Question

Basic Tutorial:
Chemical Kinetics

Answer 1

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

Answer 2

\(\huge\color{blue}{\bigstar}\color{red}{\rm Differential~Rate~Law}\color{blue}{\bigstar} \)

\(\Large\color{black}{\rm The~differential~rate~law~relates~the~rate~of\\ reaction~to~the~concentrations~of~the~various~species\\ in~the~system. }\)

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

I think It is a good introduction of chemical kinetics!

Answer 7

Once again another great tutorial from @rvc. Well done.

Answer 8

Thanks :) @Michele_Laino @sweetburger

Answer 9

\(\huge\color{blue}{\bigstar}{\rm\color{red} {Integrated~ Rate~ Law}}\color{blue}{\bigstar}\)

\(\Large\color{black}{\rm The~differential~rate~law~is~directly~proportional~\\to~conc^n~of ~the~reactants. That~ is,the~ rate~ is~\\ proportional~ to~ a~ derivative~ of~ a~ concentration. }\)

Answer 10

\(\large\color{black}{\rm Consider~ the~ reaction\\ \hspace{150pt} A \rightarrow B\\ ~\\ The ~rate~ of~ reaction,r,is ~given~ by : ~~ r=-\Large\frac{d[A]}{dt} }\)

\(\large\color{black}{\rm Suppose~this~reaction~obeys~a~first~order~rate~law:~r=k[A]}\)

\(\large\color{black}{\rm }\)

Answer 11

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

Answer 12

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

Answer 13

AWES☻ME!!!!

Answer 14

Thanks @SyedMohammed98 :)

Answer 15

You're Welcome :)

Answer 16

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.

Answer 17

https://www.youtube.com/watch?v=qo0TUPNMf_8

Tutorial 1 of 17 on chemical kinetics.

Answer 18

@ShizukaTheOtaku i noticed u and other user asking the same question regarding the unit of rate constant
So i planned making one :)

Answer 19

This is a great way to allow the other to learn.
Good job @rvc!

Answer 20

thank you @Hoslos :)

Answer 21

No problem.

Answer 22

http://openstudy.com/study#/updates/539e8e4ce4b0206eed09e15e

Answer 23

Good job :)

Answer 24

Thank you @TheSmartOne :)

Answer 25

yeah @rvc great job with this

Answer 26

thank you @Photon336 :)

Answer 27

very good tutorial

Answer 28

Thank you so much @welshfella :)

Answer 29

Always got confused by the units.

Answer 30

Nice. @rvc
Good job!

Answer 31

Good job!

Answer 32

Thank You @sammixboo