Question

Can anyone please explain me the rate equation for a first order reaction....i mean its derivation? I have a very elementary knowledge of calculus but you may use integration and differentiation

Answer 1

Alright lets assume we have a first order reaction at we set up the following equation:
\[v=k[X]\]
Now per definition we say, that the rate of a reaction is given by:
\[v=-\frac{ d[X] }{ dt }\]
Insert that into the equation before and we get the following expression:
\[-\frac{ d[X] }{ dt }=k[X]\]
This is a fist order differential equation.

Answer 2

https://www.youtube.com/watch?v=lx3Nlj4ZaJc&NR=1&feature=endscreen this video helped me in some way too....

Answer 3

but i stil need a complete explanation specific to the question

Answer 4

The solution to this equation would be:
\[\frac{ 1 }{ [X] } d[X]=-k dt\]
integrate it over the time t=0 to t=t:
\[\int\limits_{[X_{0}]}^{[X]} \frac{ 1 }{ [X] } d[X]=\int\limits_{0}^{t}-k dt\]
This is the same as the following when you calculate it:
\[\ln([X])-\ln([X_{0}])=-kt\]
Rearange it:
\[\ln([X])=-kt + \ln([X_{0}])\]
Take the exponential function on both sides and we get:
\[[X]=[X_{0}]*e ^{-kt}\]