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I was wondering how to write down the deterministic rate equation for a bimolecular reaction with similar particles.

e.g.

$$ \ce{A ->[k_+] 2B} $$

and

$$ \ce{2B ->[k_-] A} $$

Now the rate equations for the above reactions are:

$$ \mathrm{\frac{\delta A}{\delta t}= - k_{+} A + k_{-} B^2} $$

$$ \mathrm{\frac{\delta B}{\delta t}= 2 k_{+} A - 2 k_{-} B^2} $$

Now I have included a factor of 2 in the second equation to conserve $\mathrm{A + \frac{B}{2}}$.

But I don't understand the reasoning for this factor of 2 in the second equation.

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There is really only one independent equation here because

$\frac{dB}{dt} = -2\frac{dA}{dt}$

The "second equation" in the OP is just -2 times the first equation.

Every molecule of A destroyed creates 2 molecules of B.

Every 2 molecules of B destroyed creates 1 molecule of A.

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  • $\begingroup$ I understand that. but chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Rate_Laws/… They mention the rate of the bimolecular reaction with similar species without a factor of 2. $\endgroup$ – nitin Nov 19 '14 at 16:17
  • $\begingroup$ I don't see anything at that site that contradicts what you did. Certainly you could replace "2k" with a different constant "c" and still have a true equation. $\endgroup$ – DavePhD Nov 19 '14 at 16:43

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