anybody know where I can get some sample problems?

    • @marciewalters143
      @marciewalters143 7 лет назад +3

      Zeke Skulski Do you have a text book? Usually in the chapter there are practice problems.

  • @PonzooonTheGreat
    @PonzooonTheGreat 7 лет назад

    I had a question where my rate constant units were mmol^(-5/2) L^(5/2) s^(-1)
    How do I change this into the sensible mmol L^(-1) s^(-1)??

  • @crossfadez5521
    @crossfadez5521 3 года назад

    that was just amazing man, seriously you should charge us for these stuff.

  • @dazzlelight711
    @dazzlelight711 8 лет назад

    what do you do if the rate determine step is the second step?

  • @Cancer_Bgmi
    @Cancer_Bgmi 6 лет назад

    Tqq sir

    • @spyfish981
      @spyfish981 3 года назад

      9i9iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiimmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm///////////////////////////////////////////////////////////////////////////####################################~~~~~~~~~~~~~~~~~~~~~~~~~#######################~~~~~~~~~~#################################################~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~@@@@@@@@@@@@@@@@@@@@@@@@@@@'''''''''''''''''''''''''''''''''''@@'''''''''''''''''''''''''''''''''''''@''''''''''''''''''''''''''''''''''''''''''''wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
      ##############
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      ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppmmmmmmmmmmmmmmmkkkkkkkkkkkkkkk mmmmmmmmmmmmmmmmmmmmmmmmmmmkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

  • @boySomebody98
    @boySomebody98 8 лет назад

    hey, isn't 250 * 0,012^2 * 0,006 = 2,16*10^-4 ? (the answer to the last part of exercise)

  • @CassieGrey21
    @CassieGrey21 6 лет назад

    What if you get something like k [Y]^ x [Z]^ y? The data gives me 2^x = 2.5. Then x would approximately be 1.322 to get 2.500 on the calculator. If I end up with k [Y]^ 1.322 x [Z]^2, would the order of Y be 1.32 or 1?

  • @breathinginfood
    @breathinginfood 2 года назад

    I love you.(sobbing)

  • @nadeemakhtar3886
    @nadeemakhtar3886 Год назад

    What if we get a fraction?

  • @aliceofei2813
    @aliceofei2813 2 года назад

    omg thank you sm! this made 0 sense to me in class lmao

  • @opufy
    @opufy Год назад

    7:27 RIP, my teacher would dock marks with a big smile for saying ".005 M" for sig figs because it's supposed to be "0.0050 M"

  • @salemalmuhairi4284
    @salemalmuhairi4284 8 лет назад +5

    :)

  • @Jennieve_
    @Jennieve_ 6 лет назад

    Is it possible to get the rate constant if one of the concentrations was never kept constant

  • @Bulbasauros
    @Bulbasauros 2 года назад +1

    For u lazy bois out there u can do ( ln ((r1/r2) ) / ( ln ( [1] / [2] ) )
    [ ☆ ] Concentration
    1 and 2 is for experiment 1 and 2 not other substance
    r stands for initial rate
    Make sure the concentration of the other reactant stays the same and this will give u the order of reaction without any thinking necessary

  • @mohamedalalami8936
    @mohamedalalami8936 2 года назад

    BRO thank u i couldn't understand my **** professor.

  • @coolguy3513
    @coolguy3513 Год назад

    God bless

    • @coolguy3513
      @coolguy3513 Год назад

      God bless again (exam edition)

  • @missmak559
    @missmak559 6 лет назад

    Asalam-O-Alaikum.....thank u sooo much......but.....isn't there anyway to find the order of reaction without experimental values???

    • @JTurn916
      @JTurn916 6 лет назад +1

      There is not

  • @KM-wb1yd
    @KM-wb1yd 2 года назад

    omg Can you be my chemistry prof please

  • @sarakazi9451
    @sarakazi9451 6 лет назад

    But what if we have no rates to compare to?

  • @hammedyusrah4912
    @hammedyusrah4912 2 года назад

    Please you did not make use of the temperature

  • @vanessaescalera4117
    @vanessaescalera4117 6 лет назад +1

    exactly why i donate lol

  • @rosegold2969
    @rosegold2969 8 лет назад +2

    Speak louder!!!!!

    • @timcook3410
      @timcook3410 7 лет назад +9

      aderinsola abayomi clean your ears!!!!!