Chain reaction vs single reaction - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-11-13T15:25:03Z https://chemistry.stackexchange.com/feeds/question/86795 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/86795 2 Chain reaction vs single reaction Sp_J https://chemistry.stackexchange.com/users/55716 2017-12-03T21:20:57Z 2017-12-04T18:25:06Z <p>I am considering the following chain reaction: \begin{align} \ce{A_0 + B &lt;=&gt;[N\cdot k_+][k_-]A_1\\ A_1 + B &lt;=&gt;[(N-1)\cdot k_+][2\cdot k_-]A_2\\ A_2 + B &lt;=&gt;[(N-2)\cdot k_+][3\cdot k_-]A_3\\ ...\\ A_{N-2} + B&lt;=&gt;[2\cdot k_+][(N-1)\cdot k_-]A_{N-1}\\ A_{N-1} + B&lt;=&gt;[k_+][N\cdot k_-]A_N} \\ \end{align} Initial condition are $A_0 = 1, A_1=0, ... ,A_N=0$ and $B$ can vary from 0 to 1, $k_+$ is faster than $k_-$.</p> <p>I have noticed (trough extensive numerical simulations) that result of this chain: $A_N$ is identical with the product of the following reaction in the power of N. \begin{align} \ce{A_0 +B&amp;&lt;=&gt;[k_+][k_-]C} \\ \end{align}</p> <p>Meaning that<br> \begin{align} \ce{  [A_N]=[C]^N} \\ \end{align}</p> <p>I am wondering how to check this result analytically. I would like to know what to start with. </p> <p>Thanks!</p> <p>PS. I have looked at the ODEs and so far my calculations failed to prove the point, but I am not sure about my directions.</p>