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rvc

  • one year ago

Basic Tutorial: Chemical Kinetics

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  1. rvc
    • one year ago
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    \(\Large\rm To~determine~the~rate~of~the~reaction~we~have~the~\\ \Large rate~law.\\ \Large A~rate~law~is~a~mathematical~equation~that~describes~\\ \Large the~progress~of~the~reaction.Rate~law~is~determined~\\ \Large experimentally.There~are~2~different~forms~of~the~law:~ \\~\\ \Large A.Differential~rate~law \\ ~\\ \Large B.Integrated~rate~law\)

  2. rvc
    • one year ago
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    \(\huge\color{blue}{\bigstar}\color{red}{\rm Differential~Rate~Law}\color{blue}{\bigstar} \) \(\Large\color{black}{\rm The~differential~rate~law~relates~the~rate~of\\ reaction~to~the~concentrations~of~the~various~species\\ in~the~system. }\)

  3. rvc
    • one year ago
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    \( \rm Each~rate~law~contains~a~constant,~k,~called~the~rate~constant.~\\~ The~units~for~the~rate~constant~depend~upon~the~rate~law,~because~\\~the~rate~always~has~ units~of~mole~ L^{-1} sec^{-1} and~the~concentration~always~has~units\\~ of~ mole~ L^{-1}.\)

  4. rvc
    • one year ago
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    \(\begin{array}{|c|c|c|} \hline \rm{Order}&\rm{Explanation}&\rm{Differential~Rate~Law}\\ \hline \rm{Zero}~&~\rm{For~a~zero-order~reaction,~\\ the~ rate~ of~ reaction~ is~ a~ constant. }~&~\rm{ r = k} \\ \hline \rm{First}~&~\rm{Rate~of~reaction~ proportional\\~to~con~of~one~of~the~reactants.}~&~\rm{r = k [A]}\\ \hline \rm{Second}~&~\rm{Rate~of~reaction~is\\~proportional~to~square~of\\ concentration~of~one~\\of~the~ reactants}~&~\rm{r=k[A]^2}~\\ \hline \end{array} \)

  5. rvc
    • one year ago
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    \(\begin{array}{|c|c|c|} \hline \Large\rm{\color{darkblue}{Order}}&\Large\rm{\color{darkblue}{Unit~of~k}}\\ \hline \rm{Zero}&\rm{mole \cdot L^{-1} \cdot sec^{-1}. }\\ \hline \rm{First}&\rm{sec^{-1}. }\\ \hline \rm{Second}&\rm{ L \cdot mole^{-1} \cdot sec^{-1}}\\ \hline \end{array}\)

  6. Michele_Laino
    • one year ago
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    I think It is a good introduction of chemical kinetics!

  7. sweetburger
    • one year ago
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    Once again another great tutorial from @rvc. Well done.

  8. rvc
    • one year ago
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    Thanks :) @Michele_Laino @sweetburger

  9. rvc
    • one year ago
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    \(\huge\color{blue}{\bigstar}{\rm\color{red} {Integrated~ Rate~ Law}}\color{blue}{\bigstar}\) \(\Large\color{black}{\rm The~differential~rate~law~is~directly~proportional~\\to~conc^n~of ~the~reactants. That~ is,the~ rate~ is~\\ proportional~ to~ a~ derivative~ of~ a~ concentration. }\)

  10. rvc
    • one year ago
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    \(\large\color{black}{\rm Consider~ the~ reaction\\ \hspace{150pt} A \rightarrow B\\ ~\\ The ~rate~ of~ reaction,r,is ~given~ by : ~~ r=-\Large\frac{d[A]}{dt} }\) \(\large\color{black}{\rm Suppose~this~reaction~obeys~a~first~order~rate~law:~r=k[A]}\) \(\large\color{black}{\rm }\)

  11. rvc
    • one year ago
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    \(\large\color{black}{\rm This~rate~ law~ can~ also~ be~ written~ as~:~r=-\large\frac{d[A]}{dt}=k[A] \\~\\~Well~ I ~will~write~the~integrated~rate~law~directly~~ \ddot\smile }\)

  12. rvc
    • one year ago
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    \(\begin{array}{|c|c|c|} \hline \rm\Large{Reaction~order}&\rm\Large{Differential~law}&\Large\rm{Integrated~law}\\ \hline \rm\large{Zero}&\rm\large{-\large\frac{d[A]}{dt}=k}&\rm\large{[A] = [A]_0 - k t}\\ \hline \rm\large{First}&\rm\large{-\large\frac{d[A]}{dt}=k[A]}&\rm\large{[A] = [A]_0 e^{- k t}}\\ \hline \rm\large{Second}&\rm\large{-\large\frac{d[A]}{dt}=k[A]^2}&\rm\large{[A] =\large\frac{[A]_0}{1 + k t [A]_0}}\\ \hline \end{array} \)

  13. SyedMohammed98
    • one year ago
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    AWES☻ME!!!!

  14. rvc
    • one year ago
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    Thanks @SyedMohammed98 :)

  15. SyedMohammed98
    • one year ago
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    You're Welcome :)

  16. anonymous
    • one year ago
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    This would have been so useful if I read it before my exam today! And just yesterday I was struggling at this @rvc but I'll definitely remember this.

  17. abb0t
    • one year ago
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    https://www.youtube.com/watch?v=qo0TUPNMf_8 Tutorial 1 of 17 on chemical kinetics.

  18. rvc
    • one year ago
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    @ShizukaTheOtaku i noticed u and other user asking the same question regarding the unit of rate constant So i planned making one :)

  19. anonymous
    • one year ago
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    This is a great way to allow the other to learn. Good job @rvc!

  20. rvc
    • one year ago
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    thank you @Hoslos :)

  21. anonymous
    • one year ago
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    No problem.

  22. nincompoop
    • one year ago
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    http://openstudy.com/study#/updates/539e8e4ce4b0206eed09e15e

  23. TheSmartOne
    • one year ago
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    Good job :)

  24. rvc
    • one year ago
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    Thank you @TheSmartOne :)

  25. Photon336
    • one year ago
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    yeah @rvc great job with this

  26. rvc
    • one year ago
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    thank you @Photon336 :)

  27. welshfella
    • one year ago
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    very good tutorial

  28. rvc
    • one year ago
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    Thank you so much @welshfella :)

  29. Photon336
    • one year ago
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    Always got confused by the units.

  30. arindameducationusc
    • one year ago
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    Nice. @rvc Good job!

  31. sammixboo
    • one month ago
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    Good job!

  32. rvc
    • one month ago
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    Thank You @sammixboo

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